Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 10 Electrostatics Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 10 Electrostatics

1. Choose the correct option.

Question 1.
A positively charged glass rod is brought close to a metallic rod isolated from ground. The charge on the side of the metallic rod away from the glass rod will be
(A) same as that on the glass rod and equal in quantity
(B) opposite to that on the glass of and equal in quantity
(C) same as that on the glass rod but lesser in quantity
(D) same as that on the glass rod but more in quantity
Answer:
(A) same as that on the glass rod and equal in quantity

Question 2.
An electron is placed between two parallel plates connected to a battery. If the battery is switched on, the electron will
(A) be attracted to the +ve plate
(B) be attracted to the -ve plate
(C) remain stationary
(D) will move parallel to the plates
Answer:
(A) be attracted to the +ve plate

Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics

Question 3.
A charge of + 7 µC is placed at the centre of two concentric spheres with radius 2.0 cm and 4.0 cm respectively. The ratio of the flux through them will be
(A) 1 : 4
(B) 1 : 2
(C) 1 : 1
(D) 1 : 16
Answer:
(C) 1 : 1

Question 4.
Two charges of 1.0 C each are placed one meter apart in free space. The force between them will be
(A) 1.0 N
(B) 9 × 109 N
(C) 9 × 10-9 N
(D) 10 N
Answer:
(B) 9 × 109 N

Question 5.
Two point charges of +5 µC are so placed that they experience a force of 80 × 10-3 N. They are then moved apart, so that the force is now 2.0 × 10-3 N. The distance between them is now
(A) 1/4 the previous distance
(B) double the previous distance
(C) four times the previous distance
(D) half the previous distance
Answer:
(B) double the previous distance

Question 6.
A metallic sphere A isolated from ground is charged to +50 µC. This sphere is brought in contact with other isolated metallic sphere B of half the radius of sphere A. The charge on the two sphere will be now in the ratio
(A) 1 : 2
(B) 2 : 1
(C) 4 : 1
(D) 1 : 1
Answer:
(D) 1 : 1

Question 7.
Which of the following produces uniform electric field?
(A) point charge
(B) linear charge
(C) two parallel plates
(D) charge distributed an circular any
Answer:
(C) two parallel plates

Question 8.
Two point charges of A = +5.0 µC and B = -5.0 µC are separated by 5.0 cm. A point charge C = 1.0 µC is placed at 3.0 cm away from the centre on the perpendicular bisector of the line joining the two point charges. The charge at C will experience a force directed towards
(A) point A
(B) point B
(C) a direction parallel to line AB
(D) a direction along the perpendicular bisector
Answer:
(C) a direction parallel to line AB

2. Answer the following questions.

Question 1.
What is the magnitude of charge on an electron?
Answer:
The magnitude of charge on an electron is 1.6 × 10-19 C

Question 2.
State the law of conservation of charge.
Answer:
In any given physical process, charge may get transferred from one part of the system to another, but the total charge in the system remains constant”
OR
For an isolated system, total charge cannot be created nor destroyed.

Question 3.
Define a unit charge.
Answer:
Unit charge (one coulomb) is the amount of charge which, when placed at a distance of one metre from another charge of the same magnitude in vacuum, experiences a force of 9.0 × 109 N.

Question 4.
Two parallel plates have a potential difference of 10V between them. If the plates are 0.5 mm apart, what will be the strength of electric charge.
Answer:
V=10V
d = 0.5 mm = 0.5 × 10-3 m
To find: The strength of electric field (E)
Formula: E = \(\frac {V}{d}\)
Calculation: From formula,
E = \(\frac {10}{0.5×10^{-3}}\)
20 × 103 V/m

Question 5.
What is uniform electric field?
Answer:
A uniform electric field is a field whose magnitude and direction are same at all points. For example, field between two parallel plates as shown in the diagram.
Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics 1

Question 6.
If two lines of force intersect of one point. What does it mean?
Answer:
If two lines of force intersect of one point, it would mean that electric field has two directions at a single point.

Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics

Question 7.
State the units of linear charge density.
Answer:
SI unit of λ is (C / m).

Question 8.
What is the unit of dipole moment?
Answer:
i. Strength of a dipole is measured in terms of a quantity called the dipole moment.
Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics 2

ii. Let q be the magnitude of each charge and 2\(\vec{l}\) be the distance from negative charge to positive charge. Then, the product q(2\(\vec{l}\)) is called the dipole moment \(\vec{p}\).

iii. Dipole moment is defined as \(\vec{p}\) = q(2\(\vec{l}\))

iv. A dipole moment is a vector whose magnitude is q (2\(\vec{l}\)) and the direction is from the negative to the positive charge.

v. The unit of dipole moment is coulomb-metre (C m) or debye (D).

Question 9.
What is relative permittivity?
Answer:
i. Relative permittivity or dielectric constant is the ratio of absolute permittivity of a medium to the permittivity of free space.
It is denoted as K or εr.
i.e., K or εr = \(\frac {ε}{ε_0}\)

ii. It is the ratio of the force between two point charges placed a certain distance apart in free space or vacuum to the force between the same two point charges when placed at the same distance in the given medium.
i.e., K or εr = \(\frac {F_{vacuum}}{F_{medium}}\)

iii. It is also called as specific inductive capacity or dielectric constant.

3. Solve numerical examples.

Question 1.
Two small spheres 18 cm apart have equal negative charges and repel each other with the force of 6 × 10-8 N. Find the total charge on both spheres.
Solution:
Given: F = 6 × 10-8 N, r = 18 cm = 18 × 10-2 m
To find: Total charge (q1 + q2)
Formula: F = \(\frac {1}{4πε_0}\) \(\frac {q_1q_2}{r^2}\)
Calculation: From formula,
Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics 3
Taking square roots from log table,
∴ q = -4.648 × 10-10 C
….(∵ the charges are negative)
Total charge = q1 + q2 = 2q
= 2 × (-4.648) × 10-10
= -9.296 × 10-10 C

Question 2.
A charge + q exerts a force of magnitude – 0.2 N on another charge -2q. If they are separated by 25.0 cm, determine the value of q.
Answer:
Given: q1 = + q, q2 = -2q, F = -0.2 N
r = 25 cm = 25 × 10-2 m
To find: Charge (q)
Formula: F = \(\frac {1}{4πε_0}\) \(\frac {q_1q_2}{r^2}\)
Calculation: From formula,
Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics 4
[Note: The answer given above is calculated in accordance with textual method considering the given data]

Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics

Question 3.
Four charges of +6 × 10-8 C each are placed at the corners of a square whose sides are 3 cm each. Calculate the resultant force on each charge and show in direction on a diagram drawn to scale.
Answer:
Given: qA = qB = qC = qD = 6 × 10-8 C, a = 3 cm
Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics 5
∴ Resultant force on ‘A’
= FAD cos 45 + FAB cos 45 + FAC
= (3.6 × 10-2 × \(\frac {1}{√2}\)) + (3.6 × 10-2 × \(\frac {1}{√2}\)) + 1.8 × 10-2
= 6.89 × 10-2 N directed along \(\vec{F_{AC}}\)

Question 4.
The electric field in a region is given by \(\vec{E}\) = 5.0 \(\hat{k}\) N/C Calculate the electric flux through a square of side 10.0 cm in the following cases
i. The square is along the XY plane
ii. The square is along XZ plane
iii. The normal to the square makes an angle of 45° with the Z axis.
Answer:
Given: \(\vec{E}\) = 5.0 \(\hat{k}\) N/C, |E| = 5 N/C
l = 10 cm = 10 × 10-2 m = 10-1 m
A = l² – 10-2
To find: Electric flux in three cases.
1) (ø2) (ø3)
Formula: (ø1) = EA cos θ
Calculation:
Case I: When square is along the XY plane,
∴ θ = 0
ø1 = 5 × 10-2 cos 0
= 5 × 10-2 V m

Case II: When square is along XZ plane,
∴ θ = 90°
ø1 = 5 × 10-2 cos 90° = 0 V m

Case III: When normal to the square makes an angle of 45° with the Z axis.
∴ 0 = 45°
∴ ø3 = 5 × 10-2 × cos 45°
= 3.5 × 10-2 V m

Question 5.
Three equal charges of 10 × 10-8 C respectively, each located at the corners of a right triangle whose sides are 15 cm, 20 cm and 25 cm respectively. Find the force exerted on the charge located at the 90° angle.
Answer:
Given: qA = qB = qC = 10 × 10-8
Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics 6
Force on B due to A,
Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics 7

Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics

Question 6.
A potential difference of 5000 volt is applied between two parallel plates 5 cm apart. A small oil drop having a charge of 9.6 x 10-19 C falls between the plates. Find (i) electric field intensity between the plates and (ii) the force on the oil drop.
Answer:
Given: V = 5000 volt, d = 5 cm = 5 × 10-2 m
q = 9.6 × 10-19 C
To find:
i. Electric field intensity (E)
ii. Force (F)
Formula:
i. E = \(\frac {V}{d}\) \(\frac {q}{r}\)
ii. E = \(\frac {F}{q}\)
Calculation: From formula (i),
E = \(\frac {F}{q}\) = 105 N/C
From formula (ii)
F = E x q
= 105 × 9.6 × 10-19
= 9.6 × 10-14 N

Question 7.
Calculate the electric field due to a charge of -8.0 × 10-8 C at a distance of 5.0 cm from it.
Answer:
Given: q = – 8 × 10-8 C, r = 5 cm = 5 × 10-2 m
To Find: Electric field (E)
Formula: E = \(\frac {1}{4πε_0}\) \(\frac {q}{r^2}\)
Calculation: From formula,
E = 9 × 109 × \(\frac {(-8×10^{-8})}{(5×10^{-2})^2}\)
= -2.88 × 105 N/C

11th Physics Digest Chapter 10 Electrostatics Intext Questions and Answers

Can you recall? (Textbookpage no. 188)

Question 1.
Have you experienced a shock while getting up from a plastic chair and shaking hand with your friend?
Answer:
Yes, sometimes a shock while getting up from a plastic chair and shaking hand with friend is experienced.

Question 2.
Ever heard a crackling sound while taking out your sweater in winter?
Answer:
Yes, sometimes while removing our sweater in winter, some crackling sound is heard and the sweater appears to stick to body.

Question 3.
Have you seen the lightning striking during pre-monsoon weather?
Answer:
Yes, sometimes lightning striking during pre-pre-monsoon weather seen.

Can you tell? (Textbook page no. 189)

i. When a petrol or a diesel tanker is emptied in a tank, it is grounded.
ii. A thick chain hangs from a petrol or a diesel tanker and it is in contact with ground when the tanker is moving.
Answer:
i. When a petrol or a diesel tanker is emptied in a tank, it is grounded so that it has an electrically conductive connection from the petrol or diesel tank to ground (Earth) to allow leakage of static and electrical charges.

ii. Metallic bodies of cars, trucks or any other big vehicles get charged because of friction between them and the air rushing past them. Hence, a thick chain is hanged from a petrol or a diesel tanker to make a contact with ground so that charge produced can leak to the ground through chain.

Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics

Can you tell? (Textbook page no. 194)

Three charges, q each, are placed at the vertices of an equilateral triangle. What will be the resultant force on charge q placed at the centroid of the triangle?
Answer:
Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics 8
Since AD. BE and CF meets at O, as centroid of an equilateral triangle.
∴ OA = OB = OC
∴ Let, r = OA = OB = OC
Force acting on point O due to charge on point A,
\(\overrightarrow{\mathrm{F}}_{\mathrm{OA}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}^{2}}{\mathrm{r}_{i}^{2}} \hat{\mathrm{r}}_{\mathrm{AO}}\)
Force acting on point O due to charge on point B,
\(\overrightarrow{\mathrm{F}}_{O \mathrm{~B}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}^{2}}{\mathrm{r}^{2}} \hat{\mathrm{r}}_{\mathrm{BO}}\)
Force acting on point O due to charge on point C,
\(\overrightarrow{\mathrm{F}}_{\mathrm{OC}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}^{2}}{\mathrm{r}^{2}} \hat{\mathrm{r}} \mathrm{co}\)
∴ Resultant force acting on point O,
F = \(\vec{F}\)OA + \(\vec{F}\)OB + \(\vec{F}\)OC
On resolving \(\vec{F}\)OB and \(\vec{F}\)OC, we get –\(\vec{F}\)OA
i.e., \(\vec{F}\)OB + \(\vec{F}\)OC = –\(\vec{F}\)OA
∴ \(\vec{F}\) = \(\vec{F}\)OA – \(\vec{F}\)OA = 0
Hence, the resultant force on the charge placed at the centroid of the equilateral triangle is zero.

Can you tell? (Textbook page no. 197)

Why a small voltage can produce a reasonably large electric field?
Answer:

  1. Electric field produced depends upon voltage as well as separation distance.
  2. Electric field varies linearly with voltage and inversely with distance.
  3. Hence, even if voltage is small, it can produce a reasonable large electric field when the gap between the electrode is reduced significantly.

Can you tell? (Textbook page no. 198)

Lines of force are imaginary; can they have any practical use?
Answer:
Yes, electric lines of force help us to visualise the nature of electric field in a region.

Can you tell? (Textbook page no. 204)

The surface charge density of Earth is σ = -1.33 nC/m². That is about 8.3 × 109 electrons per square metre. If that is the case why don’t we feel it?
Answer:
The Earth along with its atmosphere acts as a neutral system. The atmosphere (ionosphere in particular) has nearly equal and opposite charge.

As a result, there exists a mechanism to replenish electric charges in the form of continual thunder storms and lightning that occurs in different parts of the globe. This makes average charge on surface of the Earth as zero at any given time instant. Hence, we do not feel it.

Maharashtra Board Class 11 Physics Solutions Chapter 10 Electrostatics

Internet my friend (Textbook page no. 205)

i. https://www.physicsclassroom.com/class
ii. https://courses.lumenleaming.com/physics/
iii. https://www,khanacademy.org/science
iv. https://www.toppr.com/guides/physics/
[Students are expected to visit the above mentioned websites and collect more information about electrostatics.]

Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 9 Optics Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 9 Optics

Question 1.
Express the speed of EM waves in terms of permittivity and permeability of the medium.
Answer:
In a material medium, the speed of EM waves is given by, c = \(\sqrt{\frac {1}{εµ}}\)
Where, ε = Permittivity and µ = permeability.
These constants depend on the electric and magnetic properties of the medium.

Question 2.
How can one classify commonly observed phenomena of light on the basis of nature of light?
Answer:
Commonly observed phenomena concerning light can be broadly split into three categories:

  1. Ray optics or geometrical optics: Ray optics can be used for understanding phenomena like reflection, refraction, double refraction, total internal reflection, etc.
  2. Wave optics or physical optics: Wave optics explains the phenomena of light such as, interference, diffraction, polarisation, Doppler effect etc.
  3. Particle nature of light: Particle nature of light can be used to explain phenomena like photoelectric effect, emission of spectral lines, Compton effect etc.

Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics

Question 3.
State the fundamental laws on which ray optics is based.
Answer:
Ray optics is based on the following fundamental laws:
i. Light travels in a straight line in a homogeneous and isotropic medium.

ii. Two or more rays can intersect at a point without affecting their paths beyond that point.

iii. Laws of reflection:
a. Reflected ray lies in the plane formed by incident ray and the normal drawn at the point of incidence and the two rays are on either side of the normal.
b. Angles of incidence and reflection are equal (i = r).

iv. Laws of refraction:
a. Refracted ray lies in the plane formed by incident ray and the normal drawn at the point of incidence; and the two rays are on either side of the normal.

b. Angle of incidence (90 and angle of refraction (62) are related by Snell’s law, given by,
n1 sin θ1 = n2 sin θ2
where, n1, n2 = refractive indices of medium 1 and medium 2 respectively.

Question 4.
Explain Cartesian sign conventions using a graph.
Answer:
According to Cartesian sign conventions:
i. All distances are measured from the optical centre or pole.

ii. Figures should be drawn in such a way that the incident rays travel from left to right. Thus, a real object should be shown to the left and virtual object or image to the right of pole (or optical centre).
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 1

iii. X-axis can be conveniently chosen as the principal axis with origin at the pole.

iv. Distances to the left of the pole are negative and those to the right of the pole are positive.

v. Distances above the principal axis (X-axis) are positive while those below it are negative.

Question 5.
Define and represent in a neat diagram the following terms:
i. Diverging beam
ii. Converging beam
Answer:
i. A diverging beam of light corresponds to rays of light coming from real point object.
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 2

ii. A converging beam corresponds to rays of light directed to a virtual point object or image.
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 3

Question 6.
Thickness of the glass of a spectacle is 2 mm and refractive index of its glass is 1.5. Calculate time taken by light to cross this thickness. Express your answer with most convenient prefix attached to the unit ‘second’.
Answer:
Speed of light in vacuum, c = 3 × 108 m/s
Given that:
Refractive index, (nglass) = 1.5
Thickness of the glass = 2 mm
= 2 × 10sup>-3 m
∵ Re fractive index (nglass) = \(\frac{\text { speed of light in vacuum }(\mathrm{c})}{\text { speed of light in glass (v) }}\)
∵ v = \(\frac{\mathrm{c}}{\mathrm{n}_{\text {glass }}}=\frac{3 \times 10^{8}}{1.5}\) = 2 × 108 m/s
As v = \(\frac {s}{t}\)
time taken (t) to cross the thickness (s),
t = \(\frac{\mathrm{s}}{\mathrm{v}}=\frac{2 \times 10^{-3}}{2 \times 10^{8}}\) = 1 × 10-11 s
Most convenient unit to express this small time is nano second. (1 ns = 10-9 s)
∴ t = 0.01 × 10-9 s = 0.01 ns

Question 7.
Explain the properties of the image formed after reflection of light from a plane surface.
Answer:

  1. The image of an object kept in front of a plane reflecting surface is virtual and laterally inverted.
  2. Image is of the same size as that of the object.
  3. It is situated at the same distance as that of object but on the other side of the reflecting surface.

Question 8.
Explain the formula to find the number of images formed when an object is placed in between two plane mirrors inclined at an angle θ.
Answer:

  1. If an object is kept between two plane mirrors inclined at an angle θ, multiple images (N) are formed due to multiple reflections from both the mirrors.
  2. The number of images can be calculated using formula n = \(\frac {360}{θ}\)
  3. Exact number of images seen (N) depends upon the angle between the mirrors and position of the object.
  4. When n is an even integer, for all positions of the object the number of images formed are N = n – 1.
  5. When n is an odd integer:
    a. For an object placed at the angle bisector of the mirrors: N = n- 1
    b. For an object placed off the angle bisector of the mirrors: N = n
  6. If n is not an integer, N = m, where m is integral part of n.

Question 9.
Define radius of curvature of a spherical mirror.
Answer:
Radius of the sphere of which a mirror is a part is called as radius of curvature of the mirror.

Question 10.
What is the focal length of a spherical mirror? Give its relation with the radius of curvature.
Answer:
i. For a concave mirror focal length is the distance at which parallel incident rays converge. For a convex mirror, it is the distance from where parallel rays appear to be diverging after reflection.

ii. In case of spherical mirrors, half of radius of curvature is focal length of the mirror,
f = \(\frac {R}{2}\)

Question 11.
Show with the help of a ray diagram that focal length of convex mirror is positive while that of concave mirror is negative.
Answer:
i. According to sign conventions, the incident rays are drawn from left of the mirror to the right as shown in the ray diagrams below.

ii. As the rays incident on convex mirror appear to converge at a point on the positive side of the origin, the focal length of the convex mirror is positive.
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 4

iii. However, in case of concave mirror, the rays converge at a point on negative side of the origin, the focal length of the concave mirror is negative.
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 5

Question 12.
Give relation between focal length, object distance and image distance for a small spherical mirror.
Answer:
For a point object or for a small finite object, the focal length of a small spherical mirror is related to object distance and image distance as,
\(\frac {1}{f}\) = \(\frac {1}{v}\) + \(\frac {1}{u}\)

Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics

Question 13.
What is lateral magnification? How does it vary in different types of spherical mirrors?
Answer:
i. Ratio of linear size of image to that of the object, measured perpendicular to the principal axis, is defined as the lateral magnification.
∴ m = \(\frac {h_2}{h_1}\) = \(\frac {v}{u}\) (for spherical lenses)
m = –\(\frac {v}{u}\) (for spherical mirrors)

ii. For any position of the object, a convex mirror always forms virtual, erect and diminished image. Thus, lateral magnification for convex lens is always m < 1.

iii. In the case of a concave mirror, it depends upon the position of the object.

Question 14.
Complete the following table for a concave mirror?

Position of object Position of image Lateral magnification
U = ∞ v = f m = 0
u > 2f ……………. m < 1
u = f V = ∞ ………………
…………… |v| > |u| m > 1
2f > u > f ……………… m > 1

Answer:

Position of object Position of image Lateral magnification
U = ∞ v = f m = 0
u > 2f 2f > v > f m < 1
u = f V = ∞ m = ∞
u < f |v| > |u| m > 1
2f > u > f v > 2f m > 1

Question 15.
Explain with proper diagram why parabolic mirrors are preferred over spherical one.
Answer:
i. Unlike spherical shape, every point on a parabola is equidistant from a straight line and a point.

ii. Consider given parabola having RS as directrix and F as the focus. Points A, B, C on it are equidistant from line RS and point F.
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 6

iii. Hence A’A = AF, B’B = BF, C’C = CF, and so on.

iv. If rays of equal optical path converge at a point, that point is the location of real image corresponding to that beam of rays.

v. From figure, the paths A”AA’, B”BB’. C”CC’, etc., are equal paths when mirror is neglected.

vi. If the parabola ABC is a mirror then by definition of parabola the respective optical paths,
A”AF = B”BF = C”CF

vii. Thus, F is the single point focus for entire beam of rays parallel to the axis and there is no spherical aberration.
Hence, parabolic mirrors are preferred over spherical one as there is no spherical aberration.

Question 16.
A small object is kept symmetrically between two plane mirrors inclined at 38°. This angle is now gradually increased to 41°, the object being symmetrical all the time. Determine the number of images visible during the process.
Answer:
The object is kept symmetrically between two plane mirrors. This implies the object is placed at angle bisector.
Thus, for θ = 38°,
n = \(\frac {360}{38}\) = 9.47
As it is not integral, N = 9 (the integral part of n)
∴ For going from 38° to 41°, the mirrors go through angles 39° and 40°. Number of images formed will remain 9 for all angles between 38° and 40°.
For angles > 40°, the n goes on decreasing and when θ = 41°,
n = \(\frac {360}{41}\) = 8.78 i.e., N = 8

Question 17.
A thin pencil of length 20 cm is kept along the principal axis of a concave mirror of curvature 30 cm. Nearest end of the pencil is 20 cm from the pole of the mirror. What will be the size of image of the pencil?
Answer:
For the pencil kept along the principal axis and the end of the pencil closest to pole is at 20 cm,
say, u1 = -20 cm
Flence, the other end of the stick is at distance, u2 = (u1 + 20) = -40 cm from pole of the mirror.
As R = -30 cm, F = \(\frac {R}{2}\) = -15 cm
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 7
∴ v2 = -24 cm
Here, negative signs indicate that images are formed on the left of the mirror.
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 8
The length of the image formed is given by,
v = v2 – v1 = -24 – (-60) = 36 cm.

Question 18.
An object is placed at 15 cm from a convex mirror having radius of curvature 20 cm. Find the position and kind of image formed by it.
Answer:
Given: u = – 15 cm,
f = \(\frac {R}{2}\) = + \(\frac {20}{2}\) = + 10cm
To find: Nature and position of image (v)
Formula: \(\frac {1}{v}\) + \(\frac {1}{u}\) = \(\frac {1}{f}\)
Calculation:
From formula,
∴ \(\frac {1}{v}\) = \(\frac {1}{f}\) – \(\frac {1}{u}\)
= \(\frac {1}{+10}\) – \(\frac {1}{-15}\) = \(\frac {1}{10}\) + \(\frac {1}{15}\)
= \(\frac {2+3}{30}\) = \(\frac {5}{30}\) = \(\frac {1}{6}\)
∴ v = 6 cm

Question 19.
Prove that refractive index of a glass slab is given by the formula,
n = \(\frac {Real depth}{Apparent depth}\)
Answer:
i. Consider a plane parallel slab of a transparent medium of refractive index n.

ii. A point object O at real depth R appears to be at I at apparent depth A, when seen from outside (air).

iii. Consider incident ray OA and OB as shown in figure.
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 9

iv. Assuming i and r to be very small, we can write,
tan r = \(\frac {x}{A}\) ≈ sin r and tan i = \(\frac {x}{R}\) ≈ sin i

v. According to Snell’s law, for a ray travelling from denser medium to rarer medium,
n = \(\frac{\sin \mathrm{r}}{\sin \mathrm{i}} \approx \frac{\left(\frac{\mathrm{x}}{\mathrm{A}}\right)}{\left(\frac{\mathrm{x}}{\mathrm{D}}\right)}=\frac{\mathrm{R}}{\mathrm{A}}=\frac{\text { Real depth }}{\text { Apparent depth }}\)

Question 20.
The depth of a pond is 10 m. What is the apparent depth for a person looking normally to the water surface? nwater = 4/3.
Answer:
Given: Real depth of pond, dreal = 10 m,
nw = \(\frac {4}{3}\)
To find: Apparent depth
Formula: n = \(\frac {Realdepth}{Apparent depth}\)
Calculation: From formula,
∴ Apparent depth = \(\frac {Realdepth}{n}\) = \(\frac {10}{(\frac{4}{3})}\)
= \(\frac {10×3}{4}\) = 7.5 m

Question 21.
A crane flying 6 m above a still, clear water lake sees a fish underwater. For the crane, the fish appears to be 6 cm below the water surface. How much deep should the crane immerse its beak to pick that fish?
For the fish, how much above the water surface does the crane appear? Refractive index of water = 4/3.
Answer:
For crane, apparent depth of fish = 6 cm,
Given that refractive index (nw) = \(\frac {4}{3}\)
nw = \(\frac {Realdepth}{Apparent depth}\)
∴ Apparent depth = \(\frac {4}{3}\) × 6 = 8 cm
Similarly, for fish, real height of crane = 6 m and
\(\frac{\mathrm{n}_{\mathrm{air}}}{\mathrm{n}_{\mathrm{w}}}=\frac{1}{\mathrm{n}_{\mathrm{w}}}=\frac{\text { Real height }}{\text { Apparent height }}\)
\(\frac {3}{4}\) = \(\frac {6}{Apparent height}\)
i.e., Apparent height = \(\frac {4×6}{3}\) = 8 m

Question 22.
Write a short note on Periscope.
Answer:
i. Instrument used to see the objects on the surface of a water body from inside of water is called periscope.
ii. It consists of two right angled prisms. The incident rays of light are reflected twice through these prisms.
iii. Total internal reflections occur inside these prisms and a clear view of the surface of water is obtained.
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 10

Question 23.
A ray of light passes from glass (ng = 1.52) to water (nw = 1.33). What is the critical angle of incidence?
Answer:
Given: ng = 1.52, nw = 1.33
To find: Critical angle (ic)
formula: sin ic = \(\frac {n_2}{n_1}\) = \(\frac {n_w}{n_g}\)
Calculation:
From formula,
ic = sin-1 (\(\frac {1.33}{1.52}\)) = sin-1 (0.875) = 61°2′

Question 24.
There is a tiny LED bulb at the center of the bottom of a cylindrical vessel of diameter 6 cm. Height of the vessel is 4 cm. The beaker is filled completely with an optically dense liquid. The bulb is visible from any inclined position but just visible if seen along the edge of the beaker. Determine refractive index of the liquid.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 11
As the bulb is just visible from the edge, the angle of incidence formed by ray OP must be equal to critical angle.
∴ refractive index (n) = \(\frac {1}{sin i_c}\)
From Figure,
tan ic = \(\frac {PQ}{OQ}\) = \(\frac {4}{3}\)
∴ sin ic = \(\frac {OQ}{OP}\) = \(\frac {3}{5}\)
∴ nliquid = \(\frac {5}{3}\)

Question 25.
What are convex and concave lenses? For which condition, convex lens will have negative focal length?
Answer:

  1. A lens is said to be convex if it is thicker in the middle and narrowing towards the periphery. According to Cartesian sign convention, its focal length is positive.
  2. Convex lens is visualized to be internal cross section of two spheres (or one sphere or a plane surface).
  3. A lens is concave if it is thicker at periphery and narrows down towards centre and has negative focal length.
  4. Concave lens is visualized to be external cross section of two spheres.
  5. For lenses of material optically rarer than the medium in which those are kept, convex lenses will have negative focal length and they will diverge the incident rays.

Question 26.
Which lenses can be considered as thin lenses?
Answer:
Lenses for which the maximum thickness is at least 50 times smaller than all the other distances are considered as thin lenses.

Question 27.
Give the expression for the focal length of combination of lenses when
i. Lenses are kept in contact with each other
ii. Two lenses kept at a distance d apart from each other.
Answer:
i. For thin lenses kept in contact:
\(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{f}_{1}}+\frac{1}{\mathrm{f}_{2}}+\frac{1}{\mathrm{f}_{3}}\) + ………

ii. For two lenses kept distance d apart:
\(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}-\frac{d}{f_{1} f_{2}}\)

Question 28.
An object is placed infront of a convex surface separating two media of refractive index 1.1 and 1.5. The radius of curvature is 40 cm. Where is the image formed when an object is placed at 220 cm from the refracting surface?
Solution:
Given: n1 = 1.1, n2 = 1.5, R = + 40 cm,
u = -220 cm
To find: Position of image (v)
Formula: \(\frac{n_{2}}{v}-\frac{n_{1}}{u}=\frac{\left(n_{2}-n_{1}\right)}{R}\)
Calculation: From formula,
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 12

Question 29.
A glass paper-weight (n = 1.5) of radius 3 cm has a tiny air bubble trapped inside it. Closest distance of the bubble from the surface is 2 cm. Where will it appear when seen from the other end (from where it is farthest)?
Answer:
From figure, distance OR = 2 cm
∴ Distance OP = 4 cm
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 13
According to sign conventions,
OP = u = -4 cm and CP = R = -3 cm
For refraction at curved surface,
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 14

Question 30.
Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?
Answer:
Given: n = 1.55, f = 20 cm,
R1 = R and R2 = – R
(By sign convention)
To Find: Radius of curvature (R)
Formula: \(\frac{1}{\mathrm{f}}=(\mathrm{n}-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
Calculation: From formula,
\(\frac {1}{20}\) = (1.55 – 1) \(\left[\frac{1}{R}-\left(-\frac{1}{R}\right)\right]=0.55 \times \frac{2}{R}\)
∴ R = \(\frac {1.10}{1}\) × 20 = 22 cm

Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics

Question 31.
A dense glass double convex lens (n = 2) designed to reduce spherical aberration has |R1| : |R2| = 1:5. If a point object is kept 15 cm in front of this lens, it produces its real image at 7.5 cm. Determine R1 and R2.
Answer:
Given: |R1| : |R2| = 1 : 5, u = -15 cm,
v = +7.5 cm, n = 2
To Find: Radii of curvature of double convex lens (R1) and (R2)
Formula:
i. \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\)
ii. \(\frac {1}{f}\) = (n – 1) \(\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
Calculation: From formula (i),
\(\frac{1}{f}=\frac{1}{7.5}-\frac{1}{(-15)}=\frac{1}{5}\)
∴ f = +5 cm
Substituting this value in formula (ii), we get,
\(\frac{1}{5}=(2-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
∴ \(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}=\frac{1}{5}\)
By sign conventions,
\(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\left(-\mathrm{R}_{2}\right)}=\frac{1}{5}\) ………….. (1)
Also \(\frac{\left|R_{1}\right|}{\left|R_{2}\right|}=\frac{1}{5}\)
∴ |R2| = 5 |R1| …………… (2)
Substituting in equation (1),
∴ \(\frac{1}{R_{1}}-\frac{1}{\left(-5 R_{1}\right)}=\frac{1}{5}\)
∴ \(\frac {6}{5R_1}\) = \(\frac {1}{5}\)
∴ R1 = 6 cm
Using in equation (2),
R2 = 5 × 6 = 30 cm

Question 32.
Why are prism preferred for dispersion over two parallel surfaces? Explain its construction in brief.
Answer:
i. In case of two parallel surfaces, for dispersion to be easily detectable, they must be separated over a large distance.
ii. In order to have appreciable and observable dispersion, two parallel surfaces are not useful. In such case we use prisms, in which two refracting surfaces inclined at an angle.
iii. Commonly used prisms have three rectangular surfaces forming a triangle.
iv. Two of which take part in refraction at a time. The one, not involved in refraction is called base of the prism.
v. Any section of prism perpendicular to the base is called principal section of the prism. Commonly all the rays considered during refraction lie in this plane.
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 15

Question 33.
Draw neat labelled diagrams showing refraction of a monochromatic light and white light through a prism.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 16

Question 34.
For a prism prove that i + e = A + δ where the symbols have their usual meanings.
Answer:
i. Consider a principal section ABC of a prism of absolute refractive index n kept in air as shown in figure.

ii. Let A be the refracting angle of prism and surface BC be the base.

iii. A monochromatic ray PQ obliquely strikes first reflecting surface AB such that, angle of incidence ∠PQM at Q is i.
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 17

iv. After refraction at Q, the ray deviates towards the normal and strikes second refracting surface AC at R which is the point of emergence.

v. Angles of refraction at Q (∠NQR) and at R (∠QRN) are r1 and r2 respectively.

vi. After R. the ray deviates away from normal and finally emerges along RS making e as the angle of emergence.

vii. Emergent ray RS meets an extended incident ray QT at X if traced backward. In this case, ∠TXS gives the angle of deviation.

viii. From figure,
∠AQN = ∠ARN = 90°
∴ From quadrilateral AQNR,
A + ∠QNR = 180° ………. (1)
From ∆ QNR,
r1 + r2 + ∠QNR = 180° ………. (2)
∴ A = r1 + r2 ……… (3)

ix. Angle δ forms an exterior angle for ∆ XQR.
∴ ∠XQR + ∠XRQ = δ
∴ (i – r1) + (e – r2) = δ
∴ (i + e) – (r1 + r2) = δ
From equation (3),
δ = i + e – A
∴ i + .e = A + δ

Question 35.
Explain δ versus i curve for refraction of light through a prism.
Answer:
i. Variation of angle of incidence i with angle of deviation δ is as shown in figure.
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 18

ii. It shows that, with increasing values of i, the angle of deviation δ decreases initially to a certain minimum (δm) value and then increases.

iii. The curve shown in the figure is not a symmetric parabola, but the slope in the part towards right is less.

iv. Except at δ = δm there are two values of i for any given δ. From principle of reversibility of light, we can conclude that if one of these values is i, the other must be e and vice versa. Thus at δ = δm, we have i = e.

Question 36.
Show that, at condition of minimum deviation, n = \(\frac{\sin \left(\frac{\mathbf{A}+\boldsymbol{\delta}_{\mathrm{m}}}{\mathbf{2}}\right)}{\sin \left(\frac{\mathbf{A}}{\mathbf{2}}\right)}\)
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 19
i. For every angle of deviation except angle of minimum deviation, there are two values of angle of incidence.

ii. However, at angle of minimum deviation there is only one corressponding angle of incidence.

iii. From principle of reversibility in path PQRS, the values of i and e are interchangeable for every δ. Thus, at minimum deviation, i = e.

iv. This implies the angles of refraction r1 and r2 are also equal. Also, A = r1 + r2
∴ A = 2 r i.e., r = \(\frac {A}{2}\) ……. (1)

v. In case of minimum deviation, QR is parallel to base BC and the figure is symmetric.

vi. Using i + e = A + δ,
i + i = A + δm
∴ i = \(\frac {A+δ_m}{2}\) …………….(2)

vii. According to Snell’s law,
n = \(\frac {sin i}{sin r}\)
Thus, using equations (1) and (2),
n = \(\frac{\sin \left(\frac{\mathrm{A}+\delta_{\mathrm{m}}}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)}\)
This is the prism formula.

Question 37.
For grazing emergence of a ray in a prism, find out minimum possible values for angle of incidence (i) and angle of refraction (r1) for commonly used glass prism.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 20
i. At the point of emergence in prism, the ray travels from a denser medium into rarer.
Thus, if r2 = sin-1 (\(\frac {1}{n}\)) is the critical angle, the angle of emergence e = 90°. This is called grazing emergence.

ii. Angle of prism A is constant for a given prism and A = r1 + r2. Hence the corresponding r1 and i will have their minimum possible values.

iii. For commonly used glass prisms,
n = 1.5 (r2)max = sin-1 (\(\frac {1}{n}\)) = sin-1 (\(\frac {1}{1.5}\)) = 41.49°

iv. If, prism is symmetric (equilateral),
A = 60°
∴ r1 = 60° – 41°49′ = 18°11′
∴ n = 1.5 = \(\frac{\sin \left(\mathrm{i}_{\min }\right)}{\sin \left(18^{\circ} 11^{\prime}\right)}\)
sin (imin) = 1.5 × sin (18°11′)
∴ iimin = 27°55′ ≅ 28°.

Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics

Question 38.
Derive the formula for angle of deviation for thin prisms.
OR
Show that in a thin prism, for small angles of incidence, angle of deviation is constant (independent of angle of incidence).
Answer:
For thin prisms (refracting angle < sin 10°). sin θ ≈ θ
∴ Refractive index, n = \(\frac{\sin \mathrm{i}}{\sin \mathrm{r}_{1}} \approx \frac{\mathrm{i}}{\mathrm{r}_{1}}\)
Also n = \(\frac{\sin e}{\sin \mathrm{r}_{2}} \approx \frac{\mathrm{e}}{\mathrm{r}_{2}}\)
∴ i ≈ n r1 and e ≈ nr2

ii. Substituting this in, i + e = A + δ, we get,
i + e = n (r1 + r2) = nA = A + δ
∴ S = A(n – 1)
A and n are constant for a given prism. Thus, for a thin prism, for small angles of incidence, angle of deviation is constant (independent of angle of incidence).

Question 39.
Give the expression for mean deviation for a beam of white light.
Answer:
For a beam of white light, yellow colour is practically chosen to be the mean colour for violet and red.
This gives mean deviation as,
δVR = \(\frac{\delta_{\mathrm{V}}+\delta_{\mathrm{R}}}{2}\) ≈ δY = A(nY – 1)
Where, nY = refractive index for yellow colour.

Question 40.
A fine beam of white light is incident upon the longer side of a plane parallel glass slab of breadth 5 cm at angle of incidence 60°. Calculate lateral deviation of red and violet rays and lateral dispersion between them as they emerge from the opposite side. Refractive indices of the glass for red and violet are 1.51 and 1.53 respectively.
Answer:
As shown in figure,
VM = LV = lateral deviation for violet colour,
RT = LR = lateral deviation for red colour,
∴ Lateral dispersion between these colours, LVR = LV – LR
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 21
∴ rR = sin-1 (0.5735) ≈ 35°
Similarly,
sin rV = \(\frac{\sin 60^{\circ}}{1.53}=\frac{\sqrt{3}}{2 \times 1.53}=\frac{1.732}{3.06}\)
= antilog {log (1.732) – log (3.06)}
= antilog {0.2385 – 0.4857}
= antilog 11.7528}
= 0.5659
∴ sin rV = 0.566
∴ rV = sin-1 (0.566) = 34°28′
∴ Angle of deviation for red colour
= i – rR = 60° – 35° = 25°
and that for violet colur = i – rV = 60° – 34°28′
= 25°32′
From figure, in ∆ANR,
AR = \(\frac{\mathrm{AN}}{\cos \mathrm{r}_{\mathrm{R}}}=\frac{5}{\cos \left(35^{\circ}\right)}=\frac{5}{0.8192}\) = 6.104 cm
Similarly ∆ANV,
AV = \(\frac{\mathrm{AN}}{\cos \mathrm{r}_{\mathrm{V}}}=\frac{5}{\cos \left(34^{\circ} 28^{\prime}\right)}=\frac{5}{0.8244}\) = 6.065 cm
∴ For red colour, LR = RT = AR [sin(i – rR)]
= AR [sin (25°)]
= 6.104 × 0.4226
= 2.58 cm
For violet colour, LV = VM
= AV [sin (i – rV)]
= AV × sin (25° 32′)
= 6.065 × 0.431
= 2.61 cm
∴ LVR = LV – LR = 2.61 – 2.58 = 0.03 cm
= 0.3 mm

Question 41.
For a glass (n = 1.5) prism having refracting angle 60°, determine the range of angle of incidence for which emergent ray is possible from the opposite surface and the corresponding angles of emergence. Also calculate the angle of incidence for which i = e. How much is the corresponding angle of minimum deviation?
Answer:
For an equilateral prism of glass, the minimum angle of incidence for which the emergent ray just emerges is imin = 27° 55′. Corresponding angle of emergence is, emax = 90°.
From the principle of reversibility of light, imax = 90° and emin = 27°55′
Also, for equilateral glass prism at minimum deviation,
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 22
Also, from prism formula,
i + e = A + δ
At minimum deviation,
∴ i + i = 60 + 37°10′ = 97°10′
∴ i = 48°35′

Question 42.
For a dense flint glass prism of refracting angle 10°, obtain angular deviation for extreme colours and dispersive power of dense flint glass. (nred = 1.712, nviolet = 1.792)
Answer:
Given: A = 10°, nR = 1.712, nV = 1.792
To find:
i. Angular deviation for extreme colours (δV and δR)
ii. Dispersive power of flint glass (ω)
Formulae:
i. δ = A(n – 1)
ii. ω = \(\frac{\delta_{\mathrm{V}}-\delta_{\mathrm{R}}}{\left(\frac{\delta_{\mathrm{V}}+\delta_{\mathrm{R}}}{2}\right)}\)
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 23

Question 43.
The refractive indices of the material of the prism for red and yellow colour are 1.620 and 1.635 respectively. Calculate the angular dispersion and dispersive power, if refracting angle is 8°.
Solution:
Given: nR = 1.620, nY = 1.635, A = 8°
To find:
i. Angular dispersion (δV – δR)
ii. Dispersive power (ω)
Formulae:
i. δv – δr = A(nV – nR)
ii. ω = \(\frac{\mathrm{n}_{\mathrm{V}}-\mathrm{n}_{\mathrm{R}}}{\mathrm{n}_{\mathrm{Y}}-1}\)
Calculation: Since, nY = \(\frac {n_V+n_R}{2}\)
∴ nV = 2nY – nR
nV = 2 × 1.635 – 1.620 = 3.27 – 1.620
∴ nV = 1.65
From formula (i),
δV – δR = 8(1.65 – 1.620)
= 8 × 0.03 = 0.24°
∴ δV – δR = 0.24°
From formula (ii),
ω = \(\frac{1.65-1.620}{1.635-1}=\frac{0.03}{0.635}\) = 0.0472

Question 44.
What could be the possible reasons for the upward bending of the light ray during hot days?
Answer:
Possible reasons for the upward bending at the road could be:
i. Angle of incidence at the road is glancing. At glancing incidence, the reflection coefficient is very large which causes reflection.
ii. Air almost in contact with the road is not steady. The non-uniform motion of the air bends the ray upwards and once it has bent upwards, it continues to do so.

Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics

Question 45.
State some properties of rainbow.
Answer:

  1. It is seen during rains and on the opposite side of the Sun.
  2. It is seen only during mornings and evenings and not throughout the day.
  3. In the commonly seen rainbow red arch is outside and violet is inside.
  4. In the rarely occurring concentric secondary rainbow, violet arch is outside and red is inside.
  5. It is in the form of arc of a circle.
  6. Complete circle can be seen from a higher altitude, i.e. from an aeroplane.
  7. Total internal reflection is not possible in this case.

Question 46.
Why is total internal reflection not possible during formation of a rainbow ?
Answer:
i. For total internal reflection, the angle of incidence in the denser medium must be greater than critical angle for the given pair of media.

ii. The relative refractive index of air with the water drop is just less than 1 and hence the critical angle is almost equal to 90°.

iii. Angle of incidence i in air, at the water drop, can’t be greater than 90°.
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 24

iv. As a result, angle of refraction r in water will always be less than the critical angle.

v. The figures shown indicates that this angle r itself acts as an angle of incidence at any point for one or more internal reflections. But this does not indicate the total internal reflection.

Question 47.
Rainbow is seen only for a definite angle range with respect to the ground. Justify.
Answer:
i. For clear visibility of rainbow, a beam must have enough intensity.

ii. The curve for angle of deviation and angle of incidence is almost parallel to X-axis near minimum deviation i.e., for majority of angles of incidence in this range, the angle of deviation is nearly the same and those rays form a beam of enough intensity.

iii. Rays beyond this range suffer wide angular dispersion and thus will not have enough intensity for visibility. Hence, the rainbow is seen only for a definite angle range with respect to the ground for which the intensity of the beam is enough for the visibility.

Question 48.
How is the range of angles for which rainbows can be observed calculated?
Answer:
i. Angle of deviation for the final emergent ray, can be shown to be equal to δ = π + 2i – 4r for primary rainbow and δ = 2π + 2i – 6r for the secondary rainbow.
ii. Using these relations along with Snell’s law, sin i = n sin r, derivatives of angle of deviation (δ) is obtained.
iii. Second derivative of δ comes out to be negative, which shows that it is the minima condition.
iv. Equating first derivative to zero corresponding values of i and r are obtained. Thus, from the figures shown, the corresponding angles θR and θV at the horizontal are obtained. These angles give the visible angular position for the rainbow.

Question 49.
When can one see complete circle of a rainbow? Explain in detail.
Answer:
i. Figure given below illustrates formation of primary and secondary rainbows with their common centre O. It is the point where the line joining the sun and the observer meets the Earth when extended.

ii. P is location of the observer. Different colours of rainbows are seen on arches of cones of respective angles.

iii. Smallest half angle refers to the cone of violet colour of primary rainbow, which is 41°.
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 25

iv. As the Sun rises, the relative position of common centre of the rainbows with respect to observer shifts down. Hence as the Sun comes up, smaller and smaller part of the rainbows will be seen. If the Sun is above 41°, violet arch of primary rainbow cannot be seen.

v. Beyond 53°, no rainbows will be visible. That is why rainbows are visible only during mornings and evenings and in the shape of a bow.

vi. However, if observer moves up (may be in an aeroplane), the line PO itself moves up making lower part of the arches visible. After a certain minimum elevation, entire circle for all the cones can be visible.

Question 50.
Define following terms:
i. Longitudinal chromatic aberration
ii. Circle of least confusion
iii. Transverse chromatic aberration
Answer:
i. Longitudinal chromatic aberration:
Due to different refractive indices and angle of deviations, violet and red colours of a white light converge at different focal points, fV and fR. The distance between fV and fR is measured as the longitudinal chromatic aberration.

ii. Circle of least confusion:
In presence of aberration the image is not a single point but always a circle. At particular location on the screen, this circle has minimum diameter. This is called circle of least confusion.

iii. Transverse chromatic aberration:
Radius of the circle of least confusion is called the transverse chromatic aberration.

Question 51.
After Cataract operation, a person is recommended with concavo-convex spectacles of curvatures 10 cm and 50 cm. Crown glass of refractive indices 1.51 for red and 1.53 for violet colours is used for this. Calculate the lateral chromatic aberration occurring due to these glasses.
Answer:
For a concavo-convex lens, with convex shape facing the object, both the radii of curvature are positive as shown in the figure.
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 26
= (1.53 – 1) × 0.08 = 0.0424
∴ fv = 23.58 cm
∴ Longitudinal (lateral) chromatic aberration
= fV – fR = 24.51 – 23.58 = 0.93 cm

Question 52.
Why do we need optical instruments for?
Answer:
i. Due to the limitation for focusing the eye lens it is not possible to take an object closer than a certain distance. This distance is called least distance of distance vision D. For a normal, unaided human eye D = 25cm.
ii. If an object is brought closer than this, we cannot see it clearly.
iii. If an object is too small the corresponding visual angle from 25 cm is not enough to see it and if we bring it closer than that, its image on the retina is blurred.
iv. Also, the visual angle made by cosmic objects far away from us such as stars is too small to make out minor details and we cannot bring those closer.
v. In such cases we need optical instruments like microscope and telescopes to observe these things clearly.

Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics

Question 53.
A convex lens has focal length of 2.0 cm. Find its magnifying power if image is formed at DDV.
Answer:
Given: f = 2 cm, v = D = 25 cm
To find: Magnifying power (M.P.)
Formula: M.P = 1 + \(\frac {D}{f}\)
Calculation:
From formula,
M.P = 1 + \(\frac {25}{2}\)
M.P. = 1 + 12.5 = 13.5

Question 54.
A magnifying glass of focal length 10 cm is used to read letters of thickness 0.5 mm held 8 cm away from the lens. Calculate the image size. How big will the letters appear? Can you read the letters if held 5 cm away from the lens? If yes, of what size would the letters appear? If no, why not?
Answer:
Given that, f = +10 cm, u = -8 cm,
From thin lens formula,
\(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\)
∴ \(\frac{1}{10}=\frac{1}{\mathrm{v}}-\frac{1}{-8}\)
∴ v = 40 cm
Magnification of a lens is,
m = \(\frac {v}{u}\) = \(\frac {Object size h-i}{Object size h-0}\)
∴ \(\frac {40}{8}\) = \(\frac {h_1}{0.5}\)
∴ h1 = 2.5 cm
This implies the height of the image is 5 times that of the object.
Magnifying power,
M = \(\frac {D}{u}\) = \(\frac {25}{8}\) = 3.125
∴ Image will appear to be 3.125 times bigger,
i.e., 3.125 × 0.5 = 1.5625 cm
For u = -5 cm, v will be -10 cm
For an average human being to see clearly, the image must be at or beyond 25 cm. Thus it will not possible to read the letters if held 5 cm away from the lens.

