Class 11

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 7 Modern Periodic Table Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 7 Modern Periodic Table

1. Explain the following

Question A.
The elements Li, B, Be and N have the electronegativities 1.0, 2.0, 1.5 and 3.0, respectively on the Pauling scale.
Answer:

• Li, B, Be and N belong to same period.
• As we move across a period from left to right in the periodic table, the effective nuclear charge increases steadily and therefore, electronegativity increases.

Hence, the elements Li, B, Be and N have the electronegativities 1.0, 2.0, 1.5 and 3.0, respectively on the Pauling scale.

Question B.
The atomic radii of Cl, I and Br are 99, 133 and 114 pm, respectively.
Answer:

• Cl, I and Br belong to group 17 (halogen group) in the periodic table.
• As we move down the group from top to bottom in the periodic table, a new shell gets added in the atom of the elements.
• As a result, the effective nuclear charge decreases due to increase in the atomic size as well as increased shielding effect.
• Therefore, the valence electrons experience less attractive force from the nucleus and are held less tightly resulting in the increased atomic radius.
• Thus, their atomic radii increases in the following order down the group.
  Cl (99 pm) < Br (114 pm) < I (133 pm)

Hence, the atomic radii of Cl, I and Br are 99, 133 and 114 pm, respectively.

Question C.
The ionic radii of F^– and Na⁺ are 133 and 98 pm, respectively.
Answer:

• F^– and Na⁺ are isoelectronic ions as they both have 10 electrons.
• However, the nuclear charge on F^– is +9 while that of Na⁺ is +11.
• In isoelectronic species, larger nuclear charge exerts greater attraction on the electrons and thus, the radius of that isoelectronic species becomes smaller.

Thus, F^– has larger ionic radii (133 pm) than Na⁺ (98 pm).

Question D.
₁₃Al is a metal, ₁₄Si is a metalloid and ₁₅P is a nonmetal.
Answer:

• Electronic configuration of Al is [Ne] 3s² 3p¹, ₁₄Si is [Ne] 3s² 3p² and that of ₁₅P is [Ne] 3s² 3p³.
• Metals are characterized by the ability to form compounds by loss of valence electrons.
• ‘Al’ has 3 valence electrons, thus shows tendency to lose 3 valence electrons to complete its octet. Hence, Al is a metal.
• Nonmetals are characterized by the ability to form compounds by gain of valence electrons in valence shell.
• ‘P’ has 5 valence electrons thus, shows tendency to gain 3 electrons to complete its octet. Hence, ‘P’ is a nonmetal.
• Si has four valence electrons, thus it can either lose/gain electrons to complete its octet. Hence, behaves as a metalloid.

Question E.
Cu forms coloured salts while Zn forms colourless salts.
Answer:

• Electronic configuration of ₂₉CU is [Ar] 3d¹⁰4s¹ while that of Zn is [Ar] 3d¹⁰4s².
• Electronic configuration of Cu in its +1 oxidation state is [Ar] 3d¹⁰ while that in +2 oxidation state is [Ar] 3d⁹.
• Therefore, Cu contains partially filled d orbitals in +2 oxidation state and thus, Cu²⁺ salts are coloured.
• However, Zn has completely filled d orbital which is highly stable and hence, it does not form coloured ions.

Hence, Cu forms coloured salts while Zn forms colourless salts.

2. Write the outer electronic configuration of the following using orbital notation method. Justify.
A. Ge (belongs to period 4 and group 14)
B. Po (belongs to period 6 and group 16)
C. Cu (belongs to period 4 and group 11)
Answer:
A. a. Ge belongs to period 4. Therefore, n = 4.
b. Group 14 indicates that the element belongs to the p-block of the modem periodic table.
c. The general outer electronic configuration of group 14 elements is ns² np².
d. Thus, the outer electronic configuration of Ge is 4s² 4p².

B. a. Po belongs to period 6. Therefore, n = 6.
b. Group 16 indicates that the element belongs to the p-block of the modem periodic table.
c. The general outer electronic configuration of group 16 elements is ns² np⁴.
d. Thus, the outer electronic configuration of Po is 6s² 6p⁴.

C. a. Cu belongs to period 4. Therefore, n = 4.
b. Group 11 indicates that the element belongs to the d-block of the modem periodic table.
c. The general outer electronic configuration of the d-block elements is ns^0-2(n-1)d^1-10.
d. The expected configuration of Cu is 4s²3d⁹. However, the observed configuration of Cu is 4s¹3d¹⁰. This is due to the extra stability associated with completely filled d-subshell. Thus, the outer electronic configuration of Cu is 4s¹3d¹⁰.

3. Answer the following

Question A.
La belongs to group 3 while Hg belongs to group 12 and both belong to period 6 of the periodic table. Write down the general outer electronic configuration of the ten elements from La to Hg together using orbital notation method.
Answer:
i. La and Hg both belongs to period 6. Therefore, n = 6.
ii. Elements of group 3 to group 12 belong to the d-block of the modem periodic table.
iii. The general outer electronic configuration of the d-block elements is ns^0-2 (n -1 )^1-10.
iv. Therefore, the outer electronic configuration of all ten elements from La to Hg is as given in the table below.

[Note: There are 14 elements between La and Hf which are called lanthanides. Therefore, after La, electrons are filled in 4f subshell of lanthanide elements. Once all the 14 elements of lanthanide series are filled, next electron enters 5d subshell of Hf. Hence, the outer electronic configurations of Hf to Hg often include completely filled 4f subshell. For example, the electronic configuration of Hf ‘5d²6s²’ can also be written as ‘4f¹⁴5d²6s²’.]

Question B.
Ionization enthalpy of Li is 520 kJ mol^-1 while that of F is 1681 kJ mol^-1. Explain.
Answer:

• Both Li and F belong to period 2.
• Across a period, the screening effect is the same while the effective nuclear charge increases.
• As a result, the outer electron is held more tightly and therefore, the ionization enthalpy increases across a period.
• Hence, F will have higher ionization enthalpy than Li.

Thus, ionization enthalpy of Li is 520 kJ mol^-1 while that of F is 1681 kJ mol^-1.

Question C.
Explain the screening effect with a suitable example.
Answer:
i. In a multi-electron atom, the electrons in the inner shells tend to prevent the attractive influence of the nucleus from reaching the outermost electron.
ii. Thus, they act as a screen or shield between the nuclear attraction and outermost or valence electrons. This effect of the inner electrons on the outer electrons is known as screening effect or shielding effect.
iii. Across a period, screening effect due to inner electrons remains the same as electrons are added to the same shell.
iv. Down the group, screening effect due to inner electrons increases as a new valence shell is added.
e.g. Potassium (19K) has electronic configuration 1s²2s²2p⁶3s²3p⁶4s¹.
K has 4 shells and thus, the valence shell electrons are effectively shielded by the electrons present in the inner three shells. As a result of this, valence shell electron (4s¹) in K experiences much less effective nuclear charge and can be easily removed.

Question D.
Why the second ionization enthalpy is greater than the first ionization enthalpy ?
Answer:
The second ionization enthalpy (Δ_iH₂) is greater than the first ionization enthalpy (Δ_iH₁) as it involves removal of electron from the positively charged species.

Question E.
Why the elements belonging to the same group do have similar chemical properties ?
Answer:

• Chemical properties of elements depend upon their valency.
• Elements belonging to the same group have the same valency.

Hence, the elements belonging to the same group show similar chemical properties.

Question F.
Explain : electronegativity and electron gain enthalpy. Which of the two can be measured experimentally?
Answer:
i. The ability of a covalently bonded atom to attract the shared electrons toward itself is called electronegativity (EN). Electronegativity cannot be measured experimentally. However, various numerical scales to express electronegativity were developed by many scientists. Pauling scale of electronegativity is the one used most widely.

ii. Electron gain enthalpy is a quantitative measure of the ease with which an atom adds an electron forming the anion and is expressed in kJ mol^-1. Thus, it is an experimentally measurable quantity.

4. Choose the correct option

Question A.
Consider the elements B, Al, Mg and K predict the correct order of metallic character :
a. B > Al > Mg > K
b. Al > Mg > B > K
c. Mg > Al > K > B
d. K > Mg > Al > B
Answer:
d. K > Mg > Al > B

Question B.
In modern periodic table, the period number indicates the :
a. atomic number
b. atomic mass
c. principal quantum number
d. azimuthal quantum number
Answer:
c. principal quantum number

Question C.
The lanthanides are placed in the periodic table at
a. left hand side
b. right hand side
c. middle
d. bottom
Answer:
d. bottom

Question D.
If the valence shell electronic configuration is ns²np⁵, the element will belong to
a. alkali metals
b. halogens
c. alkaline earth metals
d. actinides
Answer:
b. halogens

Question E.
In which group of elements of the modern periodic table are halogen placed ?
a. 17
b. 6
c. 4
d. 2
Answer:
a. 17

Question F.
Which of the atomic number represent the s-block elements ?
a. 7, 15
b. 3, 12
c. 6, 14
d. 9, 17
Answer:
b. 3, 12

Question G.
Which of the following pairs is NOT isoelectronic ?
a. Na⁺ and Na
b. Mg²⁺ and Ne
c. Al³⁺ and B³⁺
d. P3^– and N^3-
Answer:
b. Mg²⁺ and Ne

Question H.
Which of the following pair of elements has similar properties ?
a. 13, 31
b. 11, 20
c. 12, 10
d. 21, 33
Answer:
a. 13, 31

5. Answer the following questions

Question A.
The electronic configuration of some elements are given below:
a. 1s²
b. 1s²2s²2p⁶
In which group and period of the periodic table they are placed ?
Answer:
a. 1s²
Here n = 1. Therefore, the element belongs to the 1st period.
The outer electronic configuration 1s² corresponds to the maximum capacity of 1s, the complete duplet. Therefore, the element is placed at the end of the 1st period in the group 18 of inert gases in the modem periodic table,

b. 1s²2s²2p⁶
Here n = 2. Therefore, the element belongs to the 2nd period.
The outer electronic configuration 2s²2p⁶ corresponds to complete octet. Therefore, the element is placed in the 2nd period of group 18 in the modem periodic table.

Question B.
For each of the following pairs, indicate which of the two species is of large size :
a. Fe²⁺ or Fe³⁺
b. Mg²⁺ or Ca²⁺
Answer:
a. Fe²⁺ has a larger size than Fe³⁺.
b. Ca²⁺ has a larger size than Mg²⁺.

Question C.
Select the smaller ion form each of the following pairs:
a. K⁺, Li⁺
b. N^3-, F^–
Answer:
i. Li⁺ has smaller ionic radius than K⁺
ii. F^– has smaller ionic radius than N^3-.

Question D.
With the help of diagram answer the questions given below:

a. Which atom should have smaller ionization enthalpy, oxygen or sulfur?
b. The lithium forms +1 ions while berylium forms +2 ions ?
Answer:
Sulfur should have smaller ionization energy than oxygen.
a. Lithium has electronic configuration 1s²2s¹ while that of beryllium is 1s²2s².
b. Li can achieve a noble gas configuration by losing one electron while Be can do so by losing two electrons. Hence, lithium forms +1 ions while beryllium forms +2 ions.

Question E.
Define : a. Ionic radius
b. Electronegativity
Answer:
a. Ionic radius: Ionic radius is defined as the distance of valence shell of electrons from the centre of the nucleus in an ion.

b. Electronegativity: The ability of a covalently bonded atom to attract the shared electrons toward itself is called electronegativity (EN).

Question F.
Compare chemical properties of metals and non-metals.
Answer:
i. Metals (like alkali metals) react vigorously with oxygen to form oxides which reacts with water to form strong bases.
e. g. Sodium (Na) reacts with oxygen to form Na₂O which produces NaOH on reaction with water.

ii. Nonmetals (like halogens) react with oxygen to form oxides which on reaction with water form strong acids.
e.g. Chlorine reacts with oxygen to form Cl₂O₇ which produces HClO₄ on reaction with water.

