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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.

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).

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.

ii. A converging beam corresponds to rays of light directed to a virtual point object or image.

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, (n_glass) = 1.5
Thickness of the glass = 2 mm
= 2 × 10sup>-3 m
∵ Re fractive index (n_glass) = \(\frac{\text { speed of light in vacuum }(\mathrm{c})}{\text { speed of light in glass (v) }}\)
∵ 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.

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.

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,
\(\frac {1}{f}\) = \(\frac {1}{v}\) + \(\frac {1}{u}\)

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.

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

∴ v₂ = -24 cm
Here, negative signs indicate that images are formed on the left of the mirror.

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.

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? n_water = 4/3.
Answer:
Given: Real depth of pond, d_real = 10 m,
n_w = \(\frac {4}{3}\)
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 (n_w) = \(\frac {4}{3}\)
n_w = \(\frac {Realdepth}{Apparent depth}\)
∴ 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.

Question 23.
A ray of light passes from glass (n_g = 1.52) to water (n_w = 1.33). What is the critical angle of incidence?
Answer:
Given: n_g = 1.52, n_w = 1.33
To find: Critical angle (i_c)
formula: sin i_c = \(\frac {n_2}{n_1}\) = \(\frac {n_w}{n_g}\)
Calculation:
From formula,
i_c = 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:

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 i_c = \(\frac {PQ}{OQ}\) = \(\frac {4}{3}\)
∴ sin i_c = \(\frac {OQ}{OP}\) = \(\frac {3}{5}\)
∴ n_liquid = \(\frac {5}{3}\)

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,

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

According to sign conventions,
OP = u = -4 cm and CP = R = -3 cm
For refraction at curved surface,

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

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.

Question 33.
Draw neat labelled diagrams showing refraction of a monochromatic light and white light through a prism.
Answer:

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.

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.

ii. It shows that, with increasing values of i, the angle of deviation δ decreases initially to a certain minimum (δ_m) 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 = δ_m 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 = δ_m, we have i = e.

Question 36.
Show that, at condition of minimum deviation, n = \(\frac{\sin \left(\frac{\mathbf{A}+\boldsymbol{\delta}_{\mathrm{m}}}{\mathbf{2}}\right)}{\sin \left(\frac{\mathbf{A}}{\mathbf{2}}\right)}\)

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 + δ_m
∴ 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:

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 (i_min) = 1.5 × sin (18°11′)
∴ ii_min = 27°55′ ≅ 28°.

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,
δ_VR = \(\frac{\delta_{\mathrm{V}}+\delta_{\mathrm{R}}}{2}\) ≈ δ_Y = A(n_Y – 1)
Where, n_Y = 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 = L_V = lateral deviation for violet colour,
RT = L_R = lateral deviation for red colour,
∴ Lateral dispersion between these colours, L_VR = L_V – L_R

∴ r_R = sin^-1 (0.5735) ≈ 35°
Similarly,
sin r_V = \(\frac{\sin 60^{\circ}}{1.53}=\frac{\sqrt{3}}{2 \times 1.53}=\frac{1.732}{3.06}\)
= antilog {log (1.732) – log (3.06)}
= antilog {0.2385 – 0.4857}
= antilog 11.7528}
= 0.5659
∴ sin r_V = 0.566
∴ r_V = sin^-1 (0.566) = 34°28′
∴ Angle of deviation for red colour
= i – r_R = 60° – 35° = 25°
and that for violet colur = i – r_V = 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, L_R = RT = AR [sin(i – r_R)]
= AR [sin (25°)]
= 6.104 × 0.4226
= 2.58 cm
For violet colour, L_V = VM
= AV [sin (i – r_V)]
= AV × sin (25° 32′)
= 6.065 × 0.431
= 2.61 cm
∴ L_VR = L_V – L_R = 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 i_min = 27° 55′. Corresponding angle of emergence is, e_max = 90°.
From the principle of reversibility of light, i_max = 90° and e_min = 27°55′
Also, for equilateral glass prism at minimum deviation,