Question 55.
A compound microscope has a magnification of 15. If the object subtends an angle of 0.5° to eye, what will be the angle subtended by the image at the eye?
Answer:
Given: M.P = 15, α = 0.5°
To Find: Angle(β)
Formula: M.P = \(\frac {β}{α}\)
Calculation:
From formula,
β = M.P × α = 15 × 0.5 = 7.5°

Question 56.
A compound microscope has a magnifying power of 40. Assume that the final image is formed at DDV(25 cm). If the focal length of eyepiece 10 cm, calculate the magnification produced by objective.
Answer:
M.P = 40, D = 25 cm, fe = 10 cm
To Find: Magnification (m0)
Formula: M.P = m0 × Me
Calculation:
From the formula,
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 27

Question 57.
The pocket microscope used by a student consists of eye lens of focal length 6.25 cm and objective of focal length 2 cm. At microscope length 15 cm, the final image appears biggest. Estimate distance of the object from the objective and magnifying power of the microscope.
Answer:
Given: fe = 6.25 cm, f0 = 2 cm, L = 15 cm
As image appears biggest, Ve = -25 cm.
To find:
i. Distance of object from objective (u0)
ii. Magnifying power (M)
Formula:
i. \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\)
ii. L = |v0| + |ue|
iii. M = \(\left(\frac{v_{o}}{u_{o}}\right)\left(\frac{D}{u_{e}}\right)\)
Calculation: For eyelens, using formula (i),
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 28
∴ ue = 5 cm
From formula (ii),
|v0| = L – |ue|
= 15 – 5 = 10 cm
Using formula (i) for objective,
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 29

Question 58.
Focal length of the objective of an astronomical telescope is 1 m. Under normal adjustment, length of the telescope is 1.05 m. Calculate focal length of the eyepiece and magnifying power under normal adjustment.
Answer:
Given: f0 = 1 m, L = 1.05 m
To find:
i. Focal length of eyepiece (fe)
Magnifying power under normal adjustment (M)
Formula:
i. L = f0 + fe
ii. M = \(\frac {f_0}{f_e}\)
Calculation. From formula (i),
fe = L – f0 = 1.05 – 1 = 0.05 m
From formula (ii),
M = \(\frac {1}{0.05}\) = 20

Question 59.
Magnifying power of an astronomical telescope is 12 and the image is formed at D.D.V. If the focal length of the objective is 90 cm, what is the focal length of the eyepiece?
Answer:
Given: M.P = 12, v = D, f0 = 90 cm,
To find: Focal length of eye piece (fe)
Formula: M.P = \(\frac {f_0}{f_e}\) (1 + \(\frac {f_e}{D}\))
Calculation:
From formula.
12 = \(\frac {90}{f_e}\) (1 + \(\frac {f_e}{25}\))
∴ fe = 10.71 cm

Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics

Question 60.
Two convex lenses of an astronomical telescope have focal length 1.3 m and 0.05 m respectively. Find the magnifying power and the length of the telescope.
Answer:
Given: f0 = 1.3 m, fe = 0.05 m
To find:
i. Magnifying power of telescope (M.P.)
ii. Length of telescope (L)
Formulae:
i. M.P = \(\frac {f_0}{f_e}\)
ii. L = f0 + fe
Calculation: From formula (i),
M.P = \(\frac {1.3}{0.05}\) = 26
From formula (ii),
L = 1.3 + 0.05 = 1.35 m

Question 61.
What is the angle of deviation of reflected ray if ray of light is incident on a plane mirror at an incident angle θ?
Answer: When a ray of light is incident on a plane mirror at an angle θ, the reflected ray gets deviated by an angle of (π – 2θ).

Question 62.
Does nature of the image depend upon size of the mirror?
Answer:
No, nature of the image is independent of size of the mirror.

Question 63.
If a convex mirror is held in air and then dipped in oil, then what will be the change in its focal length?
Answer:
Focal length of spherical mirrors are independent of the medium.

Question 64.
When ray of light falls normally on a mirror, its angle of incidence is 90°. True or false? Justify your answer.
Answer:
False, when light falls normally on a mirror, its angle of incidence is zero degree.

Question 65.
In one of the performances, a magician keeps a gold ring beneath a thick glass slab (µ = \(\frac {3}{2}\)) Then he keeps a flask filled with water (µ = \(\frac {4}{3}\)), over the slab. Now when spectators one by one observe from the open end of the flask, the ring disappears at a certain angle of viewing.
i. What could be the reason behind the disappearance?
ii. At what angle of viewing the ring vanishes?
Answer:
i. The ring disappears due to total internal reflection of the light at water-air interface.

ii. aµw = \(\frac {4}{3}\)
sin ic = \(\frac {1}{µ}\)
∴ ic = sin-1 (\(\frac {1}{_aµ_w}\)) = sin-1 (\(\frac {3}{4}\)) = 48.6°
Hence, for angle of viewing for which angle of incidence of ring from water to air is greater that 48.6°, the ring will vanish.

Question 66.
Why dispersion of light is not observed in glass slab but it is observed in prism?
Answer:
When a light passes from one medium to another, at one interface, it changes its speed. The glass slab and prism both have two glass-air interfaces. Hence, the light undergoes refraction twice in both the cases. When the two interfaces are parallel to each other, although the colours are separated at first interface, they all travel the same path after refracting from second interface. However, in prism, the two interfaces are not parallel. Therefore, the colours separated at first interface do not travel the same path after second refraction but emerge out at different wavelengths producing spectrum.

Question 67.
A prism manufacturer is planning to build a dispersive prism out of following materials with the refracting angles as given.
i. Glass (µ = 1.5), A = 60°
ii. Plastic (µ = 1.4), A = 90°
iii. Fluorite (µ = 1.45), A = 64°
If he desires to give the prism following relations of i and δ, then which of the above combinations can be used to construct the prism?
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 30
Answer:
From the given values of i and δ, δm = 37°
From prism formula, µ = \(\frac{\sin \left(\frac{\mathrm{A}+\delta_{\mathrm{m}}}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)}\)

i. For Glass (µ = 1.5), A = 60°;
µ = \(\frac{\sin \left(\frac{60^{\circ}+37^{\circ}}{2}\right)}{\sin \left(30^{\circ}\right)}\)
= 1.5
Hence, this combination can be used for fabricating the desired prism.

ii. For Plastic (µ = 1.4), A = 90°;
µ = \(\frac{\sin \left(\frac{90^{\circ}+37^{\circ}}{2}\right)}{\sin \left(45^{\circ}\right)}\)
= 1.26
As Ppiastic = 1-4, this combination cannot be used.

iii. For Fluorite (µ = 1.45), A = 64°
µ = \(\frac{\sin \left(\frac{64^{\circ}+37^{\circ}}{2}\right)}{\sin \left(32^{\circ}\right)}\)
= 1.45
Hence, this combination can also be used for fabrication of prism.

Question 68.
Find the refractive index of material of following prism if the ray of light incident at angle 45° suffers minimum deviation through the prism.
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 31
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics 32
A = 60°
Also, ray of light suffers minimum deviation.
∴ 2i = A + δm
∴ δm = 2i – A = 90° – 60° = 30°
From prism formula,
µ = \(\frac{\sin \left(\frac{\mathrm{A}+\delta_{\mathrm{m}}}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)}\)
= \(\frac{\sin \left(\frac{60^{\circ}+30^{\circ}}{2}\right)}{\sin \left(30^{\circ}\right)}\)
= √2
Hence, refractive index of material of prism is √2.

Multiple Choice Questions

Question 1.
Time taken by light to cross a glass slab of thickness 4 mm and refractive index 3 is
(A) 4 × 10-11 s
(B) 2 × 10-11 ns
(C) 16 × 10-11 s
(D) 8 × 10-10 s
Answer:
(A) 4 × 10-11 s

Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics

Question 2.
If mirrors are inclined to each other at an angle of 90°, the total number of images seen for a symmetric position of an object will be
(A) 3
(B) 4
(C) 5
(D) 3 or 4
Answer:
(A) 3

Question 3.
In case of a convex mirror, the image formed is
(A) always on opposite side, virtual, erect.
(B) always on the same side, virtual, erect.
(C) always on opposite side, real, inverted.
(D) dependent on object distance.
Answer:
(A) always on opposite side, virtual, erect.

Question 4.
A glass slab is placed in the path of a beam of convergent light. The point of convergence of light
(A) moves towards the glass slab.
(B) moves away from the glass slab.
(C) remains at the same point.
(D) undergoes a lateral shift.
Answer:
(A) moves towards the glass slab.

Question 5.
For a person seeing an object placed in optically rarer medium,
(A) apparent depth of the object is more than real depth
(B) apparent depth is smaller than the real depth.
(C) apparent depthe might be smaller or greater depending on the position of the person.
(D) nothing can be concluded about the depth of object from given data.
Answer:
(A) apparent depth of the object is more than real depth

Question 6.
Light travels from a medium of refractive index µ1 to another of refractive index µ21 > µ2). For total internal reflection of light, which is NOT true?
(A) Light must travel from medium of refractive index µ1 to µ2.
(B) Angle of incidence must be greater than the critical angle.
(C) There is no refraction of light.
(D) Light must travel from the medium of refractive index µ2 to µ1.
Answer:
(D) Light must travel from the medium of refractive index µ2 to µ1.

Question 7.
Optical fibre is based on which of the following phenomenon?
(A) Reflection.
(B) Refraction.
(C) Total internal reflection.
(D) Dispersion.
Answer:
(C) Total internal reflection.

Question 8.
Commonly used glass have refractive index of 1.5. What is the critical angle for such glass?
(A) 49°
(B) 42°
(C) 45°
(D) 40°
Answer:
(B) 42°

Question 9.
If the refractive index of water is 4/3 and that of glass slab is 5/3. Then the critical angle of incidence for which a light ray tending to go from glass to water is totally reflected, is
(A) sin-1 (\(\frac {3}{4}\))
(B) sin-1 (\(\frac {3}{5}\))
(C) sin-1 (\(\frac {2}{3}\))
(D) sin-1 (\(\frac {4}{5}\))
Answer:
(D) sin-1 (\(\frac {4}{5}\))

Question 10.
While deriving prism formula, which of the following condition is NOT satisfied?
(A) I = e
(B) r1 = r2
(C) r = \(\frac {A}{2}\)
(D) δm = i + e + r
Answer:
(D) δm = i + e + r

Question 11.
If the critical angle for the material of a prism is C and the angle of the prism is A, then there will be no emergent ray when
(A) A < 2 C
(B) A = 2 C
(C) A > 2 C
(D) A < \(\frac {C}{2}\)
Answer:
(C) A > 2 C

Maharashtra Board Class 11 Physics Important Questions Chapter 9 Optics

Question 12.
Chromatic aberrations is caused due to
(A) spherical shape of lens
(B) spherical shape of mirrors
(C) angle of deviation for violet light being more than that for red light.
(D) refractive index for violet light being less than that for red light
Answer:
(C) angle of deviation for violet light being more than that for red light.

Question 13.
In normal adjustment, magnifying powser of a astronomical telescope is given by
(A) \(\frac {D}{f_0}\) \(\frac {L}{f_e}\)
(B) \(\frac {L}{D}\) \(\frac {f_e}{f_0}\)
(C) \(\frac {f_0}{f_e}\)
(D) \(\frac {f_e}{f_0}\)
Answer:
(C) \(\frac {f_0}{f_e}\)

Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 9 Optics Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 9 Optics

1. Multiple Choise Questions

Question 1.
As per recent understanding light consists of
(A) rays
(B) waves
(C) corpuscles
(D) photons obeying the rules of waves
Answer:
(D) photons obeying the rules of waves

Question 2.
Consider the optically denser lenses P, Q, R and S drawn below. According to Cartesian sign convention which of these have positive focal length?
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 1
(A) OnlyP
(B) Only P and Q
(C) Only P and R
(D) Only Q and S
Answer:
(B) Only P and Q

Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics

Question 3.
Two plane mirrors are inclined at angle 40° between them. Number of images seen of a tiny object kept between them is
(A) Only 8
(B) Only 9
(C) 8 or 9
(D) 9 or 10
Answer:
(C) 8 or 9

Question 4.
A concave mirror of curvature 40 cm, used for shaving purpose produces image of double size as that of the object. Object distance must be
(A) 10 cm only
(B) 20 cm only
(C) 30 cm only
(D) 10 cm or 30 cm
Answer:
(D) 10 cm or 30 cm

Question 5.
Which of the following aberrations will NOT occur for spherical mirrors?
(A) Chromatic aberration
(B) Coma
(C) Distortion
(D) Spherical aberration
Answer:
(A) Chromatic aberration

Question 6.
There are different fish, monkeys and water of the habitable planet of the star Proxima b. A fish swimming underwater feels that there is a monkey at 2.5 m on the top of a tree. The same monkey feels that the fish is 1.6 m below the water surface. Interestingly, height of the tree and the depth at which the fish is swimming are exactly same. Refractive index of that water must be
(A) 6/5
(B) 5/4
(C) 4/3
(D) 7/5
Answer:
(B) 5/4

Question 7.
Consider following phenomena/applications: P) Mirage, Q) rainbow, R) Optical fibre and S) glittering of a diamond. Total internal reflection is involved in
(A) Only R and S
(B) Only R
(C) Only P, R and S
(D) all the four
Answer:
(A) Only R and S

Question 8.
A student uses spectacles of number -2 for seeing distant objects. Commonly used lenses for her/his spectacles are
(A) bi-concave
(B) piano concave
(C) concavo-convex
(D) convexo-concave
Answer:
(A) bi-concave

Question 9.
A spherical marble, of refractive index 1.5 and curvature 1.5 cm, contains a tiny air bubble at its centre. Where will it appear when seen from outside?
(A) 1 cm inside
(B) at the centre
(C) 5/3 cm inside
(D) 2 cm inside
Answer:
(B) at the centre

Question 10.
Select the WRONG statement.
(A) Smaller angle of prism is recommended for greater angular dispersion.
(B) Right angled isosceles glass prism is commonly used for total internal reflection.
(C) Angle of deviation is practically constant for thin prisms.
(D) For emergent ray to be possible from the second refracting surface, certain minimum angle of incidence is necessary from the first surface.
Answer:
(A) Smaller angle of prism is recommended for greater angular dispersion.

Question 11.
Angles of deviation for extreme colours are given for different prisms. Select the one having maximum dispersive power of its material.
(A) 7°, 10°
(B) 8°, 11°
(C) 12°, 16°
(D) 10°, 14°
Answer:
(A) 7°, 10°

Question 12.
Which of the following is not involved in formation of a rainbow?
(A) refraction
(B) angular dispersion
(C) angular deviation
(D) total internal reflection
Answer:
(D) total internal reflection

Question 13.
Consider following statements regarding a simple microscope:
(P) It allows us to keep the object within the least distance of distinct vision.
(Q) Image appears to be biggest if the object is at the focus.
(R) It is simply a convex lens.
(A) Only (P) is correct
(B) Only (P) and (Q) are correct
(C) Only (Q) and (R) are correct
(D) Only (P) and (R) are correct
Answer:
(D) Only (P) and (R) are correct

2. Answer the following questions.

Question 1.
As per recent development, what is the nature of light? Wave optics and particle nature of light are used to explain which phenomena of light respectively?
Answer:

  1. As per recent development, it is now an established fact that light possesses dual nature. Light consists of energy carrier photons. These photons follow the rules of electromagnetic waves.
  2. Wave optics explains the phenomena of light such as, interference, diffraction, polarisation, Doppler effect etc.
  3. Particle nature of light can be used to explain phenomena like photoelectric effect, emission of spectral lines, Compton effect etc.

Question 2.
Which phenomena can be satisfactorily explained using ray optics?
Answer:
Ray optics or geometrical optics: Ray optics can be used for understanding phenomena like reflection, refraction, double refraction, total internal reflection, etc.

Question 3.
What is focal power of a spherical mirror or a lens? What may be the reason for using P = \(\frac {1}{f}\) its expression?
Answer:
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 2

  1. Converging or diverging ability of a lens or of a mirror is defined as its focal power.
  2. This implies, more the power of any spherical mirror or a lens, the more is its ability to converge or diverge the light that passes through it.
  3. In case of convex lens or concave mirror, more the convergence, shorter is the focal length as shown in the figure.
  4. Similarly, in case of concave lens or convex mirror, more the divergence, shorter is the focal length.
  5. This explains that the focal power of any spherical lens or mirror is inversely proportional to the focal length.
  6. Hence, the expression of focal power is given by the formula, P = \(\frac {1}{f}\).

Question 4.
At which positions of the objects do spherical mirrors produce (i) diminished image (ii) magnified image?
Answer:
i. Amongst the two types of spherical mirrors, convex mirror always produces a diminished image at all positions of the object.

ii. Concave mirror produces diminished image when object is placed:

  • Beyond radius of curvature (i.e., u > 2f)
  • At infinity (i.e., u = ∞)

iii. Concave mirror produces magnified image when object is placed:

  • between centre of curvature and focus (i.e., 2f > u > f)
  • between focus and pole of the mirror (i.e., u < f)

Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics

Question 5.
State the restrictions for having images produced by spherical mirrors to be appreciably clear.
Answer:
i. In order to obtain clear images, the formulae for image formation by mirrors or lens follow the given assumptions:

  • Objects and images are situated close to the principal axis.
  • Rays diverging from the objects are confined to a cone of very small angle.
  • If there is a parallel beam of rays, it is paraxial, i.e., parallel and close to the principal axis.

ii. In case of spherical mirrors (excluding small aperture spherical mirrors), rays farther from the principle axis do not remain parallel to the principle axis. Thus, the third assumption is not followed and the focus gradually shifts towards the pole.

iii. The relation (f = \(\frac {R}{2}\)) giving a single point focus is not followed and the image does not get converged at a single point resulting into a distorted or defective image.

iv. This defect arises due to the spherical shape of the reflecting surface.

Question 6.
Explain spherical aberration for spherical mirrors. How can it be minimized? Can it be eliminated by some curved mirrors?
Answer:
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 3

  1. In case of spherical mirrors (excluding small aperture spherical mirror), The rays coming from a distant object farther from principal axis do not remain parallel to the axis. Thus, the focus gradually shifts towards the pole.
  2. The assumption for clear image formation namely, ‘If there is a parallel beam of rays, it is paraxial, i.e., parallel and close to the principal axis’, is not followed in this case.
  3. The relation f (f = \(\frac {R}{2}\)) giving a single point focus is not followed and the image does not get converged at a single point resulting into a distorted or defective image.
  4. This phenomenon is known as spherical aberration.
  5. It occurs due to spherical shape of the reflecting surface, hence known as spherical aberration.
  6. The rays near the edge of the mirror converge at focal point FM Whereas, the rays near the principal axis converge at point FP. The distance between FM and FP is measured as the longitudinal spherical aberration.
  7. In spherical aberration, single point image is not possible at any point on the screen and the image formed is always a circle.
  8. At a particular location of the screen (across AB in figure), the diameter of this circle is minimum. This is called the circle of least confusion. Radius of this circle is transverse spherical aberration.

Remedies for Spherical Aberration:

  1. Spherical aberration can be minimized by reducing the aperture of the mirror.
  2. Spherical aberration in curved mirrors can be completely eliminated by using a parabolic mirror.

Question 7.
Define absolute refractive index and relative refractive index. Explain in brief with an illustration for each.
Answer:
i. Absolute refractive index of a medium is defined as the ratio of speed of light in vacuum to that in the given medium.

ii. A stick or pencil kept obliquely in a glass containing water appears broken as its part in water appears to be raised.

iii. As the speed of light is different in two media, the rays of light coming from water undergo refraction at the boundary separating two media.

iv. Consider speed of light to be v in water and c in air. (Speed of light in air ~ speed of light in vacuum)
∴ refractive index of water = \(\frac {n_w}{n_s}\) = \(\frac {n_w}{n_{vacuum}}\) = \(\frac {c}{v}\)

v. Relative refractive index of a medium 2 is the refractive index of medium 2 with respect to medium 1 and it is defined as the ratio of speed of light v1 in medium 1 to its speed v1 in medium 2.
∴ Relative refractive index of medium 2,
1n2 = \(\frac {v_1}{v_2}\)

vi. Consider a beaker filled with water of absolute refractive index n1 kept on a transparent glass slab of absolute refractive index n2.

vii. Thus, the refractive index of water with respect to that of glass will be,
nw2 = \(\frac {n_2}{n_1}\) = \(\frac {c/v_2}{c/v_1}\) = \(\frac {v_1}{v_2}\)

Question 8.
Explain ‘mirage’ as an illustration of refraction.
Answer:
i. On a hot clear Sunny day, along a level road, there appears a pond of water ahead of the road. Flowever, if we physically reach the spot, there is nothing but the dry road and water pond again appears some distance ahead. This illusion is called mirage.

ii. Mirage results from the refraction of light through a non-uniform medium.

iii. On a hot day the air in contact with the road is hottest and as we go up, it gets gradually cooler. The refractive index of air thus decreases with height. Hot air tends to be less optically dense than cooler air which results into a non-uniform medium.

iv. Light travels in a straight line through a uniform medium but refracts when traveling through a non-uniform medium.

v. Thus, the ray of light coming from the top of an object get refracted while travelling downwards into less optically dense air and become more and more horizontal as shown in Figure.
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 4

vi. As it almost touches the road, it bends (refracts) upward. Then onwards, upward bending continues due to denser air.

vii. As a result, for an observer, it appears to be coming from below thereby giving an illusion of reflection from an (imaginary) water surface.

Question 9.
Under what conditions is total internal reflection possible? Explain it with a suitable example. Define critical angle of incidence and obtain an expression for it.
Answer:
Conditions for total internal reflection:
i. The light ray must travel from denser medium to rarer medium.

ii. The angle of incidence in the denser medium must be greater than critical angle for the given pair of media.

Total internal reflection in optical fibre:
iii. Consider an optical fibre made up of core of refractive index n1 and cladding of refractive index n2 such that, n1 > n2.

iv. When a ray of light is incident from a core (denser medium), the refracted ray is bent away from the normal.

v. At a particular angle of incidence ic in the denser medium, the corresponding angle of refraction in the rarer medium is 90°.

vi. For angles of incidence greater than ic, the angle of refraction become larger than 90° and the ray does not enter into rarer medium at all but is reflected totally into the denser medium as shown in figure.
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 5

critical angle of incidence and obtain an expression:
i. Critical angle for a pair of refracting media can be defined as that angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90°.

ii. Let n be the relative refractive index of denser medium with respect to the rarer.

iii. Then, according to Snell’s law,
n = \(\frac {n_{denser}}{n_{rarer}}\) = \(\frac {sin r}{sin i_c}\) = \(\frac {sin 90°}{sin i_c}\)
∴ sin (ic) = \(\frac {1}{n}\)

Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics

Question 10.
Describe construction and working of an optical fibre. What are the advantages of optical fibre communication over electronic communication?
Answer:
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 6
Construction:

  1. An optical fibre consists of an extremely thin, transparent and flexible core surrounded by an optically rarer flexible cover called cladding.
  2. For protection, the whole system is coated by a buffer and a jacket.
  3. Entire thickness of the fibre is less than half a mm.
  4. Many such fibres can be packed together in an outer cover.

Working:
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 7

  1. Working of optical fibre is based on the principle of total internal reflection.
  2. An optical signal (a ray of light) entering the core suffers multiple total internal reflections before emerging out after a several kilometres.
  3. The optical signal travels with the highest possible speed in the material.
  4. The emerged optical signal has extremely low loss in signal strength.

Advantages of optical fibre communication over electronic communication:

  1. Broad bandwidth (frequency range): For TV signals, a single optical fibre can carry over 90000 independent signals (channels).
  2. Immune to EM interference: Optical fibre being electrically non-conductive, does not pick up nearby EM signals.
  3. Low attenuation loss: loss being lower than 0.2 dB/km, a single long cable can be used for several kilometres.
  4. Electrical insulator: Optical fibres being electrical insulators, ground loops of metal wires or lightning do not cause any harm.
  5. Theft prevention: Optical fibres do not use copper or other expensive material which are prone to be robbed.
  6. Security of information: Internal damage is most unlikely to occur, keeping the information secure.

Question 11.
Why are prism binoculars preferred over traditional binoculars? Describe its working in brief.
Answer:

  1. Traditional binoculars use only two cylinders. Distance between the two cylinders can’t be greater than that between the two eyes. This creates a limitation of field of view.
  2. A prism binocular has two right angled glass prisms which apply the principle of total internal reflection.
  3. The incident light rays are reflected internally twice giving the viewer a wider field of view. For this reason, prism binoculars are preferred over traditional binoculars.

Working:
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 8

  1. The prism binoculars consist of 4 isosceles, right angled prisms of material having critical angle less than 45°.
  2. The prism binoculars have a wider input range compared to traditional binoculars.
  3. The light rays incident on the prism binoculars, first get total internally reflected by the isosceles, right angled prisms 1 and 4.
  4. These reflected rays undergo another total internal reflection by prisms 2 and 3 to form the final image.

Question 12.
A spherical surface separates two transparent media. Derive an expression that relates object and image distances with the radius of curvature for a point object. Clearly state the assumptions, if any.
Answer:
i. Consider a spherical surface YPY’ of radius curvature R, separating two transparent media of refractive indices n1 and n2 respectively with ni1 < n2.

ii. P is the pole and X’PX is the principal axis. A point object O is at a distance u from the pole, in the medium of refractive index n1.

iii. In order to minimize spherical aberration, we consider two paraxial rays.

iv. The ray OP along the principal axis travels undeviated along PX. Another ray OA strikes the surface at A.
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 9

v. As n1 < n2, the ray deviates towards the normal (CAN), travels along AZ and real image of point object O is formed at I.

vi. Let α, β and γ be the angles subtended by incident ray, normal and refracted ray with the principal axis.
∴ i = (α + β) and r = (β – γ)

vii. As, the rays are paraxial, all the angles can be considered to be very small.
i.e., sin i ≈ i and sin r ≈ r
Angles α, β and γ can also be expressed as,
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 10

viii. According to Snell’s law,
n1 sin (i) = n2 sin (r)
For small angles, Snell’s law can be written
as, n1i = n2r
∴ n1 (α + β) = n2 (β – γ)
∴ (n2 – n1)β = n1α + n2γ
Substituting values of α, β and γ, we get,
(n2 – n1) \(\frac {arc PA}{R}\) = n1(\(\frac {arc PA}{-u}\)) + n2(\(\frac {arc PA}{v}\))
∴ \(\frac {(n_2-n_1)}{R}\) = \(\frac {n_2}{v}\) – \(\frac {n_1}{u}\)

Assumptions: To derive an expression that relates object and image distances with the radius of curvature for a point object, the two rays considered are assumed to be paraxial thus making the angles subtended by incident ray, normal and refracted ray with the principal axis very small.

Question 13.
Derive lens makers’ equation. Why is it called so? Under which conditions focal length f and radii of curvature R are numerically equal for a lens?
Answer:
i. Consider a lens of radii of curvature Ri and R2 kept in a medium such that refractive index of material of the lens with respect to the medium is denoted as n.

ii. Assuming the lens to be thin, P is the common pole for both the surfaces. O is a point object on the principal axis at a distance u from P.

iii. The refracting surface facing the object is considered as first refracting surface with radii R1.

iv. In the absence of second refracting surface, the paraxial ray OA deviates towards normal and would intersect axis at I1. PI1 = V1 is the image distance for intermediate image I1.
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 11
Before reaching I1, the incident rays (AB and OP) strike the second refracting surface. In this case, image I1 acts as a virtual object for second surface.

vii. For second refracting surface,
n2 = 1, n1 = n, R = R2, u = v1 and PI = v
∴ \(\frac{(1-\mathrm{n})}{\mathrm{R}_{2}}=\frac{1}{\mathrm{v}}-\frac{\mathrm{n}}{\mathrm{v}_{1}}-\frac{(\mathrm{n}-1)}{\mathrm{R}_{2}}=\frac{1}{\mathrm{v}}-\frac{\mathrm{n}}{\mathrm{v}_{1}}\) ………… (2)

viii. Adding equations (1) and (2),
(n – 1) \(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)=\frac{1}{v}-\frac{1}{u}\)
For object at infinity, image is formed at focus, i.e., for u = ∞, v = f. Substituting this in above equation,
\(\frac{1}{\mathrm{f}}=(\mathrm{n}-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\) …………. (3)
This equation in known as the lens makers’ formula.

ix. Since the equation can be used to calculate the radii of curvature for the lens, it is called the lens makers’ equation.

x. The numeric value of focal length f and radius of curvature R is same under following two conditions:
Case I:
For a thin, symmetric and double convex lens made of glass (n = 1.5), R1 is positive and R2 is negative but, |R1| = |R2|.
In this case,
\(\frac{1}{\mathrm{f}}=(1.5-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{-\mathrm{R}}\right)=0.5\left(\frac{2}{\mathrm{R}}\right)=\frac{1}{\mathrm{R}}\)
∴ f = R

Case II:
Similarly, for a thin, symmetric and double concave lens made of glass (n = 1.5), R1 is negative and R2 is positive but, |R1| = |R2|.
In this case,
\(\frac{1}{\mathrm{f}}=(1.5-1)\left(\frac{1}{-\mathrm{R}}-\frac{1}{\mathrm{R}}\right)=0.5\left(-\frac{2}{\mathrm{R}}\right)=-\frac{1}{\mathrm{R}}\)
∴ f = -R or |f| = |R|

Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics

3. Answer the following questions in detail.

Question 1.
What are different types of dispersions of light? Why do they occur?
Answer:
i. There are two types of dispersions:
a. Angular dispersion
b. Lateral dispersion

ii. The refractive index of material depends on the frequency of incident light. Hence, for different colours, refractive index of material is different.

iii. For an obliquely incident ray, the angles of refraction are different for each colour and they separate as they travel along different directions resulting into angular dispersion.

iv. When a polychromatic beam of light is obliquely incident upon a plane parallel transparent slab, emergent beam consists of all component colours separated out.

v. In this case, these colours are parallel to each other and are also parallel to their initial direction resulting into lateral dispersion

Question 2
Define angular dispersion for a prism. Obtain its expression for a thin prism. Relate it with the refractive indices of the material of the prism for corresponding colours.
Answer:
i. If a polychromatic beam is incident upon a prism, the emergent beam consists of all the individual colours angularly separated. This phenomenon is known as angular dispersion for a prism.

ii. For any two component colours, angular dispersion is given by,
δ21 = δ2 – δ1

iii. For white light, we consider two extreme colours viz., red and violet.
∴ δVR = δV – δR

iv. For thin prism,
δ = A(n – 1)
δ21 = δ2 – δ1
= A(n2 – 1) – A(n1 – 1) = A(n2 – n1)
where n1 and n2 are refractive indices for the two colours.

v. For white light,
δVR = δV – δR
= A(nV – 1) – A(nR – 1) = A(nV – nR).

Question 3.
Explain and define dispersive power of a transparent material. Obtain its expressions in terms of angles of deviation and refractive indices.
Answer:
Ability of an optical material to disperse constituent colours is its dispersive power.

It is measured for any two colours as the ratio of angular dispersion to the mean deviation for those two colours. Thus, for the extreme colours of white light, dispersive power is given by,
\(\omega=\frac{\delta_{\mathrm{V}}-\delta_{\mathrm{R}}}{\left(\frac{\delta_{\mathrm{V}}+\delta_{\mathrm{R}}}{2}\right)} \approx \frac{\delta_{\mathrm{V}}-\delta_{\mathrm{R}}}{\delta_{\mathrm{Y}}}=\frac{\mathrm{A}\left(\mathrm{n}_{\mathrm{V}}-\mathrm{n}_{\mathrm{R}}\right)}{\mathrm{A}\left(\mathrm{n}_{\mathrm{Y}}-1\right)}=\frac{\mathrm{n}_{\mathrm{V}}-\mathrm{n}_{\mathrm{R}}}{\mathrm{n}_{\mathrm{Y}}-1}\)

Question 4.
(i) State the conditions under which a rainbow can be seen.
Answer:
A rainbow can be observed when there is a light shower with relatively large raindrop occurring during morning or evening time with enough sunlight around.

(ii) Explain the formation of a primary rainbow. For which angular range with the horizontal is it visible?
Answer:
i. A ray AB incident from Sun (white light) strikes the upper portion of a water drop at an incident angle i.

ii. On entering into water, it deviates and disperses into constituent colours. The figure shows the extreme colours (violet and red).
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 12

iii. Refracted rays BV and BR strike the opposite inner surface of water drop and suffer internal reflection.

iv. These reflected rays finally emerge from V’ and R’ and can be seen by an observer on the ground.

v. For the observer they appear to be coming from opposite side of the Sun.

vi. Minimum deviation rays of red and violet colour are inclined to the ground level at θR = 42.8° ≈ 43° and θV = 40.8 ≈ 41° respectively. As a result, in the rainbow, the red is above and violet is below.

(iii) Explain the formation of a secondary rainbow. For which angular range with the horizontal is it visible?
Answer:
i. A ray AB incident from Sun (white light) strikes the lower portion of a water drop at an incident angle i.

ii. On entering into water, it deviates and disperses into constituent colours. The figure shows the extreme colours (violet and red).
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 13

iii. Refracted rays BV and BR finally emerge the drop from V’ and R’ after suffering two internal reflections and can be seen by an observer on the ground.

iv. Minimum deviation rays of red and violet colour are inclined to the ground level at θR ≈ 51° and θV ≈ 53° respectively. As a result, in the rainbow, the violet is above and red is below.

(iv) Is it possible to see primary and secondary rainbow simultaneously? Under what conditions?
Answer:
Yes, it is possible to see primary and secondary rainbows simultaneously. This can occur when the centres of both the rainbows coincide.

Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics

Question 5.
(i) Explain chromatic aberration for spherical lenses. State a method to minimize or eliminate it.
Answer:
Lenses are prepared by using a transparent material medium having different refractive index for different colours. Hence angular dispersion is present.
If the lens is thick, this will result into notably different foci corresponding to each colour for a polychromatic beam, like a white light. This defect is called chromatic aberration.
As violet light has maximum deviation, it is focussed closest to the pole.
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 14
Reducing/eliminating chromatic aberration:

  1. Eliminating chromatic aberrations for all colours is impossible. Hence, it is minimised by eliminating aberrations for extreme colours.
  2. This is achieved by using either a convex and a concave lens in contact or two thin convex lenses with proper separation. Such a combination is called achromatic combination.

(ii) What is achromatism? Derive a condition to achieve achromatism for a lens combination. State the conditions for it to be converging.
Answer:
i. To eliminate chromatic aberrations for extreme colours from a lens, either a convex and a concave lens in contact or two thin convex lenses with proper separation are used.

ii. This combination is called achromatic combination. The process of using this combination is termed as achromatism of a lens.

iii. Let ω1 and ω2 be the dispersive powers of materials of the two component lenses used in contact for an achromatic combination.

iv. Let V, R and Y denote the focal lengths for violet, red and yellow colours respectively.

v. For lens 1, let
K1 = (\(\frac {1}{R_1}\)–\(\frac {1}{R_2}\))1 and K2 = (\(\frac {1}{R_1}\)–\(\frac {1}{R_2}\))2

vi. For the combination to be achromatic, the resultant focal length of the combination must be the same for both the colours,
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 15
This is the condition for achromatism of a combination of lenses.

Condition for converging:
For this combination to be converging, fY must be positive.
Using equation (3), for fY to be positive, (fY)1 < (fY)2 ⇒ ω1 < ω2

Question 6.
Describe spherical aberration for spherical lenses. What are different ways to minimize or eliminate it?
Answer:
i. All the formulae used for image formation by lenses are based on some assumption. However, in reality these assumptions are not always true.

ii. A single point focus in case of lenses is possible only for small aperture spherical lenses and for paraxial rays.

iii. The rays coming from a distant object farther from principal axis no longer remain parallel to the axis. Thus, the focus gradually shifts towards pole.

iv. This defect arises due to spherical shape of the refracting surface, hence known as spherical aberration. It results into a blurred image with unclear boundaries.
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 16

v. As shown in figure, the rays near the edge of the lens converge at focal point FM. Whereas, the rays near the principal axis converge at point FP. The distance between FM and FP is measured as the longitudinal spherical aberration.

vi. In absence of this aberration, a single point image can be obtained on a screen. In the presence of spherical aberration, the image is always a circle.

vii. At a particular location of the screen (across AB in figure), the diameter of this circle is minimum. This is called the circle of least confusion. Radius of this circle is transverse spherical aberration.

Methods to eliminate/reduce spherical aberration in lenses:
i. Cheapest method to reduce the spherical aberration is to use a planoconvex or planoconcave lens with curved side facing the incident rays.

ii. Certain ratio of radii of curvature for a given refractive index almost eliminates the spherical aberration. For n = 1.5, the ratio is
\(\frac {R_1}{R_2}\) = \(\frac {1}{6}\) and for n = 2, \(\frac {R_1}{R_2}\) = \(\frac {1}{5}\)

iii. Use of two thin converging lenses separated by distance equal to difference between their focal lengths with lens of larger focal length facing the incident rays considerably reduces spherical aberration.

iv. Spherical aberration of a convex lens is positive (for real image), while that of a concave lens is negative. Thus, a suitable combination of them can completely eliminate spherical aberration.

Question 7.
Define and describe magnifying power of an optical instrument. How does it differ from linear or lateral magnification?
Answer:
i. Angular magnification or magnifying power of an optical instrument is defined as the ratio of the visual angle made by the image formed by that optical instrument (β) to the visual angle subtended by the object when kept at the least distance of distinct vision (α).

ii. The linear magnification is the ratio of the size of the image to the size of the object.

iii. When the distances of the object and image formed are very large as compared to the focal lengths of the instruments used, the magnification becomes infinite. Whereas, the magnifying power being the ratio of angle subtended by the object and image, gives the finite value.

iv. For example, in case of a compound microscope,
Mmin = \(\frac {D}{f}\) = \(\frac {25}{5}\) = 5 and Mmax = 1 + \(\frac {D}{f}\) = 6
Hence image appears to be only 5 to 6 times bigger for a lens of focal length 5 cm.
For Mmin = \(\frac {D}{f}\) = 5, V = ∞
∴ Lateral magnification (m) = \(\frac {v}{u}\) = ∞
Thus, the image size is infinite times that of the object, but appears only 5 times bigger.

Question 8.
Derive an expression for magnifying power of a simple microscope. Obtain its minimum and maximum values in terms of its focal length.
Answer:
i. Figure (a) shows visual angle a made by an object, when kept at the least distance of distinct vision (D = 25 cm). Without an optical instrument this is the greatest possible visual angle as we cannot get the object closer than this.
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 17

ii. Figure (b) shows a convex lens forming erect, virtual and magnified image of the same object, when placed within the focus.

iii. The visual angle p of the object and the image in this case are the same. However, this time the viewer is looking at the image which is not closer than D. Hence the same object is now at a distance smaller than D.

iv. Angular magnification or magnifying power, in this case, is given by
M = \(\frac {Visual angle of theimage}{Visual angle of the object at D}\) = \(\frac {β}{α}\)
For small angles,
M = \(\frac {β}{α}\) ≈ \(\frac {tan(β)}{tan(α)}\) = \(\frac {BA/PA}{BA/D}\) = \(\frac {D}{u}\)

v. For maximum magnifying power, the image should be at D. For thin lens, considering thin lens formula.
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 18

Question 9.
Derive the expressions for the magnifying power and the length of a compound microscope using two convex lenses.
Answer:
i. The final image formed in compound microscope (A” B”) as shown in figure, makes a visual angle β at the eye.
Visual angle made by the object from distance D is α.
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 19
From figure,
tan β = \(\frac {A”B”}{v_c}\) = \(\frac {A’B’}{u_c}\)
and tan α = \(\frac {AB}{D}\)

Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics

Question 10.
What is a terrestrial telescope and an astronomical telescope?
Answer:

  1. Telescopes used to see the objects on the Earth, like mountains, trees, players playing a match in a stadium, etc. are called terrestrial telescopes.
  2. In such case, the final image must be erect. Eye lens used for this purpose must be concave and such a telescope is popularly called a binocular.
  3. Most of the binoculars use three convex lenses with proper separation. The image formed by second lens is inverted with respect to object. The third lens again inverts this image and makes final image erect with respect to the object.
  4. An astronomical telescope is the telescope used to see the objects like planets, stars, galaxies, etc. In this case there is no necessity of erect image. Such telescopes use convex lens as eye lens.

Question 11.
Obtain the expressions for magnifying power and the length of an astronomical telescope under normal adjustments.
Answer:
i. For telescopes, a is the visual angle of the object from its own position, which is practically at infinity.

ii. Visual angle of the final image is p and its position can be adjusted to be at D. However, under normal adjustments, the final image is also at infinity making a greater visual angle than that of the object.
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 20
iii. The parallax at the cross wires can be avoided by using the telescopes in normal adjustments.

iv. Objective of focal length f0 focusses the parallel incident beam at a distance f0 from the objective giving an inverted image AB.

v. For normal adjustment, the intermediate image AB forms at the focus of the eye lens. Rays refracted beyond the eye lens form a parallel beam inclined at an angle β with the principal axis.

vi. Angular magnification or magnifying power for telescope is given by,
M = \(\frac {β}{α}\) ≈ \(\frac {tan(β)}{tan(α)}\) = \(\frac {BA/P_cB}{BA/P_0B}\) = \(\frac {f_0}{f_e}\)

vii. Length of the telescope for normal adjustment is, L = f0 + fe.

Question 12.
What are the limitations in increasing the magnifying powers of (i) simple microscope (ii) compound microscope (iii) astronomical telescope?
Answer:
i. In case of simple microscope
\(\mathrm{M}_{\max }=\frac{\mathrm{D}}{\mathrm{u}}=1+\frac{\mathrm{D}}{\mathrm{f}}\)
Thus, the limitation in increasing the magnifying power is determined by the value of focal length and the closeness with which the lens can be held near the eye.

ii. In case of compound microscope,
M = \(\mathrm{m}_{0} \times \mathrm{M}_{\mathrm{e}}=\frac{\mathrm{v}_{0}}{\mathrm{u}_{\mathrm{o}}} \times \frac{\mathrm{D}}{\mathrm{u}_{\mathrm{e}}}\)
Thus, in order to increase m0, we need to decrease u0. Thereby, the object comes closer and closer to the focus of the objective. This increases v0 and hence length of the microscope. Therefore, mQ can be increased only within the limitation of length of the microscope.

iii. In case of telescopes,
M = \(\frac {f_0}{f_e}\)
Where f0 = focal length of the objective
fe = focal length of the eye-piece
Length of the telescope for normal adjustment is, L = f0 + fe.
Thus, magnifying power of telescope can be increased only within the limitations of length of the telescope.

4. Solve the following numerical examples

Question 1.
A monochromatic ray of light strikes the water (n = 4/3) surface in a cylindrical vessel at angle of incidence 53°. Depth of water is 36 cm. After striking the water surface, how long will the light take to reach the bottom of the vessel? [Angles of the most popular Pythagorean triangle of sides in the ratio 3 : 4 : 5 are nearly 37°, 53° and 90°]
Answer:
From figure, ray PO = incident ray
ray OA = refracted ray
QOB = Normal to the water surface.
Given that,
∠POQ = angle of incidence (θ1) = 53°
Seg OB = 36 cm and nwater = \(\frac {4}{3}\)
From Snell’s law,
n1 sin θ1 = n2 sin θ2
∴ nwater = \(\frac {sinθ_1}{sinθ_2}\)
Or sin θ2 = \(\frac {sinθ_1}{n_{water}}\) = \(\frac {sin(53°)×3}{4}\)
∴ θ2 ~ 37°
ΔOBA forms a Pythagorean triangle with angles 53°, 37° and 90°.
Thus, sides of ΔOBA will be in ratio 3 : 4 : 5 Such that OA forms the hypotenuse. From figure, we can infer that,
Seg OB = 4x = 36 cm
∴ x = 9
∴ seg OA = 5x = 45 cm and
seg AB = 3x = 27 cm.
This means the light has to travel 45 cm to reach the bottom of the vessel.
The speed of the light in water is given by,
v = \(\frac {c}{n}\)
∴ v = \(\frac {3×10^8}{4/3}\) = \(\frac {9}{4}\) × 108 m/s
∴ Time taken by light to reach the bottom of vessel is,
t = \(\frac {s}{v}\) = \(\frac {45×10^{-2}}{\frac {9}{4} × 10^8}\) = 20 × 10-10 = 2 ns or 0.002 µs

Question 2.
Estimate the number of images produced if a tiny object is kept in between two plane mirrors inclined at 35°, 36°, 40° and 45°.
Answer:
i. For θ1 =35°
n1 = \(\frac {360}{θ_1}\) = \(\frac {360}{35}\) = 10.28
As ni is non-integer, N1 = integral part of n1 = 10

ii. For θ2 = 36°
n2 = \(\frac {360}{36}\) = 10
As n2 is even integer, N2 = (n2 – 1) = 9

iii. For θ3 = 40°
n3 = \(\frac {360}{36}\) = 9
As n3 is odd integer.
Number of images seen (N3) = n3 – 1 = 8
(if the object is placed at the angle bisector) or Number of images seen (N3) = n3 = 9
(if the object is placed off the angle bisector)

iv. For θ4 = 45°
n4 = \(\frac {360}{45}\) = 8
As n4 is even integer,
N4 = n4 – 1 = 7

Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics

Question 3.
A rectangular sheet of length 30 cm and breadth 3 cm is kept on the principal axis of a concave mirror of focal length 30 cm. Draw the image formed by the mirror on the same diagram, as far as possible on scale.
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 21
Answer:
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 22
[Note: The question has been modified and the ray digram is inserted in question in order to find the correct position of the image.]