Question G.
What are the valence electrons ? For s-block and p-block elements show that number of valence electrons is equal to its group number.
Answer:

• Electrons present in the outermost shell of the atom of an element are called valence electrons.
• 3Li is an s-block element and its electronic configuration is 1s²2s¹. Since it has one valence electron, it is placed in group 1.
• Therefore, for s-block elements, group number = number of valence electrons.
• However, for p-block elements, group number = 18 – number of electrons required to attain complete octet.
• 7N is a p-block element and its electronic configuration is 1s²2s²2p³. Since it has five electrons in its valence shell, it is short of three electrons to complete its octet.
• Therefore, its group number = 18 – 3 = 15.

Question H.
Define ionization enthalpy. Name the factors on which ionisation enthalpy depends? How does it vary down the group and across a period?
Answer:
i. The energy required to remove an electron from the isolated gaseous atom in its ground state is called ionization enthalpy (Δ_iH).
Ionization enthalpy is the quantitative measure of tendency of an element to lose electron and expressed in kJ mol^-1.

ii. Ionization energy depends on the following factors

• Size (radius) of an atom
• Nuclear charge
• The shielding or screening effect of inner electrons
• Nature of electronic configuration

iii. Variation of ionization energy down the group: On moving down the group, the ionization enthalpy decreases. This is because electron is to be removed from the larger valence shell. Screening due to core electrons goes on increasing and the effective nuclear charge decreases down the group. As a result, the removal of the outer electron becomes easier down the group.

iv. Variation of ionization energy across a period: The screening effect is the same while the effective nuclear charge increases across a period. As a result, the outer electron is held more tightly and hence, the ionization enthalpy increases across a period. Therefore, the alkali metal shows the lowest first ionization enthalpy while the inert gas shows the highest first ionization enthalpy across a period.

Note: First ionization enthalpy values of elements of group 1.

Note: First ionization enthalpy values of elements of period 2.

Question I.
How the atomic size vary in a group and across a period? Explain with suitable example.
Answer:
i. Variation in atomic size down the group:
a. As we move down the group from top to bottom in the periodic table, the atomic size increases with the increase in atomic number.
b. This is because, as the atomic number increases, nuclear charge increases but simultaneously the number of shells in the atoms also increases.
c. Asa result, the effective nuclear charge decreases due to increase in the size of the atom and shielding effect increases down the group. Thus, the valence electrons experience less attractive force from nucleus and are held less tightly.
d. Hence, the atomic size increases in a group from top to bottom.

e. g.

• In group 1, as we move from top to bottom i.e., from Li to Cs, a new shell gets added in the atom of the elements and the electrons are added in this new shell.
• As a result of this, the effective nuclear charge goes on decreasing and screening effect goes on increasing down a group.
• Therefore, the atomic size is the largest for Cs and is the smallest for Li in group 1.

[Note: Atomic radii of Li and Cs are 152 pm and 262 pm respectively.]

ii. Variation in atomic size across a period:
a. As we move across a period from left to right in the periodic table, the atomic size of an element decreases with the increase in atomic number.
b. This is because, as the atomic number increases, nuclear charge increases gradually but addition of electrons takes place in the same shell.
c. Therefore, as we move across a period, the effective nuclear charge increases but screening effect caused by the core electrons remains the same.
d. As a result of this, attraction between the nucleus and the valence electrons increases. Therefore, valence electrons are more tightly bound and hence, the atomic radius goes on decreasing along a period resulting in decrease in atomic size.

e. g.

• In the second period, as we move from left towards right i.e., from Li to F, the electrons are added in the second shell of all the elements in second period (except noble gas Ne).
• As a result of this, the effective nuclear charge goes on increasing from Li to F, however, screening effect remains the same.
• Therefore, the atomic size is the largest for Li (alkali metal) and is the smallest for F (halogen).

[Note: Atomic radii of Li and F are 152 pm and 64 pm respectively.]

Question J.
Give reasons.
a. Alkali metals have low ionization energies.
b. Inert gases have exceptionally high ionization energies.
c. Fluorine has less electron affinity than chlorine.
d. Noble gases possess relatively large atomic size.
Answer:
a. i. Across a period, the screening effect is the same while the effective nuclear charge increases.
ii. As a result, the outer electron is held more tightly and hence, the ionization enthalpy increases across a period.
iii. Since the alkali metals are present in the group 1 of the modem periodic table, they have low ionization energies.

b. i. Across a period, the screening effect is the same and the effective nuclear charge increases.
ii. As a result, the outer electron is held more tightly and hence, the ionization enthalpy increases across a period.
iii. Inert gases are present on the extreme right of the periodic table i.e., in group 18. Also, inert gases have stable electronic configurations i.e., complete octet or duplet. Due to this, they are extremely stable and it is very difficult to remove electrons from their valence shell.
Hence, inert gases have exceptionally high ionization potential.

c. The less electron affinity of fluorine is due to its smaller size. Adding an electron to the 2p orbital in fluorine leads to a greater repulsion than adding an electron to the larger 3p orbital of chlorine.
Hence, fluorine has less electron affinity than chlorine.

d. i. Noble gases have completely filled valence shell i.e., complete octet (except He with complete duplet).
ii. Since their valence shell contains eight electrons, they experience greater electronic repulsion and this results in increased atomic size (atomic radii) of the noble gas elements.
Hence, noble gases possess

Question K.
Consider the oxides Li₂O, CO₂, B₂O₃.
a. Which oxide would you expect to be the most basic?
b. Which oxide would be the most acidic?
c. Give the formula of an amphoteric oxide.
Answer:
a. Li₂O is the most basic oxide.
b. CO₂ is the most acidic oxide.
c. Formula of an amphoteric oxide: Al₂O₃.
[Note: Both B₂O₃ and CO₂ are acidic oxides. But CO₂ is more acidic oxide as compared to B₂O₃. Hence, CO₂ is most acidic oxide amongst the given.]

Activity :

Question 1.
Prepare a wall mounting chart of the modern periodic table.
Answer:
Students can scan the adjacent Q.R. Code to visualise the modern periodic table and are expected to prepare the chart on their own.

11th Chemistry Digest Chapter 7 Modern Periodic Table Intext Questions and Answers

Can you recall? (Textbook Page No. 93)

Question 1.
What was the basis of classification of elements before the knowledge of electronic structure of atom?
Answer:
Elements were classified on the basis of their physical properties before the knowledge of electronic structure of atom.

Question 2.
Name the scientists who made the classification of elements in the nineteenth century.
Answer:
Dmitri Mendeleev, John Newlands and Johann Doberiener were the scientists who made the classification of elements based on their atomic mass in the nineteenth century.

Question 3.
What is Mendeleev’s periodic law?
Answer:
Mendeleev’s periodic law: “The physical and chemical properties of elements are the periodic function of their atomic masses

Question 4.
How many elements are discovered until now?
Answer:
Including manmade elements, total 118 elements are discovered until now.

Question 5.
How many horizontal rows and vertical columns are present in the modern periodic table?
Answer:
The modem periodic table consists of seven horizontal rows called periods numbered from 1 to 7 and eighteen vertical columns called groups numbered from 1 to 18.

Just think. (Textbook Page No. 93)

Question 1.
How many days pass between two successive full moon nights?
Answer:
29.5 days i.e., approximately 30 days pass between two successive full moon nights.

Question 2.
What type of motion does a pendulum exhibit?
Answer:
A pendulum exhibits periodic motion since it traces the same path after regular interval of time.

Question 3.
Give some other examples of periodic events.
Answer:
Following are some other examples of periodic events:

• Motion of earth around the sun.
• Rotation of earth around its own axis.
• Day and night.

Can you recall? (Textbook Page No. 95)

Question i.
What does the principal quantum number ‘n’ and azimuthal quantum number ‘l’ of an electron belonging to an atom represent?
Answer:
The principal quantum number ‘n’ represents the outermost or valence shell of an element (which corresponds to period number) while azimuthal quantum number ‘l’ constitutes a subshell belonging to the shell for the given ‘n’.

Question ii.
Which principle is followed in the distribution of electrons in an atom?
Answer:
The distribution of electrons in an atom is according to the following three principles:

1. Aufbau principle
2. Pauli’s exclusion principle
3. Hund’s rule of maximum multiplicity

[Note: According to aufbau principle, electrons are filled in the subshells in the increasing order of their energies which follows the following order: s < p < d < f.]

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

Maharashtra State Board 11th Physics Solutions Chapter 9 Optics

1. Multiple Choise Questions

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

Question 2.
Consider the optically denser lenses P, Q, R and S drawn below. According to Cartesian sign convention which of these have positive focal length?

(A) OnlyP
(B) Only P and Q
(C) Only P and R
(D) Only Q and S
Answer:
(B) Only P and Q

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

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

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

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

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

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

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

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

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

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

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

2. Answer the following questions.

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

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

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

Question 3.
What is focal power of a spherical mirror or a lens? What may be the reason for using P = \(\frac {1}{f}\) its expression?
Answer:

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

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

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

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

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

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

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

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

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

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

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

Question 6.
Explain spherical aberration for spherical mirrors. How can it be minimized? Can it be eliminated by some curved mirrors?
Answer:

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

Remedies for Spherical Aberration:

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

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

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

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

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

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

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

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

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

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

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

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

v. Thus, the ray of light coming from the top of an object get refracted while travelling downwards into less optically dense air and become more and more horizontal as shown in Figure.

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

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

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

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

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

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

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

vi. For angles of incidence greater than ic, the angle of refraction become larger than 90° and the ray does not enter into rarer medium at all but is reflected totally into the denser medium as shown in figure.

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

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

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

Question 10.
Describe construction and working of an optical fibre. What are the advantages of optical fibre communication over electronic communication?
Answer:

Construction:

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

Working:

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

Advantages of optical fibre communication over electronic communication:

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

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

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

Working:

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

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

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

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

iv. The ray OP along the principal axis travels undeviated along PX. Another ray OA strikes the surface at A.

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

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

vii. As, the rays are paraxial, all the angles can be considered to be very small.
i.e., sin i ≈ i and sin r ≈ r
Angles α, β and γ can also be expressed as,

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

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

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

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

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

iv. In the absence of second refracting surface, the paraxial ray OA deviates towards normal and would intersect axis at I₁. PI₁ = V₁ is the image distance for intermediate image I₁.

Before reaching I₁, the incident rays (AB and OP) strike the second refracting surface. In this case, image I₁ acts as a virtual object for second surface.

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

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

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

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

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

3. Answer the following questions in detail.

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

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

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

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

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

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

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

iii. For white light, we consider two extreme colours viz., red and violet.
∴ δ_VR = δ_V – δ_R

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

v. For white light,
δ_VR = δ_V – δ_R
= A(n_V – 1) – A(n_R – 1) = A(n_V – n_R).

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

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

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

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

ii. On entering into water, it deviates and disperses into constituent colours. The figure shows the extreme colours (violet and red).

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

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

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

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

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

ii. On entering into water, it deviates and disperses into constituent colours. The figure shows the extreme colours (violet and red).

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

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

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

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

Reducing/eliminating chromatic aberration:

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

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

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

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

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

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

vi. For the combination to be achromatic, the resultant focal length of the combination must be the same for both the colours,

This is the condition for achromatism of a combination of lenses.

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

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

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

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

iv. This defect arises due to spherical shape of the refracting surface, hence known as spherical aberration. It results into a blurred image with unclear boundaries.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

v. For maximum magnifying power, the image should be at D. For thin lens, considering thin lens formula.

Question 9.
Derive the expressions for the magnifying power and the length of a compound microscope using two convex lenses.
Answer:
i. The final image formed in compound microscope (A” B”) as shown in figure, makes a visual angle β at the eye.
Visual angle made by the object from distance D is α.

From figure,
tan β = \(\frac {A”B”}{v_c}\) = \(\frac {A’B’}{u_c}\)
and tan α = \(\frac {AB}{D}\)

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

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

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

ii. Visual angle of the final image is p and its position can be adjusted to be at D. However, under normal adjustments, the final image is also at infinity making a greater visual angle than that of the object.

iii. The parallax at the cross wires can be avoided by using the telescopes in normal adjustments.