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. (n_red = 1.712, n_violet = 1.792)
Answer:
Given: A = 10°, n_R = 1.712, n_V = 1.792
To find:
i. Angular deviation for extreme colours (δ_V and δ_R)
ii. Dispersive power of flint glass (ω)
Formulae:
i. δ = A(n – 1)
ii. ω = \(\frac{\delta_{\mathrm{V}}-\delta_{\mathrm{R}}}{\left(\frac{\delta_{\mathrm{V}}+\delta_{\mathrm{R}}}{2}\right)}\)

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: n_R = 1.620, n_Y = 1.635, A = 8°
To find:
i. Angular dispersion (δ_V – δ_R)
ii. Dispersive power (ω)
Formulae:
i. δ_v – δ_r = A(n_V – n_R)
ii. ω = \(\frac{\mathrm{n}_{\mathrm{V}}-\mathrm{n}_{\mathrm{R}}}{\mathrm{n}_{\mathrm{Y}}-1}\)
Calculation: Since, n_Y = \(\frac {n_V+n_R}{2}\)
∴ n_V = 2n_Y – n_R
n_V = 2 × 1.635 – 1.620 = 3.27 – 1.620
∴ n_V = 1.65
From formula (i),
δ_V – δ_R = 8(1.65 – 1.620)
= 8 × 0.03 = 0.24°
∴ δ_V – δ_R = 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.

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°.

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 θ_R and θ_V 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°.

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, f_V and f_R. The distance between f_V and f_R 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.

= (1.53 – 1) × 0.08 = 0.0424
∴ f_v = 23.58 cm
∴ Longitudinal (lateral) chromatic aberration
= f_V – f_R = 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.

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, f_e = 10 cm
To Find: Magnification (m₀)
Formula: M.P = m₀ × M_e
Calculation:
From the formula,

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: f_e = 6.25 cm, f₀ = 2 cm, L = 15 cm
As image appears biggest, V_e = -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| + |u_e|
iii. M = \(\left(\frac{v_{o}}{u_{o}}\right)\left(\frac{D}{u_{e}}\right)\)
Calculation: For eyelens, using formula (i),

∴ u_e = 5 cm
From formula (ii),
|v₀| = L – |u_e|
= 15 – 5 = 10 cm
Using formula (i) for objective,

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 (f_e)
Magnifying power under normal adjustment (M)
Formula:
i. L = f₀ + f_e
ii. M = \(\frac {f_0}{f_e}\)
Calculation. From formula (i),
f_e = 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 (f_e)
Formula: M.P = \(\frac {f_0}{f_e}\) (1 + \(\frac {f_e}{D}\))
Calculation:
From formula.
12 = \(\frac {90}{f_e}\) (1 + \(\frac {f_e}{25}\))
∴ f_e = 10.71 cm

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, f_e = 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 + f_e
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. _aµ_w = \(\frac {4}{3}\)
sin i_c = \(\frac {1}{µ}\)
∴ i_c = 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?

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 P_piastic = 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.

Answer:

A = 60°
Also, ray of light suffers minimum deviation.
∴ 2i = A + δ_m
∴ δ_m = 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

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 µ₂ (µ₁ > 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) δ_m = i + e + r
Answer:
(D) δ_m = 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

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) \(\frac {f_0}{f_e}\)

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

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

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

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

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

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

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.

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).

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.

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.

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.

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

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?

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.

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.

(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.

Question 4.
Observe the given diagram and answer the questions based on it.

(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.

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.

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.

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.

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.

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

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:

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 (kcal^{m -2})
(c) energy in food calories
(d) calories /sq. m.
Answer:
(b) weight (g^-2) or energy (kcal^m-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.

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

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.

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).

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.

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.

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:

Question 2.
Sketch and label phosphorus cycle
Answer:

Question 3.
Sketch and label decomposition cycle
Answer:

Question 4.
Sketch and label simple grazing food chain
Answer:

Question 5.
Sketch and label ecological pyramids of energy and pyramid of biomass
Answer:

Question 6.
Observe pyramids of numbers and answer the questions below

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

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:

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

Answer:
A : Combustion
B : Respiration by plants
C : Respiration by animals
D : Photosynthesis
E : Decomposition.

Question 10.
Observe the diagram and answer the questions

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:

(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.

(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.

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

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

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.

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.

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.

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.

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:

Question 2.
Sketch and label the graph showing exponential growth curve of the population.
Answer:

Question 3.
Sketch the different types of Age Pyramids.
Answer:

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?

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.

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:

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.

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.