Question 4.
A car uses a convex mirror of curvature 1.2 m as its rear-view mirror. A minibus of cross section 2.2 m × 2.2 m is 6.6 m away from the mirror. Estimate the image size.
Answer:
For a convex mirror,
f = +\(\frac {R}{2}\) = \(\frac {1.2}{2}\) = +0.6m
Given that, a minibus, approximately of a shape of square is at distance 6.6 m from mirror.
i.e., u = -6.6 m
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 23
∴ h2 = 0.183 m
i.e., h2 0.2 m

Question 5.
A glass slab of thickness 2.5 cm having refractive index 5/3 is kept on an ink spot. A transparent beaker of very thin bottom, containing water of refractive index 4/3 up to 8 cm, is kept on the glass block. Calculate apparent depth of the ink spot when seen from the outside air.
Answer:
When observed from the outside air, the light coming from ink spot gets refracted twice; once through glass and once through water.
∴ When observed from water,
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 24
∴ Apparent depth = 2 cm
Now when observed from outside air, the total real depth of ink spot can be taken as (8 + 2) cm = 10 cm.
∴ \(\frac {n_w}{n_{air}}\) = \(\frac {Real depth}{Apparent depth}\)
∴ Apparent depth = \(\frac {10}{4/3}\)
= \(\frac {10×3}{4}\) = 7.5 cm

Question 64.
A convex lens held some distance above a 6 cm long pencil produces its image of SOME size. On shifting the lens by a distance equal to its focal length, it again produces the image of the SAME size as earlier. Determine the image size.
Answer:
For a convex lens, it is given that the image size remains unchanged after shifting the lens through distance equal to its focal length. From given conditions, it can be inferred that the object distance should be u = –\(\frac {f}{2}\)
Also, h1 = 6 cm, v1 = v2
From formula for thin lenses,
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 25

Question 7.
Figure below shows the section ABCD of a transparent slab. There is a tiny green LED light source at the bottom left corner B. A certain ray of light from B suffers total internal reflection at nearest point P on the surface AD and strikes the surface CD at point Question Determine refractive index of the material of the slab and distance DQ. At Q, the ray PQ will suffer partial or total internal reflection?
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 26
Answer:
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 27

As, the light ray undergo total internal reflection at P, the ray BP may be incident at critical angle.
For a Pythagorean triangle with sides in ratio 3 : 4 : 5 the angle opposite to side 3 units is 37° and that opposite to 4 units is 53°.
Thus, from figure, we can say, in ΔBAP
∠ABP = 53°
∠BPN = ic = 53°
∴ nglass = \(\frac {1}{sin_c}\) = \(\frac {1}{sin(53°)}\) ≈ \(\frac {1}{0.8}\) = \(\frac {5}{4}\)
∴ Refractive index (n) of the slab is \(\frac {5}{4}\)
From symmetry, ∆PDQ is also a Pythagorean triangle with sides in ratio QD : PD : PQ = 3 : 4 : 5.
PD = 2 cm ⇒ QD = 1.5 cm.
As critical angle is ic = 53° and angle of incidence at Q, ∠PQN = 37° is less than critical angle, there will be partial internal reflection at Question

Question 8.
A point object is kept 10 cm away from a double convex lens of refractive index 1.5 and radii of curvature 10 cm and 8 cm. Determine location of the final image considering paraxial rays only.
Answer:
Given that, R1 = 10 cm, R2 = -8 cm,
u = -10 cm and n = 1.5
From lens maker’s equation,
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 28

Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics

Question 9.
A monochromatic ray of light is incident at 37° on an equilateral prism of refractive index 3/2. Determine angle of emergence and angle of deviation. If angle of prism is adjustable, what should its value be for emergent ray to be just possible for the same angle of incidence.
Answer:
By Snell’s law, in case of prism,
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 29
For equilateral prism, A = 60°
Also, A= r1 + r2
∴ r2 = A – r1 = 60° – 23°39′ = 36°21′
Applying snell’s law on the second surface of
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 30
= sin-1 (0.889)
= 62°44′
≈ 63°
For any prism,
i + e = A + δ
∴ δ = (i + e) – A
= (37 + 63) – 60
= 40°
For an emergent ray to just emerge, the angle r’2 acts as a critical angle.
∴ r’2 = sin-1 (\(\frac {1}{n}\))
= sin-1 (\(\frac {2}{3}\))
= 41°48′
As, A = r’1 + r’2 and i to be kept the same.
⇒ A’ = r’1 + r’2
= 23°39′ + 41°48′
= 65°27’

Question 10.
From the given data set, determine angular dispersion by the prism and dispersive power of its material for extreme colours. nR = 1.62 nv = 1.66, δR = 3.1°
Answer:
Given: nR = 1.62, nV = 1.66, δR = 3.1°
To find:
i. Angular dispersion (δvr)
ii. Dispersive power (ωVR)
Formula:
i. δ = A (n – 1)
ii. δVR = δV – δR
(iii) ω = \(\frac{\delta_{\mathrm{V}}-\delta_{\mathrm{R}}}{\left(\frac{\delta_{\mathrm{V}}+\delta_{\mathrm{R}}}{2}\right)}\)
Calculation: From formula (i),
δR = A(nR – 1)
∴ A = \(\frac{\delta_{R}}{\left(n_{R}-1\right)}=\frac{3.1}{(1.62-1)}=\frac{3.1}{0.62}\)
= 5
δV = A(nv – 1) = 5 × (1.66 – 1) = 3.3C
From formula (ii),
δVR = 3.3 – 3.1 = 0.2°
From formula (iii),
ωVR = \(\frac{3.3-3.1}{\left(\frac{3.3+3.1}{2}\right)}=\frac{0.2}{6.4} \times 2=\frac{0.2}{3.2}=\frac{1}{16}\)
= 0.0625

Question 11.
Refractive index of a flint glass varies from 1.60 to 1.66 for visible range. Radii of curvature of a thin convex lens are 10 cm and 15 cm. Calculate the chromatic aberration between extreme colours.
Answer:
Given the refractive indices for extreme colours. As, nR < nV
nR = 1.60 and nV = 1.66
For convex lens,
R1 = 10 cm and R10 = -15 cm
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 31
= 0.11
∴ fV = 11 cm
∴ Longitudinal chromatic aberration
= fV – fR = 11 – 10 = 1 cm

Question 12.
A person uses spectacles of ‘number’ 2.00 for reading. Determine the range of magnifying power (angular magnification) possible. It is a concave convex lens (n = 1.5) having curvature of one of its surfaces to be 10 cm. Estimate that of the other.
Answer:
For a single concavo-convex lens, the magnifying power will be same as that for simple microscope As, the number represents the power of the lens,
P = \(\frac {1}{f}\) = 2 ⇒ f = 0.5 m.
∴ Range of magnifying power of a lens will be,
Mmin = \(\frac {D}{f}\) = \(\frac {0.25}{0.5}\) = 0.5
and Mmin = 1 + \(\frac {D}{f}\) = 1 + 0.5 = 1.5
Given that, n = 1.5, |R1| = 10 cm
f = 0.5 m = 50 cm
From lens maker’s equation,
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 32

Question 13.
Focal power of the eye lens of a compound microscope is 6 dioptre. The microscope is to be used for maximum magnifying power (angular magnification) of at least 12.5. The packing instructions demand that length of the microscope should be 25 cm. Determine minimum focal power of the objective. How much will its radius of curvature be if it is a biconvex lens of n = 1.5.
Answer:
Focal power of the eye lens,
Pe = \(\frac {1}{f_e}\) = 6D
∴ fe = \(\frac {1}{6}\) = 0.1667 m = 16.67 cm
Now, as the magnifying power is maximum,
ve = 25 cm,
Also (Me)max = 1 + \(\frac {D}{f_e}\) = 1 + \(\frac {25}{16.67}\) ≈ 2.5
Given that,
M = m0 × Me = 12.5
∴ m0 × 2.5 = 12.5
∴ m0 = \(\frac {v_0}{u_0}\) = 5 ……….. (1)
From thin lens formula,
Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics 33
Length of a compound microscope,
L = |v0| +|u0|
∴ 25 = |v0| + 10
∴ |v0|= 15 cm
∴ |u0| = \(\frac {v_0}{5}\) = 3 cm …………… (from 1)
From lens formula for objective,
\(\frac {1}{f_0}\) = \(\frac {1}{v_0}\) – \(\frac {1}{u_0}\)
= \(\frac {1}{15}\) – \(\frac {1}{-3}\)
= \(\frac {2}{5}\)
∴ f0 = 2.5 cm = 0.025 m
Thus, focal power of objective,
P = \(\frac {1}{f_0(m)}\)
= \(\frac {1}{0.025}\) = 40 D
Using lens maker’s equation for a biconvex lens,
\(\frac{1}{f_{o}}=(n-1)\left(\frac{1}{R}-\frac{1}{-R}\right)\)
∴ \(\frac{1}{2.5}=(1.5-1)\left(\frac{2}{R}\right)=\frac{1}{R}\)
∴ R = 2.5 cm

11th Physics Digest Chapter 9 Optics Intext Questions and Answers

Can you recall? (Textbook rage no 159)

What are laws of reflection and refraction?
Answer:
Laws of reflection:
a. Reflected ray lies in the plane formed by incident ray and the normal drawn at the point of incidence and the two rays are on either side of the normal.
b. Angles of incidence and reflection are equal (i = r).

Laws of refraction:
a. Refracted ray lies in the plane formed by incident ray and the normal drawn at the point of incidence; and the two rays are on either side of the normal.

b. Angle of incidence (θ1) and angle of refraction (θ2) are related by Snell’s law, given by, n1 sin θ1 = n2 sin θ2 where, n1, n2 = refractive indices of medium 1 and medium 2 respectively.

Maharashtra Board Class 11 Physics Solutions Chapter 9 Optics

Can you recall? (Textbook page no. 159)

Question 1.
What is refractive index?
Answer:
The ratio of velocity of light in vacuum to the velocity’ of light in a medium is called the refractive index of the medium.

Question 2.
What is total internal reflection?
Answer:
For angles of incidence larger than the critical angle, the angle of refraction is larger than 90°. Thus, all the incident light gets reflected back into the denser medium. This is called total internal reflection.

Question 3.
How does a rainbow form?
Answer:

  1. The rainbow appears in the sky after a rainfall.
  2. Water droplets present in the atmosphere act as small prism.
  3. When sunlight enters these water droplets it gets refracted and dispersed.
  4. This dispersed light gets totally reflected inside the droplet and again is refracted while coming out of the droplet.
  5. As a combined effect of all these phenomena, the seven coloured rainbow is observed.

Question 4.
What is dispersion of light?
Answer:
Splitting of a white light into its constituent colours is known as dispersion of light.

Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 8 Sound Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 8 Sound

Question 1.
State the different types of waves.
Answer:

  1. Waves which require a material medium for their propagation are called mechanical waves. Example: Sound waves, string waves, seismic waves, etc.
  2. Waves which do not require material medium for their propagation are called electromagnetic waves. Example: Light waves, radio waves, y-rays, etc.
  3. The wave associated with any object when it is in motion is called as matter wave.
  4. Waves in which a disturbance created at one place travels to distant points and keeps travelling unless stopped by an external force are known as travelling or progressive waves.
  5. Waves are also of stationary type.

Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound

Question 2.
Define the following terms. Give their SI units.
i. Period
ii. Frequency
iii. Velocity
Answer:
i. Period (T):
The time taken by the particle of a medium to complete one vibration is called period of the wave.
SI unit: second (s)

ii. Frequency (n):
The number of vibrations performed by a particle per second is called frequency of a wave.
SI unit: hertz (Hz)

iii. Velocity (v):
The distance covered by a wave per unit time is called the velocity of the wave.
SI unit: m/s

Question 3.
State the properties that should be possessed by a medium for a mechanical wave to propagate through it.
Answer:

  1. The medium must be continuous and elastic so that it can regain its original state as soon as the deforming forces are removed.
  2. The medium should possess inertia. It must be capable of storing energy and transferring it in the form of waves.
  3. The frictional resistance of the medium should be negligible to avoid damped oscillations.

Question 4.
What are two types of progressive waves? State two characteristics of progressive waves.
Answer:
Progressive waves are classified into two types:
a. Transverse progressive waves
b. Longitudinal progressive waves.

Characteristics of progressive waves:
1. All the vibrating particles of medium have same amplitude, period and frequency.
2.. State of oscillation i.e., phase changes from particle to particle.

Question 5.
A violin string emits sound of frequency 510 Hz. How far will the sound waves reach when string completes 250 vibrations? The velocity of sound is 340 m/s.
Answer:
Given: n = 510 Hz, v = 340 m/s,
number of vibrations = 250
To find: Distance
Formula: v = nλ
Calculation:
From formula,
λ = \(\frac {v}{n}\) = \(\frac {340}{510}\) = \(\frac {2}{3}\) m
Distance covered in 1 vibration = \(\frac {2}{3}\) m
∴ Distance covered in 250 vibration
= \(\frac {2}{3}\) × 250 = 166.7 m 3
Answer: The distance covered by sound waves is 166.7 m

Question 6.
The speed of sound in air is 330 m/s and that in glass is 4500 m/s. What is the ratio of the wavelength of sound of a given frequency in the two media?
Answer:
Given: vair = 330 m/s, vglass = 4500 m/s
To find: \(\frac {λ_{air}}{λ_{glass}}\)
Formula: v = nλ
Calculation: From formula,
vair = n λair
vglass = n λglass
∴ \(\frac {λ_{air}}{λ_{glass}}\) = \(\frac {v_{air}}{v_{glass}}\) = \(\frac {330}{4500}\) = 7.33 × 10-2

Question 7.
The velocity of sound in gas is 498 m/s and in air is 332 m/s. What is the ratio of wavelength of sound waves in gas to air?
Answer:
vg = 498 m/s, va = 332 m/s
To find: Ratio of wavelengths (\(\frac {λ_g}{λ_a}\))
Formula: v = nλ
Calculation:
Frequency of sound wave remains same.
From formula,
For gas λg = \(\frac {v_g}{n}\) and for air λag = \(\frac {v_a}{n}\)
∴ \(\frac {v_g}{v_a}\) = \(\frac {v_g}{v_a}\) = \(\frac {498}{332}\) = \(\frac {3}{2}\)
∴ \(\frac {v_g}{v_a}\) = 3 : 2

Question 8.
A human ear responds to sound waves of frequencies in the range of 20 Hz to 20 kHz. What are the corresponding wavelengths, if the speed of sound in air is 330 m/s? Answer:
Given: v1 = vg = 330 m/s, n1 = 20 Hz,
n2 = 20 kHz = 20 × 10³ Hz
To find: Wavelength (λ1 and λ2)
Formula: v = nλ
Calculation:
From formula,
λ1 = \(\frac {v_1}{n_1}\) = \(\frac {330}{20}\) = 16.5 m
λ2 = \(\frac {v_2}{n_2}\) = \(\frac {330}{20×10^3}\) = 16.5 × 10-3 = 0.0165 m

Question 9.
A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (i) the reflected sound, (ii) the transmitted sound? Speed of sound in air is 340 m s-1 and in water is 1486 m s-1
Answer:
Given: n = 1000 kHz = 106 Hz,
va = 340 m/s,
vw = 1486 m/s
To find: Reflected wavelength (λR),
Transmitted wavelength (λT),
Formula: v = nλ
Calculation:
In different medium, frequency of sound wave remains same.
From formula,
Sound is reflected in air,
i. ∴ λR = \(\frac {v_a}{n}\) = \(\frac {330}{10^6}\) = 3.4 × 10-4 m
Sound is transmitted in water,
ii. ∴ λT = \(\frac {v_w}{n}\) = \(\frac {1486}{10^6}\) = 1.486 × 10-3 m

Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound

Question 10.
The wavelength of a sound note is 1 m in air and 2.5 m in a liquid. Find the speed of sound in the liquid, if the speed of the sound in air is 330 m/s.
Answer:
Given: λa = 1 m, λl = 2.5 m, va = 330 m/s,
To find: Speed of sound (vl)
Formula: v = nλ
Calculation:
From formula,
Since the frequency n remains the same,
va = nλa and vl = nλl
∴ \(\frac {v_l}{v_a}\) = \(\frac {λ_l}{λ_a}\)
∴ vl = va \(\frac {λ_l}{λ_a}\) = 330 × \(\frac {2.5}{1}\) = 825 m/s

Question 11.
Define a polarised wave.
Answer:
A wave in which the vibrations of all the particles along the path of a wave are constrained to a single plane is called a polarised wave.

Question 12.
Write down the main characteristics of longitudinal waves.
Answer:
Characteristics of longitudinal waves:

  1. All the particles of medium in the path of wave vibrate in a direction parallel to the direction of propagation of wave with same period and amplitude.
  2. When longitudinal wave passes through a medium, the medium is divided into alternate compressions (high pressure zone) and rarefactions (low pressure zone).
  3. A compression and adjacent rarefaction form one cycle of a longitudinal wave. The distance between any two consecutive points having same phase (successive compression or rarefactions) is called wavelength of the wave.
  4. For propagation of longitudinal waves, the medium should possess the property of elasticity of volume (bulk modulus). Thus, longitudinal waves can travel through solids, liquids and gases. Longitudinal wave cannot travel through vacuum or empty space.
  5. The compressions and rarefactions advance in the medium and are responsible for transfer of energy.
  6. When longitudinal wave advances through medium, there is periodic variations in pressure and density along the path of wave and with time.
  7. Since the direction of vibration of particles and direction of propagation of wave are same or parallel, longitudinal waves cannot be polarised.

Question 13.
State the mathematical expression for a transverse progressive wave travelling along the positive and negative x-axis.
Answer:
i. Consider a transverse progressive wave whose particle position is described by x and displacement from equilibrium position is described by y.
Such a sinusoidal wave can be written as follows:
∴ y (x, t) = a sin (kx – ωt + ø) ……… (1)
where a, k, ω and ø are constants,
y (x, t) = displacement as a function of position (x) and time (t)
a = amplitude of the wave,
ω = angular frequency of the wave
(kx0 – ωt + ø) = argument of the sinusoidal wave and is the phase of the particle at x at time t.

ii. At a particular instant, t = t0,
y (x, t0) = a sin (kx – ωt0 + ø)
= a sin (kx + constant)
Thus at t = t0, shape of wave as a function of x is a sine wave.

iii. At a fixed location x = x0
y(x0, t) = a sin (kx0 – ωt + ø)
= a sin (constant – ωt)
Hence the displacement y, at x = x0 varies as a sine function.

iv. This means that the particles of the medium, through which the wave travels, execute simple harmonic motion around their equilibrium position.

v. For (kx – ωt + ø) to remain constant, x must increase in the positive direction as time t increases. Thus, the equation (1) represents a wave travelling along the positive x axis.

vi. Similarly, a wave travelling in the direction of the negative x axis is represented by,
y(x, t) = a sin (kx + ωt +ø) …….(2)

Question 14.
Explain the Laplace’s correction to the Newton’s formula for the velocity of sound in air.
Answer:
Laplace’s correction:
Laplace suggested that the compression or rarefaction takes place too rapidly. Heat produced during compression and lost during rarefaction does not get sufficient time for dissipation. Due to this, the whole heat content remains same. Thus, it is an adiabatic process.

According to Laplace, E is the adiabatic modulus of elasticity of air medium which is given by,
E = γP ….(1)
where P = pressure of the air medium γ = ratio of specific heat of air at constant pressure (cp) and specific heat of air at constant volume (cv). i.e., γ = cp/cv.

iii. Using equation, v = \(\sqrt{\frac {E}{ρ}}\), we have velocity of sound in air,
v = \(\sqrt{\frac {γP}{ρ}}\), …. [From equation (1)]
For air, γ = 1.41
At NTP, P = 0.76 × 13600 × 9.8 N/m²
ρ = 1.293 kg/m³.
∴ v = \(\sqrt{\frac {1.41×0.76×13600×9.8}{1.293}}\) = 332.35 m/s
This value is in close agreement with the experimental value.

Question 15.
What is the effect of temperature on the velocity of sound in air?
Answer:
Effect of temperature on velocity of sound:
i. Let v0 and v be the velocity of sound in air at T0 and T Kelvin respectively. Let ρ0 and p be the densities of gas at temperature T0 and T respectively.

ii. Considering the number of moles n = 1 for the gas, we have,
Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound 1

iii. From above formula, we can conclude that velocity of sound in air increases with increase in temperature.

Question 16.
Show that for 1 °C rise in temperature, the velocity of sound in air increases by 0.61 m/s.
Answer:
Let v0 = velocity of sound at 0 °C or 273 K
v = velocity of sound at t °C or (273 + 1) K
we have, v ∝ √T
Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound 2
Hence, velocity of sound increases by 0.61 m/s when temperature increases by 1 °C.

Question 17.
Suppose you are listening to an out-door live concert sitting at a distance of 150 m from the speakers. Your friend is listening to the live broadcast of the concert in another country and the radio signal has to travel 3000 km to reach him. Who will hear the music first and what will be the time difference between the two? Velocity of light = 3 × 108 m/s and that of sound is 330 m/s.
Answer:
Distance between me and the speakers
(s1) = 150 m, distance radio signal has to travel (S2) = 3000 km, vsound 330 m/s, vlight = 3 × 108 m/s
Time taken by sound to reach me,
= \(\frac {s_1}{v_sound}\) = \(\frac {150}{330}\) = 0.4546 s
Time taken by the broadcasted sound (done by
EM waves = \(\frac {s_2}{v_light}\) = \(\frac {3000km}{30×10^5km/s}\) = \(\frac {3×1^30}{3×10^5}\) = 10-2 s
∴My friend will hear the sound first.
The time difference will be = 0.4546 – 0.01
= 0.4446 s.

Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound

Question 18.
Consider a closed box of rigid walls so that the density’ of the air inside it is constant. On heating, the pressure of this enclosed air is increased from P0 to P. It is now observed that sound travels 1.5 times faster than at pressure P0. Calculate P/P0.
Answer:
Given: vP = 1.5 vP0
To find: Ratio of pressure (\(\frac {p}{p_0}\))
Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound 3

Question 19.
The densities of nitrogen and oxygen at NTP are 1.25 kg/m³ and 1.43 kg/m³ respectively. If the speed of sound in oxygen at NTP is 320 m/s, calculate the speed in nitrogen, under the same conditions of temperature and pressure, (γ for both the gases is 1.4)
Answer:
Given: ρN = 1.25 kg/m³, ρ = 1.43 kg/m³,
v0 = 320 m/s,
To find: Speed in nitrogen (vN)
Formula: v = (\(\sqrt{\frac {γP}{ρ}}\) )
Calculation: From formula,
Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound 4

Question 20.
Find the temperature at which the velocity of sound in air will be 1.5 times its velocity at 0 °C
Answer:
Given: \(\frac {p}{p_0}\) = 1.5, T0 = 0 °C = 273 K
To find: Temperature (T)
Formula: \(\frac {v}{v_0}\) = \(\sqrt{\frac {T}{T_0}}\)
Calculation:
From formula,
\(\frac {T}{T_0}\) = (\(\frac {v}{v_0}\))²
∴ T = T0 (\(\frac {v}{v_0}\))²
∴ T = 273 (1.5)² = 614.25 K = 341.25 °C

Question 21.
The velocity of sound in air at 27 °C is 340 m/s. Calculate the velocity of sound in air at 127 °C.
Answer:
Given: T1 = 27 °C = 27 + 273 = 300 K,
v1 = 340 m/s,
T2 = 127 °C = 127 + 273 = 400 K
To find: Velocity (v2)
Formula: \(\frac {v_1}{v_2}\) = \(\sqrt{\frac {T_1}{T_2}}\)
Calculation: From formula,
v2 = v1 \(\sqrt{\frac {T_2}{T_1}}\) = 340, \(\sqrt{\frac {400}{300}}\)
= 340 × 1.1547
∴ v2 = 392.6 m/s

Question 22.
The wavelength of a note is 27 m in air when the temperature is 27 °C. What is the wavelength when the temperature is increased to 37 °C?
Answer:
Given: λ1 = 27 m,
T1 = 27 °C = 273 + 27 = 300 K,
T2 = 37 °C = 273 + 37 = 310 K
To find: Wavelength (λ2)
Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound 5

Question 23.
We cannot hear an echo at every place. Give reason.
Answer:

  1. Echo of sound depends upon the temperature of the surrounding and distance between source and reflecting surface.
  2. To hear a distinct echo at 22 °C, the minimum distance required between the source of sound and reflecting surface should be 17.2 metre.
  3. The velocity of sound depends on the temperature of air. Thus, the minimum distance will change with temperature. Hence, we cannot hear an echo at every place.

Question 24.
Write a short note on reverberation.
Answer:

  1. Reverberation is the phenomenon in which sound waves are reflected multiple times causing a single sound to be heard more than once.
  2. Sound wave gets reflected multiple times if the distance between reflecting surface and source of sound is less than 15 m.
  3. During reverberation, the time interval between the successive reflections of a sound is small.
  4. As a result, the reflected sound waves overlap and produce a continuously increasing loud sound which is at times difficult to understand. Measures to decrease reverberation:
  5. Reverberation can be decreased by making the walls and roofs rough and by using curtains in the hall to avoid reflection of sound.
  6. Chairs and wall surfaces should be covered with sound absorbing materials.
  7. Porous cardboard sheets, perforated acoustic tiles, gypsum boards, thick curtains etc. should be used on the ceilings and walls.

Question 25.
Define acoustics.
Answer:
The branch of physics which deals with the study of production, transmission and reception of sound is called acoustics.

Question 26.
State the conditions that must be satisfied for proper acoustics in an auditorium along-with their remedies.
Answer:
i. Acoustics of an auditorium should be such that the sound is heard sufficiently loudly at all the points in the auditorium. The surface behind the speaker should be parabolic with the speaker at its focus for uniform distribution of sound in the auditorium. Reflection of sound helps to maintain good loudness through the entire auditorium.

ii. Echoes and reverberations should be reduced. More absorptive reflecting surfaces and full auditoriums help in reducing echoes.

iii. Unnecessary focusing of sound, poor audibility zone or region of silence should be avoided. Curved surface of the wall or ceiling should be avoided for this purpose.

iv. Echelon effect which arises due to the mixing of sound produced in the hall by the echoes of sound produced in front of regular structure like stairs should be reduced. Stair type construction in the hall must be avoided for this purpose.

v. To avoid outside stray sound from entering, the auditorium should be sound-proof when closed.

vi. Inside fittings, seats, etc. should not produce any sound for proper acoustics. Air conditioners instead of fans and soft action door closers should be used.

Question 27.
State the applications of acoustics observed in nature.
Answer:
Application of acoustics in nature:
i. Bats apply the principle of acoustics to locate objects. They emit short ultrasonic pulses of frequency 30 kHz to 150 kHz. The resulting echoes give them information about location of the obstacle. This helps the bats to fly in even in total darkness of caves.

ii. Dolphins navigate underwater with the help of an analogous system. They emit subsonic frequencies which can be about 100 Hz. They can sense an object about 1.4 m or larger.

Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound

Question 28.
State the medical applications of acoustics.
Answer:
i. High pressure and high amplitude shock waves are used to split kidney stones into smaller pieces without invasive surgery. A reflector or acoustic lens is used to focus a produced shock wave so that as much of its energy as possible converges on the stone. The resulting stresses in the stone causes the stone to break into small pieces which can then be removed easily.

ii. Ultrasonic imaging uses reflection of ultrasonic waves from regions in the interior of body. It is used for prenatal (before the birth) examination, detection of anomalous conditions like tumour etc. and the study of heart valve action.

iii. Ultrasound at a very high-power level, destroys selective pathological tissues which is helpful in treatment of arthritis and certain type of cancer.

Question 29.
State the underwater applications of acoustics.
Answer:

  1. SONAR (Sound Navigational Ranging) is a technique for locating objects underwater by transmitting a pulse of ultrasonic sound and detecting the reflected pulse.
  2. The time delay between transmission of a pulse and the reception of reflected pulse indicates the depth of the object.
  3. Motion and position of submerged objects like submarine can be measured with the help of this system.

Question 30.
State the applications of acoustics in environmental and geological studies.
Answer:
i. Acoustic principle has important application to environmental problems like noise control. The quiet mass transit vehicle is designed by studying the generation and propagation of sound in the motor’s wheels and supporting structures.

ii. Reflected and refracted elastic waves passing through the Earth’s interior can be measured by applying the principles of acoustics.

iii. This is useful in studying the properties of the Earth. Principles of acoustics are applied to detect local anomalies like oil deposits etc. making it useful for geological studies.

Question 31.
A man shouts loudly close to a high wall. He hears an echo. If the man is at 40 m from the wall, how long after the shout will the echo be heard? (speed of sound in air = 330 m/s)
Answer:
Given s = 40m, v = 330 m/s
To Find: time (t)
Formula: Time = distance \(\frac {distence}{speed}\)
Calculation:
The distance travelled by the sound wave
= 2 × distance from man to wall.
= 2 × 40 = 80 m.
From formula,
∴ Time taken to travel the distance
\(\frac {distence}{speed}\) = \(\frac {80}{30}\) = 0.24 s

Question 32.
Write a short note on pitch of sound note.
Answer:

  1. Pitch refers to the sharpness or shrillness of sound.
  2. Increase in frequency of sound results in increase in the pitch and the sound is said to be sharper.
  3. Tone refers to a single frequency of a wave.
  4. A note may contain single or multiple tones.
  5. High frequency is generally referred as high pitch or high tone.
  6. Generally, speech of the men is of low pitch (shrill) and that of the women is of high pitch (sharp). Tones of an acoustic guitar are sharper than that of a base guitar. Sound of table is sharper than that of a dagga.

Question 33.
Write a short note on quality (timbre) of sound note.
Answer:
i. Timbre of a sound refers to the quality of the sound which depends upon the mixture of tones and overtones in the sound. Same sound played on different musical instruments feels significantly different and the musical instrument from which the sound generated can be easily identified.

Question 34.
Write a short note on loudness of sound.
OR
Explain how loudness affects the characteristics of sound.
Answer:
Loudness:
i. Loudness depends upon the intensity of vibration.

ii. Intensity of a wave is proportional to square of the amplitude (I ∝ A²) and is measured in the (SI) unit ofW/m²

iii. The human response to intensity is not linear, i.e., a sound of double intensity is louder but not doubly loud.

iv. Under ideal conditions, for a perfectly healthy human ear, the least audible intensity is I0 = 10-12 W/m².

v. Loudness of a sound of intensity I (measured in unit bel) is given by,
L2 = log10 (\(\frac {I}{I_0}\)) ………….. (1)

vi. Decibel is the commonly used unit for loudness.

vii. As, 1 decibel or 1 dB = 0.1 bel.
∴ 1 bel = 10 dB. Thus, loudness in dB is 10 times loudness in bel.
∴ LdB = 10Lbel = 10 log10 (\(\frac {I}{I_0}\))
For sound of least audible intensity I0
LdB = 10 log10 (\(\frac {I_0}{I_0}\)) = 10 log10 (1) = 0 ………… (2)
This corresponds to threshold of hearing.

viii. For sound of 10 dB,
10 = 10 log10 (\(\frac {I}{I_0}\))
∴ (\(\frac {I}{I_0}\)) = 10 1 or I = 10 I0
For sound of 20 dB,
20 = 10 log10 (\(\frac {I}{I_0}\))
= (\(\frac {I}{I_0}\)) = 10² or I = 100 I0 and so on.

ix. This implies, loudness of 20 dB sound is felt double that of 10 dB, but its intensity is 10 times that of the 10 dB sound. Similarly, 40 dB sound is left twice as loud as 20 dB sound but its intensity is 100 times as that of 20 dB sound and 10000 times that of 10 dB sound. This is the power of logarithmic or exponential scale. If we move away from a (practically) point source, the intensity of its sound varies inversely with square of the distance, i.e., I ∝ \(\frac {1}{r^2}\).

Question 35.
When heard independently, two sound waves produce sensations of 60 dB and 55 dB respectively. How much will the sensation be if those are sounded together, perfectly in phase?
Answer:
L1 = 60 dB = 10 log10 \(\frac {I_1}{I_0}\)
∴ \(\frac {I_1}{I_0}\) = 106
∴ I1 = 106I0
Similarly, I2 = 105.5 I0
As the waves combine perfectly in phase, the vector addition of their amplitudes will be given by A² = (A1 + A2)² = A\(_1^2\) + A\(_2^2\) + 2A1, A2 As intensity is proportional to square of the amplitude.
∴ I = I1 + I2 + 2\(\sqrt {I_1I_2}\) = 105 I0 (101 +100.5 + 2\(\sqrt {10^{1.5}}\))
= 105I0(10 + 3.1623 +2 × 100.75)
= 24.41 × 105I0 = 2.441 × 106I0
∴ L = 10 log10 (\(\frac {I}{I_0}\)) = 10 log10 (2.441 × 106)
= 10[log10 (2.441) + log10(106)]
= 10(0.3876 + 6)
L = 63.876 dB ~ 64 dB

Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound

Question 36.
The noise level in a class-room in absence of the teacher is 50 dB when 50 students are present. Assuming that on the average each student outputs same sound energy per second, what will be the noise level if the number of students is increases to 100?
Answer:
Loudness of sound is given as,
Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound 6
∴ LB – LA = 0.301 × 10 = 3.01
∴ LB = LA + 3.01 = 53.01 dB

Question 37.
Calculate the decibel increase if there is a two-fold increase in the intensity of a wave. (Given: log10 2 = 0.3010)
Answer:
L = 10 log10 \(\frac {I}{I_0}\) decibel
L’ = 10 log10 \(\frac {2I}{I_0}\) decibel
L’ – L = 10 (log10 \(\frac {2I}{I_0}\) – log10 (\(\frac {I}{I_0}\))
= 10 log10 2
= 10 × 0.3010
∴ L’ – L = 3.01 dB

Question 38.
Derive the expression for apparent frequency when listener is stationary and source is moving away from the listener.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound 7
i. Consider a source of sound S moving away from a stationary listener L with velocity vs. Let the speed of sound with respect to the medium be v (always positive). The listener uses a detector for counting each wave crest that reaches it.

ii. Let at t = 0, the source at point Si which is at a distance d from the listener, emit a crest. This crest reaches the listener at time t1, given as, t1 = d/v. …………(1)

iii. Let T0 be the time period at which the waves are emitted.
At t = T0, distance travelled by the source away from the stationary listener to reach point S2 = vsT0.
∴ Distance of point S2 from the listener = d + vsT0.
At S2, The source emits second crest. This crest reaches the listener at t2, given as,
t2 = T0 + (\(\frac {d+v_sT_0}{v}\)) …………. (2)

iv. Similarly, the time taken by the (p+1)th crest (where, p is an integer, p = 1, 2, 3,…), emitted by the source at time pT0, to reach the listener is given as,
tp+1 = pT0 + (\(\frac {d+pv_sT_0}{v}\)) …………. (3)
∴ the listener’s detector counts p crests in the time interval,
tp+1 – t1 = pT0 + (\(\frac {d+pv_sT_0}{v}\)) – \(\frac {d}{v}\)
The period of wave as recorded by the listener is,
Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound 8

Where, n = frequency recorded by the listener (apparent frequency)
n0 = frequency emitted by the source (actual frequency).
This is the expression for apparent frequency when the listener is stationary and the source is moving away from the listener.

Question 39.
Derive an expression for apparent frequency when listener is stationary and source is moving towards the listener.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound 9
i. Consider a source of sound S moving towards a stationary listener L with velocity vs. Let the speed of sound with respect to the medium be v (always positive). The listener use a detector for counting each wave crest that reaches it.

ii. Let at t = 0, the source at point S1 which is at a distance d from the listener, emit a Crest. This crest reaches the listener at time t1, given as,
∴ t1 = d/v. ……….(1)

iii. Let T0 be the time period at which the waves are emitted.
At t = T0, distance travelled by the source away from the stationary listener to reach point S2 = vsT0.
Distance of S2 from the listener = d – vsT0.
At S2, The source emits second crest. This crest reaches the listener at
t2 = T0 + (\(\frac {d-v_sT_0}{v}\)) ………….. (2)

iv. Similarly, the time taken by the (p+1)th crest (where, p = 1,2,3,…), emitted by the source at time pT0, to reach the listener is given as,
tp+1 = pT0 + (\(\frac {d-pv_sT_0}{v}\)) ……………. (3)
∴ the listener’s detector counts p crests in the time interval,
tp+1 – t1 = pT0 + (\(\frac {d-pv_sT_0}{v}\)) – \(\frac {d}{v}\)
∴ the period of wave as recorded by the listener is,
Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound 10
Where, n = frequency recorded by the listener (apparent frequency)
n0 = frequency emitted by the source (actual frequency).
This is the expression for apparent frequency when the listener is stationary and the source is moving towards the listener.

Question 40.
Derive the expression for apparent frequency when the source is stationary and the listener is moving towards the source.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound 11
i. Consider a listener approaching a stationary source S with velocity vL as shown in figure. Let the speed of sound with respect to the medium be v (always positive).

ii. Let at time t = 0, the source emits the first wave when the listener L1 is at an initial distance d from the source.
At time t = t1 the listener receives the first wave at the position L2.
Distance travelled by the listener towards the stationary source during time t1 = vLt1.
Distance travelled by the sound wave during time t1 = d – vLt1
∴ time taken by the sound wave to travel this distance, t1 = \(\frac {d-v_Lt_1}{v}\)
∴ t1 = \(\frac {d}{v+v_L}\) ………….. (1)

iii. Let at time t = T0 (time period of the waves emitted by the source), the source emits a second wave.
At t = t2, the listener receives the second wave. Distance travelled by the listener towards the stationary source during time t2 = vLt2.
Distance travelled by the sound wave during time t2 = d – vLt2
∴ time taken by the sound wave to travel this distance = \(\frac {d-v_Lt_2}{v}\)
However, this time should be counted after T0, as the second wave was emitted at t = T0.
∴ t2 = T0 + \(\frac {d-v_Lt_2}{v}\)
∴ t2 = \(\frac {vT_0+d}{v+v_L}\) …………. (2)

iv. Similarly, for the third wave, we get,
t3 = 2T0 + \(\frac {d-v_Lt_3}{v}\)
∴ t3 = \(\frac {2vT_0+d}{v+v_L}\) …………. (3)

v. Extending this argument to the (p + 1)th wave, we get,
tp+1 = pT0 + \(\frac {d-v_Lt_{p+1}}{v}\)
∴ tp+1 = \(\frac {pvT_0+d}{v+v_L}\) …………. (4)

vi. The observed or recorded period T is the time duration between instances of receiving successive waves.
Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound 12
This is the expression for apparent frequency when the source is stationary and the listener is moving towards the source.

Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound

Question 41.
Derive the expression for apparent frequency when the source is stationary and the listener is moving away from the source.
Answer:
Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound 13
i. Consider a listener moving away a stationary source S with velocity VL. Let the speed of sound with respect to the medium be y (always positive).

ii. Let at time t = 0, the source emits the first wave when the listener L1 is at an initial distance d from the source.
At time t = t1 the listener receives the first wave at the position L2.
Distance travelled by the listener away from the stationary source during time t1 = vLt1.
Distance travelled by the sound wave during time t1 = d + vLt1
∴ time taken by the sound wave to travel this distance,
t1 = \(\frac {d+v_Lt_1}{v}\)
∴ t1 = \(\frac {d}{v-v_L}\) ………….. (1)

iii. Let at time t = T0 (time period of the waves emitted by the source), the source emits a second wave.
At t = t2, the listener receives the second wave. Distance travelled by the listener away from the stationary source during time t2 = vLt2.
∴ Distance travelled by the sound wave during time t2 = d + vLt2.
∴ time taken by the sound wave to travel this distance = \(\frac {d+v_Lt_2}{v}\)
However, this time should be counted after T0, as the second wave was emitted at t = T0.
∴ t2 = T0 \(\frac {d+v_Lt_2}{v}\) ………….. (2)
∴ t2 = \(\frac {vT_0+d}{v-v_L}\)

iv. Similarly for the third wave, we get
t3 = 2T0 \(\frac {d+v_Lt_3}{v}\)
∴ t3 = \(\frac {2vT_0+d}{v-v_L}\) …………..(3)

v. Extending this argument to the (p + 1)th wave, we get,
tp+1 = pT0 + \(\frac {d+v_Lt_{p+1}}{v}\)
∴ tp+1 = \(\frac {pvT_0+d}{v-v_L}\) …………..(4)

vi. The observed or recorded period T is the time duration between instances of receiving successive waves.
Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound 14
This is the expression for apparent frequency when the source is stationary and the listener is moving away from the source.

Question 42.
State the common properties between Doppler effect of sound and light.
Answer:
i. The recorded frequency is different than the emitted frequency in case of relative motion between listener (or observer) and source (of sound or light waves).

ii. In case of relative approach, recorded frequency > emitted frequency.

iii. In case of relative recede, recorded frequency < emitted frequency.

iv. For values of listener velocity (vL) or source velocity (vs) much smaller then wave speed (speed of sound or light).
n = n0 (1±\(\frac {v_r}{v}\))
Where, vr = relative velocity
n = actual frequency of the source
n0 = apparent frequency of the source
v = velocity of sound in air.
(upper sign is used during relative approach and lower sign is during relative recede.)

v. If velocities of source and observer (listener) are along different lines, their respective components along the line joining them should be chosen for longitudinal Doppler effect and the same mathematical treatment is applicable.

Question 43.
State the major difference between Doppler effects of sound and light.
Answer:

  1. Speed of light being absolute, only relative velocity between the observer and the source matter irrespective of who is in motion. However, for obtaining exact Doppler shift for sound waves, it is absolutely important to know who is in motion.
  2. In case of light, classical and relativistic Doppler effects are different while sound only has classical doppler effect.
  3. Presence of wind affects the velocity of sound which affects the Doppler shift in sound.

Question 44.
A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air. (i) What is the frequency of the whistle for a platform observer when the train (a) approaches the platform with a speed of 10 m s-1 (b) recedes from the platform with a speed of 10 m s-1? (ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 m s-1.
Answer:
Given: vs = 10 m/s, v = 340 m/s, n0 = 400 Hz
Apparent frequency (n), velocity of sound (vs) in each case
Formulae:
i. n = n0 (\(\frac {v}{v-v_s}\))
ii. n = n0 \(\frac {v}{v+v_s}\)
Calculation:
a. As the train approaches the platform, using formula (i),
n = 400 (\(\frac {340}{340-10}\)) = 421.12 Hz

b. As the train recedes from the platform, using formula (ii),
n = 400 (\(\frac {340}{340+10}\)) = 388.57 Hz

ii. The relative motion of source and observer results in the apparent change in the frequency but has no effect on the speed of sound. Hence, the speed of sound remains unchanged in both the cases.

Question 45.
A train blows a whistle of frequency 640 Hz in air. Find the difference in apparent frequencies of the whistle for a stationary observer, when the train moves towards and away from the observer with the speed of 72 km/hour. (Speed of sound in air = 340 m/s)
Answer:
Given: vs = 72 km/ hr = 20 m/s, n0 = 640 Hz,
v = 340 m/s
To find: Difference in apparent frequencies
(nA – n’A)
Formulae:
i. When the train moves towards the stationary observer then,
nA = n0 (\(\frac {v}{v-v_s}\))
ii. When the train moves away the stationary observer then,
n’A = n0 (\(\frac {v}{v+v_s}\))
Calculation: From formula (i),
nA = 640 (\(\frac {340}{340-20}\))
∴ nA = 680 Hz
From formula (ii),
n’A = 640 (\(\frac {340}{340+20}\))
∴ n’A = 604.4 Hz
Difference between nA and n’A
= nA – n’A = 75.6 Hz

Question 46.
The speed limit for a vehicle on road is 120 km/hr. A policeman detects a drop of 10% in the pitch of horn of a car as it passes him. Is the policeman justified in punishing the car driver for crossing the speed limit? (Given: Velocity of sound=340 m/s).
Answer:
Given: Speed limit, vL = 120 km/hr
n’A = nA – \(\frac {10}{100}\) nA = 0.9 nA
Velocity of sound, v = 340 m/s
To Find: Velocity of source (vs)
i. nA = (\(\frac {v}{v-v_s}\))n
ii. n’A = (\(\frac {v}{v+v_s}\))n
Calculation:
Dividing formula (i) by (ii),
Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound 15

Question 47.
A stationary source produces a note of frequency 350 Hz. An observer in a car moving towards the source measures the frequency of sound as 370 Hz. Find the speed of the car. What will be the frequency of sound as measured by the observer in the car if the car moves away from the source at the same speed? (Assume speed of sound = 340 m/s)
Answer:
Given: n0 = 350 Hz, v = 340 m/s,
nA = 370 Hz
To find: Speed (vL), Frequency (nA)
Formulae:
i. When the car moves towards the stationary source then,
nA = n0 (\(\frac {v+v_s}{v}\))

ii. When the car moves away from the stationary source then,
nA = n0 (\(\frac {v-v_L}{v}\))
Calculation: From formula (i),
370 = 35o (\(\frac {340+v_L}{340}\))
∴ 359.43 =340 +vL
∴ vL = 19.43 m/s
From formula (ii),
∴ nA = 35o (\(\frac {340-20}{340}\)) = \(\frac {35}{34}\) × 320
∴ nA = 329.41 Hz

Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound

Question 48.
A train, standing in a station-yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with a speed of 10 m s-1. What are the frequency, wavelength, and speed of sound for an observer standing on the station’s platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of 10 m s-1? The speed of sound in still air can be taken as 340 m s-1.
Answer:
Blowing of wind changes the velocity of sound. As the wind is blowing in the direction of sound, effective speed of sound ve = v + vw = 340 + 10 = 350 m/s
As the source and listener both are at rest, frequency is unchanged, i.e., n = 400 Hz.
∴ wavelength, λ = \(\frac {v_e}{n}\) = \(\frac {350}{400}\) = 0.875 m
For still air, vw = 0 and ve = v = 340 m/s
Also, as observer runs towards the stationary train vL = 10 m/s and vs = 0
The frequency now heard by the observer,
n = n0 (\(\frac {v+v_L}{v}\)) = 400 (\(\frac {340+10}{340}\))
= 411.76 Hz
As the source is at rest, wavelength does not change i.e., λ’ = λ = 0.875 m
Comparing the answers, it can be stated that, the situations in two cases are different.

Question 49.
A SONAR system fixed in a submarine operates at a frequency 40 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h-1. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be 1450 m s-1.
Answer:
Frequency of SONAR (source)
n = 40 kHz = 40 × 10³ Hz
Speed of sound waves, v = 1450 m s-1
Speed of the listener, vL = 360 km h-1
= 360 × \(\frac {5}{18}\) ms-1
= 100 m s-1
Since, the source is at rest and the observer moves towards the source (SONAR).
We have,
n = n0 (\(\frac {v+v_L}{v}\)) = 40 × 10³ × (\(\frac {1450+100}{14540}\))
∴ n = 4.276 × 10⁴ Hz
This frequency n’ is reflected by the enemy ship and is observed by the SONAR (which now acts as observer). Therefore, in this case vs = 100 m s-1
Apparent frequency,
n = n0 (\(\frac {v}{v-v_s}\))
= 4.276 × 10⁴ × (\(\frac {1450}{1450-100}\)) = 4.59 × 10⁴ Hz
∴ n = 45.9 kHz

Question 50.
A rocket is moving at a speed of 220 m s-1 towards a stationary target. While moving, it emits a wave of frequency 1200 Hz. Some of the sound reaching the target gets reflected back to the rocket as an echo. Calculate (i) the frequency of the sound as detected by the target and (ii) the frequency of the echo as detected by the rocket (velocity of sound = 330 m/s)
Answer:
Given: vs = 220 m/s, vL = 0 m/s, n = 1200 Hz
To find: Apparent frequency (n)
i. n = n0 (\(\frac {v}{v-v_s}\))
ii. n = n0 \(\frac {v+v_L}{v}\)
Calculation: For first case, observer is stationary and source i.e., rocket is moving towards the target.
Hence, using formula (i),
frequency of sound as detected by the target,
n = 1200 (\(\frac {330}{330-220}\)) = 3600 Hz
For second case, target acts as a source with frequency 3600 Hz as it is the source of echo. While rocket detector acts as an observer. This means, vs = 0 and VL = 220 m/s
Using formula (ii),
frequency of echo as detected by the rocket,
n = 3600 (\(\frac {330+220}{330}\)) = 600 Hz

Question 51.
A bat is flying about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly towards a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall?
Answer:
Here, frequency of sound emitted by bat,
n = 40 kHz
Velocity of bat, vs = 0.03 v
where v is velocity of sound.
The bat is moving towards the flat wall. This is the case of source in motion and the observer at rest.
Therefore, the frequency of sound reflected at the wall is,
n = ns (\(\frac {v}{v-v_s}\)) = n × (\(\frac {v}{v-0.03v}\))
= n × \(\frac {1}{0.97}\) = \(\frac {n}{0.97}\)
The frequency n’ is reflected by the wall and is again received by the bat moving towards the wall. This is the case of an observer moving towards the source with velocity vL = 0.03 v.
The frequency observed by bat,
Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound 16

Question 52.
A bat, flying at velocity VB = 12.5 m/s, is followed by a car running at velocity Vc = 50 m/s. Actual directions of the velocities of the car and the bat are as shown in the figure below, both being in the same horizontal plane (the plane of the figure). To detect the car, the bat radiates ultrasonic waves of frequency 36 kHz. Speed of sound at surrounding temperature is 350 m/s.
Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound 17
There is an ultrasonic frequency detector fitted in the car. Calculate the frequency recorded by this detector. The ultrasonic waves radiated by the bat are reflected by the car. The bat detects these waves and from the detected frequency, it knows about the speed of the car. Calculate the frequency of the reflected waves as detected by the bat. (sin 37° = cos 53° ~ 0.6, sin 53° = cos 37° ~ 0.8)
Answer:
The components of velocities of the bat and the car, along the line joining them, are
vc cos 53° ~ 50 × 0.6 = 30 m s-1 and
vB cos 37° ~ 12.5 × 0.8 = 10 m s-1
Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound 18
Doppler shifted frequency n = n0 (\(\frac {v±v_1}{v±v_s}\))
upper signs to be used during approach, lower signs during recede.
Case I: Frequency radiated by the bat
n0 = 36 × 10³ Hz,
The source (bat) is receding, while the listener (car) is approaching
vS = vB cos 37° = 10 m/s and
VL = vC cos 53° = 30 m/s
∴ Frequency detected by the detector in the car,
n = n0 (\(\frac {v+v_L}{v+v_s}\))
∴ n = 36 × 10³ (\(\frac {350+30}{350+10}\)) = 36 × 10³ × \(\frac {38}{36}\)
∴ n = 38 × 10³ Hz = 38 kHz

Case II: The car is the source.
Emitted frequency by the car, is given as,
n0 = 38 × 10³ Hz,
Car, the source, is approaching the listener (bat).
Now bat-the listener is receding while car the source is approaching,
∴ vs = vc cos 53° = 30 m/s
∴ vL = vB cos 37° = 10 m/s
∴ n = n0 (\(\frac {v-v_L}{v-v_s}\))
∴ n = 38 × 10³ (\(\frac {350-10}{350-30}\)) = 38 × 10³ × \(\frac {34}{33}\)
= 39.15 × 10³ Hz
∴ n = 39.15 kHz

Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound

Question 53.
Source of sound is placed at one end of a copper bar of length 1 km. Two sounds are heard at the other end at an interval of 2.75 seconds, (speed of sound in air = 330 m/s)
i. Why do we hear two sounds?
ii. Find the velocity of sound in copper.
Answer:
i. Two sounds are heard because sound travels through air as well as through copper.

ii. In air, t1 = \(\frac {distence}{time}\) = \(\frac {1000}{330}\) = 3.03 s
As the time interval is 2.75 seconds and sound travels faster in copper.
∴ In copper, t2 = 3.03 – 2.75 = 0.28 s
∴ velocity of sound in copper = \(\frac {1000}{0.28}\) = 3571 m/s

Question 54.
If all the persons mentioned in the table below are listening to a match commentary on the same channel at their respective locations positioning at same distance from television, then will they hear the same line of the commentary at same instant of time? Justify your answer.

Name of a person Location Humidity
Aijun Bangalore 65 %
Virendra Hyderabad 56%
Vikas Delhi 54%
Nilesh Mumbai 75%

Answer:
As the order of humidity for the above locations is Mumbai > Bangalore > Hyderabad > Delhi.
As velocity of sound increases with increase in humidity, the order of velocity of sound at their respective locations is Mumbai > Bangalore > Hyderabad > Delhi.
Hence, the order of persons who would listen the line of commentary first to last is Nilesh, Arjun, Virendra, Vikas.