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

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

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

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

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

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

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

4. Solve the following numerical examples

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

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

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

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

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

Question 3.
A rectangular sheet of length 30 cm and breadth 3 cm is kept on the principal axis of a concave mirror of focal length 30 cm. Draw the image formed by the mirror on the same diagram, as far as possible on scale.

Answer:

[Note: The question has been modified and the ray digram is inserted in question in order to find the correct position of the image.]

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

∴ h₂ = 0.183 m
i.e., h₂ 0.2 m

Question 5.
A glass slab of thickness 2.5 cm having refractive index 5/3 is kept on an ink spot. A transparent beaker of very thin bottom, containing water of refractive index 4/3 up to 8 cm, is kept on the glass block. Calculate apparent depth of the ink spot when seen from the outside air.
Answer:
When observed from the outside air, the light coming from ink spot gets refracted twice; once through glass and once through water.
∴ When observed from water,

∴ Apparent depth = 2 cm
Now when observed from outside air, the total real depth of ink spot can be taken as (8 + 2) cm = 10 cm.
∴ \(\frac {n_w}{n_{air}}\) = \(\frac {Real depth}{Apparent depth}\)
∴ Apparent depth = \(\frac {10}{4/3}\)
= \(\frac {10×3}{4}\) = 7.5 cm

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

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

Answer:

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

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

Question 9.
A monochromatic ray of light is incident at 37° on an equilateral prism of refractive index 3/2. Determine angle of emergence and angle of deviation. If angle of prism is adjustable, what should its value be for emergent ray to be just possible for the same angle of incidence.
Answer:
By Snell’s law, in case of prism,

For equilateral prism, A = 60°
Also, A= r₁ + r₂
∴ r₂ = A – r₁ = 60° – 23°39′ = 36°21′
Applying snell’s law on the second surface of

= sin^-1 (0.889)
= 62°44′
≈ 63°
For any prism,
i + e = A + δ
∴ δ = (i + e) – A
= (37 + 63) – 60
= 40°
For an emergent ray to just emerge, the angle r’₂ acts as a critical angle.
∴ r’₂ = sin^-1 (\(\frac {1}{n}\))
= sin^-1 (\(\frac {2}{3}\))
= 41°48′
As, A = r’₁ + r’₂ and i to be kept the same.
⇒ A’ = r’₁ + r’₂‘
= 23°39′ + 41°48′
= 65°27’

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

Question 11.
Refractive index of a flint glass varies from 1.60 to 1.66 for visible range. Radii of curvature of a thin convex lens are 10 cm and 15 cm. Calculate the chromatic aberration between extreme colours.
Answer:
Given the refractive indices for extreme colours. As, n_R < n_V
n_R = 1.60 and n_V = 1.66
For convex lens,
R₁ = 10 cm and R₁₀ = -15 cm

= 0.11
∴ f_V = 11 cm
∴ Longitudinal chromatic aberration
= f_V – f_R = 11 – 10 = 1 cm

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

Question 13.
Focal power of the eye lens of a compound microscope is 6 dioptre. The microscope is to be used for maximum magnifying power (angular magnification) of at least 12.5. The packing instructions demand that length of the microscope should be 25 cm. Determine minimum focal power of the objective. How much will its radius of curvature be if it is a biconvex lens of n = 1.5.
Answer:
Focal power of the eye lens,
P_e = \(\frac {1}{f_e}\) = 6D
∴ f_e = \(\frac {1}{6}\) = 0.1667 m = 16.67 cm
Now, as the magnifying power is maximum,
v_e = 25 cm,
Also (M_e)_max = 1 + \(\frac {D}{f_e}\) = 1 + \(\frac {25}{16.67}\) ≈ 2.5
Given that,
M = m₀ × M_e = 12.5
∴ m₀ × 2.5 = 12.5
∴ m₀ = \(\frac {v_0}{u_0}\) = 5 ……….. (1)
From thin lens formula,

Length of a compound microscope,
L = |v₀| +|u₀|
∴ 25 = |v₀| + 10
∴ |v₀|= 15 cm
∴ |u₀| = \(\frac {v_0}{5}\) = 3 cm …………… (from 1)
From lens formula for objective,
\(\frac {1}{f_0}\) = \(\frac {1}{v_0}\) – \(\frac {1}{u_0}\)
= \(\frac {1}{15}\) – \(\frac {1}{-3}\)
= \(\frac {2}{5}\)
∴ f₀ = 2.5 cm = 0.025 m
Thus, focal power of objective,
P = \(\frac {1}{f_0(m)}\)
= \(\frac {1}{0.025}\) = 40 D
Using lens maker’s equation for a biconvex lens,
\(\frac{1}{f_{o}}=(n-1)\left(\frac{1}{R}-\frac{1}{-R}\right)\)
∴ \(\frac{1}{2.5}=(1.5-1)\left(\frac{2}{R}\right)=\frac{1}{R}\)
∴ R = 2.5 cm

11th Physics Digest Chapter 9 Optics Intext Questions and Answers

Can you recall? (Textbook rage no 159)

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

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

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

Can you recall? (Textbook page no. 159)

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

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

Question 3.
How does a rainbow form?
Answer:

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

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

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 6 Redox Reactions Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 6 Redox Reactions

1. Choose the most correct option

Question A.
Oxidction numbers of Cl atoms marked as Cl^a and Cl^b in CaOCl₂ (bleaching powder) are

a. zero in each
b. -1 in Cl^a and +1 in Cl^b
c. +1 in Cl^a and -1 in Cl^b
d. 1 in each
Answer:
b. -1 in Cl^a and +1 in Cl^b

Question B.
Which of the following is not an example of redox reacton ?
a. CuO + H₂ → Cu + H₂O
b. Fe₂O₃ + 3CO₂ → 2Fe + 3CO₂
c. 2K + F₂ → 2KF
d. BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl
Answer:
d. BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl

Question C.
A compound contains atoms of three elements A, B and C. If the oxidation state of A is +2, B is +5 and that of C is -2, the compound is possibly represented by
a. A₂(BC₃)₂
b. A₃(BC₄)₂
c. A₃(B₄C)₂
d. ABC₂
Answer:
b. A₃(BC₄)₂

Question D.
The coefficients p, q, r, s in the reaction
\(\mathrm{pCr}_{2} \mathrm{O}_{7}^{2-}\) + q Fe^2⊕ → r Cr^3⊕ + s Fe^3⊕ + H₂O respectively are :
a. 1, 2, 6, 6
b. 6, 1, 2, 4
c. 1, 6, 2, 6
d. 1, 2, 4, 6
Answer:
c. 1, 6, 2, 6

Question E.
For the following redox reactions, find the correct statement.
Sn^2⊕ + 2Fe^3⊕ → Sn^4⊕ + 2Fe^2⊕
a. Sn^2⊕ is undergoing oxidation
b. Fe^3⊕ is undergoing oxidation
c. It is not a redox reaction
d. Both Sn^2⊕ and Fe^3⊕ are oxidised
Answer:
a. Sn^2⊕ is undergoing oxidation

Question F.
Oxidation number of carbon in H₂CO₃ is
a. +1
b. +2
c. +3
d. +4
Answer:
d. +4

Question G.
Which is the correct stock notation for magenese dioxide ?
a. Mn(I)O₂
b. Mn(II)O₂
c. Mn(III)O₂
d. Mn(IV)O₂
Answer:
d. Mn(IV)O₂

Question I.
Oxidation number of oxygen in superoxide is
a. -2
b. -1
c. –\(\frac {1}{2}\)
d. 0
Answer:
c. –\(\frac {1}{2}\)

Question J.
Which of the following halogens does always show oxidation state -1 ?
a. F
b. Cl
c. Br
d. I
Answer:
a. F

Question K.
The process SO₂ → S₂Cl₂ is
a. Reduction
b. Oxidation
c. Neither oxidation nor reduction
d. Oxidation and reduction.
Answer:
a. Reduction

2. Write the formula for the following compounds :
A. Mercury(II) chloride
B. Thallium(I) sulphate
C. Tin(IV) oxide
D. Chromium(III) oxide
Answer:
i. HgCl₂
ii. Tl₂SO₄
iii. SnO₂
iv. Cr₂O₃

3. Answer the following questions

Question A.
In which chemical reaction does carbon exibit variation of oxidation state from -4 to +4 ? Write balanced chemical reaction.
Answer:
In combustion of methane, carbon exhibits variation from -4 to +4. The reaction is as follows:
CH₄ + 2O₂ → CO₂ + 2H₂O
In CH₄, the oxidation state of carbon is -4 while in CO₂, the oxidation state of carbon is +4.

Question B.
In which reaction does nitrogen exhibit variation of oxidation state from -3 to +5 ?
Answer:

C. Calculate the oxidation number of underlined atoms.
a. H₂SO₄
b. HNO₃
c. H₃PO₃
d. K₂C₂O₄
e. H₂S₄O₆
f. Cr₂O₇^2-
g. NaH₂PO₄
Answer:
i. H₂SO₄
Oxidation number of H = +1
Oxidation number of O = -2
H₂SO₄ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms of H₂SO₄ = 0
∴ 2 × (Oxidation number of H) + (Oxidation number of S) + 4 × (Oxidation number of O) = 0
∴ 2 × (+1) + (Oxidation number of S) + 4 × (-2) = 0
∴ Oxidation number of S + 2 – 8 = 0
∴ Oxidation number of S in H₂SO₄ = +6

ii. HNO₃
Oxidation number of H = +1
Oxidation number of O = -2
HNO₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms of HNO₃ = 0
∴ (Oxidation number of H) + (Oxidation number of N) + 3 × (Oxidation number of O) = 0
∴ (+1) + (Oxidation number of N) + 3 × (-2) = 0
∴ Oxidation number of N + 1 – 6 = 0
∴ Oxidation number of N in HNO₃ = +5

iii. H₃PO₃
Oxidation number of O = -2
Oxidation number of H = +1
H₃PO₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 3 × (Oxidation number of H) + (Oxidation number of P) + 3 × (Oxidation number of O) = 0
∴ 3 × (+1) + (Oxidation number of P) + 3 × (-2) = 0
∴ Oxidation number of P + 3 – 6 = 0
Oxidation number of P is H₃PO₃ = +3

iv. K₂C₂O₄
Oxidation number of K = +1
Oxidation number of O = -2
K₂C₂O₄ is a neutral molecule.
∴ Sum of the oxidation number of all atoms = 0
∴ 2 × (Oxidation number of K) + 2 × (Oxidation number of C) + 4 × (Oxidation number of O) = 0
∴ 2 × (+1) + 2 × (Oxidation number of C) + 4 × (-2) = 0
∴ 2 × (Oxidation number of C) + 2 – 8 = 0
∴ 2 × (Oxidation number of C) = + 6
∴ Oxidation number of C = +\(\frac {6}{2}\)
∴ Oxidation number of C in K₂C₂O₄ = +3

v. H₂S₄O₆
Oxidation number of H = +1
Oxidation number of O = -2
H₂S₄O₆ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 2 × (Oxidation number of H) + 4 × (Oxidation number of S) + 6 × (Oxidation number of O) = 0
∴ 2 × (+1) + 4 × (Oxidation number of S) + 6 × (-2) = 0
∴ 4 × (Oxidation number of S) + 2 – 12 = 0
∴ 4 × (Oxidation number of S) = + 10
∴ Oxidation number of S = +\(\frac {10}{4}\)
∴ Oxidation number of S in H₂S₄O₆ = +2.5

vi. Cr₂O₇^2-
Oxidation of O = -2
Cr₂O₇^2- is an ionic species.
∴ Sum of the oxidation numbers of all atoms = – 2
∴ 2 × (Oxidation number of Cr) + 7 × (Oxidation number of O) = -2
∴ 2 × (Oxidation number of Cr) + 7 × (-2) = – 2
∴ 2 × (Oxidation number of Cr) – 14 = – 2
∴ 2 × (Oxidation number of Cr) = – 2 + 14
∴ Oxidation number of Cr = +\(\frac {12}{2}\)
∴ Oxidation number of Cr in Cr₂O₇^2- = +6

vii. NaH₂PO₄
Oxidation number of Na = +1
Oxidation number of H = +1
Oxidation number of O = -2
NaH₂PO₄ is a neutral molecule
Sum of the oxidation numbers of all atoms = 0
(Oxidation number of Na) + 2 × (Oxidation number of H) + (Oxidation number of P) + 4 × (Oxidation number of O) = 0
(+1) + 2 × (+1) + (Oxidation number of P) + 4 × (-2) = 0
(Oxidation number of P) + 3 – 8 = 0
Oxidation number of P in NaH₂PO₄ = +5