Question 55.
Speed of sound is greater during day than at night. True or False? Justify your answer.
Answer:
True. At night, the amount of CO2 in atmosphere increases the density of atmosphere. Since, Speed of sound is inversely proportional to the square root of density. Hence, speed of sound is greater during day than in night.

Question 56.
Case I: During summer (33 °C), Prakash was waiting for a train at the platform, train arrived tt seconds after he heard train’s whistle.
Case II: During winter (19 °C), train arrived t2 seconds after Prakash heard the sound of train’s whistle.
i. Will t2 be equal to t1? Justify your answer.
ii. Calculate the velocity of sound in both the cases.
(velocity of sound in air at 0 °C = 330 m/s)
Answer:
i. Velocity of sound is directly proportional to square root of absolute temperature.
Hence, whistle’s sound will be first heard by Prakash in summer than in winter.
Therefore, the time interval between sound and train reaching Prakash in summer will be more than in winter.
i.e.,t1 > t2

ii. When t = 33 °C
∴ v1 = v0 + 0.61t
= 330 + 0.61 × 33
∴ v1 = 350.13 m/s
When t = 19 °C
v2 = v0 + 0.61t
= 330 + 0.61 × 19
v2 = 341.59 m/s

Question 57.
You are at a large outdoor concert, seated 300 m from speaker system. The concert is also being broadcast live. Consider a listener 5000 km away who receives the broadcast. Who will hear the music first, you or listener and by what time difference? (Speed of light = 3 × 108 m/s and speed of sound in air = 343 m/s)
Answer:
s1 = 300 m,
v1 = 343 m/s,
∴ t1 = \(\frac {s_1}{v_1}\) = \(\frac {300}{343}\) = 0.8746 s
Now,
s2 = 5000 km = 5 × 106 m,
v2 = c = 3 × 108 m/s
∴ t2 = \(\frac {s_2}{c}\) = \(\frac {5×10^6}{3×10^8}\) = 0.0167 s
∴ t2 < t1
∴ Listener will hear the music first.
Time difference = t1 – t2
= 0.8746 – 0.0167
= 0.8579 s
The listener will hear the music first, about 0.8579 s before the person present at the concert.

Question 58.
When a source of sound moves towards a stationary observer, then the pitch increases. Give reason.
Answer:
When a source of sound moves towards a stationary observer, then the increase in pitch is due to actual or apparent change in wavelength. When the source of sound moves towards an observer at rest, waves get compressed and the effective velocity of sound waves relative to source becomes less than the actual velocity. Hence the wavelength of sound waves an decreases which results into increase in pitch.

Question 59.
In the examples given below, state if the wave motion is transverse, longitudinal or a combination of both?

  1. Light waves travelling from Sun to Earth.
  2. ultrasonic waves in air produced by a vibrating quartz crystal.
  3. Waves produced by a motor boat sailing in water.

Answer:

  1. Light waves (from Sun to Earth) are electromagnetic waves which are transverse in nature.
  2. Ultrasonic waves in air are basically sound waves of frequency greater than the audible frequencies. Therefore, these waves are longitudinal.
  3. The water surface is cut laterally and pushed backwards by the propeller of motor boat. Therefore, the waves produced are a mixture of longitudinal and transverse waves.

Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound

Question 60.
Internet my friend
Answer:
https://hyperphysics.phys-astr.gsu.edu/ hbase/hframe.html
[Students are expected to visit the above mentioned website and collect more information about sound.]

Multiple Choice Questions

Question 1.
Water waves are …………….
(A) longitudinal
(B) transverse
(C) both longitudinal and transverse
(D) neither longitudinal nor transverse
Answer:
(C) both longitudinal and transverse

Question 2.
Sound travels fastest in ……………..
(A) water
(B) air
(C) steel
(D) kerosene oil
Answer:
(C) steel

Question 3.
At room temperature, velocity of sound in air at 10 atmospheric pressure and at 1 atmospheric pressure will be in the ratio ……………..
(A) 10 : 1
(B) 1 : 10
(C) 1 : 1
(D) cannot say
Answer:
(C) 1 : 1

Question 4.
In a gas, velocity of sound varies directly as ………………
(A) square root of isothermal elasticity.
(B) square of isothermal elasticity.
(C) square root of adiabatic elasticity.
(D) adiabatic elasticity.
Answer:
(C) square root of adiabatic elasticity.

Question 5.
At a given temperature, velocity of sound in oxygen and in hydrogen has the ratio …………………
(A) 4 : 1
(B) 1 : 4
(C) 1 : 1
(D) 2 : 1
Answer:
(B) 1 : 4

Question 6.
With decrease in water vapour content in air, velocity of sound …………………..
(A) increases
(B) decreases
(C) remains constant
(D) cannot say
Answer:
(B) decreases

Question 7.
The temperature at which speed of sound in air becomes double its value at 0 °C is ……………….
(A) 546 °C
(B) 819 °C
(C) 273 °C
(D) 1092 °C
Answer:
(B) 819 °C

Question 8.
The velocity of sound in air at NTP is 330 m/s. What will be its value when temperature is doubled and pressure is halved?
(A) 330 m/s
(B) 165 m/s
(C) 330 √2 m/s
(D) \(\frac {330}{√2}\) m/s
Answer:
(D) \(\frac {330}{√2}\) m/s

Question 9.
A series of ocean waves, each 5.0 m from crest to crest, moving past the observer at a rate of 2 waves per second have wave velocity
(A) 2.5 m/s
(B) 5.0 m/s
(C) 8.0 m/s
(D) 10.0 m/s
Answer:
(D) 10.0 m/s

Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound

Question 10.
A radio station broadcasts at 760 kHz. What is the wavelength of the station?
(A) 395 m
(B) 790 m
(C) 760 m
(D) 197.5 m
Answer:
(A) 395 m

Question 11.
If the bulk modulus of water is 2100 MPa, what is the speed of sound in water?
(A) 1450 m/s
(B) 2100 m/s
(C) 0.21 m/s
(D) 21 m/s
Answer:
(A) 1450 m/s

Question 12.
If speed of sound in air at 0°C is 331 m/s. What will be its value at 35° C?
(A) 331 m/s
(B) 366 m/s
(C) 351.6 m/s
(D) 332 m/s.
Answer:
(C) 351.6 m/s

Question 13.
For a progressive wave, in the usual notation
(A) v = λT
(B) n = \(\frac {v}{λ}\)
(C) T = λv
(D) λ = \(\frac {1}{n}\)
Answer:
(B) n = \(\frac {v}{λ}\)

Question 14.
At normal temperature, for an echo to be heard the reflecting surface should be at a minimum distance of ………………. m.
(A) 34.4
(B) 17.2
(C) 10
(D) 20
Answer:
(B) 17.2

Question 15.
In a transverse wave, are regions of negative displacement.
(A) rarefactions
(B) compressions
(C) crests
(D) troughs
Answer:
(D) troughs

Question 16.
If pressure of air gets doubled at constant temperature then velocity of sound in air ……………….
(A) gets doubled
(B) remains unchanged
(C) √2 times initial velocity
(D) becomes half
Answer:
(B) remains unchanged

Question 17.
Wave motion has ……………………
(A) single periodicity.
(B) double periodicity.
(C) only periodicity in space.
(D) only periodicity in time.
Answer:
(B) double periodicity.

Question 18.
The speed of the mechanical wave depends upon ………………
(A) elastic properties of the medium only.
(B) density of the medium only.
(C) elastic properties and density of the medium
(D) initial speed.
Answer:
(C) elastic properties and density of the medium

Question 19.
Longitudinal waves CANNOT be …………………
(A) reflected
(B) refracted
(C) scattered
(D) polarised
Answer:
(D) polarised

Question 20.
Wavelength of the transverse wave is 30 cm. If the particle at some instant has displacement 2 cm, find the displacement of the particle 15 cm away at the same instant.
(A) 2 cm
(B) 17 cm
(C) -2 cm
(D) -17 cm
Answer:
(C) -2 cm

Question 21.
The wavelength of sound in air is 1.5 m and that in liquid is 2 m. If the velocity of sound in air is 330 m/s, the velocity of sound in liquid is
(A) 330 m/s
(B) 440 m/’s
(C) 495 m/s
(D) 660 m/s
Answer:
(B) 440 m/’s

Question 22.
The velocity of sound in a gas is 340 m/s at the pressure P, what will be the velocity of the gas when only pressure is doubled and temperature same?
(A) 170 m/s
(B) 243 m/s
(C) 340 m/s
(D) 680 m/s
Answer:
(C) 340 m/s

Question 23.
Choose the correct statement.
(A) For 1 °C rise in temperature, velocity of sound increases by 0.61 m/s.
(B) For 1 °C rise in temperature, velocity of sound decreases by 0.61 m/s.
(C) For 1 °C rise in temperature, velocity of sound decreases by \(\frac {1}{273}\) m/s.
(D) For 1 °C rise in temperature, velocity of sound increases by \(\frac {1}{273}\) m/s.
Answer:
(A) For 1 °C rise in temperature, velocity of sound increases by 0.61 m/s.

Question 24.
A sound note emitted from a certain source has a velocity of 300 m/s in air and 1050 m/s in water. If the wavelength of sound note in air is 2 m, the wavelength in water is …………
(A) 2 m
(B) 6 m
(C) 7 m
(D) 12 m
Answer:
(C) 7 m

Question 25.
A thunder clap was heard 6 seconds after a lightening flash was seen. If the speed of sound in air is 340 m/s at the time of observation, the distance of the listener from the thunder clap is ………………
(A) 56.6 m
(B) 346 m
(C) 1020 m
(D) 2040 m
Answer:
(D) 2040 m

Question 26.
The speed of sound in air at NTP is 330 m/s. The period of sound wave of wavelength 66 cm is …………………
(A) 0.2 s
(B) 0.1
(C) 0.1 × 10-2 s
(D) 0.2 × 10-2 s
Answer:
(D) 0.2 × 10-2 s

Question 27.
If the velocity of sound in hydrogen is 1248 m/s, the velocity of sound in oxygen is [Given: MO = 32 and MH = 2]
(A) 1248 m/s
(B) 624 m/s
(C) 312 m/s
(D) 300 m/s
Answer:
(C) 312 m/s

Maharashtra Board Class 11 Physics Important Questions Chapter 8 Sound

Question 28.
If the source is moving away from the observer, then the apparent frequency …………..
(A) will increase
(B) will remain the same
(C) will be zero
(D) will decrease
Answer:
(D) will decrease

Question 29.
The working of SONAR is based on …………………
(A) resonance
(B) speed of a star
(C) Doppler effect
(D) speed of rotation of sun
Answer:
(C) Doppler effect

Question 30.
The formula for speed of a transverse wave on a stretched spring is ……………… (m = linear mass density, T = tension in Spring)
(A) v = \(\sqrt{\frac {m}{T}}\)
(B) v = \(\sqrt{\frac {T}{m}}\)
(C) v = (\(\frac {m}{T}\))\(\frac {3}{2}\)
(D) v = (\(\frac {T}{m}\))\(\frac {3}{2}\)
Answer:
(B) v = \(\sqrt{\frac {T}{m}}\)

Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 8 Sound Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 8 Sound

1. Choose the correct alternatives

Question 1.
A sound carried by air from a sitar to a listener is a wave of following type.
(A) Longitudinal stationary
(B)Transverse progressive
(C) Transverse stationary
(D) Longitudinal progressive
Answer:
(D) Longitudinal progressive

Question 2.
When sound waves travel from air to water, which of these remains constant ?
(A) Velocity
(B) Frequency
(C) Wavelength
(D) All of above
Answer:
(B) Frequency

Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound

Question 3.
The Laplace’s correction in the expression for velocity of sound given by Newton is needed because sound waves
(A) are longitudinal
(B) propagate isothermally
(C) propagate adiabatically
(D) are of long wavelength
Answer:
(C) propagate adiabatically

Question 4.
Speed of sound is maximum in
(A) air
(B) water
(C) vacuum
(D) solid
Answer:
(D) solid

Question 5.
The walls of the hall built for music concerns should
(A) amplify sound
(B) Reflect sound
(C) transmit sound
(D) Absorb sound
Answer:
(D) Absorb sound

2. Answer briefly.

Question 1.
Wave motion is doubly periodic. Explain.
Answer:
i. A wave particle repeats its motion after a definite interval of time at every location, making it periodic in time.
ii. Similarly, at any given instant, the form of a wave repeats itself at equal distances making it periodic in space.
iii. Thus, wave motion is a doubly periodic phenomenon, i.e., periodic in time as well as periodic in space.

Question 2.
What is Doppler effect?
Answer:
The apparent change in the frequency of sound heard by a listener, due to relative motion between the source of sound and the listener is called Doppler effect in sound.

Question 3.
Describe a transverse wave.
Answer:
Transverse wave:
A wave in which particles of the medium vibrate in a direction perpendicular to the direction of propagation of the wave is called transverse wave.
Example: Ripples on the surface of water, light waves.

Characteristics of transverse waves:

  1. All the particles of medium in the path of wave vibrate in a direction perpendicular to the direction of propagation of wave with same period and amplitude.
  2. When transverse wave passes through the medium, the medium is divided into alternate crests i.e., regions of positive displacements and troughs i.e., regions of negative displacement, that are periodic in time.
  3. A crest and an adjacent trough form one cycle of a transverse wave. The distance between any two successive crests or troughs is called wavelength ‘λ’ of the wave.
  4. Crests and troughs advance in the medium and are responsible for transfer of energy.
  5. Transverse waves can travel only through solids and not through liquids and gases. Electromagnetic waves are transverse waves, but they do not require material medium for propagation.
  6. When transverse waves advance through a medium, there is no change of pressure and density at any point of the medium, but the shape changes periodically.
  7. Transverse wave can be polarised.
  8. Medium conveying a transverse wave must possess elasticity of shape, i.e., modulus of rigidity.

Question 4.
Define a longitudinal wave.
Answer:
A wave in which particles of medium vibrate in a direction parallel to the direction of propagation of the wave is called longitudinal wave. Example: Sound waves.

Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound

Question 5.
State Newton’s formula for velocity of sound.
Answer:
Newton’s formula for velocity of sound:
i. Sound wave travels through a medium in the form of compression and rarefaction. At compression, the density of medium is greater while at rarefaction density is smaller. This is possible only in elastic medium.

ii. Thus, the velocity of sound depends upon density and elasticity of medium. It is given by
v = \(\sqrt{\frac {E}{ρ}}\) ….(1)
Where, E is the modulus of elasticity of medium and ρ is density of medium.

Assumptions:
1. Newton assumed that during propagation of sound wave in air, average temperature of the medium remains constant. Hence, propagation of sound wave in air is an isothermal process and isothermal elasticity should be considered.

2. The volume elasticity of air determined under isothermal change is called isothermal bulk modulus.

Calculations:
1. For a gas or air, the isothermal elasticity E is equal to the atmospheric pressure P.
Substituting this value in equation (1), the velocity of sound in air or a gas is given by
v = \(\sqrt{\frac {P}{ρ}}\) ….(∵ E = P)
This is the Newton’s formula for velocity of sound in air.

2. But atmospheric pressure is given by,
P = hdg
∴ v = \(\sqrt{\frac {hdg}{ρ}\) ….(2)

3. At N.T.P., h = 0.76 m of mercury, density of mercury d = 13600 kg/m³ and acceleration due to gravity, g = 9.8 m/s², density of air ρ = 1.293 kg/m³

4. From equation (2) we have velocity of sound,
v = \(\sqrt{\frac {0.76×13600×9.8}{1.293}}\) = 279.9 m/s at N.T.P

Question 6.
What is the effect of pressure on velocity of sound?
Answer:
Effect of pressure:
i. Let v be the velocity of sound in air when the pressure is P and density is ρ.

ii. Using Laplace’s formula, we can write,
v = \(\sqrt{\frac {γP}{ρ}}\) ….(1)

iii. If V be the volume of a gas having mass M then, ρ = \(\frac {M}{V}\)

iv. Substituting ρ in equation (1), we get,
v = \(\sqrt{\frac {γPV}{M}}\) ….(2)

v. But according to Boyle’s law,
PV = constant (at constant temperature)
Also, M and γ are constant.
∴ v = constant

vi. Hence, the velocity of sound does not depend upon the change in pressure, as long as the temperature remains constant.

vii. For a gaseous medium, PV= nRT.
Substituting in equation (2), we get,
v = \(\sqrt{\frac {γnRT}{M}}\)

viii. Thus, even for a gaseous medium obeying ideal gas equation, the velocity of sound does not depend upon the change in pressure, as long as the temperature remains constant.

Question 7.
What is the effect of humidity of air on velocity of sound?
Answer:
Effect of humidity:
i. Let vm and vd be the velocities of sound in moist air and dry air respectively.
Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound 1

ii. Humid air contains a large proportion of water vapour. Density of water vapour at 0 °C is 0.81 kg/m³ while that of dry air at 0°C is 1.29 kg/m³. So, the density ρm of moist air is less than the density ρd of dry air i.e., ρm < ρd.

iii. Thus \(\frac {v_m}{v_d}\) > 1
∴ vm > vd

iv. Hence, sound travels faster in moist air than in dry air. It means that velocity of sound increases with increase in moistness (humidity) of air.

Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound

Question 8.
What do you mean by an echo?
Answer:
An echo is the repetition of the original sound because of reflection from some rigid surface at a distance from the source of sound.

Question 9.
State any two applications of acoustics.
Answer:
Application of acoustics in nature:
i. Bats apply the principle of acoustics to locate objects. They emit short ultrasonic pulses of frequency 30 kHz to 150 kHz. The resulting echoes give them information about location of the obstacle. This helps the bats to fly in even in total darkness of caves.

ii. Dolphins navigate underwater with the help of an analogous system. They emit subsonic frequencies which can be about 100 Hz. They can sense an object about 1.4 m or larger.

Medical applications of acoustics:
i. High pressure and high amplitude shock waves are used to split kidney stones into smaller pieces without invasive surgery. A reflector or acoustic lens is used to focus a produced shock wave so that as much of its energy as possible converges on the stone. The resulting stresses in the stone causes the stone to break into small pieces which can then be removed easily.

ii. Ultrasonic imaging uses reflection of ultrasonic waves from regions in the interior of body. It is used for prenatal (before the birth) examination, detection of anomalous conditions like tumour etc. and the study of heart valve action.

iii. Ultrasound at a very high-power level, destroys selective pathological tissues which is helpful in treatment of arthritis and certain type of cancer.

Underwater applications of acoustics:
i. SONAR (Sound Navigational Ranging) is a technique for locating objects underwater by transmitting a pulse of ultrasonic sound and detecting the reflected pulse.
ii. The time delay between transmission of a pulse and the reception of reflected pulse indicates the depth of the object.
iii. Motion and position of submerged objects like submarine can be measured with the help of this system.

Applications of acoustics in environmental and geological studies:
i. Acoustic principle has important application to environmental problems like noise control. The quiet mass transit vehicle is designed by studying the generation and propagation of sound in the motor’s wheels and supporting structures.

Reflected and refracted elastic waves passing through the Earth’s interior can be measured by applying the principles of acoustics. This is useful in studying the properties of the Earth.

Principles of acoustics are applied to detect local anomalies like oil deposits etc. making it useful for geological studies.

Question 10.
Define amplitude and wavelength of a wave.
Answer:
i. Amplitude (A): The largest displacement of a particle of a medium through which the wave is propagating, from its rest position, is called amplitude of that wave.
SI unit: (m)

ii. Wavelength (λ): The distance between two successive particles which are in the same state of vibration is called wavelength of the wave.
SI unit: (m)

Question 11.
Draw a wave and indicate points which are (i) in phase (ii) out of phase (iii) have a phase difference of π/2.
Answer:
Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound 2
i. In phase point: A and F; B and H; C and I; D and J
ii. Out of phase points: A and B, B and D, FI and J, E and F,
iii. Point having phase difference of π/2: A and B; B and C; D; D and F; F and H; H and I; J and I

Question 12.
Define the relation between velocity, wavelength and frequency of wave.
Answer:
i. A wave covers a distance equal to the wavelength (λ) during one period (T).
Therefore, the magnitude of the velocity (v) is given by,
Magnitude of velocity = \(\frac {Distance covered}{Corresponding time}\)

ii. v = \(\frac {22}{7}\) i.e., v = λ × (\(\frac {1}{T}\)) …………….. (1)

iii. But reciprocal of the period is equal to the frequency (n) of the waves.
∴ \(\frac {1}{T}\) = n …………… (2)

iv. From equations (1) and (2), we get
v = nλ
i.e., wave velocity = frequency × wavelength.

Question 13.
State and explain principle of superposition of waves.
Answer:
Principle:
As waves don’t repulse each other, they overlap in the same region of the space without affecting each other. When two waves overlap, their displacements add vectorially.

Explanation:
i. Consider two waves travelling through a medium arriving at a point simultaneously.

ii. Let each wave produce its own displacement at that point independent of the others. This displacement can be given as,
y1 = displacement due to first wave.
y2 = displacement due to second wave.

iii. Then according to superposition of waves, the resultant displacement at that point is equal to the vector sum of the displacements due to all the waves.
∴\(\vec{y}\) = \(\vec{y_1}\) + \(\vec{y_2}\)

Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound

Question 14.
State the expression for apparent frequency when source of sound and listener are
i) moving towards each other
ii) moving away from each other
Answer:
i. Let,
n = actual frequency of the source.
n0 = apparent frequency of the source,
v = velocity of sound in air.
vs = velocity of the source.
vl = velocity of the listener.

ii. Apparent frequency heard by the listener is given by,
n = n0(\(\frac {v±v_L}{v±v_s}\))
Where upper signs (+ ve in numerator and -ve in denominator) indicate that source and observer move towards each other. Lower signs (-ve in numerator and +ve in denominator) indicate that source and listener move away from each other.

iii. If source and listener are moving towards each other, then apparent frequency is given by,
n = n0(\(\frac {v+v_L}{v-v_s}\)) i.e., apparent frequency increases.

iv. If source and listener are moving away from each other, then apparent frequency is given by,
n = n0(\(\frac {v-v_L}{v+v_s}\)) i.e., apparent frequency decreases.

Question 15.
State the expression for apparent frequency when source is stationary and listener is
1) moving towards the source
2) moving away from the source
Answer:
Let,
n = actual frequency of the source.
n0 = apparent frequency of the source,
v = velocity of sound in air.
vs = velocity of the source.
vl = velocity of the listener.

i. If listener is moving towards source then apparent frequency is given by,
n = n0(\(\frac {v+v_L}{v}\)) i.e., apparent frequency increases.

ii. If listener is receding away from source then apparent frequency is given by,
n = n0(\(\frac {v-v_L}{v}\)) i.e., apparent frequency decreases.

Question 16.
State the expression for apparent frequency when listener is stationary and source is.

(i) moving towards the listener
(ii) moving away from the listener
Answer:
Let,
n = actual frequency of the source.
n0 = apparent frequency of the source,
v = velocity of sound in air.
vs = velocity of the source.
vl = velocity of the listener.

i. If source is moving towards observer then apparent frequency is given by,
n = n0(\(\frac {v}{v-v_s}\)) i.e., apparent frequency increases.

ii. If source is receding away from observer then apparent frequency is given by,
n = n0(\(\frac {v}{v+v_s}\)) i.e., apparent frequency decreases.

Question 17.
Explain what is meant by phase of a wave.
Answer:
Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound 3
i. The state of oscillation of a particle is called the phase of the particle.

ii. The displacement, direction of velocity and oscillation number of the particle describe the phase of the particle at a place.

iii. Particles r and t (q and u or v and s) have same displacements but the directions of their velocities are opposite.

iv. Particles having same magnitude of displacements and same direction of velocity are said to be in phase during their respective oscillations. Example: particles v and p.

v. Separation between two particles which are in phase is wavelength (λ).

vi. The two successive particles differ by ‘1’ in their oscillation number i.e., if particle v is at its nth oscillation, particle p will be at its (n + 1)th oscillation as the wave is travelling along + X direction.

vii. In the given graph, if the disturbance (energy) has just reached the particle w, the phase angle corresponding to particle is 0°. At this instant, particle v has completed quarter oscillation and reached its positive maximum (sin θ = +1). The phase angle θ of this particle v is \(\frac {π^c}{2}\) = 90° at this instant.

viii. Phase angles of particles u and q are πc (180°) and 2rcc (360°) respectively.

ix. Particle p has completed one oscillation and is at its positive maximum during its second oscillation.
∴ phase angle = 2πc + \(\frac {π^c}{2}\)
= \(\frac {5π^c}{2}\)

x. v and p are the successive particles in the same state (same displacement and same direction of velocity) during their respective oscillations. Phase angle between these two differs by 2πc.

Question 18.
Define progressive wave. State any four properties.
Answer:
i. Waves in which a disturbance created at one place travels to distant points and keeps travelling unless stopped by an external force are known as travelling or progressive waves.
Properties of progressive waves are:
Amplitude, wavelength, period, double periodicity, frequency and velocity.

Question 19.
Distinguish between traverse waves and longitudinal waves.
Answer:

Longitudinal wave Transverse wave
1. The particles of the medium vibrate along the direction of propagation of the wave. 1. The particles of the medium vibrate perpendicular to the direction of propagation of the wave.
2. Alternate compressions and rarefactions are formed. 2. Alternate crests and troughs are formed.
3. Periodic compressions and rarefactions, in space and time, produce periodic pressure and density variations in the medium. There are no pressure and density, variations in the medium.
4. For propagation of a longitudinal wave, the medium must be able to resist changes in volume. For propagation of a transverse wave, the medium must be able to resist shear or change in shape.
5. It can propagate through any material medium (solid, liquid or gas). It can propagate only through solids.
6. These waves cannot be polarised. These waves can be polarised.
7. eg.: Sound waves eg.: Light waves

Question 20.
Explain Newtons formula for velocity of sound. What is its limitation?
Answer:
Newton’s formula for velocity of sound:
i. Sound wave travels through a medium in the form of compression and rarefaction. At compression, the density of medium is greater while at rarefaction density is smaller. This is possible only in elastic medium.

ii. Thus, the velocity of sound depends upon density and elasticity of medium. It is given by
v = \(\sqrt{\frac {E}{ρ}}\) ….(1)
Where, E is the modulus of elasticity of medium and ρ is density of medium.

Assumptions:
1. Newton assumed that during propagation of sound wave in air, average temperature of the medium remains constant. Hence, propagation of sound wave in air is an isothermal process and isothermal elasticity should be considered.

2. The volume elasticity of air determined under isothermal change is called isothermal bulk modulus.

Calculations:
1. For a gas or air, the isothermal elasticity E is equal to the atmospheric pressure P.
Substituting this value in equation (1), the velocity of sound in air or a gas is given by
v = \(\sqrt{\frac {P}{ρ}}\) ….(∵ E = P)
This is the Newton’s formula for velocity of sound in air.

2. But atmospheric pressure is given by,
P = hdg
∴ v = \(\sqrt{\frac {hdg}{ρ}}\) ….(2)

3. At N.T.P., h = 0.76 m of mercury, density of mercury d = 13600 kg/m³ and acceleration due to gravity, g = 9.8 m/s², density of air ρ = 1.293 kg/m³

4. From equation (2) we have velocity of sound,
v = \(\sqrt{\frac {0.76×13600×9.8}{1.293}}\) = 279.9 m/s at N.T.P

Limitations:
1. Experimentally, it is found that the velocity of sound in air at N. T. P is 332 m/s. Thus, there is considerable difference between the value predicted by Newton’s formula and the experimental value.

2. Experimental value is 16% greater than the value given by the formula. Newton failed to provide a satisfactory explanation for the difference.

Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound

3. Solve the following problems.

Question 1.
A certain sound wave in air has a speed 340 m/s and wavelength 1.7 m for this wave, calculate
(i) the frequency
(ii) the period.
Answer:
Given: v = 340 m/s, λ = 1.7 m
To find: frequency (n), period (T)
Formulae:
i. n = \(\frac {v}{λ}\)
ii. T = \(\frac {1}{n}\)
Calculation: From formula, (i)
n = \(\frac {340}{1.7}\)
∴ n = 200 Hz
From formula, (ii)
T = \(\frac {1}{n}\) = \(\frac {1}{2×10^2}\)
= 5 × 10-3
…….. (using reciprocal Table)
∴ T = 0.005 s

Question 2.
A tuning fork of frequency 170 Hz produces sound waves of wavelength 2m. Calculate speed of sound.
Answer:
Given: n = 170 Hz, λ = 2 m
To find: velocity of sound (v)
Formula: v = nλ
Calculation: From formula,
v = 170 × 2
∴ v = 340 m/s

Question 3.
An echo-sounder in a fishing boat receives an echo from a shoal of fish 0.45s after it was sent. If the speed of sound in water is 1500 m/s, how deep is the shoal?
Answer:
Given: t = 0.45 s, v = 1500 m/s,
To Find: depth (d)
Formula: speed (v) = \(\frac {distance}{time}\)
Calculation:
For an echo distance travelled by the sound wave = 2 × (distance between echo sounder and shoal) (d)
v = \(\frac {2 × d}{t}\)
∴ d = \(\frac {1500 × 0.45}{2}\) = 337.5 m

Question 4.
A girl stands 170 m away from a high wall and claps her hands at a steady rateso that each clap coincides with the echo of the one before.
a) If she makes 60 claps in 1 minute, what value should be the speed of sound in air?
b) Now, she moves to another location and finds that she should now make 45 claps in 1 minute to coincide with successive echoes. Calculate her distance for the new position from the wall.
Answer:
i. When the girl makes 60 claps in 1 minute, the value of speed of is 340 m/s.

ii. The girl is at a distance of 226.67 m from the wall when she produces 45 claps per minute.
[Note: The answer given above is calculated in accordance with textual method considering the given data]

Question 5.
Sound wave A has period 0.015 s, sound wave B has period 0.025. Which sound has greater frequency?
Answer:
Given: TA = 0.015 s, TB = 0.025 s
To find: greater frequency (n)
Formula: n = \(\frac {1}{T}\)
Calculation: From formula,
nA = \(\frac {1}{T_A}\) = \(\frac {1}{0.025}\) = \(\frac {1}{2.5 ×10^{-2}}\)
∴ nA = 66.67
…. (using reciprocal table)
nB = \(\frac {1}{T_B}\) = \(\frac {1}{0.025}\) = \(\frac {1}{2.5 ×10^{-2}}\)
∴ nB = 40 Hz
…. (using reciprocal table)
∴ nA > nB

Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound

Question 6.
At what temperature will the speed of sound in air be 1.75 times its speed at N.T.P?
Answer:
Given:
vair = 1.75 VS.T.P = \(\frac {7}{4}\) vS.T.P
TS.T.P = 273 K
To find: temperature Tair
Formula: v ∝ √T
Calculation: From formula,
Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound 4

Question 7.
A man standing between 2 parallel eliffs fires a gun. He hearns two echos one after 3 seconds and other after 5 seconds. The separation between the two cliffs is 1360 m, what is the speed of sound?
Answer:
distance (s) = 1360 m,
time for first echo = 3 s,
time for second echo = 5 s
To Find : speed of sound (v)
Formula : speed = \(\frac {distence}{time}\)
Calculation:
Time for first echo = 3 s
∴ time taken by sound to travel given distance t1
= \(\frac {3}{2}\) = 1.5 s
Time for second echo = 5 s
∴ time taken by sound to travel given distance t2
= \(\frac {5}{2}\) = 2.5 s
∴Total time taken by sound to travel given distance, T = 1.5 + 2.5 = 4 s
From formula,
v = \(\frac {1360}{4}\)
∴v = 340 m/s

Question 8.
If the velocity of sound in air at a given place on two different days of a given week are in the ratio of 1 : 1.1. Assuming the temperatures on the two days to be same what quantitative conclusion can your draw about the condition on the two days?
Answer:
Let v1 and v2 be the velocity of sound on day 1 and day 2 respectively.
\(\frac {v_1}{v_2}\) = \(\frac {1}{1.1}\)
We know, v ∝ \(\frac {1}{√ρ}\)
Let ρ1 and ρ2 be the density of air on day 1 and day 2 respectively.
∴ \(\sqrt{\frac {ρ_2}{ρ_1}}\) = \(\frac {1}{1.1}\)
∴ \(\frac {ρ_2}{ρ_1}\) = (\(\frac {1}{1.1}\))²
∴ ρ1 = 1.1² ρ2 = 1.21 ρ²
From above equation, we can conclude,
ρ1 > ρ2
∴ v2 > v1 i.e., the velocity of sound is greater on the second day than on the first day.
We know, speed of sound in moist air (vm) is greater than speed of sound in dry air (vd).
∴ We can conclude, air is moist on second day and dry on the first day.

Question 9.
A police car travels towards a stationary observer at a speed of 15 m/s. The siren on the car emits a sound of frequency 250 Hz. Calculate the recorded frequency. The speed of sound is 340 m/s.
Answer:
Given: vs = 15 m/s, n0 = 250 Hz, v = 340 m/s
To find: Frequency (n)
Formula: n = n0(\(\frac {v}{v-v_s}\))
Calculation: As the source approaches listener, apparent frequency is given by,
n = 250 (\(\frac {340}{340-15}\)) = \(\frac {3400}{13}\)
∴ n = 261.54 Hz

Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound

Question 10.
The sound emitted from the siren of an ambulance has frequency of 1500 Hz. The speed of sound is 340 m/s. Calculate the difference in frequencies heard by a stationary observer if the ambulance initially travels towards and then away from the observer at a speed of 30 m/s.
Answer:
Given: vs = 30 m/s, n0 = 1500 Hz, v = 340 m/s
To find: Difference in apparent frequencies (nA – n’A)
Formulae:
i. When the ambulance moves towards he stationary observer then nA = n0(\(\frac {v}{v-v_s}\))

ii. When the ambulance moves away from the stationary observer then, n’A = n0(\(\frac {v}{v+v_s}\))

Calculation:
From formula (i), icon’ 340
nA = 1500(\(\frac {340}{340-30}\))
∴ nA = 1645 Hz
From (ii)
n’A = 1500(\(\frac {340}{340+30}\))
∴ nA = 1378 Hz
Difference between nA and n’A
= nA – n’A = 1645 – 1378 = 267 Hz

11th Physics Digest Chapter 8 Sound Intext Questions and Answers

Can you recall? (Textbook page no. 142)

i. What type of wave is a sound wave?
ii. Can sound travel in vacuum?
iii. What are reverberation and echo?
iv. What is meant by pitch of a sound?
Answer:
i. Sound wave is a longitudinal wave.

ii. Sound cannot travel in vacuum.

iii. a. Reverberation is the phenomenon in which sound waves are reflected multiple times causing a single sound to be heard more than once.
b. An echo is the repetition of the original sound because of reflection by some surface.

iv. The characteristic of sound which is determined by the value of frequency is called as the pitch of the sound.

Activity (Textbook page no. 144)

i. Using axes of displacement and distance, sketch two waves A and B such that A has twice the wavelength and half the amplitude of B.
ii. Determine the wavelength and amplitude of each of the two waves P and Q shown in figure below.
Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound 5
Answer:
Maharashtra Board Class 11 Physics Solutions Chapter 8 Sound 6

Wave Wavelength (λ) Amplitude (A)
A 4 m 2 m
B 2 m 4 m
Wave Wavelength (λ) Amplitude (A)
P 6 units 3 units
Q 4 units 2 units

Maharashtra Board Class 12 Biology Important Questions Chapter 15 Biodiversity, Conservation and Environmental Issues

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 15 Biodiversity, Conservation and Environmental Issues Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 15 Biodiversity, Conservation and Environmental Issues

Multiple choice questions

Question 1.
Diversity in living beings is due to ……………………
(a) mutation
(b) long term evolutionary change
(c) gradual change
(d) short term evolutionary change
Answer:
(b) long term evolutionary change

Question 2.
Diversification of plant life appeared ……………………
(a) due to long periods of evolutionary changes
(b) due to abrupt mutations
(c) suddenly on the earth
(d) by seed dispersal
Answer:
(a) due to long periods of evolutionary changes

Maharashtra Board Class 12 Biology Important Questions Chapter 15 Biodiversity, Conservation and Environmental Issues

Question 3.
Rauwolfia vomitoria shows …………………. in terms of the potency and concentration of reserpine that it produces.
(a) genetic diversity
(b) species diversity
(c) ecological diversity
(d) biodiversity
Answer:
(a) genetic diversity

Question 4.
Which of the following country has the greatest ecosystem diversity?
(a) Norway
(b) India
(c) Sweden
(d) Finland
Answer:
(b) India

Question 5.
India has deserts, rain forests, mangroves, coral reefs, wetlands, estuaries and alpine meadows. What kind of biodiversity is depicted in this statement?
(a) Geographic diversity
(b) Species diversity
(c) Ecological diversity
(d) Genetic diversity
Answer:
(c) Ecological diversity

Question 6.
Biodiversity and its conservation are vital environmental issues of international concern because ……………….
(a) it fetches more economic progress and development
(b) it can attract more international tourists
(c) biodiversity and its conservation is essential for our survival and well-being of earth
(d) all the animals and plants would be extinct if not taken care of
Answer:
(c) biodiversity and its conservation is essential for our survival and well-being of earth

Question 7.
Biodiversity of geographical region represents …………………….
(a) endangered species found in the region
(b) the diversity in the organisms living in the region
(c) genetic diversity present in the dominant species of the region
(d) species endemic to the region
Answer:
(b) the diversity in the organisms living in the region

Question 8.
The latitudinal gradient in the pattern of biodiversity shows that …………………
(a) species diversity decreases as one moves away from the equator towards the poles
(b) species diversity increases as we move away from the equator towards the poles
(c) species diversity decreases as we move away from the poles towards the equator
(d) species diversity remains constant as we move away from the poles towards the equator
Answer:
(a) species diversity decreases as one moves away from the equator towards the poles

Question 9.
Which of the following is the possible cause for greater biodiversity in the tropics?
(a) Higher productivity due to more solar energy.
(b) Lesser technological development.
(c) Traditional and religious practices for conservation of nature.
(d) Lesser natural calamities.
Answer:
(a) Higher productivity due to more solar energy.

Question 10.
Log S = log C = Z log A is the equation that depicts relation between ………………….
(a) population density and time
(b) population growth and time
(c) species richness and area
(d) area and species migrations
Answer:
(c) species richness and area

Question 11.
Regression coefficient is shown by ……………………….. in the Humboldt’s equation [Log S = log C + Z log A] of species richness.
(a) S
(b) Z
(c) A
(d) C
Answer:
(b) Z

Question 12.
Choose an incorrect statement:
(a) The relation between species richness and area for a wide variety of taxa turns out to be a rectangular hyperbola.
(b) The relation between species richness and area on a logarithmic scale, the relationship is a straight line.
(c) For the species-area relationships among very large areas like the entire continents, the slope of the line appears to be much steeper.
(d) Value of Z always keep on changing for every taxonomic group or the region.
Answer:
(d) Value of Z always keep on changing for every taxonomic group or the region.

Question 13.
Who observed that within a region, species richness increased with increasing explored area, to a certain limit ?
(a) Robert May
(b) John Muir
(c) Alexander von Humboldt
(d) David Tilman
Answer:
(c) Alexander von Humboldt

Question 14.
Who was Alexander von Humboldt ?
(a) American Population biologist
(b) German naturalist and geographer
(c) Dutch Botanist
(d) French Zoologist
Answer:
(b) German naturalist and geographer

Question 15.
Which is the most well-known pattern of biodiversity ?
(a) Species-Area relationship
(b) Latitudinal gradient
(c) Longitudinal gradient
(d) Altitudinal gradient
Answer:
(b) Latitudinal gradient

Question 16.
Name the scientist who studied ecosystem by using analogy of ‘The rivet popper hypothesis’.
(a) John Muir
(b) Alexander von Humboldt
(c) Robert May
(d) Paul Ehrlich
Answer:
(d) Paul Ehrlich

Question 17.
When is the serious threat developed to an ecosystem ?
(a) When any one or two species become extinct.
(b) When key species that drive ecosystem become extinct.
(c) When native species are replaced by exotic species.
(d) When human beings take conservation measures.
Answer:
(b) When key species that drive ecosystem become extinct.

Question 18.
Which animal group is more vulnerable to the process of extinction ?
(a) Amphibian
(b) Reptilia
(c) Aves
(d) Mammalia
Answer:
(a) Amphibian

Question 19.
Deforestation does not lead to
(a) quick nutrient cycling
(b) soil erosion
(c) alteration of local weather condition
(d) destruction of natural habitat of wild animals
Answer:
(a) quick nutrient cycling

Question 20.
What are the ‘The Evil Quartet’ for the loss of biodiversity ?
(a) Habitat loss and fragmentation, Over exploitation, Alien species invasion, Co-extinctions
(b) Pollution, Global warming, Increasing population, Reclamation
(c) Greenhouse effect, Sea level rise, Air pollution, Deforestation
(d) Agriculture, Industrialization, Urbanization, Constructing transport facilities
Answer:
(a) Habitat loss and fragmentation, Over exploitation, Alien species invasion, Co-extinctions

Question 21.
Introduction of which aquaculture fish has created threat to the indigenous catfishes in Indian rivers ?
(a) Clarias gariepinus
(b) Arius sps.
(c) Heteropneustus Jossilis
(d) Pangasius pangasius
Answer:
(a) Clarlas gariepinus

Question 22.
The phenomena of co-extinction is observed when ………………….
(a) host fish gets extinct, its parasites also meet the same fate
(b) parasites are killed, the host too suffers
(c) host-parasite relationship is terminated
(d) parasites overpower the host
Answer:
(a) host fish gets extinct, its parasites also meet the same fate

Question 23.
Which of the following suffers due to co-extinction ?
(a) Selection of mates for reproduction
(b) Plant-pollinator mutualism
(c) Feeding preferences
(d) Prey-predator relationships
Answer:
(b) Plant-pollinator mutualism

Question 24.
The region with very high levels of species richness is called
(a) Biodiversity hotspot
(b) National Park
(c) Sanctuary
(d) Biosphere
Answer:
(a) Biodiversity hotspot

Question 25.
Which of the following does not offer ex-situ conservation to the flora and fauna ?
(a) Zoological parks
(b) Botanical gardens
(c) Sanctuaries
(d) Gene banks
Answer:
(c) Sanctuaries

Question 26.
Gametes of threatened species can be preserved in viable and fertile condition for long periods using ………………… techniques.
(a) cryopreservation
(b) tissue culture
(c) formalin preservation
(d) DNA hybridization
Answer:
(a) cryopreservation

Question 27.
Find the odd one out
(a) Seed banks
(b) Gene banks
(c) In vitro fertilization
(d) Electrophoresis
Answer:
(d) Electrophoresis

Question 28.
Chipko andolan movement is to protect the ………………….
(a) flora
(b) fauna
(c) trees
(d) rivers
Answer:
(c) trees

Question 29.
Hotspots are the examples of …………………..
(a) in-situ conservation
(b) ex-situ conservation
(c) wildlife protection
(d) water conservation
Answer:
(a) in-situ conservation

Question 30.
Which of the following is not the outcome of preserving biodiversity ?
(a) To maintain the ecological processes.
(b) To build national economy.
(c) To study life in its natural habitats.
(d) To disturb ecological balance.
Answer:
(d) To disturb ecological balance.

Question 31.
Which of the following is not an example of ex-situ conservation of biodiversity ?
(a) Botanical gardens
(b) Culture collections
(c) Zoological parks
(d) Colleges teaching courses on biodiversity
Answer:
(d) Colleges teaching courses on biodiversity

Question 32.
Cryopreservation of gametes of threatened species in viable and fertile condition can be referred to as ………………..
(a) In-situ cryo-conservation of biodiversity
(b) In-situ conservation of biodiversity
(c) Advanced ex-situ conservation of biodiversity
(d) In-situ conservation by sacred groves
Answer:
(c) Advanced ex-situ conservation of biodiversity

Question 33.
In which of the following, both pairs have correct combination ?
(a) In-situ conservation : Tissue culture Ex-situ conservation : Sacred groves
(b) In-situ conservation : National Park Ex-situ conservation : Botanical Garden
(c) In-situ conservation : Cryopreservation Ex-situ conservation : Wildlife Sanctuary
(d) In-situ conservation : Seed Bank Ex-situ conservation : National Park
Answer:
(b) In-situ conservation : National Park, Ex-situ conservation : Botanical Garden

Question 34.
The organization which publishes the Red List of species is …………………
(a) UNEP
(b) WWF
(c) ICFRE
(d) IUCN
Answer:
(d) IUCN

Question 35.
The World Biodiversity Day is observed on ………………..
(a) 22nd April
(b) 5th June
(c) 3rd March
(d) 22nd May
Answer:
(d) 22nd May

Question 36.
Which of the following expanded form of the given acronyms is correct?
(a) IPCC = International Panel for Climate Change.
(b) UNEP = United Nations Environment Policy.
(c) EPA = Environmental Pollution Agency.
(d) IUCN = International Union for Conservation of Nature and Natural Resources.
Answer:
(d) IUCN = International Union for Conservation of Nature and Natural Resources

Question 37.
The Air Prevention and Control of Pollution Act came into force in …………………
(a) 1975
(b) 1981
(c) 1985
(d) 1990
Answer:
(b) 1981

Question 38.
Which is the most dangerous and common kind of environmental pollution ?
(a) Air
(b) Water
(c) Noise
(d) Radioactive
Answer:
(a) Air

Maharashtra Board Class 12 Biology Important Questions Chapter 15 Biodiversity, Conservation and Environmental Issues

Question 39.
How does carbon monoxide, a poisonous gas emitted by automobiles, prevent transport of oxygen into the body tissues ?
(a) By destroying haemoglobin.
(b) By forming a stable compound with haemoglobin.
(c) By obstructing the reaction of oxygen with haemoglobin.
(d) By changing oxygen into carbon dioxide.
Answer:
(b) By forming a stable compound with haemoglobin

Question 40.
A scrubber in the exhaust of a chemical industrial plant removes
(a) gases like ozone and methane
(b) particulate matter of the size 2.5 micrometer or less
(c) gases like sulphur dioxide
(d) particulate matter of the size 5 micrometer or above
Answer:
(c) gases like sulphur dioxide

Question 41.
Which of the following can be considered as the most hazardous effect of air pollution ?
(a) Reduction in the growth and yield of the crop.
(b) Premature death of the plants.
(c) Effect on the monuments.
(d) Deleterious effects on the respiratory system of all animals.
Answer:
(d) Deleterious effects on the respiratory system of all animals.

Question 42.
Harmful effects of air pollution does not depend on the
(a) concentration of pollutants
(b) duration of exposure
(c) the type of organism
(d) time of the day
Answer:
(d) time of the day

Question 43.
Which equipment is most widely used for filtering out particulate matter ?
(a) Scrubber
(b) Electrostatic precipitator
(c) Filters
(d) Centrifuges
Answer:
(b) Electrostatic precipitator

Question 44.
What is the percentage of particulate matter removed from the thermal power exhaust with the help of electrostatic precipitator ?
(a) 25%
(b) 50%
(c) 80%
(d) 99%
Answer:
(d) 99%

Question 45.
Scrubber removes gases like
(a) Ozone
(b) Carbon dioxide
(c) Sulphur dioxide
(d) Methane
Answer:
(c) Sulphur dioxide

Question 46.
Which part of the electrostatic precipitator is maintained at several thousand volts ?
(a) Collection plates
(b) Electrode wires
(c) Corona
(d) Water line spray
Answer:
(b) Electrode wires

Question 47.
Which particles are responsible for causing the greatest harm to human health ?
(a) Particulates with size 1.00 micrometres in diameter.
(b) Particulates with size of 10 mm.
(c) Particulates with size 2.5 micrometres or less in diameter.
(d) Particulates with size of 100 micrometres in diameter
Answer:
(c) Particulates with size 2.5 micrometres or less in diameter.

Question 48.
According to Central Pollution Control Board (CPCB), which particulate size in diameter (in micrometers) of the air pollutants is responsible for greatest harm to human health ?
(a) 2.5 or less
(b) 1.5 or less
(c) 1.0 or less
(d) Between 2.5-5.3
Answer:
(a) 2.5 or less

Question 49.
When the exhaust passes through the catalytic converters, what happens to the unburnt hydrocarbons ?
(a) They are converted to oxygen and water.
(b) They are converted to energy to run the car.
(c) They are converted to carbon dioxide and water.
(d) They are converted to carbonates.
Answer:
(c) They are converted to carbon dioxide and water.

Question 50.
When the exhaust passes through the catalytic converters, what happens to the carbon monoxide and nitric oxide ?
(a) They are changed to carbon dioxide and nitrogen gas, respectively.
(b) They are changed to oxygen and carbon monoxide respectively.
(c) They are converted into hydrocarbons.
(d) They remain unchanged.
Answer:
(a) They are changed to carbon dioxide and nitrogen gas, respectively.

Question 51.
Motor vehicles equipped with catalytic converter should use unleaded petrol because
(a) lead causes pollution
(b) lead in the petrol inactivates the catalyst
(c) lead makes automobile machinery inefficient
(d) lead causes more consumption of petrol
Answer:
(b) lead in the petrol inactivates the catalyst

Question 52.
Which expensive metals are fitted into catalytic converters of the automobiles for reducing emission of poisonous gases ?
(a) Platinum-palladium and rhodium
(b) Silver, Gold
(c) Platinum and Gold
(d) Rhodium and Silver
Answer:
(a) Platinum-palladium and rhodium

Question 53.
Which part of the electrostatic precipitator attract the charged dust particles ?
(a) Collection plates
(b) Electrode wires
(c) Corona
(d) Dust particles
Answer:
(a) Collection plates

Question 54.
Two thirds of sulphur dioxides are produced by
(a) heating plants
(b) industrial processes
(c) automobile traffic
(d) electric power plants
Answer:
(d) electric power plants

Question 55.
Which is the worst polluted city among the world with respect to air pollution ?
(a) New York
(b) Tokyo
(c) New Delhi
(d) Dubai
Answer:
(c) New Delhi

Question 56.
Which are the other equivalent norms to Euro-II norms ?
(a) Bharat stage II
(b) Euro-III
(c) Euro-IV
(d) Air (Prevention and Control of Pollution) Act
Answer:
(a) Bharat stage II

Question 57.
Which of the following statements is inaccurate ?
(a) All automobiles should have Euro-III emission norm compliant automobiles and fuels by 2010.
(b) All automobiles and fuel-petrol and diesel – were to have met the Euro-III emission specifications in major 11 cities from April 1, 2005.
(c) All automobiles should have to meet the Euro-IV norms by April 1, 2010.
(d) Quality of Delhi air has significantly deteriorated due to all the above norms.
Answer:
(d) Quality of Delhi air has significantly deteriorated due to all the above norms.

Question 58.
What was observed within a period of 1997 and 2005 in Delhi as regards to air quality ?
(a) There was net increase in all the types of air pollutants.
(b) Delhi became totally pollution-free during this period only.
(c) There was substantial fall in concentrations of CO2 and SO2.
(d) There was decrease in the concentration of H2S and CO.
Answer:
(c) There was substantial fall in concentrations of CO2 and SO2.

Question 59.
When was Air (Prevention and Control of Pollution) Act amended to include noise as an air Pollutant ?
(a) 1981
(b) 1984
(c) 1986
(d) 1987
Answer:
(d) 1987

Question 60.
Which of the following is not the measure to reduce the noise pollution ?
(a) Delimitation of horn-free zones around hospitals and schools.
(b) Permissible sound-levels of crackers and of loudspeakers.
(c) Time limits after which loudspeakers cannot be played.
(d) Complete ban on processions playing loud percussion instruments.
Answer:
(d) Complete ban on processions playing loud percussion instruments.

Question 61.
How does CO affect plant respiration ?
(a) By yellowing leaves
(b) By closing stomatal openings
(c) By reacting with cytochrome oxidase enzyme system
(d) By causing defoliation and leaf lesions
Answer:
(c) By reacting with cytochrome oxidase enzyme system

Question 62.
Acid rains are produced by
(a) excess emissions of NO2 and SO2 from burning fossil fuels
(b) excess production of NH3 by industry and coal gas
(c) excess release of carbon monoxide by incomplete combustion
(d) excess formation of CO2 by combustion and animal respiration
Answer:
(a) excess emissions of NO2 and SO2 from burning fossil fuels

Question 63.
How much decibel sound is produced by the jet plane or rocket ?
(a) 50 dB
(b) 80 dB
(c) 150 dB
(d) 200 dB
Answer:
(c) 150 dB

Question 64.
To what decibel level noise rises during festive seasons due to crackers ?
(a) 20 dB
(b) 50 dB
(c) 100 dB
(d) 150 dB
Answer:
(c) 100 dB

Maharashtra Board Class 12 Biology Important Questions Chapter 15 Biodiversity, Conservation and Environmental Issues

Question 65.
dB is the standard abbreviation used for the quantitative expression of
(a) the density of bacteria in a medium
(b) a particular pollutant
(c) the dominant Bacillus in a culture
(d) a certain pesticide
Answer:
(b) a particular pollutant

Question 66.
Sound becomes hazardous noise pollution at level
(a) above 30 dB
(b) above 80 dB
(c) above 100 dB
(d) above 120 dB
Answer:
(b) above 80 dB

Question 67.
When was Water (Prevention and Control of Pollution) Act passed by the Government of India ?
(a) 1974
(b) 1984
(c) 1986
(d) 1992
Answer:
(a) 1974

Question 68.
A mere ………………… impurities make water contaminated with domestic sewage unfit for human use.
(a) 0.8%
(b) 0.7%
(c) 0.5%
(d) 0.1%
Answer:
(d) 0.1%

Question 69.
What is the outcome of algal bloom ?
(a) Lots of algae available for fodder.
(b) Deterioration of water quality and fish mortality.
(c) Cleaning up of the ambient water.
(d) Decrease in BOD amount.
Answer:
(b) Deterioration of water quality and fish mortality

Question 70.
When there are excessive microorganisms in the water that cause biodegradation there is
(a) sharp rise in the dissolved oxygen content
(b) sharp decline of dissolved oxygen content
(c) refreshing odour to the water
(d) loss of algal population
Answer:
(b) sharp decline of dissolved oxygen content

Question 71.
Which is world’s most problematic aquatic weed ?
(a) Hydrilla
(b) Pistia
(c) Eichhornia crassipes
(d) Duckweed
Answer:
(c) Eichhornia crassipes

Maharashtra Board Class 12 Biology Important Questions Chapter 15 Biodiversity, Conservation and Environmental Issues

Question 72.
Which two substances are well-known for biomagnification ?
(a) Mercury and DDT
(b) Cadmium and Lead
(c) Petroleum hydrocarbons and sewage
(d) Paper manufacturing effluents and copper
Answer:
(a) Mercury and DDT

Question 73.
Choose the correct statement
(a) Concentration of DDT in the water declined with the passing time.
(b) Concentration of DDT in the water remains the same over many years without any effect.
(c) If concentration of DDT starts at 0.003 ppb in water, it can ultimately reach 25 ppm in fish-eating birds.
(d) If concentration of DDT is 25 ppm in water, in fish-eating bird population it reduces to 0.003 ppb later.
Answer:
(c) If concentration of DDT starts at 0.003 ppb in water, it can ultimately reach 25 ppm in fish-eating birds

Question 74.
Which major pollutant is released from electricity generating units ?
(a) heated water
(b) arsenic
(c) cadmium
(d) ammonia
Answer:
(a) heated water

Question 75.
Which of the statement is incorrect with reference to thermal waste water ?
(a) Thermal wastewater eliminates or reduces the number of organisms sensitive to high temperature.
(b) Thermal wastewater may enhance the growth of plants and fish in extremely cold areas.
(c) Thermal wastewater causes damage to the indigenous flora and fauna.
(d) Thermal waste water is not an important category of pollutants.
Answer:
(d) Thermal waste water is not an important category of pollutants.

Question 76.
Choose the incorrect statement from the following statements
(a) Ecological sanitation is a sustainable system for handling human excreta.
(b) One can save lot of water if flush is not used for sanitation.
(c) Ecological sanitation is a practical, hygienic, efficient and cost-effective solution to human waste disposed.
(d) Human excreta cannot be recycled into any resource or natural and safe fertilizer.
Answer:
(d) Human excreta cannot be recycled into any resource or natural and safe fertilizer.

Question 77.
Eutrophication is caused by
(a) acid rain
(b) nitrates and phosphates
(c) sulphates and carbonates
(d) CO2 and CO
Answer:
(b) nitrates and phosphates

Question 78.
Measuring Biochemical Oxygen Demand (BOD) is a method used for
(a) estimating the amount of organic matter in sewage water.
(b) working out the efficiency of oil driven automobile engines.
(c) measuring the activity of Saccharomyces cerevisae in producing curd on a commercial scale.
(d) working out the efficiency of RBCs about their capacity to carry oxygen.
Answer:
(a) estimating the amount of organic matter in sewage water.

Question 79.
High value of BOD (Biochemical Oxygen Demand) indicates that
(a) consumption of organic matter in the water is higher by the microbes
(b) water is pure
(c) water is highly polluted
(d) water is less polluted
Answer:
(c) water is highly polluted

Question 80.
When huge amount of sewage is dumped into a river, its BOD will
(a) increase
(b) decrease
(c) sharply decrease
(d) remain unchanged
Answer:
(a) increase

Question 81.
Which river of India is considered as an unending sewer ?
(a) Mula in Pune
(b) Panchaganga in Kolhapur
(c) Ganga from Haridwar to Kolkata
(d) Patalganga in Panvel
Answer:
(c) Ganga from Haridwar to Kolkata

Question 82.
Sewage drained into water bodies kill fishes because
(a) excessive carbon dioxide is added to water
(b) it gives off a bad smell
(c) it removes the food eaten by the fish
(d) it increases competition fishes for dissolved oxygen
Answer:
(d) it increases competition fishes for dissolved oxygen

Question 83.
Which method was commonly practised for managing solid waste generated by municipal bodies ?
(a) Open dumps
(b) Open burning dumps
(c) Accumulation in trenches
(d) Accumulation in water bodies
Answer:
(b) Open burning dumps

Question 84.
Biochemical Oxygen Demand (BOD) may not be a good index for pollution of water bodies receiving effluents from
(a) domestic sewage
(b) dairy industry
(c) petroleum industry
(d) sugar industry
Answer:
(c) petroleum industry

Question 85.
What is the appropriate scientific method for waste disposed ?
(a) Land fill
(b) Open burning dump
(c) Sanitary landfill
(d) Open dumps (Junk yards)
Answer:
(c) Sanitary landfill

Question 86.
Which statement correctly describes the process of waste disposal in sanitary landfill ?
(a) Solid wastes are dumped in a trench or depression.
(b) Solid wastes are dumped in a depression and compacted.
(c) Solid wastes are dumped in a trench, compacted and covered by soil.
(d) Solid wastes are dumped in a trench and burnt to reduce volume.
Answer:
(c) Solid wastes are dumped in a trench, compacted and covered by soil.

Question 87.
Which is adverse effect of sanitary land fill noticed occasionally?
(a) Breeding place for rats and flies.
(b) Seepage of chemicals polluting ground water.
(c) Burning of hazardous waste.
(d) Accumulation of biodegradable materials.
Answer:
(b) Seepage of chemicals polluting ground water.

Question 88.
From the following, which is not a category of sorting of waste ?
(a) Recyclable
(b) Biodegradable
(c) Non-biodegradable
(d) Explosive
Answer:
(d) Explosive

Question 89.
From the following, which is a recyclable waste ?
(a) Food waste
(b) Newspaper
(c) Leather
(d) Rubber
Answer:
(b) Newspaper

Question 90.
Which is appropriate method for disposal of hospital wastes ?
(a) Sanitary landfills
(b) Open dumps
(c) Use of incinerators
(d) Composting
Answer:
(c) Use of incinerators

Question 91.
Which is valid suggestion for responsible citizen to deal with managing non- biodegradable waste ?
(a) Mixing of biodegradable and recyclable waste material.
(b) Mixing of biodegradable and non- biodegradable waste material.
(c) Use of more and more disposable material.
(d) Use of cloth bags and avoid plastic carry bags.
Answer:
(d) Use of cloth bags and avoid plastic carry bags.

Question 92.
For treatment of e-waste, which is the most suitable solution ?
(a) Recycling and recovery
(b) Buried in landfills
(c) Incineration
(d) Disposal and storage in open
Answer:
(a) Recycling and recovery

Question 93.
Which metals are recovered from recycling of E-waste ?
(a) Copper, iron, silicon, nickel and gold.
(b) Platinum, aluminium, silicon, silver and rhodium.
(c) Sodium, iron, silicon, uranium and potassium.
(d) Copper, radium, zinc, cobalt and titanium.
Answer:
(a) Copper, iron, silicon, nickel and gold.

Question 94.
Greenhouse effect is warming due to
(a) infra-red rays reaching earth
(b) moisture layer in the atmosphere
(c) increase in temperature due to increase in carbon dioxide concentration of the atmosphere
(d) ozone layer in the atmosphere
Answer:
(c) increase in temperature due to increase in carbon dioxide concentration of the atmosphere

Question 95.
Which one of the following is the correct percentage of the two (out of the total of 4) greenhouse gases that contribute to the total global warming ?
(a) CFCs 14%, Methane 20%
(b) CO2 40%, CFCs 30%
(c) N2O 6%, CO2 86%
(d) Methane 20%, N2O 18%
Answer:
(a) CFCs 14%, Methane 20%

Question 96.
The two gases making highest relative contribution to the greenhouse gases are
(a) CO2 and CH4
(b) CH4 and N2O
(c) CFCs and N2O
(d) CO2 and N2
Answer:
(a) CO2 and CH4

Maharashtra Board Class 12 Biology Important Questions Chapter 15 Biodiversity, Conservation and Environmental Issues

Question 97.
Which is the chief reason of global warming ?
(a) Greenhouse effect
(b) Absorption of UV radiations by ozone
(c) Effect of visible light
(d) Trapping of radio waves
Answer:
(a) Greenhouse effect

Question 98.
Why naturally occurring greenhouse effect is important?
(a) It maintains the temperature of earth at average 15 °C.
(b) It maintains the temperature of earth at 18 °C.
(c) It maintains lot of greenery on the surface of the earth.
(d) It maintains level of ozone in atmosphere.
Answer:
(a) It maintains the temperature of earth at average 15 °C.

Question 99.
Clouds and gases reflect about ………………… of the incoming solar radiation.
(a) one half
(b) one-fourth
(c) one tenth
(d) three-fourth
Answer:
(b) one-fourth

Question 100.
Montreal Protocol aims at ……………………
(a) Biodiversity conservation
(b) Control of water pollution
(c) Control of CO2 emission
(d) Reduction of ozone depleting substances
Answer:
(d) Reduction of ozone depleting substances

Question 101.
Which are the four most affected aspects due to climate change caused due to global warming?
(a) Energy, Agricultural research, Sewage disposed, Entertainment.
(b) Food supply, Water, Health, Infrastructure.
(c) Political views, Equality, Natural Resources, Safety of women.
(d) Education, Empowerment of people, Shelter, Processed food.
Answer:
(b) Food supply, Water, Health, Infrastructure.

Question 102.
What is exactly measured in Dobson units or DU?
(a) The thickness of the ozone in a column of air from the ground to the top of the Atmosphere.
(b) The noise level in the circumscribed area.
(c) The amount of chlorofluorocarbons.
(d) The hole in the ozone umbrella.
Answer:
(a) The thickness of the ozone in a column of air from the ground to the top of the Atmosphere.

Question 103.
‘Good ozone’ is found in the while the bad ozone is in ………………….
(a) Mesosphere, Ionosphere
(b) Mesosphere, Troposphere
(c) Stratosphere, Troposphere
(d) Stratosphere, Ionosphere
Answer:
(c) Stratosphere, Troposphere

Question 104.
UV-B does not cause ……………………
(a) aging of skin
(b) damage to skin cells
(c) various types of skin cancers
(d) albinism or lightning of the skin
Answer:
(d) albinism or lightning of the skin

Question 105.
What is snow-blindness cataract ?
(a) Cataract noticed in people living in snow clad areas.
(b) Cataract that shows symptom of white dense patch.
(c) Cataract resulted due to inflammation of cornea.
(d) Cataract that causes total blindness.
Answer:
(c) Cataract resulted due to inflammation of cornea.

Question 106.
Which policy is introduced by Government of India to conserve forests effectively with local people ?
(a) Wildlife protection
(b) Chipko movement
(c) Joint forest movement
(d) Joint tree plantation
Answer:
(c) Joint forest movement

Question 107.
Name the award declared by Government of India to motivate people for protecting wildlife.
(a) Amrita Devi-Bishnoi Tree Protection Award.
(b) Amrita Devi – Bishnoi Wildlife Protection Award.
(c) Amrita Devi-Chipko Movement Award.
(d) Bahuguna – Chipko Movement Award.
Answer:
(b) Amrita Devi – Bishnoi Wildlife Protection Award.

Match the columns

Question 1.

Column A Column B
(1) Walter Rosen (a) Popularisation of term biodiversity
(2) David Tillman (b) Rivet Popper Hypothesis
(3) Paul Ehrlich (c) Productivity Stability Hypothesis
(4) Edward Wilson (d) Coined

Answer:

Column A Column B
(1) Walter Rosen (d) Coined
(2) David Tillman (c) Productivity Stability Hypothesis
(3) Paul Ehrlich (b) Rivet Popper Hypothesis
(4) Edward Wilson (a) Popularisation of term biodiversity

Question 2.

Column I (Phenomena) Column II (Effect)
(1) Eutrophication (a) Soil erosion
(2) Biomagnification (b) Prevention of extinction
(3) Conservation (c) Accumulation of non-biodegradable substance
(4) Deforestation (d) Death of aquatic ecosystem

Answer:

Column I (Phenomena) Column II (Effect)
(1) Eutrophication (d) Death of aquatic ecosystem
(2) Biomagnification (c) Accumulation of non-biodegradable substance
(3) Conservation (b) Prevention of extinction
(4) Deforestation (a) Soil erosion

Maharashtra Board Class 12 Biology Important Questions Chapter 15 Biodiversity, Conservation and Environmental Issues

Question 3.

Indian region shows Number
(1) Biosphere reserves (a) 448
(2) National parks (b) 14
(3) Wildlife sanctuaries (c) 90

Answer:

Indian region shows Number
(1) Biosphere reserves (b) 14
(2) National parks (c) 90
(3) Wildlife sanctuaries (a) 448

Classify the following to form Column B as per category given in Column A.

Question 1.
Karnataka, Chanda, Khasi, Rajasthan, Sarguja, Jaintia, Maharashtra, Bastar.

Region Sacred groves seen at
(1) Meghalaya —————
(2) Western ghat regions —————
(3) Aravali hills —————
(4) Madhya Pradesh ————–

Answer:

Region Sacred groves seen at
(1) Meghalaya Khasi, Jaintia
(2) Western ghat regions Karnataka,
(3) Aravali hills Maharashtra
(4) Madhya Pradesh Rajasthan, Bastar

Question 2.
Dust, Carbon monoxide, Lead, Smog, Methane, Mercury, DDT, Cadmium.

Column A Column B (Examples)
(1) Particulate pollutant —————
(2) Gaseous pollutant —————
(3) Biomagnification —————
(4) Heavy metals ————–

Answer:

Column A Column B (Examples)
(1) Particulate pollutant Dust, Smog
(2) Gaseous pollutant Carbon monoxide, Methane
(3) Biomagnification Mercury, DDT
(4) Heavy metals Lead, Cadmium

Very short answer questions

Question 1.
Which are the biodiversity hotspots in India?
Answer:
India has three of world’s biodiversity viz. Western Ghats, Indo-Burma and Eastern- Himalayas.

Question 2.
How many national parks and sanctuaries are present in Maharashtra?
Answer:
In Maharashtra, there are 5 national parks and 11 sanctuaries.

Question 3.
What is included under biodiversity?
Answer:
Biodiversity includes a vast array of species of microorganisms – viruses, algae, fungi, plants and animals occurring in all the different habitats on Earth and forming the ecological complexes.

Question 4.
Who coined the biodiversity?
Answer:
The term biodiversity was coined by Walter Rosen in 1982.

Question 5.
Who popularised the term biodiversity?
Answer:
Edward Wilson popularized the term biodiversity to describe combined diversity at all the levels of biological organisation.

Question 6.
How long it has taken for Biodiversity to form on the earth?
Answer:
Biodiversity which is currently present took over 3.5 billion of years of evolutionary history to form on earth.

Question 7.
Where does India stand as far as species diversity is concerned?
Answer:
Answer:
India is one among 15 nations that are rich in species diversity.

Question 8.
Which habitats do not show latitudinal and altitudinal gradients?
Answer:
Arid and Semiarid habitats and aquatic habitat do not show latitudinal and altitudinal gradient.

Question 9.
What is the relationship between species richness and latitudinal gradient?
Answer:
Species richness is high at lower latitudes and there is a steady decline towards the poles, i.e. species richness for plants and animals decreases as we move away from equator to the poles.

Question 10.
What is the relationship between species diversity and altitude?
Answer:
Species diversity is more at lower altitudes than at the higher altitude.

Question 11.
At which latitude there is maximum diversity?
Answer:
Biodiversity is maximum in tropical rain forests at equator.

Question 12.
Enumerate the biodiversity in Amazon rain forest
Answer:
The world’s largest tropical rainforest of Amazon, there are around 40,000 plant species, nearly 1,300 bird species, 3,000 types of fish, 427 species of mammals and more than 1,25,000 invertebrates.

Question 13.
How many species have been documented as per IUCN data of 2004?
Answer:
Over 1.5 million species have been documented as per IUCN data (2004).

Question 14.
How much of global biodiversity wealth has been recorded as per May’s estimate?
Answer:
As per May’s estimate of global biodiversity, there are 22% of our natural wealth recorded.

Question 15.
Which are the activities of humans that could cause loss of biodiversity even before they are recorded?
Answer:
Human activities like reclamation and deforestation can cause loss of varieties even before they are identified and recorded.

Question 16.
Which one is considered the sixth extinction? Why?
Answer:
The current loss of biodiversity is considered to be the sixth extinction. It is considered so because the loss of biodiversity is progressing at an alarming rate of 100 to 1000 times faster than pre-human times.

Question 17.
Enlist the causes of Biodiversity losses.
Answer:
There are four major causes of biodiversity losses which are popularly known as, ‘The Evil Quartet’ which are habitat loss and fragmentation, over exploitation, alien species invasion and co-extinction.

Maharashtra Board Class 12 Biology Important Questions Chapter 15 Biodiversity, Conservation and Environmental Issues

Question 18.
Why alien species of animals become invasive and harmful to local species?
Answer:
There is lack of local predator for the invasive species, therefore, the invasive species cannot be controlled.

Question 19.
What is the work of The International Union for Conservation of Nature and Natural Resources (IUCN)?
Answer:
The International Union for Conservation of Nature and Natural Resources or IUCN maintains a Red Data Book or Red list which includes records of conservation status of plant and animal species.

Question 20.
What was the main reason of loss of natural resources in last ten decades?
Answer:
Human population has grown exponentially along with industrial development, both of these have resulted in the rampant loss of natural resources in last ten decades.

Question 21.
Which Act was passed to protect and improve quality of environment?
Answer:
In order to protect and improve the quality of our environment, the Government of India has passed the Environment Protection Act 1986.

Question 22.
Which is the Indian law that includes noise as an air pollutant?
Answer:
In India, the Air (Prevention and Control of Pollution) Act 1981, Amendment 1987, includes noise as an air pollutant.

Question 23.
What are the sources of noise pollution?
Answer:
The common sources of noise pollution are machines, transportation, construction sites, industries, etc.

Question 24.
How do we identify polluted water?
Answer:
Polluted water is usually turbid, foul smelling, coloured and contains number of pathogens, heavy metals, oils, etc.

Question 25.
What is the most common source of water pollution?
Answer:
Domestic sewage is one of the most common sources of water pollution.

Question 26.
What is algal bloom?
Answer:
Algal bloom is excessive growth of planktonic freely floating blue-green algae caused due to presence of large amount of nutrients in water.

Question 27.
Why is algal bloom considered to be bad?
Answer:
Algal bloom makes the water coloured and unpotable, releasing toxins in water which can kill the fish.

Question 28.
Why water pollution act was passed in India? When was it passed?
Answer:
When Government realised the importance of maintaining the cleanliness of the water bodies, the Water (Prevention and Control of Pollution) Act was passed in year 1974 to safeguard our water resources.

Question 29.
Which substances can cause biomagnification ?
Answer:
Those substance like some pesticides which are non-biodegradable and those which get accumulated in the tissues of living organisms without metabolism or excretion, cause biomagnification.

Question 30.
Which method of recycling of sewage is used in Tirumala hills?
Answer:
At Tirumala hills there are reverse osmosis units set up to recycle sewage water, which helps in solving the huge water demand.

Question 31.
What is the advantage of recycling of sewage water by reverse osmosis (RO)?
Answer:
Recycling sewage water by RO System helps to solve the problem of scarcity of water and also disposal of sewage water.

Question 32.
Which systems are now made mandatory by Municipal Corporation for new constructions ?
Answer:
Rainwater harvesting is made mandatory for new constructions by Municipal Corporation.

Question 33.
Which ban was imposed by Maharashtra Government on 23rd June 2018?
Answer:
Maharashtra government sent a notification to ban use, sale, distribution and storage of plastic material to fight pollution caused due to extensive use of plastic.

Question 34.
What are biomedical wastes?
Answer:
The harmful wastes generated by the hospitals that contains disinfectants, harmful chemicals, discarded body parts, blood and also pathogenic microorganisms are called biomedical wastes.

Question 35.
Why recycling of e-wastes is considered dangerous?
Answer:
In developing countries, due to lack of facilities for recycling of e-waste, it is done by manual methods, which is harmful as the workers are exposed to toxic substances from e-waste.

Question 36.
Which are commonly called greenhouse gases?
Answer:
CO2 and methane are commonly called greenhouse gases.

Question 37.
What are Dobson units?
Answer:
Thickness of the ozone in a column of air from the ground to the top of the atmosphere is measured as Dobson units (DU).

Question 38.
If ozone layer is intact, which UV radiations are absorbed by earth’s atmosphere?
Answer:
UV radiations of wavelength shorter than UV-B i.e. 100-280 nm are almost completely absorbed by earth’s atmosphere, given that the ozone layer is intact.

Question 39.
What is the aim of National Forest Policy?
Answer:
National Forest Policy (NFP) aims at maintaining 33% forest cover in the country.

Question 40.
How much forest cover is being lost due to deforestation?
Answer:
Almost 40% tropical forests and 1% temperate forests are lost due to deforestation.

Give definition of the following

Question 1.
Biodiversity
Answer:
Biodiversity is part of nature which includes the differences in the genes among the individuals of a species; the variety and richness of all plants and animal species at different scales in a space – local regions, country and the world; and the types of ecosystem, both terrestrial and aquatic, within a defined area.

Question 2.
Extinct species
Answer:
The species which gets totally eliminated from the earth is called extinct species.

Question 3.
Endangered species
Answer:
When the number of members of a species starts dwindling, it is said to be endangered species.

Question 4.
Invasive species
Answer:
The species that does not belong to the region or locality but is introduced accidentally or intentionally which causes harmful effects to the already existing local species, are called invasive species.

Question 5.
Bioprospecting
Answer:
Bioprospecting is systematic search for development of new sources of chemical compounds, genes, microorganisms, macroorganism and other valuable products from nature.

Question 6.
Biochemical Oxygen Demand (BOD)
Answer:
BOD is the amount of dissolved oxygen required by microorganisms for decomposing the organic matter present in water which is expressed in milligram of oxygen per litre (mg/L) of water.

Question 7.
Natural Eutrophication
Answer:
Natural eutrophication is the process of aging of a lake due to nutrient enrichment of water.

Question 8.
Cultural or Accelerated eutrophication
Answer:
The process of aging of the water body due to pollutants from human activities such as effluents from agricultural lands, industries and homes (household).

Maharashtra Board Class 12 Biology Important Questions Chapter 15 Biodiversity, Conservation and Environmental Issues

Question 9.
Biological Magnification (Biomagnification) :
Answer:
Biological magnification is the phenomenon through which certain pollutants get accumulated in tissues in increasing concentration along the food chains in successive trophic levels.

Question 10.
Deforestation
Answer:
Deforestation is conversion of forest area into non-forest area.

Question 11.
Reforestation
Answer:
Reforestation is the natural process of restoring a forest that once existed but was destroyed or removed at some time in past.

Question 12.
Chipko Movement
Answer:
Chipko Movement is people’s participation for the protection of trees in which people hug the trees and save it from the axe of tree-cutters. Initially it happened in 1974 in Garhwal region of Himalayas and now it is spread world-wide.

Question 13.
Joint Forest Management (JFM)
Answer:
Joint Forest Management is an attempt to conserve forests in a sustainable matter which has been introduced by the Government of India in 1980s for working with the local communities for protection and management of the forests.

Name the following/Give examples

Question 1.
Levels of biodiversity
Answer:

  1. Genetic diversity
  2. Species diversity or community diversity
  3. Ecosystem diversity or ecological diversity

Question 2.
Two patterns of biodiversity
Answer:

  1. Latitudinal and Altitudinal gradient
  2. Species-area relationship.

Question 3.
Three types of extinction
Answer:

  1. Natural extinction
  2. Mass extinction
  3. Manmade (anthropogenic) extinction

Question 4.
Three examples of animals that are now extinct due to over-exploitation
Answer:

  1. Dodo
  2. Stellar sea cow
  3. Passenger pigeon

Question 5.
Types of air pollutants
Answer:

  1. Particulate pollutants
  2. Gaseous pollutants.

Question 6.
Name the different types of pollution.
Answer:

  1. Water pollution
  2. Air pollution
  3. Radioactive pollution
  4. Noise pollution
  5. Soil pollution.

Question 7.
Main types of wastes
Answer:

  1. Biodegradable
  2. Recyclable
  3. Non-biodegradable.

Distinguish between the following

Question 1.
Genetic diversity and Species diversity.
Answer:

Genetic diversity Species diversity
(1) Genetic diversity is intraspecific diversity. (1) Species diversity is interspecific diversity.
(2) Genetic diversity is due to number and types of genes and chromosomes present in different species. (2) Species diversity is due to number of species of plants and animals that are present in a region.
(3) Genetic diversity includes variations in genes, alleles and chromosomes (3) Species diversity includes species richness and species evenness.

Question 2.
In-situ and ex-situ conservation.
Answer:

In-situ conservation Ex-situ conservation
(1) In-situ conservation is a onsite conservation. (1) Ex-situ conservation is done outside the habitat of plants and animals.
(2) Plant and animal species are conserved in their natural habitat for protecting endangered species. (2) Plant and animal species are conserved in artificial or manmade place.
(3) It is done in natural environment. (3) It is done in manmade environment.
(4) National parks, Sanctuaries, biosphere reserve, etc. are set up for in-situ conservation. (4) Zoo, aquarium, seed banks are the examples of ex-situ conservation
(5) It is a dynamic process. Cheap and convenient to conduct. (5) It is static process. Its expensive and commercial process.
(6) Captive breeding is not successful in all cases of in-situ conservation method. (6) Captive breeding is successful and can help in increasing the number of endangered organisms.

Give reasons

Question 1.
There is decrease in species diversity at higher altitude.
Answer:

  1. At higher altitudes, there are different climatic conditions such as drastic season variations and lesser ambient temperature.
  2. Survival of organisms thus becomes difficult. Therefore, at such altitudes, species diversity decreases.

Question 2.
Loss of biodiversity leads to the overall imbalance in the ecosystem.
Answer:

  1. Loss of biodiversity in any area leads to the decline in plant production.
  2. There is lower resilience to environmental disturbance like flood.
  3. It may also lead to alteration in environmental processes like disease cycles, plant productivity, etc.
  4. The food chains and food webs are disturbed.
  5. The productivity of the ecosystem is reduced and this results into overall imbalance in the ecosystem.

Question 3.
Motor vehicles equipped with catalytic converter should use unleaded petrol.
Answer:

  1. There is lead in the petrol which can inactivate the catalyst present in catalytic converter.
  2. Due to this inactivation catalytic converter will not work properly. Therefore motor vehicles equipped with catalytic converter should use unleaded petrol.

Question 4.
Using CNG as a fuel is more eco-friendly.
Answer:

  1. Diesel or petrol cause air pollution.
  2. On the other hand, CNG is advantageous because it is cheaper and thus people find it economical to use CNG.
  3. CNG burns efficiently and causes lesser pollution.
  4. CNG cannot be adulterated. Thus, it is, the only adulteration-proof fuel which is eco-friendly.

Question 5.
High BOD indicates intense level of microbial pollution.
Answer:

  1. When BOD of a water sample is high, it denotes that the amount of dissolved oxygen in the water which is required by the microorganisms to decompose the organic matter in that water is high.
  2. This shows that there are many microbes in the water body. Thus, high BOD indicates intense level of microbial pollution.

Question 6.
If microorganisms are more in a water body, other aquatic creatures find it difficult to survive.
Answer:

  1. Microorganisms involved in biodegradation of organic matter in water body consume lot of dissolved oxygen.
  2. Due to this consumption there is sharp decline in oxygen level of water which leads to mortality of fish and other aquatic creatures. Therefore, other aquatic creatures find it difficult to survive.

Question 7.
Lake can literally get choked to death due to eutrophication.
Answer:

  1. Eutrophication means enrichment due to nutrients.
  2. Pollution due to human activities releases effluents through agricultural lands, industries and houses.
  3. Many of these contain phosphates and sulphates which cause excessive eutrophication.
  4. This leads to non-availability of oxygen for other aquatic organisms, mainly causing death of fish.
  5. The processes of decomposition of these dead fish adds to further depletion of oxygen and hence lake can literally get choked to death due to eutrophication.

Question 8.
Water hyacinth is called ‘Terror of Bengal’.
Answer:

  1. Water hyacinth (Eichhornia crassipes) is native plant of amazon basin which was introduced in India for its beautiful, purple flowers.
  2. It has become an invasive species.
  3. This plant is a nuisance as it grows excessively and covers entire water body in which it is present.
  4. It grows faster than our ability to remove it, so it is commonly called ‘Terror of Bengal’.

Question 9.
Most of the water pollution is manmade.
Answer:

  1. There Eire many activities of human beings which dump the wastes into water bodies.
  2. The industrial processes also cause dumping of hazardous waste into surrounding water bodies.
  3. Human activities is the only factor that causes water pollution and there are no natural causes for water pollution.

Question 10.
There is steady concentration of ozone in the stratosphere.
Answer:

  1. Ozone is a form of oxygen which is photo-dissociated and is generated by absorption of short wavelength UV radiations.
  2. Both generation and dissociation of ozone is in equilibrium.
  3. This is the equation which shows these conversions.
    O3 → O2 + [O]
    O2 + [O] → UV RAYS → O3
  4. Therefore, there is steady concentration of ozone in the stratosphere.

Question 11.
The CO2 has crucial role in global warming.
Answer:
(1) Carbon dioxide is produced by many activities of human beings such as destruction of forests, combustion of fossil fuels, cement plants and other industries, burning and respiration by all living organisms.

(2) This CO2 forms a layer in the upper atmosphere.

(3) When solar energy reaches the earth surface, the infrared radiations from this are trapped by the layer of CO2.

(4) Along with CO2 other gases such as methane, CFCs and nitrogen oxides also form a blanket in the atmosphere which traps the reflected infrared rays.

(5) This results in greenhouse effect and global warming. CO2 alone can increase the temperature by about 50%. During the last 50 years the average temperature of the earth has increased due to steadily increasing CO2 concentration. Thus, CO2 plays a crucial role in global warming.

Maharashtra Board Class 12 Biology Important Questions Chapter 15 Biodiversity, Conservation and Environmental Issues

Question 12.
Global warming is caused by ‘greenhouse effect’.
Answer:

  1. Carbon dioxide along with methane, nitrogen oxides and CFCs can absorb infrared radiations reflected from the earth’s surface.
  2. The blanket formed by these gases in the atmosphere traps the reflected infrared rays and produces heat on the earth’s surface which results in greenhouse effect.
  3. The greenhouse effect in turn causes global warming.

Question 13.
Ozone present in the stratosphere is called good ozone.
Answer:

  1. The ozone present in the upper atmospheric region, i.e. in the stratosphere, absorbs the ultraviolet radiations present in the sunlight.
  2. These radiations are harmful for living organisms.
  3. Since ozone protects the living organisms from such dangerous UV radiations, therefore, the ozone present in the stratosphere is called good ozone.

Question 14.
The UV radiations are injurious.
Answer:

  1. When ultraviolet (UV) radiation falls on the cells of living organisms it is absorbed in the DNA and proteins present in the nucleus.
  2. The high energy of UV radiations break the chemical bonds within these molecules.
  3. This results in damage to the skin cells and cause skin cancers of various types.
  4. High doses of UV-B radiations also cause inflammation of cornea called snow blindness, cataract, etc. The UV radiations therefore are injurious.

Question 15.
Plastic ban in Maharashtra is an essential step.
Answer:
1. Plastic is non-biodegradable and man-made substance which cannot be decomposed naturally.

2. If burnt it causes toxic fumes. If buried it will contaminate the soil. If thrown anyhow, it can damage other animals like cattle. If thrown in water bodies, it kills the aquatic organisms. It clogs the water outlets and can cause flooding of cities during rainy season. Therefore, disposal of plastic has become a major issue.

3. On the contrary, the users of plastic have increased tremendously due to ease in using the plastic items.

4. Because plastic causes damage to ecosystem, it was very essential to curtail its use. Banning its use reduces is helpful in managing the excessive and unnecessary use, therefore, plastic ban in Maharashtra is an essential step.

Write short notes

Question 1.
Ecological (Ecosystem) diversity.
Answer:

  1. Different types of ecosystems or habitats within a given geographical area forms ecosystem diversity.
    On Earth there are a large variety of ecosystems.
  2. Each ecosystem has its own complement of distinctive interlinked species, based on the differences in the habitat. It is also specific for particular geographical region.
  3. In one region, generally, there may be one or many different types of ecosystems.
  4. In India, there are varieties of ecosystems such as deserts, rain forests, deciduous forests, estuaries, wetlands, grasslands, etc.
  5. In India, the Western Ghats show great ecosystem diversity. However, regions like Ladakh and Rann of Kutch have less ecosystem diversity.

Question 2.
Habitat loss and fragmentation.
Answer:

  1. Habitat loss and fragmentation is the prime cause of destruction of biodiversity.
  2. Due to degradation and pollution there is reduction in vast natural habitats. This creates crisis situation for resident living organisms.
  3. This is largely due to human activities.
  4. There is also a threat to migratory birds and for animals which need larger territories.
  5. Reduction in tropical rain forests has been reduced from 14% to 6% over the years, which has surely destroyed many species.

Question 3.
Over-exploitation.
Answer:

  1. Humans have exploited natural resources beyond their needs.
  2. The excessive consumption and accumulation have resulted into problem of over¬exploitation.
  3. Overexploitation of resources has caused threats to various organisms.
  4. Dodo bird, stellar sea cow and passenger pigeon are extinct due to overexploitation.
  5. Over exploitation of fish from sea has also resulted into dearth of fish.

Question 4.
Alien species invasion.
Answer:

  1. When invasive species are accidentally or intentionally introduced into a particular region which causes extinction of local and already existing species, it is called alien species invasion.
  2. Examples of such invasive plant species are
    (a) the carrot grass (Parthenium)
    (b) Lantana(c) Water hyacinth (Eichhornia).
  3. Invasive animal species is African catfish Clarias gariepinus, introduced for aquaculture purpose. This has caused harm to endemic catfish varieties.
  4. Invasion by such species is one of the major reasons for extinction of local species.

Question 5.
Co-extinctions.
Answer:

  1. When one organism is associated with other one in an obligatory way, then, if one is extinct, the other also gets extinct.
  2. Extinction of one variety leads to loss of associate variety from the ecosystem. Such phenomena is called co-extinction.
  3. Extinction of host fish causes extinction of unique parasites.
  4. Coevolved plant-pollinator also will have such a threat.

Question 6.
Write a note on BD Act 2002.
Answer:

  1. The law broadly defines biodiversity.
  2. It includes plants, animals and microorganisms and their parts, their genetic materials and by-products.
  3. It excludes value added products and human genetic material.
  4. Regulation of access to Indian biological resources as well as scientific cataloguing of traditional knowledge about ethnobiological materials were the main objectives for proposing this Act.
  5. There is three-tier system, viz. National Biodiversity Authority (NBA) at the national level, the State Biodiversity Boards (SBBs) at the state level and Biodiversity Management Committees (BMCs) at the local level that gives approval of utilization of any biological resource for commercial or research purpose.
  6. It is mandatory for foreigners, NRIs as well as Indian citizens and institutions to seek permission from NBA before exploiting local resource.
  7. NBA has powers of civil court. Not seeking approval of NBA, can incur jail and fine up to 10 lakh rupees.

Question 7.
Particulate air pollutants.
Answer:

  1. Particulate air pollutants are either solids or liquids.
  2. Particles having larger diameter of 10 pm settle in the soil but finer particles with 1 pm or less remain suspended in the air.
  3. Central Pollution Control Board (CPCB) has declared that, particulate matter of size 2.5 pm or less in diameter (PM 2.5) are responsible for causing the greatest harm to humans.
  4. These fine particulates can be inhaled deep into the lungs and are responsible for irritation, inflammation and damage to lungs.
  5. In addition to this, it causes breathing and respiratory disorders and premature deaths.
  6. Examples of particulate pollutants are : Smoke, smog, pesticides, heavy metals, dust and radioactive elements.

Question 8.
Gaseous pollutants.
Answer:
(1) Gaseous pollutants are gases which cause air pollution.

(2) Some common gaseous pollutants are CO2, CO, SO2, NO, NO2, etc.

(3) Carbon dioxide (CO2) : It is a greenhouse gas which is continuously produced due to human activities such as burning of fossil fuels and rampant deforestation. Photosynthesis carried out by plants can balance CO2 : O2 ratio in the air, provided there is good green cover. CO2 is also removed from the air by weathering of silicate rocks forming limestone. Aeroplane traffic especially release lot of CO2. CO2 is the main cause for global warming and climate change.

(4) Carbon monoxide (CO) : CO is a toxic gas produced by incomplete combustion of different fuels. Therefore, vehicular exhausts are the largest source of CO.

(5) Nitrogen dioxide (NO2) and nitrogen monoxide (NO) : Nitrogen oxides are released through automobiles and chemical industries as waste gases. NOa can form nitric acid after reacting with water vapour, this causes irritation to eyes and lungs. Injury to lungs, liver and kidneys is caused due to these gases.

Question 9.
Thermal pollution.
Answer:

  1. Thermal pollution of water is caused when heated water is added to the water body which results into rise in temperature of water.
  2. Thermal and nuclear power plants cause thermal pollution of adjoining water bodies.
  3. The power plants use water as coolant and release back this hot water.
  4. Many resident organisms which are sensitive to temperature die due to sudden rise in temperature.
  5. This leads to loss of flora and fauna of the water body.

Question 10.
Ecosan.
Answer:

  1. Ecological sanitation (Ecosan) is a sanitation provision that safely reuses excreta in agriculture as a manure.
  2. By Ecosan the need for chemical fertilizers is reduced. Ecosan toilet is a closed system without water and it is an alternative to leach pit toilets.
  3. They are useful in places of water scarcity or places with risk of ground water contamination.
  4. The principle of recovery and recycling of nutrients from excreta to create a valuable resource for agriculture is used here in Ecosan.
  5. The pit of an ecosan toilet fills up after some time, then it is closed and sealed for about 8-9 months.
  6. In this time the faeces gets completely composted to organic manure.
  7. It is a practical, efficient and cost-effective solution for human waste disposal.
  8. There are working Ecosan toilets in many areas of Gujarat, Kerala, Tamil Nadu and Sri Lanka.

Question 11.
Sanitary landfills.
Answer:

  1. Sanitary landfills are the places where wastes are dumped in depression or trench. Wastes are dumped here on everyday basis.
  2. In large metro cities, landfills are over-filled rapidly. Landfills are unhealthy and they may emit foul odour.
  3. Also there is a danger of chemicals percolating and reaching down to ground water and contaminate this water source.

Maharashtra Board Class 12 Biology Important Questions Chapter 15 Biodiversity, Conservation and Environmental Issues

Question 12.
Greenhouse effect.
Answer:
1. Greenhouse effect is caused due to heating up of earth’s surface and atmosphere. This heating is due to trapped infrared rays that are reflected from the earth’s surface by atmospheric gases.

2. The solar energy reaches the earth in the form of ultraviolet radiations, visible light and infrared and radio waves. Clouds and gases reflect about Vith radiations and absorb some of it.

3. Harmful UV radiations are absorbed by the ozone layer of the stratosphere and thus do not reach the earth’s surface.

4. Infrared radiation has heating effect thus it warms up the earth’s atmosphere and various objects. A part of the infrared radiations falling on the earth surface which have longer wavelength is reflected back into the outer space.

5. But there is a layer of carbon dioxide in the lower region of the atmosphere along with other atmospheric gases such as methane, nitrogen oxide, etc.

6. Due to these gases radiations cannot escape out. Carbon dioxide, along with methane, nitrogen oxides and CFCs can absorb infrared radiations reflected from the earth’s surface. They are therefore called greenhouse gases.

7. The blanket of these greenhouse gases in the atmosphere traps the reflected infrared rays and produces heat on the earth’s surface. This results in greenhouse effect which in turn causes global warming.

Question 13.
Mission Harit Maharashtra.
Answer:

  1. Government of Maharashtra in 2016, undertook an ambitious project of planting 50 crore trees in four years.
  2. Yearly targets were given to each district for plantations.
  3. The plantations are under the guidelines of National Forest Policy (NFP).
  4. For information about plantation, protection and mass awareness, a 24-hour toll free helpline number 1926 called ‘Hello Forest’ has been set up.
  5. There is also a mobile application called ‘My Plants’ which is set up by Forest Department. It records details of the plantation such as numbers, species and location.
  6. Following number of saplings were planted. 2.87 crore saplings in 2016, 5.17 crore saplings in 2017, 15.17 crore saplings in 2018, 33 crore saplings in 2019. Authorities are taking care of these plantations.
  7. Also Japanese Miyawaki method of plantation is adapted by State Forest Department and Social Forestry Department. Such plantations are in districts of Beed, Hingoli, Pune, Jalgaon, Aurangabad, etc.

Short answer questions

Question 1.
Why and how did biodiversity evolve on the earth?
Answer:

  1. There is great diversity with respect to size from microscopic to macroscopic, shape, colour, form, mode of nutrition, type of habitat, reproduction, motility, duration of life cycle span, etc.
  2. This diversity evolved in living beings for surviving and perpetuating to accommodate with different environmental conditions such as climatic, edaphic, topographic, geographic, etc. and different situations.
  3. For achieving this, living organisms adapted to different conditions and various habitats.
  4. This lead in formation of different features which lead to diversity in them.
  5. These adaptations in different environments serve as basis for diversity.

Question 2.
Explain species diversity with suitable examples.
Answer:

  1. Species diversity is interspecific diversity due to number of species of plants and animals that are present in a region.
  2. It can be expressed by variety of species i.e. species richness as well as number of individuals of different species i.e. species evenness.
  3. E.g. Species diversity is more in Western Ghats than in Eastern Ghats.
  4. Natural undisturbed tropical forests have much greater species richness than monoculture plantation of timber plant, developed by forest plantation.

Question 3.
What are latitudes and longitudes ? Which of these imaginary lines are more significant with reference to diversification of living beings?
Answer:

  1. Latitude is a geographic coordinate that specifies the north-south position of a point on the Earth’s surface. It is an angle which ranges from 0° at the Equator to 90° (North or South) at the poles.
  2. Longitude is a geographic coordinate that specifies the east-west position of a point on the Earth’s surface, or the surface of a celestial body. It is an angular measurement, usually expressed in degrees.
  3. Latitude is the imaginary line that is more significant to diversification of living beings.
  4. Species richness shows latitudinal gradient for many plants and animal species. Species richness is high at lower latitudes and there is a steady decline towards the poles.

Question 4.
Explain Productivity Stability Hypothesis.
Answer:

  1. Productivity Stability Hypothesis emphasises the importance of species diversity to the ecosystem. This hypothesis was given by David Tillman.
  2. It states that rich diversity leads to lesser variation in biomass production over a period of time and species richness is not needed for maintaining the stability of an ecological community.
  3. If average biomass production remains fairly constant over a period of time, then that community remains stable.
  4. The stable community remains strong to withstand disturbances and also recover quickly. Such community is resistant to invasive species.

Question 5.
What are the shortcomings of the graphic representation diagrams that give data of existing organisms?
Answer:

  1. In the diagrams of graphic representation of known animal and plant groups there is no data about prokaryotes.
  2. Several moneran species which are not cultivable under laboratory conditions are also not included.
  3. Conventional taxonomic methods are not suitable for identification of prokaryotic species. Therefore, such species are not included in these diagrams.

Question 6.
What is India’s wealth in biodiversity?
Answer:

  1. India has a share of 8.1% of total biodiversity wealth of the earth.
  2. India is one of the 12 megadiversity countries of the globe.
  3. India has 2.4% of total land area of the world but there are around 45000 identified plant species and nearly double the number of animal varieties in India.

Question 7.
What are the reasons for loss of biodiversity?
Answer:
Loss of biodiversity occurs due to two main causes:

  1. Natural reasons : E.g. Forest fires, earthquakes, volcanic eruptions, etc.
  2. Manmade reasons : E.g. Habitat destruction, hunting, settlement, overexploitation and reclamation.

Question 8.
When did major mass extinction events occurred ?
Answer:
In the following geological time scale, plants as well as animal groups underwent major mass extinctions.

  1. Between Cretaceous and Coenozoic period.
  2. Between Triassic and Jurassic period.
  3. Between Permian and Triassic period.
  4. Between Devonian and Carboniferous period.
  5. Between Ordovician and Silurian period.

Question 9.
What do you understand by invasive species? How does it affect local population?
Answer:

  1. New species are introduced into any ecosystem either accidentally or intentionally.
  2. Such introduction proves harmful for existing species.
  3. Sometimes even local species get extinct.
  4. If such extinction happens, then this new species is called an invasive species. E.g. Parthenium or carrot grass, Lantana and water hyacinth (Eichhornia) are such invasive plant species.
  5. Nile perch which is a predator fish in Lake Victoria cause harm to 200 local species of Cichlid fish.
  6. Clarias gariepinus (African catfish) was brought to India for aquaculture purpose. This catfish species has proved harmful to endemic catfish varieties.
  7. Since there is lack of local predator, this alien species survives and cause harmful effect on local species.