Question D.
Justify that the following reactions are redox reaction; identify the species oxidized/reduced, which acts as an oxidant and which act as a reductant.
a. 2Cu₂O_(s) + Cu₂S_(s) → 6Cu_(s) + SO_2(g)
b. HF_(aq) + OH^–_(aq) → H₂O_(l) + F^–_(aq)
c. I_2(aq) + 2 S₂O₃^2-_(aq) → S₄O₆^2-_(aq) + 2I^–_(aq)
Answer:
i. 2Cu₂O_(s) + Cu₂S_(s) → 6Cu_(s) + SO_2(g)
a. Write oxidation number of all the atoms of reactants and products.

b. Identify the species that undergoes change in oxidation number.

c. The oxidation number of S increases from -2 to +4 and that of Cu decreases from +1 to 0. Because oxidation number of one species increases and that of the other decreases, the reaction is a redox reaction.
d. The oxidation number of S increases by loss of electrons and therefore, S is a reducing agent and it itself is oxidised. On the other hand, the oxidation number of Cu decreases by gain of electrons and therefore, Cu is an oxidising agent and itself is reduced.

Result:

1. The given reaction is a redox reaction.
2. Oxidant/oxidising agents (Reduced species): Cu₂O/ Cu₂S
3. Reductant/reducing agent (Oxidised species): Cu₂S

[Note: Cu in both Cu₂O and Cu₂S undergoes reduction. Hence, both Cu₂O and Cu₂S can be termed as oxidising agents in the given reaction.]

ii. HF_(aq) + OH^–_(aq) → H₂O_(l) + F^–_(aq)
a. Write oxidation number of all the atoms of reactants and products.

b. Since, the oxidation numbers of all the species remain same, this is NOT a redox reaction. Result:
The given reaction is NOT a redox reaction.

iii. I_2(aq) + 2 S₂O₃^2-_(aq) → S₄O₆^2-_(aq) + 2I^–_(aq)
a. Write oxidation number of all the atoms of reactants and products.

b. Identify the species that undergoes change in oxidation number.

c. The oxidation number of S increases from +2 to +2.5 and that of I decreases from 0 to -1. Because oxidation number of one species increases and that of the other decreases, the reaction is a redox reaction.
d. The oxidation number of S increases by loss of electrons and therefore, S is a reducing agent and itself is oxidised. On the other hand, the oxidation number of I decreases by gain of electrons and therefore, I is an oxidising agent and itself is reduced.

Result:

1. The given reaction is a redox reaction.
2. Oxidant/oxidising agent (Reduced species): I₂
3. Reductant/reducing agent (Oxidised species): S₂O₃^2-

Question E.
What is oxidation? Which one of the following pairs of species is in its oxidized state ?
a. Mg / Mg²⁺
b. Cu / Cu²⁺
c. O₂ / O^2-
d. Cl₂ / Cl^–
Answer:
a. Mg / Mg²⁺
Here, Mg loses two electrons to form Mg²⁺ ion.
\(\mathrm{Mg}_{(\mathrm{s})} \longrightarrow \mathrm{Mg}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-}\)
Hence, Mg / Mg²⁺ is an oxidized state.

b. Cu/Cu²⁺
Here, Cu loses two electrons to form Cu²⁺ ion.
\(\mathrm{Cu}_{(\mathrm{s})} \longrightarrow \mathrm{Cu}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-}\)
Hence, Cu/Cu²⁺ is in an oxidized state.

c. O₂ / O^2-
Here, each O gains two electrons to form O^2- ion.
\(\mathrm{O}_{2(\mathrm{~g})}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{O}_{(\mathrm{aq})}^{2-}\)
Hence, O₂ / O^2- is in a reduced state.

d. Cl₂ / Cl^–
Here, each Cl gains one electron to form Cl^– ion.
\(\mathrm{Cl}_{2(\mathrm{~g})}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cl}_{(\mathrm{aq})}^{-}\)
Hence, Cl₂ / Cl^– is in a reduced state.

Question F.
Justify the following reaction as redox reaction.
2 Na_2(s) + S_(s) → Na₂S_(s)
Find out the oxidizing and reducing agents.
Answer:
i. Redox reaction can be described as electron transfer as shown below:
2Na_(s) + S_(s) → 2Na⁺ + S^2-
ii. Charge development suggests that each sodium atom loses one electron to form Na⁺ and sulphur atom gains two electrons to form S^2-. This can be represented as follows:

iii. When Na is oxidised to Na₂S, the neutral Na atom loses electrons to form Na⁺ in Na₂S while the elemental sulphur gains electrons and forms S^2- in Na₂S.
iv. Each of the above steps represents a half reaction which involves electron transfer (loss or gain).
v. Sum of these two half reactions or the overall reaction is a redox reaction.
vi. Oxidising agent is an electron acceptor and hence, S is an oxidising agent. Reducing agent is an electron donor and hence, Na is a reducing agent.

Question G.
Provide the stock notation for the following compounds : HAuCl₄, Tl₂O, FeO, Fe₂O₃, MnO and CuO.
Answer:

Question H.
Assign oxidation number to each atom in the following species.
a. Cr(OH)₄^–
b. Na₂S₂O₃
c. H₃BO₃
Answer:
i. Cr(OH)₄^–
Oxidation number of O = -2
Oxidation number of H = +1
Cr(OH)₄^– is an ionic species.
∴ Sum of the oxidation numbers of all atoms = – 1
∴ Oxidation number of Cr + 4 × (Oxidation number of O) + 4 × (Oxidation number of H) = – 1
∴ Oxidation number of Cr + 4 × (-2) + 4 × (+1) = – 1
∴ Oxidation number of Cr – 8 + 4 = – 1
∴ Oxidation number of Cr – 4 = – 1 –
∴ Oxidation number of Cr = – 1 + 4
∴ Oxidation number of Cr in Cr(OH)₄^– = +3

ii. Na₂S₂O₃
Oxidation number of Na = +1
Oxidation number of O = -2
Na₂S₂O₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 2 × (Oxidation number of Na) + 2 × (Oxidation number of S) + 3 × (Oxidation number of O) = 0
∴ 2 × (+1) + 2 × (Oxidation number of S) + 3 × (-2) = 0
∴ 2 × (Oxidation number of S) + 2 – 6 = 0
∴ 2 × (Oxidation number of S) = + 4
∴ Oxidation number of S = +\(\frac {4}{2}\)
∴ Oxidation number of S in Na₂S₂O₃ = +2

iii. H₃BO₃
Oxidation number of H = +1
Oxidation number of O = -2
H₃BO₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 3 × (Oxidation number of H) + (Oxidation number of B) + 3 × (Oxidation number of O) = 0
∴ 3 × (+1) + (Oxidation number of B) + 3 × (-2) = 0
∴ Oxidation number of B + 3 – 6 = 0
∴ Oxidation number of B in H₃BO₃ = +3

Question I.
Which of the following redox couple is stronger oxidizing agent ?
a. Cl₂ (E⁰ = 1.36 V) and Br₂ (E⁰ = 1.09 V)
b. \(\mathrm{MnO}_{4}^{\Theta}\) (E0 = 1.51 V) and \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2 \Theta}\) (E⁰ = 1.33 V)
Answer:
a. Cl₂ has a larger positive value of E⁰ than Br₂. Thus, Cl₂ is a stronger oxidizing agent than Br₂.
b. \(\mathrm{MnO}_{4}^{\Theta}\) has larger positive value of E⁰ than \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2 \Theta}\). Thus, \(\mathrm{MnO}_{4}^{\Theta}\) is stronger oxidizing agent than \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2 \Theta}\)

Question J.
Which of the following redox couple is stronger reducing agent ?
a. Li (E⁰ = – 3.05 V) and Mg(E⁰ = – 2.36 V)
b. Zn(E⁰ = – 0.76 V) and Fe(E⁰ = – 0.44 V)
Answer:
a. Li has a larger negative value of E⁰ than Mg. Thus, Li is a stronger reducing agent than Mg.
b. Zn has a larger negative value of E⁰ than Fe. Thus, Zn is a stronger reducing agent than Fe.

4. Balance the reactions/equations :

Question A.
Balance the following reactions by oxidation number method

Answer:
i. \(\mathrm{Cr}_{2} \mathrm{O}_{7(\mathrm{aq})}^{2-}+\mathrm{SO}_{3(\mathrm{aq})}^{2-} \longrightarrow \mathrm{Cr}_{(\mathrm{aq})}^{3+}+\mathrm{SO}_{4(\mathrm{aq})}^{2-} \quad(\text { acidic })\)
Step 1: Write skeletal equation and balance the elements other than O and H.
\(\mathrm{Cr}_{2} \mathrm{O}_{7(a q)}^{2-}+\mathrm{SO}_{3(a)}^{2-} \longrightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{3+}+\mathrm{SO}_{4(\mathrm{aq})}^{2-}\)
Step 2: Assign oxidation number to Cr and S. Calculate the increase and decrease in the oxidation number and make them equal.

To make the net increase and decrease equal, we must take 3 atoms of S and 2 atoms of Cr. (There are already 2 Cr atoms.)
Step 3: Balance ‘O’ atoms by adding 4H₂O to the right-hand side.
\(\mathrm{Cr}_{2} \mathrm{O}_{7(\mathrm{aq})}^{2-}+3 \mathrm{SO}_{3(\mathrm{aq})}^{2-} \longrightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{3+}+3 \mathrm{SO}_{4(\mathrm{aq})}^{2-}+4 \mathrm{H}_{2} \mathrm{O}_{(l)}\)
Step 4: The medium is acidic. To make the charges and hydrogen atoms on the two sides equal, add 8H on the left-hand side.
\(\mathrm{Cr}_{2} \mathrm{O}_{7(\mathrm{aq})}^{2-}+3 \mathrm{SO}_{3(\mathrm{aq})}^{2-}+8 \mathrm{H}_{(\mathrm{aq})}^{+} \longrightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{3+}+3 \mathrm{SO}_{4(\mathrm{aq})}^{2-}+4 \mathrm{H}_{2} \mathrm{O}_{(l)}\)
Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation:

ii. \(\mathrm{MnO}_{4(\mathrm{aq})}^{-}+\mathrm{Br}_{(\mathrm{aq})}^{-} \longrightarrow \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{BrO}_{3}^{-}{(a q)} \quad \text { (basic) }\)
Step 1: Write skeletal equation and balance the elements other than O and H.
\(\mathrm{MnO}_{4(\mathrm{aq})}^{-}+\mathrm{Br}_{(\mathrm{aq})}^{-} \longrightarrow \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{BrO}_{3}^{-}{ }_{(\mathrm{aq})}\)
Step 2: Assign oxidation number to Mn and Br. Calculate the increase and decrease in the oxidation number and make them equal.