Question 10.
Explain how loss of species diversity can harm ecosystem?
Answer:

  1. When species diversity is lost, the ecosystem enters into imbalance.
  2. The biodiversity loss results into lesser plant production.
  3. This causes disruption of further food chains and food webs.
  4. The environmental processes such as disease cycles, plant productivity, etc. are also adversely affected.
  5. The productivity of the ecosystem is reduced and this results into overall imbalance in the ecosystem.

Question 11.
What is the basic cause of pollution?
Answer:

  1. Exponential growth of human population coupled with industrial development is the main cause of imbalance in all the ecosystems.
  2. There is also excessive utilization and production of synthetic materials and construction activities.
  3. These together have caused several undesired substances in ecosphere.
  4. This dumping of substances has resulted in severe pollution.

Question 12.
What are the effects of air pollution?
Answer:

  1. Air pollutants affect the surfaces of the respiratory system of all living beings. Therefore, any type of air pollution affects the process of respiration and respiratory system.
  2. The concentration of pollutants decide the severity of damage caused to body.
  3. Duration of exposure and the type of the organism also decide the effects of air pollution.
  4. In plants, air pollution results in poor yield of crops and premature death of plants.
  5. Particulate pollutants are very harmful for human beings. Fine particulates which enter the depths of lungs are responsible for irritation, inflammation and damage to lungs.
  6. This causes breathing and respiratory disorders and premature deaths.

Question 13.
Which is the major cause of atmospheric air pollution and how can it be avoided?
Answer:

  1. Automobiles are the major cause for atmospheric (air) pollution.
  2. Regular maintenance of vehicles and use of lead-free petrol or diesel can reduce air pollution as lesser pollutants are released from the exhausts.
  3. Using public transport and carpooling can be done to reduce number of automobiles on the road.

Question 14.
Does particulate matter help to reduce atmospheric temperature?
Answer:
Particulate matter plays a complicated role when it comes to influencing the temperature of the earth. The particles are light-absorbing and consequently contribute to the rise in global temperatures, but they also reflect a portion of the sunlight and so play a role in increasing the albedo, which moderates the temperature increase. This is what is known as negative radiative forcing.

Question 15.
Give any norms for reducing sulphur and aromatic contents of petrol and diesel.
Answer:
Euro I, II, III and IV norms were suggested in world to reduce sulphur and aromatic contents of petrol and diesel. In India, the aim is to reduce sulphur emission to 50 ppm in petrol and diesel along with aromatic hydrocarbons to 35%. Government has directly adapted BS VI norm.

Question 16.
What are the ill effects of noise pollution on human health?
Answer:

  1. Noise causes psychological and physiological changes in human beings. It causes sleeplessness, increased heartbeat, altered breathing pattern and psychological stress.
  2. Extremely high sound level of more than 150 decibels or more generated during a take-off of a jet plane or rocket, may damage ear drums, resulting into permanent hearing loss.
  3. Noise negatively interferes with child’s learning and behaviour pattern.

Question 17.
What are the ways to reduce noise pollution?
Answer:

  1. Using sound absorbent materials or by muffling the noise, especially in the industrial areas.
  2. Laws to reduce noise pollution should be strictly implemented.
  3. Blowing of horns should be discouraged in the areas of schools and hospitals.
  4. Firecrackers and loudspeakers should be completely banned. Government rules regarding this should be strictly followed.
  5. Supreme Court of India has banned loudspeakers at public gatherings after 10 pm.
  6. Awareness about noise pollution caused during festivals and processions should be spread among masses.

Question 18.
Explain BOD and its effects on aquatic ecosystem.
Answer:

  1. BOD is the amount of dissolved oxygen required by microorganisms for decomposing the organic matter present in water.
  2. This can be measured by chemical testing and is expressed in milligram of oxygen per litre (nigT) of water.
  3. If BOD of water is high, it denotes that the water is highly polluted with decomposing organic matter.
  4. High BOD shows that water is not potable but is polluted with microorganisms and organic debris.
    Waters with higher BOD will also cause death of resident aquatic organisms.
  5. Lower BOD on the other hand can vouch for cleaner water.
  6. BOD is an indicator of polluted ecosystem.

Maharashtra Board Class 12 Biology Important Questions Chapter 15 Biodiversity, Conservation and Environmental Issues

Question 19.
Why do you think the amount of DDT is maximum in birds?
Answer:

  1. Birds like hawk or kingfisher occupy the higher trophic levels.
  2. They feed on smaller prey like small birds, frogs, fishes, snakes, etc. These smaller animals feed on insects which already may have accumulated some amount of DDT due to their feeding on plants having DDT content.
  3. Since non-biodegradable DDT remains in tissues of organisms, larger birds will receive the maximum dose of DDT.
  4. As the phenomenon of biomagnification occurs in case of non-biodegradable DDT, the amount of DDT will go on rising from plants → insects → smaller animals → then to larger birds. Birds of prey will thus have maximum DDT in their tissues.

Question 20.
Ecological sanitation is the need of the day. Justify.
Answer:

  1. Large megacities have excessive human population.
  2. The problem of water shortage is also severe in some cities.
  3. Many slum-dwelling people do not have access to clean and safe toilet too.
  4. Open-air defecation results into outbreak of many communicable diseases.
  5. It also results into unclean and unhygienic atmosphere. In such case, ecological sanitation is a much better option.

Question 21.
What is solid waste? How is it disposed?
Answer:

  1. Solid waste means all the trash which is not needed by a person who throws it. Wastes from home, offices, stores, schools, hospitals, etc. form the wastes which is collected and disposed by municipality.
  2. Municipal solid waste is composed of paper, food, plastic, glass, metals, rubber, leather, textile, etc.
  3. Burning reduces volume of the waste. But it produces toxic air pollution.
  4. The incomplete burning also causes release of CO.
  5. Open dumps of wastes are breeding ground for rats and flies, so sanitary landfills are created.

Question 22.
Why is solid waste management facing endless problems?
Answer:

  1. Disposal of solid wastes is huge problem and personnel looking after this are overburdened.
  2. Open burning causes obnoxious gases which lead to air pollution.
  3. Open dumps serve as breeding grounds for rats and flies causing contamination of the environment.
  4. Epidemics of infectious diseases can spread if waste is kept unattended.
  5. Method of landfills is also inadequate. Such sites in large metros are getting filled. From such landfills, dangerous chemicals also seep and pollute underwater ground resources.
  6. Plastic and other synthetic items in the solid waste cause further problems in decomposition and hence they damage the ecosystem.

Question 23.
How pollution by domestic garbage can be controlled?
Answer:

  1. Everybody should learn to segregate garbage at source. The wet i.e. biodegradable and dry i.e. non-biodegradable materials should be separated, instead of throwing them callously in trash.
  2. Non-biodegradable material can be recycled or reused. It should be disposed off in proper way.
  3. Biodegradable material should be composted at home by simple methods of vermicomposting.
  4. Every household should practise these methods which will reduce the volume of garbage that is sent out to land filling or burning.
  5. The excessive load on Municipalities can also be reduced if each one takes the responsibility of his or her own garbage.
  6. Larger establishments like offices, schools or industries should provide space for solid waste and other garbage management.

Question 24.
What is e-waste? How is it disposed?
Answer:

  1. Any useless part of electronic origin or irreparable computers and other electronic goods as well as electrical waste are known as e-wastes.
  2. e-wastes are buried in landfills or are completely burnt.
  3. e-waste is also exported to developing countries like China, India and Pakistan from developed countries for recycling.
  4. Recycling is done for e-wastes in which metals like copper, iron, silicon, nickel and gold are recovered.

Question 25.
How citizens should sort out solid wastes?
Answer:

  1. All the collected wastes should be categorized into three types, viz. biodegradable, non- biodegradable and recyclable.
  2. Each person should sort out the waste. The biodegradable wastes should be put into deep pits and allowed to undergo slow degradation. Every building should have pit for biodegradable domestic wastes.
  3. Recyclable material like newspaper, plastic bags, empty milk pouches should be sold off.
  4. Every citizen should reduce his or her garbage generation and should also minimize the use of non-biodegradable products. Simple change such as refusing a plastic bag and using a reusable cloth bag can solve the problem of plastic garbage.

Question 26.
What are the preventive measures against Global warming?
Answer:

  1. Reducing use of fossil fuel.
  2. Efficient use of energy. Use of alternative energy sources like solar or wind energy.
  3. Reducing deforestation.
  4. Tree plantation and afforestation activities.
  5. Reducing the rate of growth of human population.
  6. International initiatives for reduction of greenhouse gases.

Question 27.
What is global warming? On what does it depend?
Answer:

  1. Global warming is increased temperature of the earth. During past century, the temperature of the
  2. Earth has increased by 0.6 °C, most of it during last three decades.
  3. This is mainly caused by greenhouse effect.
  4. Global warming depends upon the amount of CO2 and other greenhouse gases present in the atmosphere.
  5. An increase in CO2 concentration increases earth’s temperature by retaining more heat.
  6. Carbon dioxide increases temperature by about 50%, CFCs increase it by 20% and methane increases it by 15% whereas other pollutants increase it by 10%.
  7. Atmospheric air pollution, industrialization and greenhouse effect cause global warming.

Question 28.
Give an account of possible effects of global warming.
Answer:

  1. This rise in temperature leads to unfavourable changes in environment and resulting in odd climatic changes. (E.g. El Nino effect).
  2. Global warming results in melting of polar ice caps and Himalayan snow caps which may be the cause for submerging of the coastal areas.
  3. The frequency and severity of cyclones, hurricanes and unseasonal rains has increased tremendously causing series of natural disasters all over the world. This results into loss of infrastructure.
  4. Outbreak of various vector borne diseases has increased in last two decades due to global warming.
  5. The marine environment is worst affected due to excessive heat being absorbed into oceans.

Question 29.
Why greenhouse gases are increasing in their proportion?
Answer:

  1. Greenhouse gases are increasing due to excessive burning of fossil fuels in industries, by automobiles and air transportation, by burning of agricultural wastes, etc.
  2. All the combustion is causing levels of CO2 to rise.
  3. Biogas plants, paddy fields, cattle sheds add methane to atmosphere.
  4. More amount of chlorofluorocarbons are emitted due to fire extinguishers and air conditioners.
  5. Due to loss of forest cover, there are few trees left to absorb atmospheric CO2.
  6. Man is making development and progress due to which green house gases are increasing in proportion.

Question 30.
How is ozone hole formed?
OR
Give effect of CFC on ozone shield.
Answer:

  1. The ozone molecules are attacked by chlorofluorocarbon molecules. This action causes disturbances in the ozone shield due to increased rate of ozone degradation by Chlorofluorocarbon (CFC).
  2. CFCs can move upwards to reach stratosphere. Here UV rays act on them and release Cl atoms. The released Cl-degrades ozone, which subsequently releasing molecular oxygen.
  3. Cl atoms act as catalyst. So, they remain in the stratosphere and continue the effect of ozone degradation.
  4. This results in ozone depletion. This leads to the formation of ozone hole which is a large area of thinned ozone layer. One such ozone hole was observed over the Antarctic region.

Question 31.
What is the effect of UV-B radiation on body?
Answer:

  1. UV-B radiations of wavelength 280-322nm cause following deleterious effects:
  2. Damaging effect to DNA.
  3. Formation of mutation.
  4. Aging of skin and damage to skin cells including various types of skin cancers.
  5. Cornea of human eye absorbs UV-B radiations. High dose of UV-B causes inflammation of cornea called snow blindness, cataract, etc. leading to permanent damage to cornea.

Question 32.
What is Montreal Protocol?
Answer:

  1. Montreal Protocol was signed agreement of an international treaty among different nations who had recognised the harmful effects of ozone depletion.
  2. It was signed at Montreal (Canada) in 1987 to control emission of ozone depleting substances.
  3. After signing of Montreal protocol and its subsequent execution in 1989, the ozone depletion has been reduced worldwide.

Question 33.
‘There is a hole in the ozone layer.’ What do you understand by this ?
Answer:

  1. The area in Antarctica region with a thin ozone layer is known as ozone hole.
  2. CFCs (chlorofluorocarbons) which -are widely used as refrigerants disturb the balance between the production and degradation of ozone.
  3. Ultraviolet (UV) rays and chlorofluorocarbons (CFCs) release chlorine (Cl) ions which act as catalyst in the degradation of ozone layer.
  4. Chloride ions also degrade the ozone layer.
  5. The degradation of ozone layer results in the depletion of ozone causing a hole in the ozone layer.

Question 34.
What is the scenario of deforestation in India?
Answer:

  1. The scenario of deforestation is grim in India because we have cut down many trees from forests due to various developmental activities.
  2. At the beginning of 20th century, 30% was the forest cover.
  3. By the end of the 20th century, it became 19.4%.
  4. The National Forest Policy 1988 of India has recommended 33% forest cover for the plains and 67% for the hills.

Question 35.
What are the causes of deforestation?
Answer:
Causes of deforestation are as follows:

  1. Conversion of forest to agricultural land for growing food for ever-increasing human population.
  2. Trees are cut for timber, firewood, for keeping cattle in farm and for other purposes.
  3. For constructions of dams, road, railways, metros, residential complexes, etc.
  4. For any kind of developmental activities due to Government Policies.

Question 36.
What is Jhum cultivation?
Answer:

  1. Jhum cultivation is practised in north eastern India.
  2. This is also called slash and burn agriculture in which farmers cut down trees of the forest and burn the plant remains.
  3. This ash from burnt trees is used as fertilizer. The land is used for farming and cattle grazing.
  4. When cultivation is harvested, the area is left for several years so as to allow its recovery.

Question 37.
How Jhum cultivation has lead to deforestation in recent years?
Answer:

  1. Because in the Jhum cultivation, the forest trees are burnt to make space for farming and for obtaining ash as fertilizer, it leads to loss of precious forest cover.
  2. Once the trees are destroyed, farmers use this land for farming. After the cultivation and harvest, the land is left barren.
  3. It is used for cattle grazing.
  4. Since long period is required for the recovery of land back into forest patch, the deforestation results.

Question 38.
What are the major effects of deforestation ?
Answer:
Major effects of deforestation:

  1. Increased concentration of COa in the atmosphere.
  2. Trees hold lot of carbon in their biomass which is lost with deforestation.
  3. Loss of biodiversity due to habitat destruction.
  4. Disturbances in hydrologic cycle.
  5. Soil erosion and desertification in extreme cases.

Question 39.
Why Amrita Devi Bishnoi Wildlife Protection Award is started by Government of India? To whom is it given?
Answer:

  1. In 1731, a Bishnoi woman Amrita Devi hugged the trees to save them but was killed by King of Jodhpur who wanted the wood for his palace. For protecting the forest Amrita Devi and her three daughters along with hundreds of other Bishnois lost their lives.
  2. The Government of India in memory of this sacrifice has started Amrita Devi Bishnoi Wildlife Protection Award.
  3. This award is given to individuals or community from rural areas who protect wildlife.

Question 40.
What is Joint Forest Management?
Answer:

  1. Government of India has introduced the concept of Joint Forest Management (JFM) for working closely with local communities who protect and manage forests.
  2. In return, for their services to the forest, the communities can get benefit of various forest products (Fruits, gum, rubber, medicine, etc.).
  3. This is done with the idea to conserve the forest in a sustainable manner.
  4. JFM has started from 1980.

Question 41.
Floods in Sangli and Kolhapur in August 2019, were responsible for many problems during and after the floods. Think and enlist different types of problems faced by flood affected areas.
Answer:

  1. Floods of Western Maharashtra were mainly due to unseasonal and extremely heavy rain caused by climate change events.
  2. The floods had hit the districts of Kolhapur and Sangli hard. Sangli got completely marooned: There was no electric supply for extended period.
  3. Over two lakh people were living without electricity in affected areas.
  4. Lack of food and drinking water was a main problem.
  5. Several water-supply schemes became dysfunctional.
  6. Fields were completely ruined as crops were damaged. In Kolhapur district alone crops over 67,000 hectares were damaged.
  7. Large number of cattle were dead.
  8. Houses were destroyed completely. Thousands of people were shifted to safer places.
  9. About 223 villages in district of Kolhapur suffered a lot. 18 out of these have been completely marooned.
  10. About 28,897 persons were affected out of which 8,923 people were shifted.
  11. 813 houses were affected in the district, out of which 89 are completely damaged.
    There was also the scarcity of petrol and diesel.
  12. The Mumbai-Bengaluru National Highway, which passes through Kolhapur was dysfunctional and hence transport was also affected.

Chart based/Table based questions

Question 1.
Complete the given table:

Vehicle Norms Cities of Implementation
—————- Bharat Stage II ——————-
4 wheelers Bharat Stage III ——————
4 wheelers ————— 13 mega cities (Delhi and NCR, Mumbai, Kolkata, Chennai, Bengaluru, Surat, Kanpur, Agra, Lucknow, Solapur) since April 2010.
2 wheelers Bharat Stage III ——————–
————— Bharat Stage III Throughout the country since October 2010

Answer:

Vehicle Norms Cities of Implementation
4 wheelers Bharat Stage II All metro cities
4 wheelers Bharat Stage III Throughout the country since October 2010
4 wheelers Bharat Stage IV 13 mega cities (Delhi and NCR, Mumbai, Kolkata, Chennai, Bengaluru, Surat, Kanpur, Agra, Lucknow, Solapur) since April 2010.
2 wheelers Bharat Stage III Throughout the country since October 2010
3 wheelers Bharat Stage III Throughout the country since October 2010

Maharashtra Board Class 12 Biology Important Questions Chapter 15 Biodiversity, Conservation and Environmental Issues

Question 2.
Complete the given table

Name of polluting gas Source Effect
—————— ————– Combining with haemoglobin and causing respiratory problems
NO2 Chemical industries —————–
CO2 ————— ——————
——————- Biogas plants Greenhouse effect

Answer:

Name of polluting gas Source Effect
CO Vehicular exhaust Combining with haemoglobin and causing respiratory problems
NO2 Chemical industries Irritation to lungs and eyes
CO2 Combustion of any kind Greenhouse effect
CH4 Biogas plants Greenhouse effect

Diagram based questions

Question 1.
What can you say about species diversity A and B?
Maharashtra Board Class 12 Biology Important Questions Chapter 15 Biodiversity, Conservation and Environmental Issues 1
Answer:

  1. Size of species A and B are same. The number of species shown in both A and B are also same as both A and B have 4 different species each.
  2. Group A : In this species 4 different species are seen, but the density of the plants is not much. Group B also shows 4 different species. But two of these are in less number.
  3. Group A is more diverse that species B, but species B is showing more population of one species.

Question 2.
Observe the figure and answer the following questions.
(a) Identify the given instrument.
(b) What is its significance?
(c) Explain its working in brief.
Maharashtra Board Class 12 Biology Important Questions Chapter 15 Biodiversity, Conservation and Environmental Issues 2
Answer:
(a) Electrostatic precipitator.

(b) This is an important equipment which is used to remove particulate matter like soot and dust present in industrial exhaust. It is capable of removing almost 99% particulate matter present in exhaust of a thermal power plant.

(c) (1) In Electrostatic precipitator, high voltage is applied which produces electric discharge.
(2) This discharge causes ionisation of air in the smokestack.
(3) As a result, the free electrons are formed.
(4) These electrons in the ionised air attach to the gaseous or dust particles moving up the stack.
(5) Negatively charged particles move towards the positive electrode and settle down there.
(6) They are then removed by vibrations of the electrodes and collected in the reservoir.

Question 3.
Observe the given diagram and answer the questions given below.
Maharashtra Board Class 12 Biology Important Questions Chapter 15 Biodiversity, Conservation and Environmental Issues 3
(a) What is the name of this equipment?
(b) What is the function of such equipment?
(c) Explain its working in brief.
Answer:
(a) Exhaust gas scrubber.
(b) Exhaust gas scrubbers are used to clean air by removing both dust and gases.
(c) The exhaust is passed through dry or wet packing material. When it is done, gases like SO2 are removed. For this purpose, the exhaust is passed through a spray of water or lime.

Maharashtra Board Class 12 Biology Important Questions Chapter 15 Biodiversity, Conservation and Environmental Issues

Question 4.
Observe the given diagram and answer the questions based on it.
Maharashtra Board Class 12 Biology Important Questions Chapter 15 Biodiversity, Conservation and Environmental Issues 4
(a) Which apparatus is shown in the given diagram ?
(b) What is the function of this apparatus?
(c) What are the reactions that can take place in blocks 1, 2, and 3?
Answer:
(a) Catalytic converter.

(b) The harmful gases like CO and nitrogen oxides which are present in the automobile exhausts are removed by catalytic converters. Thereby, harmful effects of air pollution are reduced.

(c) In block 1 : Nitrogen oxides are present in the exhaust gases. They enter into reduction block of catalyst. The oxides of nitrogen react forming nitrogen and oxygen.
In block 2 : The exhaust gases enter the next block called oxidation block of the catalyst. Here, hydrocarbons and the newly formed oxygen react to form carbon dioxide.
In block 3 : The exhaust gases enter into last block from here the least harmful gases are released out.

Question 5.
Observe the graph and explain Alexander von Humboldt’s views about species richness and area relationship.
Maharashtra Board Class 12 Biology Important Questions Chapter 15 Biodiversity, Conservation and Environmental Issues 5
Answer:

  1. Scientists have tried to establish relationship between species diversity and the size of the habitat. It is considered that number of species present is directly proportional to the area.
  2. It is understood that larger areas may have more resources that can be distributed amongst the inhabitant species.
  3. Alexander von Humboldt observed that species richness does increase with the increase in area but only till a certain limit.
  4. For many species this curve is a rectangular hyperbola.
  5. If we consider S to be species richness, A as area under study, C as the Y intercept and Z as the slope of the line, this relationship can be described by the equation, log S = log C + Z log A.
  6. On logarithmic scale this relationship is a straight line, as observed in the figure above. For smaller areas, value of Z ranges between 0.1 to 0.2 regardless of species or region under study.
  7. But for the larger areas like the entire continents, slopes are closer to vertical axis i.e. steeper.
  8. This observation indicates that in very large areas, number of species found, increase faster than the area explored.

Question 6.
Explain the phenomenon of biomagnification by observing the diagram given below.
OR
Explain the phenomenon of biological magnification.
Maharashtra Board Class 12 Biology Important Questions Chapter 15 Biodiversity, Conservation and Environmental Issues 6
Answer:

  1. Some non-biodegradable substances like pesiticides or heavy metals have the tendency to accumulate in the tissues of living organisms.
  2. When such organisms are eaten by their predators, these pollutants enter the bodies of predators.
  3. At lower trophic level the concentration of such pollutant may be low, but when they are fed upon by their predator the amount of pollutant goes on increasing.
  4. As shown in the diagram there is only 0.000003 ppm DDT in the water. This DDT level is meagre but when zooplankton survive in this water, DDT concentration increases in their body and becomes 0.04 ppm.
  5. When many of these zooplankton are eaten by small fish, it rises to 0.5 ppm.
  6. In turn, the several smaller fish are eaten by a large fish and in it the concentration rises to 2 ppm.
  7. When such larger fishes are consumed by a bird, it receives maximum amount of DDT which might kill this bird.
  8. In this way, the DDT level shows biomagnification. Biomagnification is thus the phenomena of increase in the concentration of non-biodegradable substances according to the food chain or trophic relationships.

Long answer questions

Question 1.
What is biodiversity? Explain genetic diversity with suitable example.
Answer:
1. Biodiversity is the part of nature which includes the differences in the genes among the individuals of a species; the variety and richness of all plants and animal species at different scales in a space – local regions, country and the world; and the types of ecosystem, both terrestrial and aquatic, within a defined area.

2. Genetic diversity:
Genetic diversity is the intraspecific diversity in the number and types of genes and chromosomes present in different species.

  • It also includes variation in the genes and their alleles in the same species. Variation within a population and diversity between populations that are associated with adaptation to local conditions.
  • Genetic diversity or variability is essential for a healthy breeding population of a species.
  • Genetic variations are changes in the allelic genes which lead to individual differences within species.
  • Such variations help in the evolution. The chances of continuation of species in the changing environmental conditions are caused due to such variation and it allows the best organisms to get adapted to survive. Races and subspecies are formed due to genetic diversity.
  • Examples of genetic diversity : (a) There are 1000 varieties of mangoes and 50.000 varieties of rice or wheat in India, (b) Rauwolfia vomitoria is a medicinal plant that secretes reserpine.
  • This plant is inhabitant of different Himalayan ranges. There is variations in terms of potency and concentration of reserpine, from different locations.

Question 2.
Species richness goes on decreasing as we move from equator to pole. Explain.
Answer:

  1. In tropical regions, there are lesser climatic changes throughout the year and availability of plenty of sunlight.
  2. Moreover, in tropical areas there are lesser disturbances like periodic glaciations as compared to those seen in the polar regions.
  3. In tropical regions, there is a stability over millions of years which favoured speciation and hence there is more species richness.
  4. Also in tropical regions, there are lesser migrations which reduce gene flow between geographically isolated regions. This too favoured speciation.
  5. There is more availability of intense sunlight, warmer temperatures and higher annual rainfall in tropics. These factors have brought higher species richness in tropics.
  6. Constant climatic conditions and abundance of resources in tropical regions provide more food preferences for animals species.
  7. E.g. fruits are available throughout the year in rain forests, therefore variety of frugivorous animals are seen here, as compared to the temperate regions.

In short, species richness or diversity for plants and animals decreases as we move away from equator to the poles.

Question 3.
Explain Rivet Popper Hypothesis.
Answer:

  1. Rivet Popper hypothesis was given by Paul Ehrlich to emphasise significance of diversity.
  2. For explaining the hypothesis, he gave an analogy between aeroplane and ecosystem.
  3. As the rivets keep all parts of the aeroplane together, similarly, all species keep the diversity of an ecosystem in functional.
  4. Just as if one species gets extinct, initially not much of a problem will take place in an ecosystem, just as in case of a single rivet mission cannot cause problem in flight. However, if the same damage is continued, the turbulence will be experienced.
  5. When more rivets are popped out gradually, there will be a serious threat to the safety of the aeroplane.
  6. Also the rivets in key positions can cause serious situation. With same analogy he explained that if loss of species occurs, initially the problem will not be obvious but later if similar damage continues, there will be a threat to the ecosystem.
  7. Thus, there is a relationship between diversity and well-being of ecosystem which is not linear.
  8. Loss of key species causes threat in very short span of time by affecting food chains, food web, energy flow, natural cycles, etc. This will disturb the balance of the ecosystem.

Question 4.
Give various categories of endangered species explained by IUCN.
Answer:
Following are the categories of endangered species as explained by IUCN.

  1. Extinct (EX) : EX is added to species in which the last individual has died or is not recorded. So now on the earth not a single organism of this kind can be seen.
  2. Extinct in the Wild (EW) : This category contains those species whose members survive only in captivity.
  3. Critically Endangered (CR) : Critically endangered is a category containing those species that possess an extremely high risk of extinction with very few surviving members around 50 or so.
  4. Endangered (EN) : EN is added to a species that possess a very high risk of extinction as a result of rapid population decline of 50 to more than 70% over past three generations or the previous 10 years.
  5. Vulnerable (VU) : VU is a category containing those species that possess a very high risk of extinction as a result of rapid population decline of 30 to more them 50% over last three generations or the previous 10 years.
  6. Near Threatened (NT) : NT are species that are close to becoming threatened or may meet the criteria for threatened status in the near future.
  7. Least Concern (LC) : LC is a category containing species that are pervasive and abundant after careful assessment.
  8. Data Deficient (DD) : DD is a condition applied to species in which the amount of available data related to its risk of extinction, is lacking in some way.
  9. Not Evaluated (NE) In NE category any of the nearly 1.9 million species described by scientists are included, but not assessed by the IUCN.

Maharashtra Board Class 12 Biology Important Questions Chapter 15 Biodiversity, Conservation and Environmental Issues

Question 5.
What were the measures taken by Delhi Government to combat air pollution in Delhi?
Answer:
Following measures were taken to combat air pollution by Delhi Government:

  1. All the city buses were converted to run on CNG (i.e. compressed natural gas) by year 2002. CNG causes less pollution and is less expensive fuel.
  2. There was new fuel policy drafted by the Government.
  3. The norms were set to reduce sulphur and aromatic content of petrol and diesel.
  4. Engines of the automobiles were upgraded.
  5. Bharat stage emission standards (BS) were set. These standards are equivalent to Euro norms and have evolved on similar lines as Bharat Stage II (BS II) to BS VI from 2001 to 2017.

Question 6.
Why is Delhi worst polluted city as far as air pollution is concerned?
Answer:

  1. Delhi has colder weather during winters.
  2. There are stagnant winds which trap smoke from various sources like automobile exhausts and firecrackers.
  3. Many farms surrounding Delhi resort to burning crop stubbles.
  4. Even in the city garbage is lit.
  5. There is more dust on the roads.
  6. Automobile traffic is heavy and public transport systems were not efficient, till the metro was started.

Effects caused by this air pollution are as follows:

  1. Citizens suffered breathlessness and chest muscle contraction.
  2. The irritation in eyes, asthma and allergy were frequently reported.

Question 7.
What were the control measures taken by Delhi Government for combating air pollution ?
Answer:
Following measures were taken by Delhi Government:

  1. By 2002, all the city buses of Delhi were converted to CNG buses which now do not run on diesel.
  2. The new fuel policy was introduced and the norms were set to reduce sulphur and aromatic content of petrol and diesel.
  3. Upgradation of engines was done.
  4. Bharat stage emission standards (BS) were set which were equivalent to Euro norms.
  5. Bharat Stage II (BS II) to BS VI norms were given from 2001 to 2017.
  6. Administration took certain measures like closing educational institutions, suspending of construction or demolition work, undertaking vacuum cleaning of roads, etc.
  7. The polluting industries were penalized and Badarpur thermal power plant was temporarily closed down.

Question 8.
How is water polluted due to domestic sewage and Industrial Effluents?
Answer:

  1. When water is impure, it cannot be used for human consumption.
  2. Small amount about 0.1% impurities in water also makes the water polluted.
  3. In domestic sewage, there are dissolved salts such as nitrates, phosphates, other nutrients and toxic metal ions as well as organic compounds.
  4. Sewage also contains biodegradable organic matter and harmful bacterial and virus.
  5. Organic matter can be decomposed by bacteria and other microorganisms. But such water is not potable.
  6. Industrial effluents also contain harmful heavy metals and other solids.
  7. Solids can be easily removed from water but the dissolved salts cannot be separated.
  8. Biodegradable organic matter in sewage water is calculated by measuring Biochemical Oxygen Demand (BOD).

Question 9.
What is eutrophication? Describe the phenomena of eutrophication.
Answer:

  1. Eutrophication is the phenomena caused when a body of water becomes overly enriched with minerals and nutrients which induce excessive growth of algae.
  2. This process may result in oxygen depletion of the water body.
  3. Planktonic algae and algal bloom are the result of such nutrient-enriched water.
  4. Due to excessive growth of these plant species, the light in the lower layer of water is reduced.
  5. These plants perform photosynthesis during daytime but at night they compete with animal species for oxygen.
  6. There organic load of water body increases causing reduction in dissolved oxygen content.
  7. This results in death of fishes and other aquatic organisms. Once these dead bodies start decomposing in the water, there is more oxygen depletion.
  8. It then results in loss of species diversity.
  9. The water body which was once eutrophic now turns into stinking and turbid water body with coloured water. This is death of an ecosystem.

Question 10.
What are the two types of eutrophication? What is the main difference in them?
Answer:

  1. Natural Eutrophication and Cultural or Accelerated Eutrophication are two types of eutrophication.
  2. Natural eutrophication is caused due to natural processes. It is also called aging of a lake due to nutrient enrichment of water.
  3. It is a very slow and gradual process.
  4. Natural aging of lakes may take thousands of years, depending on the size of the lake, climatic conditions and other factors.
  5. Cultural or Accelerated eutrophication is caused due to human activities and chiefly due to pollution. Effluents from agricultural lands, industries and homes (household) cause such type of eutrophication.
  6. This phenomenon is called Cultural or Accelerated eutrophication.
  7. Due to excessive growth of algae there is lesser amount of dissolved oxygen for aquatic organisms, especially during night time. Due to this death of fish and other aquatic organisms take place.
  8. The dead bodies of aquatic organisms decompose causing further depletion of the dissolved oxygen.
  9. This results into complete collapse of the ecosystem of water body.

Question 11.
Comment on deforestation status of the world and its major effects.
Answer:
I. Deforestation status of the world:

  1. Forest area has declined all across the world in the past three decades. But in last decade, it is the reverse trend as the rate of forest loss has declined due to the growth of sustainable management.
  2. Global Forest Resources Assessment 2020 (FRA 2020) has given the figures of the rate of forest loss in 2015-2020. It is said to be declined to 10 million hectares (mha), reducing from 12 million hectares (mha) in 2010-2015.
  3. The FRA 2020 was released by the United Nations Food and Agriculture Organization (FAO) on May 13, 2020. It has examined the status of, and trends in, more than 60 forest-related variables in 236 countries and territories in the period 1990-2020.
  4. The world lost 178 mha of forest since 1990. However, the rate of net forest loss decreased substantially during 1990-2020 due to a reduction in deforestation in some countries, plus increases in forest area in others through afforestation and the natural expansion of forests.
  5. The rate of net forest loss declined from 7.8 mha per year in the decade 1990-2000 to 5.2 mha per year in 2000-2010 and 4.7 mha per year in 2010-2020.
  6. Among the world’s regions, Africa had the largest annual rate of net forest loss in 2010-2020, at 3.9 mha, followed by South America, at 2.6 mha.
  7. On the other hand, Asia had the highest net gain of forest area in 2010-2020, followed by Oceania and Europe.
  8. The world’s total forest area was 4.06 billion hectares (bha), which was 31% of the total land area. This area was equivalent to 0.52 ha per person, the report noted.
  9. The largest proportion of the world’s forests were tropical (45%), followed by boreal, temperate and subtropical.
  10. More than 54% of the world’s forests were in only five countries – the Russian Federation, Brazil, Canada, the United States of America and China. The area of naturally regenerating forests worldwide decreased since 1990, but the area of planted forests increased by 123 mha.

Maharashtra Board Class 12 Biology Important Questions Chapter 15 Biodiversity, Conservation and Environmental Issues

The rate of increase in the area of planted forest slowed in the last ten years.
(Source : https : //www.downtoearth.org.in/ news/forests/deforestation-rate-globally- declined-between-2015-and-2020-fao- report-71107)

II. The effects of deforestation:

  1. Increased concentration of COa in the atmosphere.
  2. Trees hold lot of carbon in their biomass which is lost with deforestation.
  3. Loss of biodiversity due to habitat destruction.
  4. Disturbances in hydrologic cycle.
  5. Soil erosion and desertification in extreme cases.

Maharashtra Board Class 12 Biology Solutions Chapter 15 Biodiversity, Conservation and Environmental Issues

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 15 Biodiversity, Conservation and Environmental Issues Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 15 Biodiversity, Conservation and Environmental Issues

1. Multiple choice questions

Question 1.
Observe the graph and select correct option.
Maharashtra Board Class 12 Biology Solutions Chapter 15 Biodiversity, Conservation and Environmental Issues 1
(a) Line A represents, S = CA²
(b) Line B represents, log C = log A + Z log S
(c) Line A represents, S = CAZ
(d) Line B represents, log S = log Z + C log A
Answer:
(c) Line A represents, S = CAZ

Maharashtra Board Class 12 Biology Solutions Chapter 15 Biodiversity, Conservation and Environmental Issues

Question 2.
Select odd one out on the basis of Ex situ conservation.
(a) Zoological park
(b) Tissue culture
(c) Sacred groves
(d) Cryopreservation
Answer:
(a) Zoological park

Question 3.
Which of the following factors will favour species diversity?
(a) Invasive species
(b) Glaciation
(c) Forest canopy
(d) Co-extinction
Answer:
(a) Invasive species

Question 4.
The term “terror of Bengal’ is used for
(a) algal bloom
(b) water hyacinth
(c) increased BOD
(d) eutrophication
Answer:
(b) water hyacinth

Question 5.
CFC are air polluting agents which are produced by
(a) Diesel trucks
(b) Jet planes
(c) Rice fields
(d) Industries
Answer:
(b) Jet planes

2. Very short answer type questions.

Question 1.
Give two examples of biodegradable materials released from sugar industry.
Answer:

  1. Molasses
  2. Bagasse.

Question 2.
Name any two modern techniques of protection of endangered species.
OR
Two modern methods of ex-situ conservation of species
Answer:

  1. Tissue culture
  2. In vitro fertilization of eggs
  3. Cryopreservation.

Question 3.
Where was ozone hole discovered?
Answer:
Ozone hole was discovered in Antarctica.

Question 4.
Give one example of natural pollutant.
Answer:
Volcanic ash is a natural pollutant.

Maharashtra Board Class 12 Biology Solutions Chapter 15 Biodiversity, Conservation and Environmental Issues

Question 5.
What do you understand by EW category of living being?
Answer:
A species which becomes extinct in the wild (EW) is called EW category, their members are seen only in captivity or as a naturalized population outside its historic range due to massive habitat loss.

3. Short answer type questions.

Question 1.
Dandiya raas is not allowed after 10.00 pm. Why?
Answer:
Dandiya rass involves blaring loudspeakers which cause noise pollution. It is undesired loud sound which could be hazardous for ears and general health. In India, the Air (Prevention and Control of Pollution) Act 1981, Amendment 1987, includes noise as an air pollutant. As per law noise after 10 pm is not allowed as many people may be resting. Therefore, Dandiya Raas is not allowed after 10 pm.

Question 2.
Tropical regions exhibit species richness as compared to polar regions. Justify.
Answer:

  1. Tropical regions are bestowed by thicker vegetation and ample food due to available sunlight and humidity.
  2. Polar regions are covered over with snow, with almost no vegetation.
  3. Only handful species of animals can survive here due to their adaptations.
  4. Species richness always shows latitudinal gradient for many plants and animal species. It is high at lower latitudes and there is a steady decline towards the poles. Therefore, tropical regions show more species richness.

Question 3.
How does genetic diversity affect sustenance of a species?
Answer:

  1. Genetic diversity develops the capability of the species to adapt to the varying changes in the environment.
  2. The large variation of the different gene sets allows an individual or the whole population to have the capacity to endure environmental stress in any form.
  3. Some individuals have, a better capacity to endure the increasing pollution in the environment whereas some do not have it.
  4. Those that do not have show infertility or even death from the same conditions.
  5. Those who are able to endure and adapt to this change survive and live in a better way.
  6. This is called natural selection which leads to a loss of genetic diversity in particular habitats.
  7. Thus, due to genetic diversity can affect sustenance of some species.

Question 4.
Greenhouse effect is boon or bane? Give your opinion.
Answer:
(1) The natural greenhouse effect is good, it is a boon but human enhanced greenhouse effect is a bane.

(2) In the absence of an atmosphere, Earth’s surface temperature would be about -18 °C, or 0 °F, which is too cold for sustaining life.

(3) Earth is habitable because of the natural greenhouse effect. Heating of Earth’s atmosphere due to the presence of greenhouse gases such as water vapour, carbon dioxide (CO2), methane (CH4) and oxides of nitrogen (NO2).

(4) Greenhouse gases have just the right molecular structure to absorb infrared radiation that the Earth emits. It re-emits most of that infrared energy in all directions, warming the atmosphere to its comfortable average temperature of 15 °C (60 °F). So, the greenhouse effect was a boon in olden days before industrialization and invention of automobiles.

(5) However, due to human impact, the proportion of greenhouse gases has increased tremendously causing global warming. Thus, now greenhouse effect has become a bane.

Maharashtra Board Class 12 Biology Solutions Chapter 15 Biodiversity, Conservation and Environmental Issues

Question 5.
State the effects of CO in human body.
OR
How does CO cause giddiness and exhaustion?
Answer:
Effects of Carbon monoxide:

  1. Carbon monoxide is tasteless, colourless and odourless gas, therefore its presence goes unnoticed.
  2. It can inhibit the blood’s ability to carry oxygen to body tissues.
  3. Supply of oxygen to vital organs such as.the heart and brain is affected due to presence of CO.
  4. When CO is inhaled, it combines with the oxygen carrying haemoglobin of the blood to form carboxyhaemoglobin. Once combined with the haemoglobin, that haemoglobin is no longer available for transporting oxygen.
  5. The symptoms of CO poisoning are headache, nausea, giddiness, etc.

Question 6.
Name two types of particulate pollutants found in air. Add a note on ill effects of the same on human health.
OR
Describe any 2 particulate and gaseous pollutants.
Answer:
I. Types of gaseous pollutants include CO2, CO, SO2, NO, NO2, etc.
(1) Carbon dioxide : It is a greenhouse gas. It is produced in excess due to human activities such as burning of fossil fuels. It is also rising due to increasing deforestation. The natural cycle of Carbon dioxide is disturbed due to human interference. Otherwise, the process of photosynthesis can balance CO2 : O2 ratio of the air. Aeroplane traffic such as a jet plane also emits lots of CO2.

(2) Carbon monoxide (CO) : CO is produced due to incomplete combustion of fuels. It is a toxic gas. Vehicular exhausts produce lot of CO.

II. Types of particulate pollutants are mist, dust, fume and smoke particles, smog, pesticides, heavy metals and radioactive elements, etc.
(1) Dust are fine particles which enter the respiratory passage and can cause damage to delicate tissues in the lungs. Various processes such as construction work, demolition of buildings and traffic can cause dust pollution. There are natural causes of release of dust too, through wind or volcanic eruption.

(2) Smoke and smog are worst type of particulate air pollutants which can cause many respiratory problems like emphysema or asthma.

4. Long answer type questions.

Question 1.
Montreal Protocol is an essential step. Why is it so?
Answer:

  1. Montreal Protocol was an international treaty signed at Montreal in Canada in 1987.
  2. Later many more efforts have been made and protocols have laid down definite roadmaps separately for developing and developed countries.
  3. All these efforts were for reducing emission of CFCs and other ozone depleting chemicals.
  4. All nations realized that ozone depletion can cause penetration of harmful UV radiations to the earth’s surface. This is very hazardous, for flora, fauna and for mainly human beings. Therefore, urgent action was needed to combat this effect.
  5. Montreal Protocol was a very positive move because after 1987, there have been much better condition of ozone layer.

Maharashtra Board Class 12 Biology Solutions Chapter 15 Biodiversity, Conservation and Environmental Issues

Question 2.
Name any 2 personalities who have contributed to control deforestation in our country. Elaborate on importance of their work.
Answer:
Two personalities who have contributed to control deforestation in our country are:
Saalumara Thimmakka from Karnataka and Moirangthem Loiya from Manipur.
1. Saalumara Thimmakka :

  • Saalumara Thimmakka is the best example of peoples’ participation in reforestation.
  • She is an Indian environmentalist from Karnataka. She has taken up work of planting and tending to 385 banyan trees along a 4 km stretch of highway between Hulikal and Kudur. Other 800 trees are also planted by her.
  • She is honoured with the National Citizens Award of India and Padma Shri in 2019.

2. Moirangthem Loiya :

  • Moirangthem Loiya is from Manipur who has restored Punshilok forest. For last 17 years he is planting trees after leaving his job.
  • He brought the lost glory back for the 300 acres forest land. He planted a variety of trees like, bamboo, oak Ficus, teak, jackfruit and Magnolia.
  • This forest now has over 250 varieties of plants including 25 varieties of bamboo along with many animals making the forest rich in biodiversity.

Question 3.
How BS emission standards changed over time? Why is it essential?
Answer:

  1. BS emission standards changed over the time due to changing city life and more vehicular traffic on the road, especially in the megacities.
  2. Since capital city of Delhi was declared as worst polluted city as far as its air quality is concerned, various measures were taken by the Government of India. There was new fuel policy declared, in which Bharat stage emission standards (BS) were set.
  3. These norms were set to reduce sulphur and aromatic content of petrol and diesel. Also the vehicular engines were upgraded.
  4. Bharat stage emission standards (BS) are standards which are equivalent to Euro norms and have evolved on similar lines as Bharat Stage II (BS II) to BS VI from 2001 to 2017.
  5. Since population of Delhi was to be saved, in 2001, Bharat stage II emission norms were set for CNG and LPG vehicles.
  6. This helped in reduced emission of sulphur which was controlled at 50 ppm in diesel and 150 ppm in petrol. Also aromatic hydrocarbons were reduced at 42% in concerned fuel according to norms.
  7. Because, in spite of all the efforts, Delhi was declared as worst air-polluted city in the world in 2016, therefore, Government of India directly adapted BS VI in the year 2018, skipping BS V These efforts decreased the levels of CO2 and SO2 in Delhi.

Question 4.
During large public gatherings like Pandharpur vari, mobile toilets are deployed by the government. Explain how this organic waste is disposed.
Answer:

  1. The toilets deployed at Pandharpur at the time of vari are of the Ecosan type.
  2. Ecosan toilet is a closed system without water and it is an alternative to leach pit toilets.
  3. When the pit of an Ecosan toilet fills up after some time, then it is closed and sealed for about 8-9 months.
  4. In this time the faeces get completely composted to organic manure. In this way the organic waste can be disposed.
  5. It is a practical, efficient and cost-effective solution for human waste disposal.
  6. Also, open-air defecation is prohibited which can cause health problems. Therefore, during large public gatherings like Pandharpur vari mobile toilets like Ecosan are deployed by the government.

Question 5.
How Indian culture and traditions helped in bio-diversity conservation? Give importance of conservation in terms of utilitarian reasons.
Answer:
In Indian culture and traditions in different religions, biodiversity is protected and conserved. Few examples of worship of animals and plants can be given here.

  1. Nagpanchami festival is towards the respect of snakes. They are worshipped on that day and the local people are aware of their role in ecosystem of control of rat population.
  2. Vatapournima festival is worshipping a banyan tree.
  3. Various other festivals teach the value of plants and animals surrounding us. Even the cattle are worshipped on a particular day as a tradition.
  4. Jain religion strongly advocates protection of all animals through vegetarianism.

Maharashtra Board Class 12 Biology Solutions Chapter 15 Biodiversity, Conservation and Environmental Issues

Conservation in terms of utilitarian reasons:
The conservation of biodiversity can be done in utilitarian way or for ethical reasons. Utilitarian reasons are further classified into narrowly utilitarian and broadly utilitarian reasons:

I. Narrowly utilitarian reasons:

  1. Humans always reap material benefits from biodiversity in the form of resources for basic needs such as food, clothes, shelter.
  2. Industrial products like resins, tannins, perfume base, etc. are also obtained through biodiversity resources.
  3. For making ornaments or artefacts for aesthetic purpose, again biodiversity is sacrificed.
  4. Many medicines are also obtained through biodiversity resources which shares 25% of global medicine market.
  5. Around 25000 species are used for traditional medicines by tribal population worldwide.
  6. Bioprospecting which is a systematic search for development of new sources of chemical compounds, genes, microorganisms, macroorganisms, and other valuable products from nature which is of economically important species is also due to biodiversity.

II. Broadly utilitarian reasons:

  1. Production of oxygen done by all green plants helps human beings to thrive. Amazon forest alone gives 25% of the oxygen to the entire world.
  2. Insects carry out pollination and seed dispersal.
  3. If insects do not carry out pollination and seed dispersal, man would go hungry without crops and fruits.
  4. Biodiversity also is useful in recreation of human beings.

III. Taking all these aspects in consideration, conservation of biodiversity becomes essential. Therefore, to protect and conserve our rich biodiversity on the planet, we have to remember all the utilitarian reasons.

Maharashtra Board Class 12 Biology Important Questions Chapter 14 Ecosystems and Energy Flow

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 14 Ecosystems and Energy Flow Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 14 Ecosystems and Energy Flow

Multiple choice questions

Question 1.
What is a true about ecosystem?
(a) Primary consumers are least dependent upon producers.
(b) Primary consumers outnumber the producers.
(c) Producers are more than primary consumers.
(d) Secondary consumers are the largest and most powerful.
Answer:
(c) Producers are more than primary consumers.

Question 2.
In an ecosystem, which shows one way passage?
(a) Free energy
(b) Carbon
(c) Nitrogen
(d) Potassium
Answer:
(a) Free energy

Maharashtra Board Class 12 Biology Important Questions Chapter 14 Ecosystems and Energy Flow

Question 3.
How many are biotic components from the following? Climate, carbohydrates, microbes, green plants, lipids, water, proteins, photosynthetic bacteria, chemosynthetic bacteria, herbivores, carnivores.
(a) 6
(b) 5
(c) 4
(d) 7
Answer:
(a) 6

Question 4.
From the following, which is the basic requirement for any type of ecosystem to function and sustain?
(a) Constant output of solar energy
(b) Constant input of solar energy
(c) Organic substances
(d) Organic substances dissolved in water
Answer:
(b) Constant input of solar energy

Question 5.
Which type of spatial patterns noticed in ecosystem structure?
(a) Zonation
(b) Stratification
(c) Zonation and stratification
(d) Zonation, stratification and distribution
Answer:
(c) Zonation and stratification

Question 6.
In which strata of any aquatic body there will be maximum photosynthesis?
(a) Bottom deposits
(b) Middle strata of a water body
(c) Near coastal region
(d) The depth till where sunlight can reach
Answer:
(d) The depth till where sunlight can reach

Question 7.
Which spatial pattern occurs vertically in an ecosystem?
(a) Zonation
(b) Stratification
(c) Y-shaped pattern
(d) Pyramid
Answer:
(b) Stratification

Question 8.
Which of the following is one of the characteristics of a biological community?
(a) Sex ratio
(b) Stratification
(c) Natality
(d) Mortality
Answer:
(b) Stratification

Question 9.
Identify the likely organisms (1), (2), (3) and (4) in the food web given below:
Maharashtra Board Class 12 Biology Important Questions Chapter 14 Ecosystems and Energy Flow 1
Answer:
(a) deer, rabbit, frog, rat

Question 10.
Which of the following is the basic requirement for any ecosystem to function and sustain?
(a) A constant input of solar energy
(b) Ample water availability
(c) Cool temperatures
(d) Enough winds and currents around
Answer:
(a) A constant input of solar energy

Question 11.
Which one is the most accurate definition of primary production?
(a) The amount of the food produced by plants.
(b) The amount of food available to herbivores.
(c) The amount of biomass or organic matter produced per unit area over a time period by plants during photosynthesis.
(d) The amount of crop produced by farmer.
Answer:
(c) The amount of biomass or organic matter produced per unit area over a time period by plants during photosynthesis.

Question 12.
Primary production is expressed in terms of
(a) kg/area
(b) weight (g-2) or energy (kcalm -2)
(c) energy in food calories
(d) calories /sq. m.
Answer:
(b) weight (g-2) or energy (kcalm-2)

Question 13.
Considerable amount of GPP is utilized by plants in
(a) metabolism
(b) respiration
(c) homeostasis
(d) excretion
Answer:
(b) respiration

Question 14.
Which one is the most appropriate definition of secondary productivity?
(a) Secondary productivity is the rate of formation of new organic matter by consumers.
(b) Secondary productivity is the amount of food consumed by the carnivores.
(c) Secondary productivity is the amount of food consumed by the producers.
(d) Secondary productivity is the amount of energy lost in the food chain.
Answer:
(a) Secondary productivity is the rate of formation of new organic matter by consumers.

Question 15.
Choose the factor on which primary productivity does not depend.
(a) The plant species inhabiting a particular area
(b) Amount of primary consumers dependent on plants
(c) Availability of nutrients
(d) Photosynthetic capacity of plants
Answer:
(b) Amount of primary consumers dependent on plants.

Question 16.
The annual net primary productivity of the whole biosphere is approximately billion tonnes (dry weight) of organic matter.
(a) 10
(b) 50
(c) 100
(d) 170
Answer:
(d) 170

Question 17.
The rate at which light energy is changed into chemical energy of organic molecules in ecosystems is
(a) net primary productivity
(b) gross primary productivity
(c) net secondary productivity
(d) gross secondary productivity
Answer:
(b) gross primary productivity

Question 18.
What is net primary productivity?
(a) Total rate of photosynthesis
(b) Rate of energy storage by consumer
(c) Amount of organic matter stored by plants
(d) Rate of energy used
Answer:
(c) Amount of organic matter stored by plants

Question 19.
Which is the correct sequence of the steps decomposition process?
(a) Fragmentation → Mineralization → Catabolism → Humification → Leaching
(b) Leaching → Catabolism → Humification → Fragmentation → Mineralization
(c) Fragmentation → Leaching → Catabolism → Humification → Mineralization
(d) Catabolism → Humification → Fragmentation → Leaching → Mineralization
Answer:
(c) Fragmentation → Leaching → Catabolism → Humification → Mineralization

Question 20.
Match the columns:

1. Fragmentation i. Release of inorganic nutrients
2. Leaching ii. Bacterial and fungal enzymes
3. Catabolism iii. Accumulation of dark amorphous substance
4. Humification iv. Precipitated as unavailable salts
5. Mineralization v. Break down into smaller pieces

(a) 1-v, 2-ii, 3-iii, 4-i, 5-iv
(b) 1-v, 2-iv, 3-ii, 4-iii, 5-i
(c) 1-ii, 2-iii, 3-iv, 4-v, 5-i
(d) 1-iii, 2-v, 3-iv, 4-i, 5-ii
Answer:
(b) 1-v, 2-iv, 3-ii, 4-iii, 5-i

Question 21.
The slow rate of decomposition of fallen logs in nature is due to their
(a) low moisture content
(b) poor nitrogen content
(c) anaerobic environment around them
(d) low cellulose content
Answer:
(a) low moisture content

Question 22.
Small amount of energy sustains the entire living world is only ……………………… of the PAR.
(a) 20-25%
(b) 10-15%
(c) 2-10%
(d) 1-2%
Answer:
(c) 2-10%

Question 23.
Choose incorrect statement out of the following
(a) Much larger fraction of energy flows through the DFC than through the GFC.
(b) Some of the organisms of DFC are prey to the GFC animals.
(c) In an aquatic ecosystem, GFC is the major conduit for energy flow.
(d) Much less fraction of energy flows through the GFC than through the DFC.
Answer:
(d) Much less fraction of energy flows through the GFC than through the DFC

Question 24.
In an ecosystem, bacteria are considered as
(a) primary consumers
(b) microconsumers
(c) macroconsumers
(d) secondary consumers
Answer:
(b) microconsumers

Question 25.
Which of the following can be a top carnivore of marine ecosystem?
(a) Zooplankton
(b) Sea cucumber
(c) Sea horse
(d) Kingfisher
Answer:
(d) Kingfisher

Question 26.
Identify the possible link ‘A’ in the following food chain.
Plant → Insect → Frog → A’ → Eagle
(a) Rabbit
(b) Wolf
(c) Cobra
(d) Parrot
Answer:
(c) Cobra

Question 27.
Why is the pyramid of biomass inverted in sea?
(a) Because plants are absent.
(b) Because fishes have less biomass.
(c) Because phytoplankton which are primary producers have less biomass.
(d) Because primary producers have more biomass.
Answer:
(c) Because phytoplankton which are primary producers have less biomass.

Question 28.
Identify the wrong statement in relation to concept of pyramids.
(a) Pyramid of energy is always upright, can never be inverted.
(b) In most ecosystems, all the pyramids of number of energy and biomass are upright.
(c) Inverted pyramid of biomass is observed in pond ecosystem as small standing crop of phytoplankton supports larger zooplanktons.
(d) Pyramid of biomass in sea is upright because fishes feed on standing crop of planktons.
Answer:
(d) Pyramid of biomass in sea is upright because fishes feed on standing crop of planktons.

Question 29.
Identify A, B, C and D in the following simplified model of phosphorus cycling in a terrestrial ecosystem.
Maharashtra Board Class 12 Biology Important Questions Chapter 14 Ecosystems and Energy Flow 2
Answer:
(a) Detritus, Rock minerals, producer, litter fall

Question 30.
From the following, which is not a fate of carbon in plants?
(a) Released in transpiration. atmosphere through
(b) Liberated in respiration. atmosphere through
(c) Consumed by animal in the form of food.
(d) Remains as it is as organic matter when plant dies.
Answer:
(a) Released in atmosphere through transpiration.

Question 31.
Bacterial role in carbon cycle is ………………
(a) photosynthesis
(b) chemosynthesis
(c) assimilation
(d) breakdown of organic matter
Answer:
(d) breakdown of organic matter

Question 32.
Which out of the following is the major reservoir of carbon?
(a) Animal bodies
(b) Fruits
(c) Ocean
(d) Coal mines
Answer:
(c) Ocean

Maharashtra Board Class 12 Biology Important Questions Chapter 14 Ecosystems and Energy Flow

Question 33.
Which act of human beings has great impact on carbon cycle?
(a) Breaking of limestone
(b) Building up of limestone reefs
(c) Burning of fossil fuels
(d) Volcanic eruption
Answer:
(c) Burning of fossil fuels

Question 34.
Rock phosphates are brought into circulation by the process of ………………….
(a) combustion
(b) sedimentation
(c) weathering
(d) absorption
Answer:
(c) weathering

Question 35.
In animals this structure is not containing phosphorus.
(a) Bones
(b) Shells
(c) Teeth
(d) Hairs
Answer:
(d) Hairs

Question 36.
Phosphorus is not a major constituent of
(a) DNA
(b) Proteins
(c) RNA
(d) ATP
Answer:
(b) Proteins

Question 37.
Secondary Succession takes place on/in
(a) Newly cooled lava
(b) Bare rock
(c) Degraded forest
(d) Newly created pond
Answer:
(c) Degraded forest

Question 38.
Vertical distribution of different species occupying different levels in a biotic community is known as
(a) Pyramid
(b) Divergence
(c) Stratification
(d) Zonation
Answer:
(c) Stratification

Question 39.
The developmental stages of ecological succession are called
(a) serial stages
(b) serai stages
(c) cereal stages
(d) trophic levels
Answer:
(b) serai stages

Question 40.
What is the peculiarity of pioneers in succession?
(a) They are always heterotrophic components.
(b) They always face favourable growth conditions.
(c) They are the most successful terminal occupants of the area.
(d) They face adverse conditions and get established there.
Answer:
(d) They face adverse conditions and get established there.

Question 41.
What are pioneers?
(a) First serai stage
(b) Heterotrophic serai stage
(c) Terminal (last) serai stage
(d) Final climax community
Answer:
(a) First serai stage

Question 42.
The stable community established in an area is known as
(a) pioneer community
(b) climax community
(c) equilibrium community
(d) autotrophic community
Answer:
(b) climax community

Question 43.
The entire sequence of communities that successively change in a given area are called
(a) climax
(b) sere
(c) pioneer
(d) standing population
Answer:
(b) sere

Question 44.
The primary succession refers to the development of communities on a ………………….
(a) freshly cleared crop field
(b) forest clearing after devastating fire
(c) pond, freshly filled with water after a dry phase
(d) Newly-exposed habitat with no record of earlier vegetation
Answer:
(d) Newly-exposed habitat with no record of earlier vegetation

Question 45.
Which is the most important service provided by environment?
(a) Release of oxygen
(b) Formation of ozone layer
(c) Carbon assimilation in photosynthesis
(d) Agents of pollination
Answer:
(c) Carbon assimilation in photosynthesis

Match the columns

Question 1.

Column A Column B
(1) Tansley (a) 10% law
(2) C. Elton (b) Ecosystem services
(3) R. Lindemann (c) Ecological pyramids
(4) Millennium Ecosystem Assessment report (d) Ecosystem

Answer:

Column A Column B
(1) Tansley (d) Ecosystem
(2) C. Elton (b) Ecosystem services
(3) R. Lindemann (a) 10% law
(4) Millennium Ecosystem Assessment report (c) Ecological pyramids

Question 2.

Column A Column B
(1) Rooted floating angiosperm (a) Cyperus
(2) Free-floating plant (b) Typha
(3) Reed swamp (c) Pistia
(4) Marsh-meadow (d) Lotus

Answer:

Column A Column B
(1) Rooted floating angiosperm (d) Lotus
(2) Free-floating plant (c) Pistia
(3) Reed swamp (b) Typha
(4) Marsh-meadow (a) Cyperus

Question 3.

Column A Column B
(1) Pioneer species (a) Entire gradient of communities
(2) Climax species (b) Spatial pattern
(3) Succession (c) Quercus
(4) Sere (d) Crustose lichen

Answer:

Column A Column B
(1) Pioneer species (d) Crustose lichen
(2) Climax species (c) Quercus
(3) Succession (b) Spatial pattern
(4) Sere (a) Entire gradient of communities

Classify the following to form Column B as per the category given in Column A.

Question 1.
Estuarine waters, Taiga, Aquarium tank, Lake, Evergreen forest, Flower garden.

Types of ecosystems Examples
(1) Aquatic ————–
(2) Terrestrial —————
(3) Artificial ————–

Answer:

Types of ecosystems Examples
(1) Aquatic (a) Estuarine waters, Lake
(2) Terrestrial (b) Taiga, Evergreen forest
(3) Artificial (c) Aquarium tank, Flower garden

Question 2.
Xerarch, Epipelagic, Hydrosere, Sublittotral, Intertidal, Benthic.

Phenomena Examples
(1) Stratification ————–
(2) Zonation —————
(3) Succession ————–

Answer:

Phenomena Examples
(1) Stratification (a) Epipelagic, Benthic
(2) Zonation (b) Sublittotral, Intertidal
(3) Succession (c) Xerarch, Hydrosere

Question 3.
Nature trails, Pollination, Sea food, Carbon sequestration, Animal therapy, Pest control, Health care, Nutrient cycling.

Column A Column B
(1) Supporting services ————–
(2) Provisioning services —————
(3) Regulating services ————–
(4) Cultured services —————

Answer:

Column A Column B
(1) Supporting services (a) Pollination, Nutrient services cycling
(2) Provisioning services (b) Sea food, Health care
(3) Regulating services (c) Carbon sequestration, services Pest control
(4) Cultured services (d) Nature trails, Animal services therapy

Very short answer questions

Question 1.
What forms the physical structure of the ecosystems?
Answer:
Interaction of biotic and abiotic components, results in a physical structure of ecosystems.

Question 2.
How is species composition of an ecosystem decided?
Answer:
By identification and enumeration of plant and animal species of a given ecosystem, its species composition can be decided.

Question 3.
What does base of each pyramid represent?
Answer:
The base of each pyramid represents the first trophic level of producers.

Question 4.
How is stratification seen in the forested land?
Answer:
Stratification seen in forested land is as follows trees occupying top vertical strata or layer of a forest, shrubs the second strata and herbs and grasses occupying the bottom layer.

Question 5.
What is meant by nutrient cycling?
Answer:
The cyclic movement of nutrient elements through the various components of an ecosystem, is called nutrient cycling.

Question 6.
Why nutrient cycling is called biogeochemical cycle?
Answer:
The nutrients are cycled from biotic organisms (bio) to abiotic components in the earth (geo) and all these nutrients are in the form of chemicals, therefore, nutrient cycling is called biogeochemical cycle.

Question 7.
What is the function of reservoirs in nutrient cycling?
Answer:
The function of the reservoir is to meet with the deficit, which occurs due to imbalance in the rate of influx and efflux in any ecosystem.

Question 8.
Till what time does the climax community remain stable?
Answer:
The climax community remains stable as long as the environment remains unchanged.

Question 9.
Which are the pioneers in the aquatic habitat during primary succession?
Answer:
Small phytoplankton are the pioneers in the aquatic habitat during primary succession.

Maharashtra Board Class 12 Biology Important Questions Chapter 14 Ecosystems and Energy Flow

Question 10.
What are serai communities?
Answer:
The individual transitional communities formed during succession are termed serai communities.

Question 11.
What are the benefits given by the healthy ecosystems?
Answer:
Healthy ecosystems give the benefits from wide range of economic, environmental and aesthetic goods and services.

Question 12.
Which are the major producers in a terrestrial ecosystem?
Answer:
The major producers in a terrestrial ecosystem are herbaceous and woody plants.

Question 13.
Which are the major producers in an aquatic ecosystem?
Answer:
Major producers in an aquatic ecosystem are phytoplankton and algae.

Question 14.
Which event is the beginning of the detritus food chain or web?
Answer:
Death of any organism is the beginning of the detritus food chain or web.

Question 15.
State the 10% law. Who gave this law?
Answer:
R. Lindermann gave the 10% law which states that only 10% of the energy is transferred to each trophic level as net energy, from the previous trophic level. B

Give definitions of the following

Question 1.
Ecosystem
Answer:
A self-regulatory and self-sustaining structural and functional unit of biosphere in which there are interacting biotic and abiotic components is called an ecosystem.

Question 2.
Stratification
Answer:
Vertical distribution of different species of plants and animals occupying different levels, is known as stratification.

Question 3.
Zonation
Answer:
Horizontal distribution of plants and animals on land or in water, is called zonation.

Question 4.
Productivity
Answer:
Conversion of inorganic chemicals into organic material with the help of the radiant energy of the sun by the autotrophs and consumption of the autotrophs by heterotrophs.

Question 5.
Decomposition
Answer:
Decomposition is the break-down of dead organic material and mineralization of the dead matter.

Question 6.
Energy flow
Answer:
Energy flow is the unidirectional flow of energy from producers to consumers and finally dissipation and loss as heat.

Question 7.
Saprotrophs
Answer:
Organisms which can degrade the detritus and obtain their energy and nutrient requirements are called saprotrophs

Question 8.
Trophic level
Answer:
A specific place occupied by the organisms in the food chain is called their trophic level.

Question 9.
Ecological pyramid
Answer:
Ecological pyramid is the graphic representation showing relationship between the organisms of different successive trophic levels with respect to energy, biomass and number.

Question 10.
Leaching
Answer:
The precipitation of water-soluble inorganic nutrients into the soil horizon in the form of salts is called leaching.

Question 11.
Humification
The formation of humus is humification.

Question 12.
Mineralization
Answer:
The process in which humus is degraded by some microbes to release inorganic nutrients is called mineralization.

Question 13.
Eutrophication
Answer:
Eutrophication is the sudden influx of phosphorus in water bodies due to agricultural runoff or industrial effluents which are rich in phosphate content.

Question 14.
Ecological succession
Answer:
The gradual and predictable change in the species composition of a given area is called ecological succession.

Question 15.
Sere
Answer:
The entire sequence of communities that successively change in a given area is called a sere.

Name the following

Question 1.
Functional aspects of ecosystem.
Answer:

  1. Productivity
  2. Decomposition
  3. Nutrient cycling
  4. Energy flow.

Question 2.
Two types of spatial patterns.
Answer:

  1. Stratification
  2. Zonation.

Question 3.
Stratification seen in open seas.
Answer:
Epipelagic, mesopelagic, bathy-pelagic and benthic zones.

Question 4.
Zonation seen at the junction of aquatic and terrestrial ecosystems.
Answer:
Inter-tidal, Littoral, Sub-littoral zones.

Question 5.
Common herbivores in terrestrial ecosystem.
Answer:
Insects (grasshopper, aphids), birds (parrot) and some mammals (sheep, cattle, goat, donkey).

Maharashtra Board Class 12 Biology Important Questions Chapter 14 Ecosystems and Energy Flow

Question 6.
Secondary consumers.
Answer:
All carnivorous animals such as frogs, lizards, birds of prey, hunting animals like tiger, lion, wolf, etc.

Question 7.
Three types of food chains.
Answer:

  1. Grazing
  2. Detritus
  3. Arasitic.

Question 8.
Different trophic levels.
Answer:
Producer, herbivore, primary carnivore, secondary carnivore, tertiary carnivore and ultimate carnivore are different trophic levels.

Question 9.
Three types of ecological pyramids.
Answer:

  1. Pyramid of biomass
  2. Pyramid of numbers
  3. Pyramid of energy

Question 10.
The important steps in the process of decomposition.
Answer:

  1. Fragmentation
  2. Leaching
  3. Catabolism
  4. Humification
  5. Mineralization

Question 11.
Sequential steps in process of succession.
Answer:

  1. Nudation
  2. Invasion
  3. Ecesis
  4. Aggregation
  5. Competition and co-action
  6. Reaction and stabilization

Question 12.
Types of Nutrient cycles.
Answer:

  1. Gaseous and
  2. Sedimentary.

Question 13.
Reservoir for the sedimentary cycle.
Answer:
Earth’s crust.

Question 14.
Reservoir for gaseous cycles.
Answer:
Atmosphere.

Distinguish between the following

Question 1.
Natural ecosystem and artificial ecosystem.
Answer:

Natural ecosystem Artificial ecosystem
1. Natural ecosystems are naturally formed. 1. Artificial ecosystems are man made.
2. There are no human inputs in natural ecosystems. 2. Artificial ecosystem is based on all human inputs.
3. Natural ecosystems are self-sustainable. 3. Artificial ecosystems are not self-sustainable.
4. In natural ecosystem, energy and materials are used and reused in cyclic manner. E.g. Ocean, forest, wetlands, estuary. 4. In artificial ecosystem, Energy and materials have to be given by human intervention which requires constant inputs. E.g. Farm land, aquaculture ponds, aquarium in the house.

Question 2.
Primary and secondary succession.
Answer:

Primary succession Secondary succession
1. The primary succession starts in the area where no living organisms ever existed. 1. The secondary succession starts in an area which has lost all the living organisms once existed.
2. Areas where primary succession starts are bare rock, newly formed pond, newly cooled lava, etc. 2. Abandoned farm, cut or burnt forest, flooded land’, etc. are areas where secondary succession begins.
3. Primary succession is a very slow process. 3. Secondary succession is comparatively a faster process.

Question 3.
Carbon cycle and phosphorus cycle.
Answer:

Carbon cycle Phosphorus cycle
1. Carbon cycle is a gaseous cycle. 1. Phosphorus cycle is a sedimentary cycle.
2. Carbon cycle has higher speed. 2. Phosphorus cycle is very slow.
3. There is respiratory release of CO2 in carbon cycle. 3. There is no respiratory release of phosphorus.
4. Atmospheric inputs of carbon through rainfall are larger. 4. Atmospheric inputs of phosphorus through rainfall are much smaller.
5. Exchanges of phosphorus between organism and environment are negligible in carbon cycle due to photosynthesis and respiration. 5. Exchanges of phosphorus between organism and environment are negligible in phosphorus cycle.

Given reasons

Question 1.
All animals are called consumers.
Answer:

  1. All animals depend on plants for food either directly as in case of herbivorous animals or indirectly as in case of carnivorous animals.
  2. Animals cannot perform photosynthesis as they are not autotrophic.
  3. They are heterotrophic and hence all are called consumers.

Question 2.
Food chains do not exist in isolation, but are always interconnected to form food web.
Answer:

  1. Food chains start from producers and end with consumers.
  2. But beyond secondary carnivores, the amount of energy available is too less.
  3. Thus, there is no tertiary carnivore that feeds exclusively on secondary carnivore.
  4. The secondary carnivore, however, many times will feed on herbivores directly.
  5. Therefore, food chains do not exist in isolation, but are always interconnected to form food web, so that the stability of ecosystem is maintained.

Question 3.
The pyramid of biomass in the sea is inverted.
Answer:

  1. The food chain in the sea is dependent on producers i.e. phytoplankton.
  2. The biomass of phytoplankton is always. lesser to the biomass of fishes which are dependent upon these phytoplankton.
  3. The pyramid of biomass, therefore, in the sea is inverted.

Question 4.
The pyramid of energy is always upright.
Answer:

  1. When the energy is moving from one trophic level to the next one, there is loss of some energy in the form of heat.
  2. Therefore, the energy is always more on the lower trophic levels as compared to the higher trophic levels.
  3. Due to this reason, the pyramid of energy is always upright and never inverted.

Maharashtra Board Class 12 Biology Important Questions Chapter 14 Ecosystems and Energy Flow

Question 5.
Only a fraction of sunlight is used for photosynthesis.
Answer:

  1. When sunlight falls on the earth surface about 34% of this is reflected back.
  2. About 10% is held by the ozone layer, water vapour and other atmospheric gases.
  3. Out of the total solar energy, about 56% reaches to the earth’s atmosphere.
  4. Only 0.02% of the sunlight is used for photosynthesis. This shows that only a fraction of sunlight is used for photosynthesis.

Question 6.
Usually crustose lichens form a pioneer species.
Answer:

  1. The pioneer species is the one which invades a bare area.
  2. When primary succession occurs it takes place on rocks.
  3. The crustose lichens can secrete acids which can dissolve rocks. This starts weathering of rocks and soil formation.
  4. Further, bryophytes and mosses can take hold of such soil. Therefore, as a pioneer on the barren rocks only crustose lichens can grow.

Write short notes

Question 1.
Zonation in wetland.
Answer:
There are four zones in wetland, viz. subtidal channels, mudflats, low marsh and high marsh. High marsh is more in terrestrial ecosystem while the subtidal channels are more of aquatic nature.

  1. Subtidal channels : These are important habitat for fish during low tide. This region allows good drainage and flooding in mudflats.
  2. Mudflats : This area is very rich in invertebrate life. Therefore, many wading birds are dependent for food in this area. Algal mat is also observed on mud flats.
  3. Low marsh : This area is a good habitat for cordgrass, insects, herons and egrets and the clapper rails.
  4. High marsh : This region supports pickleweed and patches of cordgrass. Sparrow and clapper rails are seen here.

Question 2.
Pond as a small ecosystem.
Answer:

  1. For every ecosystem there are four important functional aspects viz. productivity, decomposition, nutrient cycling and energy flow.
  2. In a small pond ecosystem, too, there are complex interactions.
  3. Pond is a shallow water body in which all the above four basic processes of an ecosystem are observed.
  4. The abiotic component is water along with dissolved inorganic and organic substances and rich soil deposit at the bottom of the pond.
  5. The sunlight acts as a source of solar energy, the cycle of temperature, day-length and other climatic conditions regulate the rate of function of the entire pond.
  6. Phytoplankton are the main producers along with algae and other aquatic plants.
  7. Zooplankton, aquatic insects and fish are the consumers.
  8. The decomposers are the fungi and bacteria present in the bottom soil.

Question 3.
Primary succession.
Answer:

  1. Primary succession means the initial development of life on barren piece of land.
  2. Primary succession occurs on newly cooled lava, rocks and newly created pond or reservoir. In a newly formed volcanic island, the life starts and takes up millions of years to develop the whole biomass.
  3. In the successive serai stages, there is a change in the diversity of species of organisms, increase in the number of species and organisms as well as an increase in the total biomass.
  4. The establishment of a new biotic community is generally very slow and takes millions of years.
  5. Primary succession depends on climatic condition, soil characteristics and natural processes.

Question 4.
Secondary succession.
Answer:

  1. Secondary succession begins in areas where previously natural biotic communities were present. But later were destroyed due to causes such as abandoned farm lands, burnt or cut forests, flooded lands, etc.
  2. Secondary succession is faster than primary succession because already there is some soil or sediment present.
  3. It occurs in the form of changed vegetation. But due to changed vegetation, there is also change in the resident animals that depend for food and shelter.
  4. Thus, with secondary succession, the numbers and types of animals and decomposers also change.
  5. During succession, natural or human induced disturbances (fire, deforestation, etc.), can convert a particular serai stage of succession to an earlier previous / preceding stage.
  6. Sometimes, disturbances like this can create new conditions that encourage some species and eliminate other species.

Question 5.
Succession of Plants.
Answer:

  1. Succession of plants is of two types based on the nature of the habitat. These are hydrarch or hydrosere which is based on water or very wetland areas and xerarch or xerosere based on very dry areas.
  2. In wetter areas hydrarch succession begins and then successional series progress from hydric to the mesic conditions.
  3. Xerarch succession starts in dry areas and the series progress from xeric to mesic conditions.
  4. Both hydrarch and xerarch successions lead to mesic or medium water condition which is neither too xeric nor too hydric.

Question 6.
Pioneer species.
Answer:

  1. Pioneer species are the ones that invade a bare area during primary succession.
  2. Crustose lichens which are able to secrete acids to dissolve rock usually start as pioneer species. They bring about weathering of rocks and soil formation.
  3. Later here bryophytes, mosses are settled as they can take hold in the small amount of soil. They are then followed by herbaceous plants, and after several more stages, ultimately a stable climax forest community is formed.

Question 7.
Succession in aquatic habitat.
Answer:

  1. In aquatic habitats the pioneer species in primary succession are the small phytoplankton.
  2. Phytoplankton are replaced by rooted- submerged plants (e.g. Hydrilla), rooted- floating angiosperms (e.g. Lotus) followed by free-floating plants (e.g.Pistia), then reed swamp (e.g. Typha), marsh-meadow (e.g. Cyperus), scrub (e.g. Alnus) and finally the trees (e.g. Quercus) in a very systematic and gradual way.
  3. The climax again would be a forest. With passage of time, the water body is converted into land.

Short Answer Questions

Question 1.
Describe the functions of the ecosystem.
Answer:
Functions of an ecosystem:

  1. Biological energy flow.
  2. Productivity of the ecosystem.
  3. Photosynthesis and respiration that take place in the ecosystem.
  4. Nutrient cycles operating in the ecosystem.
  5. Regulation of the environment by the organisms and the regulation of the organisms by the environment in turn.
  6. Interactions of biotic and abiotic components of the ecosystem.

Question 2.
What are the types of ecosystems? Give their suitable examples.
Answer:
Ecosystems are of two types, viz. natural ecosystem and artificial ecosystem.
(1) Natural ecosystems : The ecosystems which operate under natural conditions without any much major interference by man are called natural ecosystems. E.g. terrestrial ecosystems such as grasslands, forests, deserts, etc. or aquatic ecosystems such as lakes, river, wetland, etc.

(2) Artificial ecosystems : The man engineered ecosystems which are maintained artificially by man by the addition of energy are called artificial ecosystems. These ecosystems are dependent upon manipulations from human beings. E.g. croplands, aquarium, aquaculture, etc. are the types of artificial ecosystems.

Question 3.
What are the different components of ecosystem?
Answer:
(1) There are two main components of the ecosystem, viz. abiotic components and biotic components.

(2) Abiotic components include all the inorganic substances such as B S, C, N, H, etc., their distribution and amount available in an ecosystem and the organic compounds such as proteins, carbohydrates, lipids, etc. along with the climate of that region.

(3) Biotic components include all the living organisms living in an ecosystem. The living organisms may be autotrophic or producers and converters or transducers. These are either green plants having photosynthetic abilities or chemosynthetic microorganisms. The heterotrophic organisms are called consumers.

They can either be macroconsumers or microconsumers.

  1. Macroconsumers are herbivores, carnivores or omnivores as per the order in which they appear in the food chain.
  2. Herbivores are primary consumers while the carnivores are secondary consumers.
  3. Omnivores can be secondary or tertiary consumers.
  4. Microconsumers are decomposing organisms and hence they are also called decomposers. They have saprophytic mode of nutrition. Bacteria, actinomycetes and fungi are some of the microconsumers.
  5. Microconsumers decompose complex organic compounds from dead and living protoplasm and release the inorganic nutrients obtained from them, back to the environment.

Question 4.
What are the two spatial patterns in an ecosystem? Describe them briefly.
Answer:

  1. There are two spatial patterns recognized in an ecosystem, viz. zonation and stratification.
  2. Zonation is the spatial pattern which occurs horizontally along the ground. Along a horizontal gradient, the density and distribution of the species keep on varying.
  3. Stratification is the spatial pattern which occurs vertically. This is determined by the height of organisms. Such stratification is seen in forest community where trees of different species grow to different heights.

Question 5.
Describe the concept of energy flow.
OR
In an ecosystem the flow of energy is unidirectional. Explain.
Answer:
(1) For considering the energy flow in an ecosystem the following aspects have to be taken into account:

  • The efficiency of the producers in the absorption and conversion of the solar energy.
  • The quantity of converted energy is used by the consumers.
  • The total input of energy in the form of food and its efficiency of assimilation by the consumers.
  • The amount of energy is lost through respiration, heat, excretion, etc.
  • The ultimate gross net production.

(2) The energy captured by autotrophs never returns back to the sun. The energy obtained by the herbivores will never go back to autotrophs. Hence energy flow is always unidirectional.

(3) The energy flow through different trophic levels is progressive and hence previous trophic level cannot get this back.

(4) The amount of energy keeps on decreasing as it travels to further trophic level. This loss of energy is due to dissipation as heat formed during various metabolic activities of the organism. The energy loss is measured as respiration coupled with unutilized energy.

(5) If the food chain is shorter there is greater amount of available food energy. When the length of food chain increases, there is corresponding more loss of energy.

Question 6.
What could be the reason for the low productivity of ocean?
Answer:

  1. The conditions of land and in ocean are not same. Primary productivity which is the biomass or dry weight produced by the plants per unit area during the photosynthesis is different for oceanic ecosystem.
  2. In ocean, sunlight does not penetrate uniformly at all depths. It is thus the main limiting factor which decreases the rate- of photosynthesis.
  3. Decrease in rate of photosynthesis decreases the growth of phytoplankton and then that of the zooplankton. Aquatic plants such as algae are also affected due to lack of sunlight.
  4. Minerals and nutrients is also a limiting factor based on location of the oceans.
  5. Therefore, there is less productivity of the oceans to about 55 billion tons as compared to productivity on land which is about 170 billion tonnes.

Question 7.
What could be the connecting points between the GFC and DFC?
Answer:

  1. In grazing food chain, the energy is transferred from producers to consumers.
  2. The autotrophs convert inorganic matter into organic compounds which is then transferred in food chain or food web.
  3. In detritus food chain energy is obtained from organic matter or detritus generated in trophic levels of the grazing food chain.
  4. Detritus such as dead bodies of animals or fallen leaves, which are then eaten by decomposers or detritivores.
  5. These detritivores can also be consumed in grazing food chain by secondary consumers or predators.
  6. When decomposers convert organic matter back into inorganic substances, the autotrophs use these minerals again.
  7. In this way GFC and DFC are connected to each other in cyclic and mutual exchanges.

Maharashtra Board Class 12 Biology Important Questions Chapter 14 Ecosystems and Energy Flow

Question 8.
How will you classify man as carnivore (primary/ secondary) or omnivore? Why?
Answer:
Autotrophic producers or plants make the first trophic level. They synthesize their own food. Herbivores are primary consumers. They eat producers. Carnivores are secondary consumers. They eat primary consumers. The trophic level of man is omnivore as he eats both plants and animals. Non vegetarian person can be called carnivore while totally vegetarian person can be herbivore. However, the most appropriate placement of man is omnivorous.

Question 9.
How many trophic levels human beings function in a food chain?
Answer:
(1) All food chains and webs have at least two dr three trophic levels. Generally, there are a maximum of four tropic levels. Many consumers feed at more than one tropic level.

(2) Man is a primary consumer when he eat plants such as vegetables. Vegans and vegetarians who do not consume any animal product, fish or meat of any kind are primary consumers. They belong to the second trophic level.

(3) But some humans have different dietary choices. Many humans are omnivores, meaning they consume both plant and animal material. Thus, they may be on the third or even fourth tropic level.

(4) For example, if man consumes goat meat he is a part of the third tropic level.

(5) But when he eats bigger fish (and considering fish eats smellier fish) he becomes tertiary consumer on the fourth tropic level.

Question 10.
What is climax community? What leads to such a community?
Answer:

  1. Climax community is the community which is near equilibrium with the environment.
  2. The sequential changes in the structure and composition of the community, to adjust with the changing environment is called a climax community.

Question 11.
From algae to forest, explain in relation with the succession.
Answer:

  1. When there is a succession from algae to forest, it depends upon the amount of water available.
  2. The succession begins with small phyto-planktons followed by submerged and free floating and then rooted hydrophytes, sages, grasses and finally the trees.
  3. Similarly, there is also a transformation from a pool of water to swamp then marsh and then mesic which means neither too dry nor too wet conditions.
  4. Then small plants like mosses can inhabit followed by herbs, shrubs and then trees. Such succession ultimately leads to a stable climax forest community.

Question 12.
Explain the following terms with reference to ecological succession.
(1) Serai stages.
Answer:
The developmental stages of the ecological succession are known as serai stages.

(2) Pioneers.
Answer:
The organisms belonging to first serai stage in the ecological succession are known as pioneers.

(3) Hydrosere.
Answer:
Hydrosere or hydrarch succession is a type of ecological succession which is determined by the amount of water available during succession. Hydrosere occurs when there is abundant water available in the area where organisms reside.

(4) Xerosere.
Answer:
Xerosere or xerarch succession is a type of ecological succession which is determined by the amount of water available during succession. Xerosere occurs when there is very little water available in the area where organisms reside. Such succession is observed in desert regions.

Question 13.
Why is zonation more pronounced at the edges of habitat?
Answer:

  1. Edge of the habitat is an abruptly changing region. E.g. the shore or littoral zone is the edge between aquatic and terrestrial habitats.
  2. At such places there is variety of environmental factors, such as temperature, wind exposure, light intensity, wave action, and salinity, etc.
  3. These abiotic factors are never constant in such areas and they show tremendous variation.
  4. Therefore, the intertidal communities show differences in regions that are occupied by them. In this way zonation is more pronounced in such areas when edges of habitat are present.

Question 14.
What is the maximum number of trophic levels in a food chain?
Answer:

  1. There are maximum four trophic levels in an ecosystem.
  2. Rarely five levels are seen where it is occupied by apex consumer.
  3. But as the trophic levels are moving from producers to consumers, lesser and lesser energy remains.
  4. Through heat loss, lot of amount of energy is dissipated, therefore no ecosystem can sustain fifth trophic level.

Question 15.
How would you explain inverted pyramid of biomass in oceanic ecosystem?
Answer:

  1. Pyramid of biomass is inverted in an oceanic ecosystem because in oceans the biomass of fishes is much more than the biomass of phytoplanktons.
  2. The biomass of phytoplanktons which are the producers in the ocean is smaller than that of zooplanktons.
  3. Zooplankton are primary consumers in oceans. The biomass of zooplanktons is smaller than that of secondary consumers which are fish.
  4. This results in the inverted pyramid of biomass is an oceanic ecosystem.
  5. Also the life span of phytoplankton is very small, which are consumed almost as rapidly as they are formed.
  6. The phytoplankton have high reproductive potential by which they reproduce rapidly. All these facts make the pyramid inverted in case of oceanic ecosystem.

Question 16.
What is secondary productivity? What is annual net primary productivity of biosphere?
Answer:

  1. Secondary productivity is defined as the rate of formation of new organic matter by consumers.
  2. It is the rate of assimilation of food energy by the consumers.
  3. This amount of energy available to consumer for transfer to the next trophic level.
  4. The annual net primary productivity of the whole biosphere is approximately 170 billion tonnes (dry weight) of organic matter. Of 170 billion tonnes, the oceans create about 55 billion tonnes, and rest is by land ecosystems.

Complete the given chart

Trophic levels Type of organisms Status Example
…………….…. …………………….. Secondary consumer ……………………..
……………..… Herbivore, Heterotrophs …………………….. …………………….
……………..… Autotrophs ……………………. Phytoplankton, grass, trees

Answer:

Trophic levels Type of organisms Status Example
Third trophic level Carnivore, Heterotrophs Secondary consumer Lion, Frog
Second trophic level Herbivore, Heterotrophs Primary consumer Cattle, Sheep, Deer
First trophic level Autotrophs Producer Phytoplankton, grass, trees

Diagram based questions

Question 1.
Sketch and label Carbon cycle
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 14 Ecosystems and Energy Flow 3

Question 2.
Sketch and label phosphorus cycle
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 14 Ecosystems and Energy Flow 4

Question 3.
Sketch and label decomposition cycle
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 14 Ecosystems and Energy Flow 5

Question 4.
Sketch and label simple grazing food chain
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 14 Ecosystems and Energy Flow 6

Maharashtra Board Class 12 Biology Important Questions Chapter 14 Ecosystems and Energy Flow

Question 5.
Sketch and label ecological pyramids of energy and pyramid of biomass
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 14 Ecosystems and Energy Flow 7

Question 6.
Observe pyramids of numbers and answer the questions below
Maharashtra Board Class 12 Biology Important Questions Chapter 14 Ecosystems and Energy Flow 8
Questions:
1. In the above pyramid of numbers how many primary consumers are supporting secondary consumers? What will happen if the numbers are reversed?
2. When does pyramid of numbers get inverted in case of a single tree ecosystem ?
3. What will happen, if in the above example, we substitute larger bird of prey feeding on small insect eating birds?
Answer:
(1) 500,000 primary consumers are supporting 5000 secondary consumers. If the numbers are reversed, i.e. if primary consumers are lesser than secondary consumers, then the secondary consumers will have fierce competition and will lead to decline in their number too. The pyramid of numbers will get inverted in such case.

(2) If we plot the number of insects on a single tree, smaller birds feeding on insects, and parasites on those birds, we get an inverted pyramid.

(3) If large birds are feeding on smaller insect eating birds, their population will decrease. This will result into increased number of insects as there will be no check on the insect population due to loss of smaller predator birds. If larger birds keep on feeding constantly and unchecked on smaller birds, the smaller ones will eventually become lesser and lesser in their population and this in turn will starve the larger birds too, leading in decrease of their population too.

Question 7.
Observe the given food web and answer the questions
Maharashtra Board Class 12 Biology Important Questions Chapter 14 Ecosystems and Energy Flow 9
Questions
1. Name the producers in the above web.
2. Name the primary consumers in the above web.
3. Which is an omnivore animal in the above web? Why?
4. Why frog is occupying an important ecological position in this web?
5. From the given food web diagram, give the trophic levels where the eagle is present.
Answer:
1. Corn crop, flowering plant, lavender plant, mango tree
2. Grasshopper, butterfly, fruit fly
3. Rat, because it feeds on grains of corn as well as on insects.

4. Frog feeds on insects and keeps the insect population under check. It also falls prey to snakes and birds of prey like eagle. Thus, supports their population by providing food.

5. (1) Eagle is apex consumer in the following cases : Lavender/Producer → Butterfly/ Primary consumer → Dragon fly /Secondary consumer → Thrush/Tertiary consumer → Eagle/Apex consumer
Or
Corn/Producer → grasshopper/Primary consumer → frog/Secondary consumer → python/Tertiary consumer → eagle/Apex consumer.
(2) Eagle is tertiary consumer in the following case : Rat/Primary consumer → python/ Secondary consumer → eagle/Tertiary consumer
(3) Secondary consumer in the following case : Rat/Primary consumer → eagle/ Secondary consumer

Question 8.
Observe the given figure and answer the questions:
Maharashtra Board Class 12 Biology Important Questions Chapter 14 Ecosystems and Energy Flow 10
Questions:
1. Which trophic level has maximum energy?
2. Why carnivores have least energy shown in this diagram?
Answer:
1. Producers which form first trophic level has maximum energy.

2. When energy flows from one trophic level to the next, some amount is lost as heat. As it can be seen that producers had 1000 joules of energy, out of which 900 joules are lost when energy flowed from producers to primary consumers. Further, it was lost again by 90 joules while herbivore to carnivore energy flow was taking place. Therefore, the carnivore gets least energy as it is the higher trophic level.

Question 9.
Identify A, B, C, D and E given in sketch of carbon cycle
Maharashtra Board Class 12 Biology Important Questions Chapter 14 Ecosystems and Energy Flow 11
Answer:
A : Combustion
B : Respiration by plants
C : Respiration by animals
D : Photosynthesis
E : Decomposition.

Question 10.
Observe the diagram and answer the questions
Maharashtra Board Class 12 Biology Important Questions Chapter 14 Ecosystems and Energy Flow 12
1. Fill in the empty boxes with proper words in the above figure.
2. Which cycle is it depicting?
3. What kind of cycle is it?
Answer:
1. A : geological uplifting, B : Weathering of phosphate from rocks, C : Phophaste in solution. D : Detritus settling at bottom, E : Sedimentation forming New rocks, F : Phosphate in soil, G : Leaching
2. Phosphorus cycle
3. Sedimentary cycle

Long answer questions

Question 1.
Describe the pathway followed by phosphorus in an ecosystem.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 14 Ecosystems and Energy Flow 13
(1) The phosphorus cycle is the simplest of all the nutrient cycles, which is a type of sedimentary cycle. It constitutes cyclic movement of phosphorus through hydrosphere, lithosphere and biosphere.

(2) Since phosphorus is a heavy molecule it never goes into the atmosphere. Phosphorus remains in the bodies of organisms, dissolved in water or in the form of rock. The natural reservoir of phosphorus is rock in which phosphorus is present in the form of phophates.

(3) Upon weathering of the rock due to action of mildly acidic water, the phosphates in the rock go into the solution.

(4) Plants take up phosphorus in the form of phosphate. The roots of plants can absorb phosphate ions from the soil. Animals obtain phosphorus through food which they consume. Thus autotrophs supply phosphorus to heterotrophs. Phosphorus is a major biological constituent of all the living organisms.
It is found in biological membranes, nucleic acids e.g. DNA, RNA, cellular energy transfer systems such as ATE Phosphorus is thus an essential element.

(5) The requirement of phosphorus in animals is much more as it is the component of bones, hooves, teeth, shells.

(6) The waste products formed through defecation and the dead organisms are decomposed by phosphate-solubilizing bacteria releasing phosphorus. This in turn is used up by other growing plants. Phosphorus is always in short supply and hence acts as limiting factor for the plant growth.

Sudden influx of phosphorus in the form of agricultural runoff or industrial effluents rich in phosphate content, leads to eutrophication in water bodies. Eutrophication is due to overgrowth of algae at the instance of high phosphorus dissolved in water. The overgrowth of algae kills or harms the aquatic life.

Question 2.
What are the most important ecological services whose value cannot be determined?
Answer:
(1) The main ecological services are fixation of atmospheric CO2 and release of O2 are the most important services provided by an ecosystem.

(2) Photosynthetic activity of photoautotrophs helps in carbon sequestration in the form of CO2 from the atmosphere. At the same time it releases O2 as a by-product.

(3) O2 is needed for respiration by all aerobic organisms. Oxygen also purifies air.

(4) For human consumption, crops and fruits are continuously required. These can be produced only after pollination of plants. Wind, water or other biotic agent such as insects therefore play an important role as ecosystem service. Without pollination there would be no crops and no fruits.

Question 3.
What are the four categories of ecosystem services as given by Millennium Ecosystem Assessment Report in 2005? What are the services included in each?
Answer:
Ecosystem services are of following four categories:
(1) Supporting services : Support to the life on earth such as nutrient cycling, primary production, soil formation, habitat provision, pollination and overall maintenance of balance of ecosystem.

(2) Provisioning services : Provides necessities such as food in the form of crops, fruits and seafood, raw materials such as timber, skins, fuel wood, genetic resources in the form of seeds, crop improvement genes, and health care, other resources such as water, medicinal resources in the form of test and assay organisms and ornamental resources such as furs, feathers, ivory, orchids, butterflies, etc.

Maharashtra Board Class 12 Biology Important Questions Chapter 14 Ecosystems and Energy Flow

(3) Regulating services : Regulation of processes on the earth such as carbon sequestration, prey-predation regulation, waste decomposition and detoxification, purification of water and air and pest control.

(4) Cultural services : Under this category, humans get services from nature in the form of cultural, spiritual and historical, recreational experiences, opportunity to learn science and indulge in education, and pets and animal therapy.

Maharashtra Board Class 12 Biology Solutions Chapter 14 Ecosystems and Energy Flow

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 14 Ecosystems and Energy Flow Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 14 Ecosystems and Energy Flow

1. Multiple choice questions

Question 1.
Which one of the following has the largest population in a food chain?
(a) Producers
(b) Primary consumers
(c) Secondary consumers
(d) Decomposers
Answer:
(a) Producers

Maharashtra Board Class 12 Biology Solutions Chapter 14 Ecosystems and Energy Flow

Question 2.
The second trophic level in a lake is ……………………
(a) Phytoplankton
(b) Zooplankton
(c) Benthos
(d) Fishes
Answer:
(b) Zooplankton

Question 3.
Secondary consumers are …………………….
(a) Herbivores
(b) Producers
(c) Carnivores
(d) Autotrophs
Answer:
(c) Carnivores

Question 4.
What is the % of photosynthetically active radiation in the incident solar radiation?
(a) 100%
(b) 50%
(c) 1-5%
(d) 2-10%
Answer:
(b) 50%

Question 5.
Give the term used to express a community in its final stage of succession?
(a) End community
(b) Final community
(c) Climax community
(d) Dark community
Answer:
(c) Climax community

Question 6.
After landslide which of the following type of succession occurs?
(a) Primary
(b) Secondary
(c) Tertiary
(d) Climax
Answer:
(a) Primary

Question 7.
Which of the following is most often a limiting factor of the primary productivity in any ecosystem?
(a) Carbon
(b) Nitrogen
(c) Phosphorus
(d) Sulphur
Answer:
(c) Phosphorus

2. Very short answer question.

Question 1.
Give an example of ecosystem which shows inverted pyramid of numbers.
Answer:
Number of insects dependent on a single tree, is an example of ecosystem having inverted pyramid of numbers.

Question 2.
Give an example of ecosystem which shows inverted pyramid of biomass.
Answer:
Oceanic ecosystem has inverted pyramid of biomass.

Question 3.
Which mineral acts as limiting factor for productivity in an aquatic ecosystem?
Answer:
Phosphorus acts as limiting factor for productivity in an aquatic ecosystem.

Maharashtra Board Class 12 Biology Solutions Chapter 14 Ecosystems and Energy Flow

Question 4.
Name the reservoir and sink of carbon in carbon cycle.
Answer:
Atmosphere is the reservoir of carbon cycle, while fossil fuels embedded in ocean and oceanic waters are the sink of carbon in carbon cycle.

3. Short answer questions.

Question 1.
Upright and inverted pyramid of biomass.
Answer:

Upright pyramid Inverted pyramid
1. In upright pyramid, the number and biomass of the organisms which are at first trophic level of producers is high. 1. In inverted pyramid, the number and biomass of organisms at first trophic levels of producers is lowest.
2. The biomass goes on decreasing at each trophic level. 2. The biomass foes on increasing at each trophic level.
3. The base of the pyramid is always in large number of producers. 3. The base of pyramid is always in small numbers of producers.
4. Pyramid is always upright. 4. Pyramid is always inverted.