To make the net increase and decrease equal, we must take 2 atoms of Mn.
\(2 \mathrm{MnO}_{4(\mathrm{aq})}^{-}+\mathrm{Br}_{(\mathrm{aq})}^{-} \longrightarrow 2 \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{BrO}_{3(\mathrm{aq})}^{-}\)
Step 3: Balance ‘O’ atoms by adding H₂O to the right-hand side.
\(2 \mathrm{MnO}_{4(a q)}^{-}+\mathrm{Br}_{(2 q)}^{-} \longrightarrow 2 \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{BrO}_{3 \text { (aq) }}^{-}+\mathrm{H}_{2} \mathrm{O}_{(l)}\)
Step 4: The medium is basic. To make the charges and hydrogen atoms on the two sides equal, add 2H⁺ on the left-hand side.

iii. H₂SO_4(aq) + C_(s) → CO_2(g) + SO_2(g) + H₂O_(l) (acidic)
Step 1: Write skeletal equation and balance the elements other than O and H.
H₂SO_4(aq) + C_(s) → CO_2(g) + SO_2(g) + H₂O_(l)
Step 2: Assign oxidation number to S and C. Calculate the increase and decrease in the oxidation number and make them equal.


To make the net increase and decrease equal, we must take 2 atoms of S.
2H₂SO_4(aq) + C_(s) → CO_2(g) + 2SO_2(g) + H₂O_(l)
Step 3: Balance ‘O’ atoms by adding H2O to the right-hand side.
2H₂SO_4(aq) + C_(s) → CO_2(g) + 2SO_2(g) + H₂O_(l) + H₂O_(l)
Step 4: The medium is acidic. There is no charge on either side. Hydrogen atoms are equal on both side.
2H₂SO_4(aq) + C_(s) → CO₂ + 2SO_2(g) + H₂O_(l)
Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation: 2H₂SO_4(aq) + C_(s) → CO_2(g) + 2SO_2(g) + H₂O_(l)

iv. \(\mathrm{Bi}(\mathrm{OH})_{3(\mathrm{~s})}+\mathrm{Sn}(\mathrm{OH})_{3(\mathrm{aq})}^{-} \longrightarrow \mathrm{Bi}_{(\mathrm{s})}+\mathrm{Sn}(\mathrm{OH})_{6(\mathrm{aq})}^{2-}\) (basic)
Step 1: Write skeletal equation and balance the elements other than O and H.
\(\mathrm{Bi}(\mathrm{OH})_{3(\mathrm{~s})}+\mathrm{Sn}(\mathrm{OH})_{3(\mathrm{aq})}^{-} \longrightarrow \mathrm{Bi}_{(\mathrm{s})}+\mathrm{Sn}(\mathrm{OH})_{6(\mathrm{aq})}^{2-}\)
Step 2: Assign oxidation numbers to Bi and Sn. Calculate the increase and decrease in the oxidation number and make them equal.

To make the net increase and decrease equal, we must take 3 atoms of Sn and 2 atoms of Bi.

Step 4: The medium is basic. To make hydrogen atoms on the two sides equal, add 3W on the right-hand side.

Question B.
Balance the following redox equation by half reaction method

Answer:
i. H₂C₂O_4(aq) + \(\mathrm{MnO}_{4(a q)}^{-}\) → CO_2(g) + \(\mathrm{Mn}_{(\mathrm{aq})}^{2+}\)
Step 1: Write unbalanced equation for the redox reaction. Assign oxidation number to all the atoms in reactants and products. Divide the equation into two half equations.

Step 2: Balance the atoms except O and H in each half equation. Balance half equation for O atoms by adding 4H₂O to the right side of reduction half equation.

Step 3: Balance H atoms by adding H⁺ ions to the side with less H. Hence, add 2H⁺ ions to the right side of oxidation half equation and 8H⁺ ions to the left side of reduction half equation.

Step 4: Now add 2 electrons to the right side of oxidation half equation and 5 electrons to the left side of reduction half equation to balance the charges.

Step 5: Multiply oxidation half equation by 5 and reduction half equation by 2 to equalize number of electrons in two half equations. Then add two half equation.

ii. \(\mathrm{Bi}(\mathrm{OH})_{3(\mathrm{~s})}+\mathrm{SnO}_{2(\mathrm{aq})}^{2-} \longrightarrow \mathrm{SnO}_{3(\mathrm{aq})}^{2-}+\mathrm{Bi}_{(\mathrm{s})}\)
Step 1: Write unbalanced equation for the redox reaction. Assign oxidation number to all the atoms in reactants and products. Divide the equation into two half equations.

Step 2: Balance half equations for O atoms by adding H₂O to the side with less O atoms. Add 1H₂O to left side of oxidation half equation and 3H₂O to the right side of reduction half equation.

Step 3: Balance H atoms by adding H⁺ ions to the side with less H. Hence, add 2H⁺ ions to the right side of oxidation half equation and 3H⁺ ions to the left side of reduction half equation.

Step 4: Now add 2 electrons to the right side of oxidation half equation and 3 electrons to the left side of reduction half equation to balance the charges.

Step 5: Multiply oxidation half equation by 3 reduction half equation by 2 to equalize number of electrons in two half equations. Then add two half equation.

Reaction occurs in basic medium. However, H⁺ ions cancel out and the reaction is balanced. Hence, no need to add OH^– ions. The equation is balanced in terms of number of atoms and the charges.
Hence, balanced equation:

5. Complete the following table :

Assign oxidation number to the underlined species and write Stock notation of compound

Compound	Oxidation number	Stock notation
AuCl3	……………..	……………..
SnCl2	……………..	……………..
\(\underline{\mathrm{V}}_{2} \mathrm{O}_{7}^{4-}\)	……………..	……………..
\(\underline{\mathrm{Pt}} \mathrm{Cl}_{6}^{2-}\)	……………..	……………..
H3AsO3	……………..	……………..

Answer:

Compound	Oxidation number	Stock notation
AuCl3	+3	Au(III)Cl3
SnCl2	+2	Sn(II)Cl2
\(\underline{\mathrm{V}}_{2} \mathrm{O}_{7}^{4-}\)	+5	V2(V)\(\mathrm{O}_{7}^{4-}\)
\(\underline{\mathrm{Pt}} \mathrm{Cl}_{6}^{2-}\)	+4	Pt(IV)\(\mathrm{Cl}_{6}^{2-}\)
H3AsO3	+3	H3As(III)O3

11th Chemistry Digest Chapter 6 Redox Reactions Intext Questions and Answers

Can you tell? (Textbook Page No. 81)

Question i.
Why does cut apple turn brown when exposed to air?
Answer:
Cut apple turns brown when exposed to air because polyphenols are released. These polyphenols undergo oxidation in the presence of air and impart brown colour.

Question ii.
Why does old car bumper change colour?
Answer:
Car bumper is made of iron which undergoes rusting over a period of time. Hence, old car bumper changes colour.

Question iii.
Why do new batteries become useless after some days?
Answer:
Batteries generate electricity by redox reactions. Once the chemicals taking part in redox reaction are used up, the battery cannot generate power. Hence, new batteries become useless after some days.

Can you recall? (Textbook Page No. 81)

Question i.
What is combustion reaction?
Answer:
Combustion is a process in which a substance combines with oxygen.

Question ii.
Write an equation for combustion of methane.
Answer:
Combustion of methane: CH₄ + 2O₂ → CO₂ + 2H₂O + Heat + Light

Question iii.
What is the driving force behind reactions of elements?
Answer:
The ability of element to combine with other element or the ability of element to replace other element in compound is the driving force behind the reactions. This may involve formation of precipitates, formation of water, release of gas, etc.

Try this. (Textbook Page No. 82)

Question 1.
Complete the following table of displacement reactions. Identify oxidising and reducing agents involved.

Reactants	Products
Zn(s) + ————(aq)	————-(aq) + Cu(s)
Cu(s) + 2Ag+(aq)	—————– + ————–
———– + ————-	\( \mathrm{Co}_{(\mathrm{aq})}^{2+}\) + Ni(s)

Answer:

Try this (Textbook Page No. 88)

Question 1.
Classify the following unbalanced half equations as oxidation and reduction.

Answer:

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 12 Photosynthesis Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Biology Solutions Chapter 12 Photosynthesis

1. Choose correct option

Question (A)
A cell that lacks chloroplast does not
(a) evolve carbon dioxide
(b) liberate oxygen
(c) require water
(d) utilize carbohydrates
Answer:
(b) liberate oxygen

Question (B)
Energy is transferred from the light reaction step to the dark reaction step by
(a) chlorophyll
(b) ADP
(c) ATP
(d) RuBP
Answer:
(c) ATP

Question (C)
Which one is wrong in photorespiration?
(a) It occurs in chloroplasts
(b) It occurs in day time only
(c) It is characteristic of C₄-plants
(d) It is characteristic of C₃-plants
Answer:
(c) It is characteristic of C₄-plants

Question (D)
Non-cyclic phosphorylation differs from cyclic photophosphorylation in that former
(a) involves only PS
(b) Include evolution of 02
(c) involves formation of assimilatory power
(d) both (b) and (c)
Answer:
(d) both (b) and (c)

Question (E)
For fixation of 6 molecules of C0₂ and formation of one molecule of glucose in Calvin cycle, requires
(a) 3 ATP and 2 MADPPE
(b) 18 ATP and 12 NADPH₂
(c) 30 ATP and 18 NADPH₂
(d) 6 ATP and 6 NADPIT₂
Answer:
(b) 18 ATP and 12 NADPH₂

Question (F)
In maize and wheat, the first stable products formed in bundle sheath cells respectively are
(a) OAA and PEPA
(b) OAA and OAA
(c) OAA and 3PGA
(d) 3PGA and OAA
Answer:
(c) OAA and 3PGA

Question (G)
The head and tail of chlorophyll are made up of
(a) porphyrin and phytin respectively
(b) pyrrole and tetrapyrrole respectively
(c) porphyrin and phytol respectively
(d) tetrapyrole and pyrrole respectively
Answer:
(c) porphyrin and phytol respectively

Question (H)
The net result of photo-oxidation of water is release of ……………. .
(a) electron and proton
(b) proton and oxygen
(c) proton, electron and oxygen
(d) electron and oxygen
Answer:
(c) proton, electron and oxygen

Question (I)
For fixing one molecule of C0₂ in Calvin cycle are required.
(a) 3ATP + 1NADPFE
(b) 3ATP + 2NADPH₂
(c) 2ATP + 3NADPH₂
(d) 3ATP + 3NADPFE
Answer:
(b) 3ATP + 2NADPH₂

Question (J)
In presence of high concentration of oxygen, RuBP carboxylase converts RuBP to …………… .
(a) Malic acid and PEP
(b) PGA and PEP
(c) PGA and malic acid
(d) PGA and phosphoglycolate
Answer:
(d) PGA and phosphoglycolate

Question (K)
The sequential order in electron transport from PSII to PSI of photosynthesis is
(a) FeS, PQ, PC and Cytochrome
(b) FeS, PQ, Cytochrome and PC
(c) PQ, Cytochrome, PC and FeS
(d) PC, Cytochrome, FeS, PQ
Answer:
(c) PQ, Cytochrome, PC and FeS

2. Answer the following questions

Question (A)
Describe the light-dependent steps of photosynthesis. How are they linked to dark reactions?
Answer:
The light dependent steps of photosynthesis include cyclic and non-cyclic photophosphorylation,
1. Cyclic photophosphorylation:
a. Illumination of photosystem-I causes electrons to move continuously out of the reaction center of photosystem-I and back to it.
b. The cyclic electron-flow is accompanied by the photophosphorylation of ADP to yield ATP. This is termed as Cyclic photophosphorylation.
c. Since this process involves only pigment system I, photolysis of water and consequent evolution of oxygen does not take place.

2. Non-cyclic photophosphorylation:
a. It involves both photosystems- PS-I and PS-II.
b. In this case, electron transport chain starts with the release of electrons from PS-II.
c. In this chain high energy electrons released from PS-II do not return to PS-II but, after passing through an electron transport chain, reach PS-I, which in turn donates it to reduce NADP to NADPH.
d. The reduced NADP⁺ (NADPH) is utilized for the reduction of CO₂ in the dark reaction.
e. Electron-deficient PS-II brings about oxidation of water-molecule. Due to this, protons, electrons and oxygen atom are released.
f. Electrons are taken up by PS-II itself to return to reduced state, protons are accepted by NADP⁺ whereas oxygen is released.
g. As in this process, high energy electrons released from PS-II do not return to PS-II and it is accompanied with ATP formation, this is called Non-cyclic photophosphorylation.

3. Link between light-dependent and dark reactions:

1. The light reaction gives rise to two important products, a reducing agent NADPH₂ and an energy rich compound ATP. Both these are utilized in the dark phase of photosynthesis.
2. ATP and NADPH₂ molecules function as vehicles for transfer of energy of sunlight into dark reaction leaving to carbon fixation. In this reaction C0₂ is reduced to carbohydrate.
3. During dark reaction, ATP and NADPH₂ are transformed into ADP, iP and NADP which are transferred to the grana in which light reaction takes place.

(B)

Question (a)
Distinguish between Respiration and Photorespiration
Answer:

Respiration	Photorespiration
1. Occurs in all aerobic and anaerobic organisms.	Occurs in C3 plants under high temperature, bright light, high oxygen and low C02 concentration.
2. A light independent process, occurs in both presence and absence of light.	A light dependent process, occurs in presence of Jight.
3. Produce energy rich molecules like ATP, GTP, FADH2, NADH2	Do not produce energy rich molecules such as ATP.
4. Respiration is an energy-producing process.	Photorespiration is an energy wastage process.

Question (b)
Distinguish between Cyclic photophosphorylation and Non-cyclic photophosphorylation
Answer:

Cyclic photophosphorylation	Non – cyclic photophosphorvlution
1. Electrons emitted by chlorophyll return back to the same chlorophyll.	The electrons emitted by chlorophyll do not return back to the same chlorophyll.
2. First electron acceptor is FRS.	First electron acceptor is CO – Q
3. It forms only ATP.	NADPH2 and ATP are formed.
4. Does not involve photolysis of H20.	Involves photolysis of H20.
5. No evolution of 02.	There is evolution of 02.
6. Only Photosystem-I (P700) is involved in this cycle.	Both Photosystem PS-I (P700) as well as PS-II (P680) are involved.

Question (C)
Answer the following questions.
1. What are the steps that are common to C₃ and C₄ photosynthesis?
2. Differentiate between C₃ and C₄ plants.
Answer:
Steps that are common to C₃ and C₄ photosynthesis are Carboxylation, Reduction, Glucose synthesis, Regeneration.
[Note: Students are expected to refer the given Q.R code for detail understanding the common steps between C₁ and C₄ plants.]

Question (D)
Are the enzymes that catalyze the dark reactions of carbon fixation located inside the thylakoids or outside the thylakoids?
Answer:
Carbon fixation occurs in the stroma by series of enzyme catalyzed steps. The enzymes that catalyze the dark reactions of carbon fixation are located outside the thylakoids.

Question (E)
Calvin cycle consists of three phases, what are they? Explain the significance of each of them.
Answer:
The entire process of dark reaction was traced by Dr. Melvin Calvin along with his co-worker, Dr. Benson. Hence, the process is called as Calvin cycle or Calvin- Benson cycle. Since the first stable product formed is a 3-carbon compound, it is also called as C₃ pathway and the plants are called C₁₄ plants.
Calvin carried out experiments on unicellular green algae (Chlorella), using radioactive isotope of carbon, C₁₄ as a tracer.
It is also called synthesis phase or second phase of photosynthesis.

The cycle is divided into the following phases:
1. Carboxylation phase:
a. Carbon dioxide reduction starts with a five-carbon sugar ribulose-l,5-bisphosphate (RuBP). It is a 5- carbon sugar with two phosphate groups attached to it.
b. RuBP reacts with CO₂ to produce an unstable 6 carbon intermediate in the presence of Rubisco.
c. It immediately splits into 3 carbon compounds called 3-phosphoglyceric acid.
d. RuBisCO is a large protein molecule and comprises 16% of the chloroplast proteins.

2. Glycolytic reversal:
a. 3-phosphoglyceric acid form 1,3-diphosphoglyceric acid by utilizing ATP molecule.
b. These are then reduced to glyceraldehyde-3-phosphate (3-PGA) by NADPH supplied by the light reactions of photosynthesis.
c. In order to keep Calvin cycle continuously running there must be sufficient number of RuBP and regular supply of ATP and NADPH.
d. Out of 12 molecules of 3-phosphoglyceraldehyde, two molecules are used for synthesis of one glucose molecule.

3. Regeneration of RuBP:
a. 10 molecules of 3-phosphoglyceraldehyde are used for the regeneration of 6 molecules of RuBP at the cost of 6 ATP.
b. Therefore, six turns of Calvin cycle are needed to get one molecule of glucose.
Significance:
1. Carboxylation: RuBisCO is the most abundant enzyme in the world. It is responsible for fixing carbon in the form of C0₂ into sugar. As a result of Carboxylation, the first stable product of carbon fixation i.e. 3- PGA is synthesized.
2. Reduction/Glycolytic reversal: NADPH2 donates electrons to 1, 3-Bisphoshoglycerate to form 3- phosphoglyceraldehyde molecules. During this process ADP and NADP are generated which are used in light reaction.
3. Regeneration of RuBP: Some 3-phosphoglyceraldehyde molecules are involved in production of glucose while others are recycled to regenerate the 5-carbon compound RuBP which used to accept new carbon molecules. Thus, regeneration of RuBP is required for Calvin cycle to run continuously.

Question (F)
Why are plants that consume more than the usual 18 ATP to produce 1 molecule of glucose favoured in tropical regions?
Answer:

1. C₄ plants are favoured in tropical regions as they require 30 ATP to produce 1 molecule of glucose.
2. High temperature in tropical regions leads to closure of stomata to reduce rate of transpiration. Due to this availability of C0₂ decreases.
3. PEP carboxylase present in mesophyll cells can fix C0₂ even at low concentration. This helps the plant in efficient assimilation of atmospheric carbon dioxide.
4. C₄ plants contain a special leaf anatomy called Kranz anatomy which minimizes the losses due to photorespiration.
5. It helps C₄ plants to survive in conditions of high daytime temperatures, intense sunlight and low moisture.

Question (G)
What is the advantage of having more than one pigment molecule in a photocenter?
Answer:
Advantages of having more pigment molecules in a photocenter are as follows:
1. Having more than one pigment molecule in photocenter means more sunlight being captured and thus facilitating more effective light reaction.
2. It will provide protection to chlorophyll molecule against photo-oxidation.
3. More pigments will capture more energy to start the initial reactions, which is not possible by single pigment.

Question (H)
Why does chlorophyll appear green in reflected light and red transmitted light? Explain the significance of these phenomena in terms of photosynthesis.
Answer:
1. Chlorophyll is a light absorbing pigment. It absorbs light in red and blue regions of the visible light spectrum. Absorption spectrum of chlorophyll for red light is maximum so chlorophyll appears red in transmitted light. Green light is not absorbed but reflected so chlorophyll appear green in reflected light.
2. Chlorophyll predominantly absorbs red and violet-blue light and it allows plants to use this light as a form of energy for photosynthesis process.
3. It is most effective wavelength of light in photosynthesis as it has exactly right amount of energy to excite electrons of chlorophyll and boost them out of their orbits to higher energy level.

Question (I)
Explain why photosynthesis is considered the most important process in the biosphere.
Answer:
Photosynthesis is considered to be the most important process in the biosphere due to following reasons:
1. Photosynthesis is the biochemical process through which all plants (primary producers) produce food.
2. It is responsible for release of oxygen in the atmosphere.
3. Heterotrophs are directly or indirectly dependent on autotrophs for energy and other related resources. Therefore, photosynthesis is considered the most important process in the biosphere.

Question (J)
Why is photolysis of water accompanied with non-cyclic photophosphorylation?
Answer:
1. Photolysis of water provides new electrons to Photosystem – II.
2. The water molecule is lysed into three components:
a. Protons (H⁺) which are used as part of reactions that makes NADPH.
b. Second component formed is electrons which replaces the electrons lost by PS-II.
c. The third component is oxygen (0₂) which is released into the atmosphere.
3. Photosystem I sends electrons to reduce NADP+.
4. Then, Photosystem II sends replacement electrons to Photosystem I.
5. Finally, photolysis of water replaces the electrons lost by Photosystem II.
6. Water is the ultimate source of electrons for photosynthesis.
Therefore, photolysis of water is accompanied with non – cyclic photophosphorylation.

Question (K)
In C-4 plants, why is C-3 pathway operated in bundle sheath cells only?
Answer:

1. Decarboxylation of malic acid occurs in bundle sheath cells of C₄ plants. Due to which concentration of C0₂ increases in bundle sheath cells.
2. The enzymes required for Calvin cycle i.e. RuBisCO is present in bundle sheath cells.
3. In presence of high concentration of C0₂, RuBisCO acts as carboxylase and bring about carboxylation of RuBP.
4. Hence, in C-4 plants, C-3 pathway is operated in bundle sheath cells only.

Question (L)
What would have happened if C-4 plants did not have Kranz anatomy?
Answer:
Photorespiration would occur if C₄ plants did not have Kranz anatomy.

Question (M)
Why does RuBisCO carry out preferentially carboxylation than oxygenation in C₄ plants?
Answer:

1. In C₄ plants, C0₂ taken from the atmosphere is accepted by a 3-carbon compound, phosphoenolpyruvic acid in the chloroplasts of mesophyll cells.
2. This leads to the formation of 4-carbon compound oxaloacetic acid with the help of enzyme phosphoenolpyruvate carboxylase.
3. It is converted to another 4-carbon compound called malate.
4. Malate is transported to chloroplasts of bundle sheath cells where malate is converted to pyruvate and releases C0₄ in the cytoplasm thus increasing the concentration of C0₂ in the bundle sheath cells.
5. Chloroplasts of bundle sheath cells contains enzymes of Calvin cycle.
6. Thus, due to high concentration of C0₂, RuBisCO participates in carboxylation and not in oxygenation.

Question (N)
What would have happened if plants did not have accessory pigments?
Answer:

1. Accessory pigments are light absorbing molecules which are found in photosynthetic organisms.
2. They transfer the absorbed light to chlorophyll-a and thus increasing the photosynthetic rate.
3. In absence of accessory pigments less amount of light will be absorbed and also there would be no protection provided to chlorophyll molecule from photo-oxidation.

Question (O)
How can you identify whether the plant is C₃ or C₄? Explain / Justify.
Answer:

1. By observing the cross section of a leaf we can identify whether the plant is a C₃ plant or a C₃ plant.
2. C₄ plants possess a special anatomy of leaves called Kranz anatomy. In Kranz anatomy two types of chloroplasts are present, agranal in bundle sheath cells and granal in mesophyll cells.
3. In C₃ plants Kranz anatomy is absent.

Question (P)
In C₄ plants, bundle sheath cells carrying out Calvin cycle are very few in number. Then also, C₄ plants are highly productive. Explain.
Answer:

1. C₄ plants have special type of leaf anatomy called Kranz anatomy.
2. In C₄ plants, C0₂ fixation occurs twice.
3. In these plants, chloroplasts of mesophyll cells contain enzyme PEP carboxylase which fixes atmospheric C0₂.
4. Thus, first C0₂ fixation occurs in mesophyll cells.
5. Decarboxylation of malic acid in bundle sheath cells results in increase in C0₂ concentration.
6. Thus, RuBisCO acts as carboxylase and brings about carboxylation of RuBP.
7. Due to this oxygenation of RuBP and photorespiration is prevented.
8. Thus, despite of having less number of bundle sheath cells carrying out Calvin cycle, C₄ plants are highly productive.

Question (Q)
What is functional significance of Kranz anatomy?
Answer:

1. Leaves of C₄ plants show some structural peculiarities called Kranz anatomy.
2. The chloroplast of mesophyll cells contain enzyme PEP Carboxylase, which can fix C0₂ at low concentration.
3. Thus, light reaction and evolution of 0₂ occurs in mesophyll cells.
4. Decarboxylation of malate occurs in bundle sheath cells, which results in release of C0₂, due to which concentration of C0₂ in bundle sheath cells increases.
5. Enzyme RuBisCO present in bundle sheath cells acts as carboxylase in presence of high C0₂ concentraion and catalyses carboxylation of RuBP.
6. Thus, possibility of oxygenation of RuBP is avoided and photorespiration does not take place.