Question 2.
Food chain and Food web.
Answer:

Food chain Food web
1. Food chain is the linear sequence of organisms for feeding purpose. 1. Food web is interconnections between many small food chains.
2. In food chain the flow of energy is through a single straight pathway from the lower trophic level to the higher trophic level. 2. In food web, the energy flow is interconnected through numerous food chains in the ecosystem.
3. In a food chain, members present at higher trophic level feeds on only single type of organisms. 3. In a food web, one organism can feed on multiple types of organisms.
4. Energy flow can be easily calculated in food chain. 4. Energy flow is difficult to calculate in a food web.
5. In food chain there is increased instability due to increasing number of separate and confined food chains. 5. In food web there is increased stability due to the presence of the complex food chains.
6. The whole food chain gets affected even if one group of an organism is disturbed. 6. The food web does not get disturbed by the removal of one group of organisms.
7. Member of higher trophic level depends or feed upon the single type of organisms of the lower trophic level. 7. The members of higher trophic level depend or feed upon many different types of the organism of the lower trophic level.
8. Food chain consists of only 4-6 trophic levels of different species. 8. Food web contains numerous trophic levels and also of different populations of species.
9. Competition is seen in members of same trophic level. 9. Competition is seen in members of same as well as different trophic levels.
10. Food chains are of two types:

1. Grazing food chain 2. Detritus food chain.

10. In food web there are no types.

4. Long answer questions

Question 1.
Define ecological pyramids and describe with examples, pyramids of number and biomass.
Answer:
1. Ecological Pyramids : Ecological Pyramids are the representation of relationships between different components of ecosystem at successive trophic levels.

2. Pyramid of numbers:

  • Pyramid of numbers is the diagrammatic representation which shows the relationship between producers, herbivores and carnivores at successive trophic levels in terms of their numbers.
  • As we go up the trophic levels, the interdependent organisms keep on reducing in their numbers.
  • For example, the number of grasses are more than the number of herbivores which eat them. The number of herbivores such as rabbits would be lesser than grass but greater than the carnivores that are dependent upon the population of rabbits.
  • Thus, the producers would be more than primary consumers and primary consumers would be more than secondary consumers. The top level consumers would be least in their numbers. This pyramid shows upright nature.

Maharashtra Board Class 12 Biology Solutions Chapter 14 Ecosystems and Energy Flow

3. Pyramid of biomass:
(1) Pyramid of biomass are constructed by taking into consideration the different biomass in every successive trophic level.
(2) Pyramid of biomass in seas in inverted as the biomass of fishes is more than the biomass of phytoplankton.
Maharashtra Board Class 12 Biology Solutions Chapter 14 Ecosystems and Energy Flow 1

Question 2.
What is primary productivity? Give brief description of factors that affect primary productivity.
Answer:
(1) Primary Productivity : The rate of generation of biomass in an ecosystem which is expressed in units of mass per unit surface (or volume) per unit time, for instance grams per square metre per day (g/m²/day) is called primary productivity.

(2) Primary productivity is described as gross primary productivity (GPP) and net primary productivity (NPP).

(3) The rate of production of organic matter during photosynthesis is called gross primary productivity of an ecosystem. Of this the amount of energy lost through respiration of plants is called respiratory losses.

(4) Gross primary productivity minus respiratory losses gives the net primary productivity (NPP).

(5) Net primary productivity is the available biomass for the consumption to heterotrophs (herbivores, carnivores and decomposers).

(6) Factors affecting primary productivity: Gross primary productivity (GPP) depends on the following factors:

  • Plant species inhabiting a particular area.
  • Variety of environmental factors such as temperature, sunlight, salinity, oxygen and carbon dioxide content, etc.
  • Availability of nutrients and
  • Photosynthetic capacity of plants.

Question 3.
Define decomposition and describe the processes and products of decomposition.
Answer:

  1. Decomposition is the process carried out by the decomposer organisms.
  2. Most of the bacteria, actinomycetes and fungi are decomposers. They convert the dead and decaying organic matter into simpler compounds. These simpler inorganic substances return back to the environment.
  3. Decomposition takes place through detritus food chain. It starts from the dead organic matter. Detritus eating organisms called detritivores like earthworm, etc. breakdown the detritus into smaller fragments. Therefore, this first step of decomposition is called fragmentation.
  4. Water soluble inorganic nutrients seep into the soil after fragmentation. These nutrients get precipitated as salts. Therefore, this second step of decomposition is called leaching.
  5. The third step of decomposition is called catabolism. In this step, fungal and bacterial enzymes degrade the detritus into simple inorganic substances.
  6. The partially decomposed organic matter is called humus which is formed by the process of humification. Humus is a dark coloured amorphous substance which is the reservoir of nutrients.
  7. Humus too undergoes decomposition by bacterial action at a very slow rate and ultimately releases inorganic matter. This process is therefore called mineralization.
  8. Decomposition requires oxygen in greater amount. The rate of decomposition is dependent upon the temperature and the humidity of the environment.

Question 4.
Write important features of a sedimentary cycle in an ecosystem.
Answer:

  1. Reservoir of sedimentary cycles is earth’s crust.
  2. The nutrients such as phosphorus which show sedimentary cycle, moves through hydrosphere, lithosphere and biosphere.
  3. There is no respiratory release of nutrients into the atmosphere which show sedimentary cycle.
  4. Natural reservoir of such nutrients are usually in the form of rocks. The rocks upon weathering release such nutrients into circulation.
  5. Sedimentary cycles are very slow in their reactions.

Question 5.
Describe carbon cycle and add a note on the impact of human activities on carbon cycle.
Answer:
I. Carbon cycle:
(1) The entire carbon cycle has following basic processes viz. Photosynthesis, Respiration, Decomposition, Sedimentation and Combustion.

(2) Carbon is an important element as it forms 49% of the dry weight of all organisms. 71% of global carbon is present in the oceans. Therefore, ocean is the major reservoir of carbon. Carbon is also present in all fossil fuels. This is long term storage places or sinks for carbon which is in the form of coal, natural gas, etc.

Maharashtra Board Class 12 Biology Solutions Chapter 14 Ecosystems and Energy Flow

(3) Respiration and photosynthesis are the two events that keep the carbon in cyclic circulation. During respiration, oxygen is used for combustion of carbohydrates as a result of which carbon dioxide and water are formed with the release of energy. The process of photosynthesis utilizes carbon dioxide and water vapour liberating oxygen and producing carbohydrates at the same time.

(4) Solar energy is stored in the carbon-carbon bonds of carbohydrates during photosynthesis whereas respiration releases the same stored energy.

(5) The main reservoirs for carbon dioxide are in the oceans and in rocks. Carbon dioxide is highly soluble in water and forms mild carbonic acid upon dissolving. This dissolved carbon dioxide precipitate as a solid rock or limestone which is calcium carbonate. This reaction in the seas is aided by corals and algae which in turn builds the coral reefs made up of limestone.

(6) Carbon moves through food chains. Autotrophic green plants on land and in water take up carbon dioxide and manufacture carbohydrates by the process of photosynthesis. The carbon stored in plants has three different fates, viz. liberation into atmosphere, consumption by animals upon feeding, storage in the plant till the plant dies.

(7) Animals get their carbon requirement through their food. When autotrophs are consumed, the heterotrophs obtain carbon. Carbon in animals also has three fates, viz. release back into the atmosphere in the process of respiration, release of stored carbon from the body by the action of decomposers or conversion into fossil fuels if buried intact.

(8) Fossil fuels such as coal, oil, natural gas, etc. can be mined and burned for energy purposes. This burning releases carbon dioxide back into the atmosphere.

(9) Carbon from limestone can also be released if pushed to the surfaces and slowly weathered away. Subducting and volcanic eruptions can also release the stored carbon from sediments.
Maharashtra Board Class 12 Biology Solutions Chapter 14 Ecosystems and Energy Flow 2

II. Impact of human activities on carbon cycle:
(1) Excessive burning of fossils fuels for power plants, industrial processes and vehicular traffic, adds excessive carbon dioxide into atmosphere. When fossil fuels burn to run factories, power plants, motor vehicles, most of the carbon quickly enters the atmosphere as carbon dioxide gas.

(2) Each year, 5.5 billion tonnes of carbon is released through combustion of fossil fuels. Of this massive amount, 3.3 billion tonnes stays in the atmosphere.

(3) Rapid deforestation also increases carbon dioxide. Since plants absorb carbon dioxide for their photosynthesis, they always reduce the concentration of CO2. But deforestation upsets this balance.

(4) Massive burning of fossil fuel for energy and transport, have significantly increased the rate of release of carbon dioxide into the atmosphere which is causing global warming and resultant climate change.

Maharashtra Board Class 12 Biology Important Questions Chapter 13 Organisms and Populations

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 13 Organisms and Populations Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 13 Organisms and Populations

Multiple choice questions

Question 1.
All ecosystems on the earth together form
(a) biosphere
(b) biome
(c) living world
(d) environment
Answer:
(a) biosphere

Question 2.
What is the mean annual temperature in the region of Arctic and Alpine tundra ?
(a) About -10 to 2 °C
(b) About -2 to 2 °C
(c) About 0 to 5 °C
(d) About 5 to 10 °C
Answer:
(a) About -10 to 2 °C

Maharashtra Board Class 12 Biology Important Questions Chapter 13 Organisms and Populations

Question 3.
Which out of the following are the major biomes in India ?
(I) Desert (II) Grassland (III) Tropical rain forest (IV) Temperate forest (V) Coniferous forest (VI) Deciduous forest (VII) Sea coast (VIII) Tundra
(a) (II) (IV) (VI) (VII)
(b) (I) (II) (V) (VIII)
(c) (I) (III) (VI) (VII)
(d) (II) (III) (VI) (VIII)
Answer:
(c) (I) (III) (VI) (VII)

Question 4.
Important key elements that bring about variations in the different habitats are
(a) temperature, water, light and soil
(b) salinity, pollutants, water
(c) modern developmental parameters in the region
(d) scientific progress and technological innovations
Answer:
(a) temperature, water, light and soil

Question 5.
Which out of the following is the most ecologically relevant factor?
(a) Water
(b) Temperature
(c) Salinity
(d) Wind speed
Answer:
(b) Temperature

Question 6.
Animals that can tolerate a narrow range of temperature are
(a) stenothermal
(b) eurythermal
(c) poikilothermic
(d) homeothermic
Answer:
(a) stenothermal

Question 7.
Animals that can tolerate a narrow range of salinity are
(a) euryhaline
(b) anadromous
(c) catadromous
(d) stenohaline
Answer:
(d) stenohaline

Question 8.
What is the source of energy in the depths of more than 500 m in the oceans?
(a) Sunlight
(b) Wind energy
(c) Dead and decaying matter
(d) Phytoplankton
Answer:
(c) Dead and decaying matter

Question 9.
Which factor does not determine percolation and water holding capacity of the soil?
(a) Soil composition
(b) Grain size
(c) Aggregation of soil particles
(d) Vegetation on that soil
Answer:
(d) Vegetation on that soil

Question 10.
Which factors of soil does not determine the vegetation in any area?
(a) pH
(b) mineral composition
(c) topography
(d) types of microorganisms
Answer:
(d) types of microorganisms

Question 11.
Find the odd one out
(a) Dormancy
(b) Hibernation
(c) Aestivation
(d) Migration
Answer:
(d) Migration

Question 12.
Which of the following ability is present in the desert animals?
(a) Ability to concentrate urine.
(b) Ability to remain inside the shelters and escape need of water.
(c) Ability to derive water from all the fruits.
(d) Ability to absorb water from air.
Answer:
(a) Ability to concentrate urine.

Question 13.
Which of the statement does not describe the adaptation of the desert plants ?
(a) Desert plants have a thick cuticle on their leaf surfaces.
(b) Desert plants have their stomata arranged in sunken pits.
(c) Desert plants have a special photosynthetic pathway called CAM.
(d) Desert plants have soft stems and large leaves.
Answer:
(d) Desert plants have soft stems and large leaves.

Question 14.
What is the use of CAM type of photosynthetic pathway for the desert plants ?
(a) It enables their stomata to remain closed during day time.
(b) It requires less sunlight for the photosynthesis.
(c) It requires less amount of chlorophyll during photosynthesis.
(d) The water absorbed from the soil by the plants during CAM photosynthetic pathway is less.
Answer:
(a) It enables their stomata to remain closed during day time.

Question 15.
In which plant photosynthetic function is taken over by the flattened stems?
(a) Nephrolepis
(b) Cycas
(c) Opuntia
(d) Zea mays
Answer:
(c) Opuntia

Question 16.
What is Allen’s rule?
(a) Mammals from colder climates generally have shorter ears and limbs to minimise heat loss.
(b) Mammals have constant temperature of the body in spite of their varied habitats.
(c) Mammals can be oviparous at times.
(d) The ecosystem of cold climate regions is equally occupied with mammals, birds and reptiles.
Answer:
(a) Mammals from colder climates generally have shorter ears and limbs to minimise heat loss.

Question 17.
Sunken stomata is the characteristic feature of
(a) hydrophyte
(b) mesophyte
(c) xerophyte
(d) halophyte
Answer:
(c) xerophyte

Question 18.
Which of the following pairs is correctly matched ?
(a) Uricotelism-aquatic habitat
(b) Parasitism-intra-specific relationship
(c) Excessive perspiration-xeric adaptation
(d) Stream-lined body-aquatic adaptation
Answer:
(d) Stream-lined body-aquatic adaptation

Question 19.
World population day is observed on
(a) 11th July
(b) 16th September
(c) 23rd October
(d) 1st December
Answer:
(a) 11th July

Question 20.
World Environment day is celebrated on
(a) 22nd April
(b) 5th June
(c) 1st October
(d) 16th November
Answer:
(b) 5th June

Question 21.
World Earth day is observed on
(a) 22nd April
(b) 5th June
(c) 21st October
(d) 26th November
Answer:
(a) 22nd April

Question 22.
World ozone day is celebrated on
(a) 5th June
(b) 16th September
(c) 1st October
(d) 23rd October
Answer:
(b) 16th September

Question 23.
A group of individuals belonging to the same species within an ecosystem is called a
(a) community
(b) habitat
(c) population
(d) specific group
Answer:
(c) population

Question 24.
The populations of different species that live and interact together in the ecosystem are called
(a) community
(b) habitat
(c) population
(d) interspecies
Answer:
(a) community

Question 25.
If do not occur in an ecosystem, the survival of organisms may not take place and functioning of ecosystem is lost.
(a) interactions
(b) struggle
(c) reproduction
(d) none of the above
Answer:
(a) interactions

Question 26.
If in a pond there were 200 lotus plants last year and through reproduction 20 new plants are added, taking the current population to 220, what is the natality rate of the lotus for that year ?
(a) 0.2
(b) 0.4
(c) 0.1
(d) 1.0
Answer:
(c) 0.1

Question 27.
During laboratory experiments, 30 fishes died from an aquarium tank having 150 fishes during one month. What is the rate of mortality of fishes per month ?
(a) 0.2
(b) 0.3
(c) 0.4
(d) 0.5
Answer:
(a) 0.2

Question 28.
Which of the following is correct statement?
(a) Natality can never be controlled in any population.
(b) If mortality is more than natality, the density of population declines.
(c) Natality and mortality are always same for every population.
(d) If natality is more than mortality the population size declines.
Answer:
(b) If mortality is more than natality, the density of population declines.

Question 29.
What is the apt definition of density?
(a) The capacity of a population to sustain.
(b) The total number of genes in a population.
(c) The total number of individuals in a population per unit area.
(d) The total number of births taking place.
Answer:
(c) The total number of individuals in a population per unit area.

Question 30.
On which of the following factors is growth rate of a population dependent ?
(a) Density and natality
(b) Natality and age structure
(c) Mortality and density
(d) Natality and mortality
Answer:
(a) Density and natality

Question 31.
When pre-reproductive age group is large in a population, what will be the growth rate of that population?
(a) Steady
(b) Rapid
(c) Declining
(d) None of these
Answer:
(b) Rapid

Question 32.
When pre-reproductive and post-reproductive age group is same in structure, the population is
(a) declining
(b) increasing
(c) steady
(d) disappearing
Answer:
(c) steady

Question 33.
In which type of interactions, interacting species do not live closely together ?
(a) Competition
(b) Parasitism
(c) Commensalism
(d) Predation
Answer:
(d) Predation

Maharashtra Board Class 12 Biology Important Questions Chapter 13 Organisms and Populations

Question 34.
Name the interaction in which one species is harmed but the other remains unaffected,
(a) Commensalism
(b) Parasitism
(c) Amensalism
(d) Competition
Answer:
(c) Amensalism

Question 35.
Choose the correct statement from the following
(a) Carnivores are the only predators in any ecosystem
(b) Herbivores are in a broad ecological context not very different from predators.
(c) Sparrow eating grain is not a predator.
(d) Predation and parasitism are one and the same.
Answer:
(b) Herbivores are in a broad ecological context not very different from predators.

Question 36.
What is the meaning of symbiosis?
(a) Living together
(b) Breeding together
(c) Fighting with each other
(d) Feeding together
Answer:
(a) Living together

Question 37.
Gause’s ‘Competitive Exclusion Principle’ can work only when
(a) the resources are limited
(b) the resources are abundant
(c) the inferior species and superior species do not need same resources
(d) only when there is interspecific competition
Answer:
(a) the resources are limited

Question 38.
The word commensalism means
(a) on the either side of the table
(b) sharing the table
(c) eating at the table of other
(d) on the top of the table
Answer:
(b) sharing the table

Question 39.
Which of the following is not a classical example of commensalism?
(a) An orchid growing as an epiphyte on a mango branch.
(b) Barnacles growing on the back of a whale.
(c) The cattle egret and grazing cattle.
(d) Lichen seen on the tree.
Answer:
(d) Lichen seen on the tree

Question 40.
Calotropis plant is poisonous for herbivores as it is rich in
(a) opium
(b) cardiac glycosides
(c) quinine
(d) strychnine
Answer:
(b) cardiac glycosides

Question 41.
Term ecology was first used by
(a) Grinnell
(b) Reiter
(c) Haeckel
(d) Darwin
Answer:
(b) Reiter

Question 42.
Who introduced subject ecology to the world.
(a) Reiter
(b) Grinnell
(c) Darwin
(d) E. Haeckel
Answer:
(d) E. Haeckel

Question 43.
The term niche was first time used by
(a) Grinnell
(b) Haeckel
(c) Mendel
(d) Darwin
Answer:
(a) Grinnell

Match the columns

Question 1.

Column A Column B
(1) Organism (a) Large unit with specific climatic zone
(2) Population (b) Different species in particular area
(3) Community (c) Same species in a geographical area
(4) Biome (d) Basic unit of ecological hierarchy

Answer:

Column A Column B
(1) Organism (d) Basic unit of ecological hierarchy
(2) Population (c) Same species in a geographical area
(3) Community (b) Different species in particular area
(4) Biome (a) Large unit with specific climatic zone

Question 2.

Water body A Salinity values B
(1) Streams (a) 340 ppt
(2) Indian Ocean (b) 5 ppt
(3) Hyper saline lagoon (c) 30-35 ppt
(4) Dead sea (d) 100 ppt

Answer:

Water body A Salinity values B
(1) Streams (b) 5 ppt
(2) Indian Ocean (c) 30-35 ppt
(3) Hyper saline lagoon (d) 100 ppt
(4) Dead sea (a) 340 ppt

Classify the following to form Column B as per the category given in Column A

Question 1.
Emigration, Immigration, More mortality, Unlimited resources, More natality, Continuous reproduction.

Column A Column B
(1) Decrease in population density ————–
(2) Increase in population density —————
(3) Exponential growth ————–

Answer:

Column A Column B
(1) Decrease in population density Emigration, More mortality
(2) Increase in population density Immigration, More natality
(3) Exponential growth Unlimited resources, Continuous reproduction.

Question 2.
Hungary, United States, Denmark, Italy, Australia, Kenya, Nigeria, Germany

Pattern of population growth Name of the countries
(1) Rapid growth ————–
(2) Slow growth —————
(3) Zero growth ————–
(4) Negative growth ————–

Answer:

Pattern of population growth Name of the countries
(1) Rapid growth Kenya, Nigeria
(2) Slow growth United States, Australia
(3) Zero growth Denmark, Italy
(4) Negative growth Hungary, Germany

Question 3.
Epiphytic orchid and mango tree, Ascaris and human, Egret and cattle, Lichen having alga and fungus, Bumblebee and flowering plant, Plasmodium vivax and man.

Column A Column B
(1) Mutualism ————–
(2) Commensalism —————
(3) Parasitism ————–

Answer:

Column A Column B
(1) Mutualism Lichen having alga and fungus, Bumblebee and flowering plant.
(2) Commensalism Epiphytic orchid and mango tree, Egret and cattle.
(3) Parasitism Ascaris and human, Plasmodium vivax and man.

Very short answer questions

Question 1.
What are the key abiotic factors that influence all habitats?
Answer:
Key abiotic factors that influence all habitats are ambient temperature, availability of water, light and type of soil.

Question 2.
What is homeostasis?
Answer:
Homeostasis is the state of steady interned, physical and chemical conditions maintained by living systems.

Question 3.
What is meant by eurythermal?
Answer:
Eurythermal are those organisms that can tolerate and thrive in a wide range of temperatures.

Question 4.
What is meant by stenothermal?
Answer:
Stenothermal are those organisms which are restricted to only narrow range of temperatures.

Maharashtra Board Class 12 Biology Important Questions Chapter 13 Organisms and Populations

Question 5.
What is euryhaline?
Answer:
Organisms which can tolerate wide range of salinities are called euryhaline.

Question 6.
What is stenohaline?
Answer:
Organisms which are restricted only to a narrow range of salinity are called stenohaline.

Question 7.
What are the various characteristics of soil?
Answer:
Characteristics of the soil are soil composition, grain size, the percolation and water holding capacity, pH, mineral composition of the soil.

Question 8.
What decides the vegetation of an area?
Answer:
The soil characteristics along with pH, mineral composition and topography, and climatic factors determine the vegetation of an area.

Question 9.
What is Population ecology?
Answer:
Population ecology is an important area of ecology because it links ecology to population dynamics, genetics and evolution.

Question 10.
In which interaction of species, both the species are at a loss?
Answer:
Competition is the type of interaction where . both the species are at a loss.

Question 11.
What is parasitism?
Answer:
Parasitism is the type of interaction between two species in which parasitic species is benefited and the host species is harmed.

Question 12.
What are ectoparasites?
Answer:
Parasites that feed on the external surface of the host organism are called ectoparasites.

Question 13.
Name blood sucking ectoparasites.
Answer:
Lice, mosquito feeding on human blood and ticks parasitic on dogs.

Question 14.
Name the malarial parasite and its vector.
Answer:
The malarial parasite is Plasmodium vivax which needs a vector anopheles mosquito.

Question 15.
Name the secondary metabolites that act as defensive substances against grazers and browsers.
Answer:
Nicotine, caffeine, quinine, strychnine, opium, etc. are secondary metabolites which act as defensive substances produced by plants against grazers and browsers.

Question 16.
Name the ectoparasites which infest the marine fish.
Answer:
Copepods are the ectoparasites which infest the marine fish.

Question 17.
What do you mean by ‘evolutionary arms race’? In which kind of interactions is it observed?
Answer:
In order to save their lives, prey species : have evolved various defence mechanisms. This are called ‘evolutionary arms race’. These defence mechanisms are seen in prey-predator interactions.

Give Definitions of the following

Question 1.
Habitat
Answer:
The physical space of an organism with the other living or non-living factors is called its habitat.

Question 2.
Microhabitat
Answer:
The immediate surrounding of an organism is called microhabitat.

Question 3.
Niche
Answer:
Niche is defined as the processes about how that organism is linked with its physical and biological environment.

Question 4.
Fundamental niche
Answer:
Fundamental niche is the niche in the absence of all competitors, this is highly improbable in nature.

Question 5.
Realized niche
Realized niche is more realistic approach, in the presence of competition for the resources available in the habitat.

Question 6.
Competition
Answer:
Competition is defined as a process in which the fitness of one species is significantly lowered in the presence of another species.

Give one or two examples of the following

Question 1.
Intraspecific competition
Answer:
Two dogs fighting for same food. Two tomcats fighting for their territory.

Question 2.
Commensalism
Answer:
An orchid growing as an epiphyte on a branch of mango tree, is benefited due to support offered by mango tree but the mango tree does not get any benefit.

Maharashtra Board Class 12 Biology Important Questions Chapter 13 Organisms and Populations

Question 3.
Ectoparasites
Answer:
Mosquito, louse sucks blood from human host.

Question 4.
Endoparasites
Answer:
Round worm. Ascaris and tape worm are endoparasites in the intestine of humans.

Name the type of association

Question 1.
Cattle egret birds with buffalo.
Answer:
Commensalism

Question 2.
Tiger and the deer.
Answer:
Predator and prey relationship

Question 3.
Visiting flamingos and fishes in the estuarine water.
Answer:
Interspecific competition.

Distinguish between the following

Question 1.
Natality and Mortality.
Answer:

Natality Mortality
1. Birth rate of any population is called its natality. 1. Death rate of any population is called its mortality.
2. Rise in natality increase the population density. 2. Rise in mortality decrease the population density.
3. Decline in natality decrease the population. 3. Decline in mortality increase the population.
4. Natality is the positive factor for population growth. 4. Mortality is the negative factor for population growth.
5. Absolute natality will always be more than realized natality. 5. Absolute mortality will always be less them realized mortality.

Question 2.
Eurythermal and stenothermal.
Answer:

Eurythermal Stenothermal
1. Animals which can tolerate wide range of temperatures are called eurythermal. 1. Animals which can tolerate only narrow range of temperature fluctuations are called stenothermal.
2. Eurythermal animals show reduced temperature sensitivity. 2. Stenothermal animals show high temperature sensitivity.
3. Body functions of eurythermal animals can occur at wide range of temperature range.

E.g. Goat, man, cat, dog, tiger, cow, sheep, monkey, crab, etc.

3. Body functions of stenothermal animals can occur at only narrow range of temperature range.

E.g. Insects, fishes, reptiles, snakes, etc.

Give scientific reasons

Question 1.
Temperature is said to be the most ecologically relevant environmental factor.
Answer:

  1. Temperature fluctuations on the earth are quite marked.
  2. The distribution of plants and animals on the earth depends upon temperature range.
  3. For the organisms ambient temperature affects their enzyme kinetics of the cell.
  4. Entire metabolism, activity and other physiology of the organism is dependent on temperature. Therefore, it is said to be the most ecologically relevant environmental factor.

Question 2.
Adaptation is an important attribute of the organism.
Answer:

  1. Organisms adapt to their surrounding environment by showing physiological, behavioural or morphological changes which are called adaptations.
  2. Due to adaptations, organisms can survive and reproduce in its environment. Therefore, adaptation is said to be am important attribute of the organisms.

Question 3.
Both host and the parasite tend to co¬evolve, against each other.
Answer:

  1. Host and parasitic relationship is most specific. This means that for a particular parasite there is specific host.
  2. Many parasites have evolved to be host-specific as they can parasitize only a single species of host. Therefore, during evolution, they both co-evolve together, against each other.

Question 4.
Cuscuta plant does not have chlorophyll.
Answer:

  1. Cuscuta is a parasitic plant. It is commonly found growing on hedge plant.
  2. It derives its nutrition directly from the host plant on which it thrives by parasitizing it.
  3. Since it does not prepare its own food by photosynthesis, it loses chlorophyll from the leaves.

Question 5.
Predators in nature are called prudent.
Answer:

  1. Predators control the prey population but if a predator over exploits its prey, then the prey might become extinct.
  2. If prey species is not available, the predator will also starve and become extinct. Predators, therefore, do not kill the prey unnecessarily. They act as prudent.

Write short notes

Question 1.
Temperature fluctuations on the earth.
Answer:

  1. The temperatures vary from subzero levels in polar areas and high altitudes, to about 50 °C in tropical deserts during summer.
  2. There are also seasonal changes in the temperature.
  3. Temperature also shows progressive decrease from the equator towards the poles and from plains to the mountain tops.
  4. Some unique habitats such as hot springs may show very high temperatures of about 80 to 100 °C.
  5. In deep-sea hydrothermal vents average temperatures may rise up to 400 °C.

Question 2.
Adaptations of mammals in colder regions.
Answer:

  1. Mammals inhabiting colder regions have shorter snout, ears, tail and limbs to minimize the loss of body heat. This is called Allen’s Rule.
  2. Aquatic mammals such as whales and seals living in the polar seas, have a thick layer of fat which is called a blubber below their skin.
  3. Blubber acts as an insulator and thus helps to keep the body warm by reducing loss of body heat.
  4. Some animals like polar bears undergo hibernation and thus tide over the stressful winters.

Question 3.
Natality.
Answer:

  1. Natality is the birth rate of a population. Due to increased natality the population density rises.
  2. Natality is a crude birth rate or specific birth rate.
  3. Crude birth rate : Number of births per 1000 population/year gives crude birth rate. Crude birth rate is helpful in calculating population size.
  4. Specific birth rate : Crude birth rate is relative to a specific criterion such as age. E.g. If in a pond, there were 200 carp fish and their population rises to 800. Then, taking the current population to 1000, the birth rate becomes 800/200 = 4 offspring per carp per year. This is specific birth rate.
  5. Absolute Natality : The number of births under ideal conditions when there is no competition and the resources such as food and water are abundant, then it give absolute natality.
  6. Realized Natality : The number of births under different environmental pressures give realized natality. Absolute natality will be always more than realized natality.

Maharashtra Board Class 12 Biology Important Questions Chapter 13 Organisms and Populations

Question 4.
Mortality.
Answer:

  1. Mortality is the death rate of a population. It gives a measure of the number of deaths in a particular population, in proportion to the size of that population, per unit of time.
  2. Mortality rate is typically expressed in deaths per 1,000 individuals per year.
    A mortality rate of 9.5 (out of 1,000) in a population of 1,000 would mean 9.5 deaths per year in that entire population or 0.95% out of the total.
  3. Absolute Mortality : The number of deaths under ideal conditions when there is no competition, and all the resources such as food and water are abundant, then it gives absolute mortality.
  4. Realized Mortality : The number of deaths under environmental pressures come into play gives realized mortality.
  5. It must be remembered that absolute mortality will always be less than realized mortality.

Question 5.
Populations Interactions
Answer:

  1. In nature, every species requires interactions with at least one other species for its food.
  2. Even autotrophic plant species needs soil microbes to break down the organic matter in soil and return the inorganic nutrients for absorption.
  3. The plants need animal agents . for pollination.
  4. Animals, plants and microbes cannot live in isolation but interact in various ways to form a biological community.
  5. Such interactions can be interspecific (within two different species) or intraspecific (within the same species).
  6. Interspecific interactions are of broad four types viz, neutralism, negative or harmful, positive or beneficial, and both positive and negative interactions.
  7. Interacting species live closely together in interactions such as predation, parasitism and commensalism.
  8. Neutralism interaction have no significant effect on either species. Negative interactions are of competition or amensalism type. Positive interactions occur in the form of mutualism, protocooperation and commensalism. Parasitism and predation are both positive and negative type of interaction.

Question 6.
Mutualism
Answer:

  1. Mutualism is an obligatory and interdependent interaction. It is an association of two species in which both of them are benefited.
  2. The classic example of mutualism is lichens. Lichen is an intimate, mutualistic relationship between a fungus and photosynthetic algae or cyanobacteria.
  3. Most of the plant and animal interactions are of mutualistic type.
  4. For pollination and seed dispersal, plants depend on the animals.
  5. Animals in turn feed on pollen and nectar during pollination. During seed dispersal juicy and nutritious fruits are used by the animals.
  6. In animal-animal interactions also mutualism is seen in many instances.

Question 7.
Brood parasitism
Answer:

  1. Brood parasitism is a type of parasitic behaviour shown by Asian Koel. Koel lays its eggs in the nest of crow.
  2. Crow acts as a host bird and incubate the eggs of koel.
  3. The eggs of koel show resemblance to the host’s egg in size and colour. This reduces the chances of the crow detecting koel’s eggs and ejecting them from the nest.
  4. Eggs of koel hatch before the host’s eggs and hence parasitic bird is in advantage.

Short answer questions

Question 1.
What are the three main types of niches?
Answer:

  1. Spatial or habitat niche : Spatial or habitat niche means the physical space occupied by the organisms.
  2. Trophic niche : This kind of niche is based on the trophic level of an organism in a food chain.
  3. Multidimensional or hypervolume niche : In multidimensional niche, number of abiotic and biotic environmental factors are considered. The resulting space by the niche is called hypervolume. Therefore it is also called hypervolume niche.
  4. It shows the position of an organism in the environmental gradient.

Question 2.
Why do animals need to maintain homeostasis?
Answer:

  1. Homeostasis keeps the body in equilibrium.
  2. All the internal functions are maintained due to homeostasis.
  3. Survival, growth and reproduction can be achieved due to this steady state.
  4. The external environment changes constantly but by homeostasis, organisms can cope up with this change.
  5. Thus homeostasis is a way of adaptation for survival.

Question 3.
How is sunlight important for every ecosystem ?
Answer:

  1. Sunlight is essential for the plants for photosynthesis.
  2. It is the only source of energy for the entire ecosystem.
  3. Without sunlight the food chains will not exist.
  4. Survival of plants is therefore dependent on sunlight.
  5. In case of animals diurnal and seasonal variations in light intensity and duration decide the feeding, foraging and reproductive activities.
  6. Migrations shown by certain animals also depend on light.
  7. Almost all animals have behaviour based on photoperiod. The proportion of sunlight on land also decides the ambient temperature. Thus, life is dependent on light.

Question 4.
What can be the causes of deviation from 1 : 1 sex ratio in natural habitat?
Answer:

  1. In nature, when there is chromosomal sex determination, usually 1 : 1 sex ratio is obtained. But there are other causes for deviation from 1 : 1 sex ratio.
  2. There are many other environmental sex determination methods where 1 : 1 sex
    ratio becomes skewed. This is due to mere chance of fertilization.
  3. Also some lower animals show parthenogenesis e.g. as in honey bees. In such cases, offspring produced may not be in 1 : 1 proportion.

Question 5.
What are the special adaptations that endoparasites show?
Answer:

  1. Endoparasites show loss of unnecessary sense organs as these are not needed for the parasite.
  2. There are adhesive organs or suckers always present in the endoparasites which are needed to cling on to the host.
  3. Endoparasites show loss of digestive system.
  4. They have very high reproductive capacity.
  5. The complex life cycles are seen in such parasites which involve intermediate hosts or vectors to facilitate transfer to the host.

Question 6.
What are the effects of parasite on the host?
Answer:

  1. Most of the parasites cause harm to the host.
  2. Host is affected by reducing its survival, growth and reproduction.
  3. Some parasites can also be fatal to the host causing death of the host.
  4. The population density of host species is reduced by parasites.
  5. The host species become more vulnerable to predation by making it physically weak.

Question 7.
What is the role of predator in balancing the ecosystem?
Answer:

  1. Predators keep prey population under control. If predators are lacking from the ecosystem, the prey population will rise without any control. Their high density may cause instability in ecosystem.
  2. Predators also help in maintaining the species diversity in a community. This is done by reducing the intensity of competition among competing prey species.
  3. Predators control the pest species and thus can be used for natural biological control measures in an ecosystem. E.g. frog controlling the locust population.
  4. Predators also control the invading exotic species and stop their rapid spread of such species.

Question 8.
How does prey species defend themselves against predators?
Answer:
Prey species show following defence mechanisms:

  1. Showing camouflage for concealment.
  2. Moving at faster speed for escape.
  3. Cryptic colouration to avoid the detection. This is seen in some insects. Also predators display such cryptic colouration to avoid detection. E.g. Colouration in frog.
  4. Bad taste due to accumulated chemicals.
    E.g. The Monarch butterfly is highly distasteful to its predator bird as it stores a special chemical in the body during its caterpillar stage by feeding on a poisonous weed.

Question 9.
With suitable examples describe commensalism.
Answer:
I. Commensalism : Commensalism is the interaction between two species in which one species derives benefit and the other one is neither harmed nor benefited.
II. Examples of commensalism:

  1. Orchid grows as epiphyte on other big trees. The tree do not get any benefit but is neither harmed. But orchid gets support.
  2. Cattle egret is the insectivorous bird which forage close to cattle. When cattle move, the hidden insects in the grass are flushed out. These insects are then captured by egrets. Cattle do not get benefit but birds do.
  3. Sea anemone has stinging cells on the tentacles which offer protection to clown fish. Clown fish gets the protection from other predators, whereas, sea anemone does not derive any benefit from this association.

Maharashtra Board Class 12 Biology Important Questions Chapter 13 Organisms and Populations

Question 10.
What are different population attributes?
Answer:

  1. Basic physical attributes of population are population size and population density.
  2. Number of individuals in a population is its size whereas number of individuals present per unit space, in a given time is called its density.
  3. The other attributes are natality or birth rate, mortality or death rate, immigration means coming into the population, emigration means leaving from the population, age pyramids, sex ratio and biotic potential etc.

Question 11.
What are the decisive factors for population density?
Answer:
The density of a population in a given habitat during a given period fluctuates due to changes in four basic processes, viz.

  1. Natality i.e. birth rate (The number of births during a given period in the population that are added to the initial density).
  2. Mortality i.e. death rate (The number of deaths in the population during a given period).
  3. Immigration i.e. number of individuals of the same species that have come into the habitat from elsewhere during the time period under consideration.
  4. Emigration i.e. the number of individuals of the population who left the habitat and gone elsewhere during the time period under consideration.
  5. Natality and immigration increase in population density whereas mortality and emigration decrease it.

Question 12.
Should an ideal parasite be able to thrive within the host without harming it?
Answer:
A parasite that resides inside the body of host, will certainly cause discomfort to the host. But if the host dies, the parasite also will diminish. Therefore, Ideal parasite will not extract more benefits from the host. But ideal parasite cannot stay alive without association with host. They need host for various purposes like nutrition, food. Some need the host for completing their life cycle. So parasite will always harm the host in order to thrive.

Question 13.
Why didn’t natural selection lead to the evolution of such totally harmless parasites?
Answer:
Being a parasite means causing trouble or harm to the host. When host-parasite relationship is concerned both of them evolve together which is also called co¬evolution. If totally harmless parasite has to be evolved through natural selection, then such parasite will not remain as a parasite but will become commensal or can show amensalism. Therefore, natural selection does not lead to evolution of totally harmless parasite.

Question 14.
What will happen when carrying capacity of any habitat is exceeded?
Answer:
When the carrying capacity of any habitat is exceeded there is severe resource crunch. This may be in the form of food, shelter or availability of mates. This results into struggle for survival. This will result into death of those who cannot cope up with the struggle. They may die due to starvation or due to interactions such as fierce competitions. This causes the population again to come back to equilibrium. Thus when carrying capacity is exceeded, the natural way is to reduce the population in the habitat.

Question 15.
What could be the reasons behind enormous increase in human population?
Answer:

  1. Human population has increased enormously in last century due to increased natality and reduced mortality.
  2. Due to advances in medical sciences, vaccinations and better life expectancy mortality rate has been considerably reduced. With the exception of Covid 19 Pandemic, all the mortalities due to epidemics were controlled due to modern medical facilities.
  3. Due to better food production and advances in agriculture methods, people are not starving anymore. This has considerably reduced the deaths due to starvations.
  4. However, natality rate has not been reduced especially in developing countries. Lack of family planning measures, ignorance, illiteracy and traditional thinking about birth of male child, all such factors have caused tremendous increase in the human population.

Question 16.
What can be reason behind the different reproductive strategies adopted by monocot plants like cereals / pulses and dicot plants like mango?
Answer:
The monocot plants such as cereals and pulses and also mango are commercially important cash crops. Therefore, in order to obtain profitable harvest, different reproductive strategies are adopted.

Chart based / Table based questions

Question 1.
Complete the following table by placing the categories and the given signs + : Benefited ; – : Inhibited ; 0 : Not affected

Interactions Species
A B
1. Neutralism – …………………..
2. Negative interactions …………………………
3. Positive interactions …………………….
4. Both positive and negative interactions …………………………..

Answer:

Interactions Species
A B
1. Neutralism – no significant effect O O
2. Negative interactions

a. Competition – direct interference type

b. Competition – resource – use type

c. Amemsalism





O
3. Positive interactions

a. Symbiosis (Mutualism)

b. Commensalism

c. Protocooperation

+
+
+
+
O
+
4. Both positive and negative interactions

a. Parasitism

b. Predation

+
+

Diagram based questions

Question 1.
Sketch and label the graph showing logistic growth curve of the population.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 13 Organisms and Populations 1

Question 2.
Sketch and label the graph showing exponential growth curve of the population.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 13 Organisms and Populations 2

Question 3.
Sketch the different types of Age Pyramids.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 13 Organisms and Populations 3

Question 4.
From the age pyramids given in figure what will be your forecast for 15 years from now for the populations of 1. Kenya, 2. Australia, 3. Italy and 4. Hungary?
Maharashtra Board Class 12 Biology Important Questions Chapter 13 Organisms and Populations 4
Answer:
(1) Forecast for Kenya : Kenya is showing rapid growth as its pre-reproductive group is large. Post-reproductive group is much reduced. Therefore, 15 years from now, the population will be large.

(2) Forecast for Australia : Australian age pyramid is showing larger post-reproductive group. In the years to come, they will be removed from the population. Also the pre-reproductive group is not much large. This will not add to the Australian population in the future. Thus, this country will show slow growth.

(3) Forecast for Italy : In Italy the reproductive and post-reproductive groups are almost similar. The pre-reproductive group is also limited. Hence, in future, the population will not expand. At the same time the post-reproductive group here is large which means that the population growth will be reduced in the next 15 years.

(4) Forecast for Hungary : Hungary is showing : typical age pyramid where the pre-reproductive and reproductive groups are small. On the contrary, the post-reproductive group is more, which shows that in the next 15 years, the old people will leave the population, causing negative growth.

Maharashtra Board Class 12 Biology Important Questions Chapter 13 Organisms and Populations

Long answer questions

Question 1.
What are the characteristics of ecological niche ?
Answer:

  1. A niche describes how that organism is linked with its physical and biological environment.
  2. Niche is described as a position of a species in the environment. It gives the idea about how the organisms are surviving and fulfilling their needs of shelter and food.
  3. By studying niche one can get idea of the flow of energy from one organism to another. This helps to understand the feeding habits and interactions involving food chain and food web.
  4. If any niche is left vacant, other organisms\fill that position.
  5. The niche is specific to each species. Two species can never share the same niche. By having specific niche, every organism tries to reduce competition for resources.
  6. E.g. In birds, each one is specific in their eating habits, some are insectivorous, while some are frugivorous. Some are omnivorous, in this way birds living in the same habitat differ in their niches because of different eating habits.

Question 2.
What are the different ways in which organisms adapt to the changes in the environment?
Answer:
To survive and propagate further in any environment, organisms show one of the four possible ways, viz. regulate, conform, migrate and suspend.
(1) Regulate : In this method, organisms maintain homeostasis by physiological and behavioural changes. Due to homeostatic regulation, they can perform thermoregulation or osmoregulation. E.g. All birds and mammals show constant body temperature and osmotic concentration irrespective of external temperature.

(2) Conform : Most of the animals and plants are unable to maintain a constant internal environment. Their body parameters change according to outside environment. E.g. Poikilothermic animals cannot maintain body temperature but they are simple conformers. In few aquatic animals, the osmotic concentration of the body fluids changes according to surrounding osmotic concentration. Few conformers can regulate the parameters in limited range.

(3) Migrate : When organism is unable to cope up with surrounding temperatures, they migrate temporarily from such stressful habitat to a more favourable habitat. After the stressful period is over, they return back. Birds show long-distance migrations during severe winter.

(4) Suspend : Suspending the life activities for particular period is one of the methods to cope up with stressful conditions. Seeds of plants remain dormant over unfavourable period and once favourable conditions are resumed they start growing. This state is called dormancy during which metabolic activities are suspended. Hibernation and aestivation seen in some animals is also for escaping severe winter or summer respectively. E.g. Polar bear shows hibernation while snails and fish show aestivation. These are also suspension measures.

Question 3.
What are the adaptations in animals living under crushing pressure at great depths of ocean?
Answer:

  1. Environment of depths of ocean are characterized by high pressure, low temperature, absence of light, calmness of water, absence of phytoplankton and other producers, scarcity of food and thus animals staying here show many adaptations.
  2. Due to extreme pressure the bodies of deep- sea fish and other animals are very much compressed.
  3. The bony skeletons are much reduced except for jaws. They have watery muscles. Some deep-sea fishes exhibit greatly enlarged eyes which act like telescope.
  4. They are highly effective as in depths there is less light. Their retina is composed of a number of tiers of rods, presumably arranged to absorb all the limited light that enters the eye. However, the eye-size is small.
  5. Some benthic fishes have eyes located on only one side of the body. E.g. Sole fish.
  6. Many deep-sea animals are bioluminescent, i.e. they produce their own light by means of luminous organs.
  7. In anglerfish, the light is used as a bait to attract prey and also for species and sex recognition.
  8. The mouth of deep-sea fish is the enormous, which enables them to gulp large sized prey.
  9. Many of the deep-sea animals have long appendages, abundant spines, stalks or other

Question 4.
With the help of suitable diagram describe the Exponential population growth curve.
Answer:
Maharashtra Board Class 12 Biology Important Questions Chapter 13 Organisms and Populations 5

  1. When the resources are abundant, organisms show continuous growth of a population without any hindrance. With unlimited resources, each species has the ability to realise fully its innate potential to grow in number.
  2. Such growth of a population is called an exponential or geometric growth.
  3. Any species growing exponentially under unlimited resource conditions can reach enormous population densities in a short time. E.g. Human population shows such exponential growth.
  4. Exponential growth shows J-shaped curve.

Question 5.
Explain the ‘Competitive Exclusion Principle’ given by Gause.
Answer:
Gause’s ‘Competitive Exclusion Principle’:
(1) This principle states that two closely related species competing for the same resources cannot co-exist indefinitely and the competitively inferior one will be eliminated eventually.

(2) The Gause’s principle may be true if resources are limiting, but not otherwise. More recent studies do not support such gross generalisations about competition. The species during competition also show resource partitioning.

(3) In resource partitioning, the species facing competition might evolve mechanisms that promote co-existence rather than exclusion. If two species compete for the same resource, they could avoid competition by choosing, for instance, different times for feeding or different foraging patterns. E.g. Five closely related species of warblers living on the same tree were able to avoid competition and co-exist due to behavioural differences in their foraging activities. If there are two competing species and one is comparatively superior than the other, then the inferior one remains restricted to smaller geographical area. If this superior species is removed then only the inferior species expands its range.

Question 6.
What are the key differences that make such a great variation in the physical and chemical conditions of different habitats?
Answer:
Different abiotic factors make great variations in the physical and chemical conditions of different habitats.
The most important abiotic factors are temperature, water, light and soil. These abiotic factors act with the resident biotic factors in that habitat. Biotic components such as pathogens, parasites, predators and competitors keep on interacting continuously. All such interactions cause variations in different habitats.

Maharashtra Board Class 12 Biology Important Questions Chapter 13 Organisms and Populations

Question 7.
Give names of eurythermal and stenothermal animals and plants?
Answer:

  1. Eurythermal animals : Goat, man, cow, crab, bivalves, etc.
  2. Stenothermal animals : Insects, reptiles, snakes, fishes, etc.
  3. Eurythermal plants : Roses, daisies, oak trees, some fruits and vegetables.
  4. Stenothermal plants : Croton, Bougainvillea, Frangipani, vines and orchids, some other fruits and vegetables

Question 8.
What will be the effect of increasing global temperatures on the different habitats and the organisms found in those habitats?
Answer:

  1. Global temperature rise or global warming is having great impact on natural habitats.
  2. Aquatic habitats like oceans are worst affected as 90% of extra heat enters marine waters. These elevated temperatures cause adverse effects on marine habitat.
  3. Coral reefs are also affected due to increased temperature. It causes bleaching of coral reefs. Fish populations are adversely affected.
  4. Due to rising temperature, the tundra regions and polar regions are showing melting of snow. These habitats are disappearing and the animals from these areas such as polar bears are on the verge of extinction. Increased temperatures also cause desertification of the once green habitats.
  5. Nearly 50% of the species are under threat of extinction due to climate change. Global warming and climate change thus reduces biodiversity.
  6. Every plant and animal species is adapted to a specific temperature conditions. But due to global warming and associated climate change, these species are affected. Some species show migrations to cooler places.
  7. Temperatures also alter the life cycles of plants and animals. When temperatures rise, many plants grow rapidly and bloom earlier in the spring and survive longer into the fall. Some animals leave hibernation sooner.

Question 9.
Give examples of an animal and plant that can survive in fresh water as well as sea water.
Answer:
Few fish species such as salmons, crustacean prawns and crabs are able to survive in fresh water as well as sea water. Among plants, few phytoplankton, some algal species and mangorves show similar phenomena.

Question 10.
What is the source of energy for the life in deep ocean trenches where sunlight does not reach?
Answer:
In deep ocean trenches, the sunlight does not penetrate. Hence photosynthesis is not possible here. Most of the organisms that live here depend upon subsistence on falling organic matter. This organic matter falls off down from the upper photic zones, where phytoplankton can perform photosynthesis. This organic matter acts as source of energy for the life in deep oceanic trenches.