3. Correct the pathway and name it.

Question 1.
Correct the pathway and name it.
Answer:

1. The pathway shown is C₄ pathway.
2. M. D. Hatch and C. R. Slack while working on sugarcane found four carbon compounds (dicarboxylic acid) as the first stable product of photosynthesis.
3. It occurs in tropical and sub-tropical grasses and some dicotyledons.
4. The first product of this cycle is a 4-carbon compound oxaloacetic acid. Hence it is also called as C₄ pathway and plants are called C₄ plants.

Mechanism:

1. C0₂ taken from atmosphere is accepted by a 3-carbon compound, phosphoenolpyruvic acid in the chloroplasts of mesophyll cells, leading to the formation of 4-C compound, oxaloacetic acid with the help of enzyme phosphoenolpyruvate carboxylase.
2. It is converted to another 4-C compound, malic acid.
3. It is transported to the chloroplasts of bundle sheath cells.
4. Malic acid (4-C) is converted to pyruvic acid (3-C) with the release of C0₂ in the cytoplasm.
5. Thus, concentration of C0₂ increases in the bundle sheath cells.
6. Chloroplasts of these cells contain enzymes of Calvin cycle.
7. Because of high concentration of C0₂, RuBP carboxylase participates in Calvin cycle and not photorespiration.
8. Sugar formed in Calvin cycle is transported into the phloem.
9. Pyruvic acid generated in the bundle sheath cells re-enter mesophyll cells and regenerates
   phosphoenolpyruvic acid by consuming one ATP.
10. Since this conversion results in the formation of AMP (not ADP), two ATP are required to regenerate ATP from AMP.
11. Thus, C₄ pathway needs 12 additional ATP.
12. The C₃ pathway requires 18 ATP for the synthesis of one glucose molecule, whereas C4 pathway requires 30 ATP.
13. Thus, C₄ plants are better photosynthesizers as compared to C₃ plants as there is no photorespiration in these plants.

4. Is there something wrong in following schematic presentation? If yes, correct it so that photosynthesis will be operated.

Question 1.
Is there something wrong in following schematic presentation? If yes, correct it so that photosynthesis will be operated.
Answer:

Non-cyclic photophosphorylation:
a. It involves both photosystems- PS-I and PS-II.
b. In this case, electron transport chain starts with the release of electrons from PS-II.
c. In this chain high energy electrons released from PS-II do not return to PS-II but, after passing through an electron transport chain, reach PS-I, which in turn donates it to reduce NADP to NADPH.
d. The reduced NADP⁺ (NADPH) is utilized for the reduction of CO₂ in the dark reaction.
e. Electron-deficient PS-II brings about oxidation of water-molecule. Due to this, protons, electrons and oxygen atom are released.
f. Electrons are taken up by PS-II itself to return to reduced state, protons are accepted by NADP⁺ whereas oxygen is released.
g. As in this process, high energy electrons released from PS-II do not return to PS-II and it is accompanied with ATP formation, this is called Non-cyclic photophosphorylation.

Practical/ Project:

Question 1.
Draw schematic presentation of different processes/ cycles/ reactions related to photosynthesis.
Answer:
Cyclic photophosphorylation:
a. Illumination of photosystem-I causes electrons to move continuously out of the reaction center of photosystem-I and back to it.
b. The cyclic electron-flow is accompanied by the photophosphorylation of ADP to yield ATP. This is termed as Cyclic photophosphorylation.
c. Since this process involves only pigment system I, photolysis of water and consequent evolution of oxygen does not take place.

Non-cyclic photophosphorylation::
a. It involves both photosystems- PS-I and PS-II.
b. In this case, electron transport chain starts with the release of electrons from PS-II.
c. In this chain high energy electrons released from PS-II do not return to PS-II but, after passing through an electron transport chain, reach PS-I, which in turn donates it to reduce NADP to NADPH.
d. The reduced NADP⁺ (NADPH) is utilized for the reduction of CO₂ in the dark reaction.
e. Electron-deficient PS-II brings about oxidation of water-molecule. Due to this, protons, electrons and oxygen atom are released.
f. Electrons are taken up by PS-II itself to return to reduced state, protons are accepted by NADP⁺ whereas oxygen is released.
g. As in this process, high energy electrons released from PS-II do not return to PS-II and it is accompanied with ATP formation, this is called Non-cyclic photophosphorylation.

Interdependence of light and dark reactions:

1. The light reaction gives rise to two important products, a reducing agent NADPH₂ and an energy rich compound ATP. Both these are utilized in the dark phase of photosynthesis.
2. ATP and NADPH₂ molecules function as vehicles for transfer of energy of sunlight into dark reaction leaving to carbon fixation. In this reaction C0₂ is reduced to carbohydrate.
3. During dark reaction, ATP and NADPH₂ are transformed into ADP, iP and NADP which are transferred to the grana in which light reaction takes place.

Calvin cycle: The entire process of dark reaction was traced by Dr. Melvin Calvin along with his co-worker, Dr. Benson. Hence, the process is called as Calvin cycle or Calvin- Benson cycle. Since the first stable product formed is a 3-carbon compound, it is also called as C₃ pathway and the plants are called C₁₄ plants.
Calvin carried out experiments on unicellular green algae (Chlorella), using radioactive isotope of carbon, C₁₄ as a tracer.
It is also called synthesis phase or second phase of photosynthesis.
The cycle is divided into the following phases:
1. Carboxylation phase:
a. Carbon dioxide reduction starts with a five-carbon sugar ribulose-l,5-bisphosphate (RuBP). It is a 5- carbon sugar with two phosphate groups attached to it.
b. RuBP reacts with CO₂ to produce an unstable 6 carbon intermediate in the presence of Rubisco.
c. It immediately splits into 3 carbon compounds called 3-phosphoglyceric acid.
d. RuBisCO is a large protein molecule and comprises 16% of the chloroplast proteins.

2. Glycolytic reversal:
a. 3-phosphoglyceric acid form 1,3-diphosphoglyceric acid by utilizing ATP molecule.
b. These are then reduced to glyceraldehyde-3-phosphate (3-PGA) by NADPH supplied by the light reactions of photosynthesis.
c. In order to keep Calvin cycle continuously running there must be sufficient number of RuBP and regular supply of ATP and NADPH.
d. Out of 12 molecules of 3-phosphoglyceraldehyde, two molecules are used for synthesis of one glucose molecule.

3. Regeneration of RuBP:
a. 10 molecules of 3-phosphoglyceraldehyde are used for the regeneration of 6 molecules of RuBP at the cost of 6 ATP.
b. Therefore, six turns of Calvin cycle are needed to get one molecule of glucose.

Photorespiration: Mechanism:
1. Photorespiration involves three organelles chloroplast, peroxisomes and mitochondria and occurs in a series of cyclic reactions which is also called PCO cycle. (Photosynthetic Carbon Cycle)
2. Enzyme Rubisco acts as oxygenase at higher concentration of O₂ and photorespiration begins.
3. When RuBP reacts with 0₂ rather than C0₂ to form a 3-carbon compound (PGA) and 2-carbon compound phosphoglycolate.
4. Phosphoglycolate is then converted to glycolate which is shuttled out of the chloroplast into the peroxisomes.
5. In Peroxisomes, glycolate is converted into glyoxylate by enzyme glycolate oxidase.
6. Glyoxylate is further converted into amino acid glycine by transamination.
7. In mitochondria, two molecules of glycine are converted into serine (amino acid) and C0₂ is given out.
8. Thus, it loses 25% of photosynthetically fixed carbon.
9. Serine is transported back to peroxisomes and converted into glycerate.
10. It is shuttled back to chloroplast to undergo phosphorylation and utilized in formation of 3-PGA, which get utilized in C₃ pathway.
Hatch-Slack pathway: M. D. Hatch and C. R. Slack while working on sugarcane found four carbon compounds (dicarboxylic acid) as the first stable product of photosynthesis.
It occurs in tropical and sub-tropical grasses and some dicotyledons.
The first product of this cycle is a 4-carbon compound oxaloacetic acid. Hence it is also called as C₄ pathway and plants are called C₄ plants.
Mechanism:

1. C0₂ taken from atmosphere is accepted by a 3-carbon compound, phosphoenolpyruvic acid in the chloroplasts of mesophyll cells, leading to the formation of 4-C compound, oxaloacetic acid with the help of enzyme phosphoenolpyruvate carboxylase.
2. It is converted to another 4-C compound, malic acid.
3. It is transported to the chloroplasts of bundle sheath cells.
4. Malic acid (4-C) is converted to pyruvic acid (3-C) with the release of C0₂ in the cytoplasm.
5. Thus, concentration of C0₂ increases in the bundle sheath cells.
6. Chloroplasts of these cells contain enzymes of Calvin cycle.
7. Because of high concentration of C0₂, RuBP carboxylase participates in Calvin cycle and not photorespiration.
8. Sugar formed in Calvin cycle is transported into the phloem.
9. Pyruvic acid generated in the bundle sheath cells re-enter mesophyll cells and regenerates
   phosphoenolpyruvic acid by consuming one ATP.
10. Since this conversion results in the formation of AMP (not ADP), two ATP are required to regenerate ATP from AMP.
11. xi. Thus, C₄ pathway needs 12 additional ATP.
12. The C₃ pathway requires 18 ATP for the synthesis of one glucose molecule, whereas C4 pathway requires 30 ATP. Thus, C₄ plants are better photosynthesizers as compared to C₃ plants as there is no photorespiration in these plants.
13. CAM Pathway: In CAM plants, malic acid accumulates during night, which is formed from Oxaloacetic acid in presence of the enzyme malate dehydrogenase.

Question 2.
Check the effects of different factors on photosynthesis under the guidance of teacher.
Answer:
External factors which affect photosynthesis are as follows:
1. Light:
a. It is an essential factor as it supplies the energy necessary for photosynthesis.
b. Quality and intensity of light affects the photosynthesis.
c. Highest rate of photosynthesis takes place in red light followed by blue light.
d. The rate of photosynthesis considerably decreases in plants which are growing under a forest canopy.
e. In most of the plants, photosynthesis is maximum in bright diffused sunlight.
f. Uninterrupted and continuous photosynthesis for a very long period of time may be sustained without any visible damage to the plant.

2. Carbon dioxide:
The main source of C0₂ in land plants is the atmosphere, which contains only 0.3% of the gas.
b. Under normal conditions of temperature and light, carbon dioxide acts as a limiting factor in photosynthesis.
c. Increase in concentration of CO₂ increases the photosynthesis.
d. Increase in C0₂ to about 1% is advantageous to most of the plants.
e. Higher concentration of the gas has an inhibitory effect on photosynthesis.

3. Temperature:
a. Like all other physiological processes, photosynthesis also needs a suitable temperature.
b. The optimum temperature at which the photosynthesis is maximum is 25-30 °C. Except in plants like Opuntia, photosynthesis takes place at as high as 55 °C.
c. This is the maximum temperature. Minimum temperature is temperature at which photosynthesis process just starts.
d. In the presence of sufficient light and CO₂, photosynthesis increases with the rise of temperature till it becomes maximum. After that there is a decrease or fall in the rate of the process.

4. Water:
a. Water is necessary for photosynthetic process.
b. An increase in water content of the leaf results in the corresponding increase in the rate of photosynthesis.
c. Thus, the limiting effect of water is not direct but indirect.
d. It is mainly due to the fact that it helps in maintaining the turgidity of the assimilatory cells and the proper hydration of their protoplasm.
[Students can refer the given information and perform this activity on their own]

11th Biology Digest Chapter 12 Photosynthesis Intext Questions and Answers

Can you recall? (Textbook Page No. 138)

(i) Why energy is essential in different life processes?
Answer:
a. Energy is the basic requirement of life.
b. Without energy no work can be done.
c. All living organisms need energy for reproduction and survival.
d. Sun is the main source of energy, and that energy should be transformed into the usable forms for living organisms to carry out life processes.
Therefore, energy is essential in different life processes.

(ii) How do we get energy?
Answer:
a. Sun is the main source of energy.
b. Plants utilize sunlight, carbon dioxide and water for the process called photosynthesis to produce sugars.
c. Animals make use of these sugars provided by the plants in their own cellular energy factories called mitochondria. Thus, energy is obtained.

Use your brainpower (Textbook Page No. 138)

Justify: All life on earth is ‘bottled solar energy’.
Answer:

1. Life on earth is dependent on solar energy directly or indirectly.
2. Plants by carrying out photosynthesis converts solar energy into chemical energy by producing carbohydrates.
3. Humans and animals depend on plants for food. Basically, life on earth depends totally on photosynthesis for food and energy.
4. Therefore, all life on earth is bottled solar energy.

Can you tell? (Textbook Page No. 140)

Draw well labeled diagram of ultrastructure of chloroplast.
Answer:

1. The chloroplasts are discoid and lens shaped in higher plants. Chloroplast is bounded by a double membrane.
   System of chlorophyll bearing a double-membrane sac is present inside the stroma.
2. These are stacked one above the other to form grana.
3. Individual sacs in each granum is are known as thylakoid.
4. All the pigments chlorophylls, carotenes and xanthophylls are located in thylakoid membranes.
5. These pigments are fat soluble and are present in lipid part of membrane also they absorb light of specific spectrum in the visible regions.

Use your brainpower (Textbook Page No. 140)

The photosynthetic lamellae taken out from a chloroplast and suspended in a nutrient medium in the presence of C0₂ and light. Will they synthesize sugar or not?
Answer:
Photosynthetic ladmellae will not synthesize sugar because sugar synthesis occurs only in stroma, as all the enzymes required for sugar synthesis are present there. In photosynthetic lamellae only light reactions occur. Thus, lamellae cannot synthesize sugar even when C0₂, light and other nutrients are provided.

Internet my friend (Textbook Page No. 139)

Collect information: Why does chlorophyll appear red in reflected light and green in transmitted light?
Answer:
Chlorophyll is a light absorbing pigment. It absorbs light in red and blue regions of the visible light spectrum. Absorption spectrum of chlorophyll for red light is maximum so chlorophyll appears red in transmitted light. Green light is not absorbed but reflected so chlorophyll appear green in reflected light. [Note: Chlorophyll appear red in transmitted light and green in reflected light.[

Activity 1 (Textbook Page No. 139)

Grind the spinach leaves in small quantity of acetone / nail paint remover. Mix the contents properly and filter with filter paper in test tube. Test tube contains green filtrate. Take the test tube in dark-room and put a flash of torch on it. Now, solution appears red. Why does this occur? Which phenomenon is this? Discuss this with your physics, chemistry and biology teachers.
Answer:
Chlorophyll is the green pigment present in chloroplast. It absorbs light in red and blue region of visible spectrum. It does not absorb green light and thus the green light is reflected which is why it appears green. In this experiment, the chlorophyll in test tube appears red when a flash torch is put on it in the dark room.

This is because when the electrons of the chlorophyll molecule are excited in dark in the absence of electron transport chain the electrons release their energy in the form of red light as they return to ground state. This phenomenon observed here is transmission of light.

Activity 2 (Textbook Page No. 139)

To separate the chloroplast pigments by paper chromatography. Concentrate the extracted chlorophyll solution by evaporation. Apply a drop of it at one end, 2cm away from edge of a strip of chromatography paper and allow it to dry thoroughly. Take a mixture of petroleum ether and acetone in the ratio of 9:1 at temperature of 40 to 60°C. Hang the strip in the jar with its loaded end dipping in the solvent. Close the jar tightly and keep it for an hour. The pigments separate into distinct green and yellow bands of chlorophyll and carotenoid respectively.
Answer:
Pigments are the molecules which reflects only certain wavelengths of visible light. Chromatography is the technique used to separate the chloroplast pigments. Carotenes form yellow-orange band, chlorophyll forms blue-green band, chlorophyll b forms yellow-green bands.

Can you tell? (Textbook Page No. 139)

Tomatoes, carrots and chillies are red in colour due to the presence of pigments. Name the pigment. Answer:
Red colour pigment present in tomatoes, carrots and chillies is lycopene.

Think about it (Textbook Page No. 140)

Large number of gas bubbles are evolved during day time in a pond of water.
Answer:
Photosynthesis occurs in the presence of sunlight. During photosynthesis, plants give out oxygen and take in carbon dioxide. The plants present underwater carryout photosynthesis and release oxygen. Hence, large number of gas bubbles are evolved during day time in a pond.

Think about it (Textbook Page No. 141)

Does moonlight support photosynthesis?
Answer:
The reactions of photosynthesis take place in the presence of sunlight. The intensity of moonlight is several thousand times less than that of direct sunlight which is insufficient for the light dependent phase of photosynthesis. As the sun sets, rate of photosynthesis also decreases. Therefore, moonlight does not support photosynthesis.

Can you tell? (Textbook Page No. 145)

How chlorophyll – a is excited? Show it with a diagram.
Answer:

1. Chlorophyll-a is an essential photosynthetic pigment as it converts light energy into chemical energy and acts as a reaction centre.
2. Initially, it lies at ground state or singlet state but when it absorbs or receives photons (solar energy), it gets activated and goes in excited state or excited second singlet state.
3. In the excited state, chlorophyll-a emits an electron. The emitted electron is energy rich, i.e. has extra amount of energy.
4. Due to the loss of electron (e_–), chlorophyll-a becomes positively charged. This is the ionized state.
5. Chlorophyll-a molecule cannot remain in the ionized state for more than 10‘9 seconds. Hence the photo-chemical reaction or electron transfer occurs very fast.
6. The energy rich electron is then transferred through various electron acceptors and donors (carriers).
7. During the transfer, the electron emits energy which is utilized for the synthesis of ATP. This shows that light energy is converted into chemical energy in the form of ATP.

Can you tell? (Textbook Page No. 140)

What made Hill to perform his experiment?
Answer:
Robert Hill proved that the source of oxygen evolved during photosynthesis is water and not carbon dioxide. Hence, it is called Hill’s Reaction.

1. In this experiment, Hill cultured isolated chloroplasts in a medium containing C0₂ free water, haemoglobin and ferric compound.
2. Ferric salts and haemoglobin were added in the medium as hydrogen and oxygen acceptors respectively.
3. When the suspension was illuminated, he observed that haemoglobin turned into oxyhaemoglobin (red colour).
4. This confirmed that water must have oxidized releasing 0₂, that reacted with haemoglobin. Reduction of ferric compound was also indicated by change in colour.
5. The H₂O molecule oxidized to evolve 0₂ as a by-product. Thus, Hill proved that the source of evolving 0₂ is H₂0 and not C0₂.
6. This process of splitting up of water molecules under the influence of light in the presence of chlorophyll is called Photolysis of water or Hill Reaction.
7. Hill’s reaction can be represented as follows:

Can you tell? (Textbook Page No. 145)

Draw a flowchart of non-cyclic photophosphorylation.
Answer:
Non-cyclic photophosphorylation:
a. It involves both photosystems- PS-I and PS-II.
b. In this case, electron transport chain starts with the release of electrons from PS-II.
c. In this chain high energy electrons released from PS-II do not return to PS-II but, after passing through an electron transport chain, reach PS-I, which in turn donates it to reduce NADP to NADPH.
d. The reduced NADP⁺ (NADPH) is utilized for the reduction of CO₂ in the dark reaction.
e. Electron-deficient PS-II brings about oxidation of water-molecule. Due to this, protons, electrons and oxygen atom are released.
f. Electrons are taken up by PS-II itself to return to reduced state, protons are accepted by NADP⁺ whereas oxygen is released.
g. As in this process, high energy electrons released from PS-II do not return to PS-II and it is accompanied with ATP formation, this is called Non-cyclic photophosphorylation.

Can you tell? (Textbook Page No. 145)

Describe Calvin’s cycle.
Answer:
The entire process of dark reaction was traced by Dr. Melvin Calvin along with his co-worker, Dr. Benson. Hence, the process is called as Calvin cycle or Calvin- Benson cycle. Since the first stable product formed is a 3-carbon compound, it is also called as C₃ pathway and the plants are called C₁₄ plants.
Calvin carried out experiments on unicellular green algae (Chlorella), using radioactive isotope of carbon, C₁₄ as a tracer.
It is also called synthesis phase or second phase of photosynthesis.
The cycle is divided into the following phases:
1. Carboxylation phase:
a. Carbon dioxide reduction starts with a five-carbon sugar ribulose-l,5-bisphosphate (RuBP). It is a 5- carbon sugar with two phosphate groups attached to it.
b. RuBP reacts with CO₂ to produce an unstable 6 carbon intermediate in the presence of Rubisco.
c. It immediately splits into 3 carbon compounds called 3-phosphoglyceric acid.
d. RuBisCO is a large protein molecule and comprises 16% of the chloroplast proteins.

2. Glycolytic reversal:
a. 3-phosphoglyceric acid form 1,3-diphosphoglyceric acid by utilizing ATP molecule.
b. These are then reduced to glyceraldehyde-3-phosphate (3-PGA) by NADPH supplied by the light reactions of photosynthesis.
c. In order to keep Calvin cycle continuously running there must be sufficient number of RuBP and regular supply of ATP and NADPH.
d. Out of 12 molecules of 3-phosphoglyceraldehyde, two molecules are used for synthesis of one glucose molecule.

3. Regeneration of RuBP:
a. 10 molecules of 3-phosphoglyceraldehyde are used for the regeneration of 6 molecules of RuBP at the cost of 6 ATP.
b. Therefore, six turns of Calvin cycle are needed to get one molecule of glucose.

Can you tell? (Textbook Page No. 147)

Summarise the photosynthetic reaction.
Answer:
6C0₂ + 12H₂0 → C₂H₁₂0₆ + 60₂ + 6H₂0
1. Photosynthesis is a two step process.
The light dependent reactions convert the light energy from the sun into chemical energy.
The light independent reactions convert the chemical energy to synthesize carbohydrates.
2. Light dependent reactions: Light is absorbed by chlorophyll which results in the production of ATP. Photolysis of water produce oxygen and hydrogen. The hydrogen and ATP are used in the light independent reactions and the oxygen is released from stomata.
3. Light independent reactions: ATP and hydrogen are transferred to the site of light independent reactions. The hydrogen is combined with carbon dioxide to form complex organic compounds.
The ATP provides the required energy to power these anabolic reactions and fix the carbon molecules.

Can you tell? (Textbook Page No. 147)

Summarise the photosynthetic reaction.
Answer:
6C0₂ + 12H₂0 → C₂H₁₂0₆ + 60₂ + 6H₂0
1. Photosynthesis is a two step process.
The light dependent reactions convert the light energy from the sun into chemical energy.
The light independent reactions convert the chemical energy to synthesize carbohydrates.
2. Light dependent reactions: Light is absorbed by chlorophyll which results in the production of ATP. Photolysis of water produce oxygen and hydrogen. The hydrogen and ATP are used in the light independent reactions and the oxygen is released from stomata.
3. Light independent reactions: ATP and hydrogen are transferred to the site of light independent reactions. The hydrogen is combined with carbon dioxide to form complex organic compounds.
The ATP provides the required energy to power these anabolic reactions and fix the carbon molecules.

Can you tell? (Textbook Page No. 147)

C₄ plants are more productive. Why?
Answer:

1. Photorespiration is considered to be a wasteful process in plants. It is an energy consuming process in plants which ultimately leads to reduction in final yield of plants.
2. During C₃ photosynthesis, 25% of the carbon dioxide fixed has to pass through photorespiratory process.
3. This decreases the photosynthetic productivity.
4. In C₄ plants, photorespiration is absent and hence they have better productivity.

Can you tell? (Textbook Page No. 147)

Xerophytic plants survive in high temperature. How?
Answer:

1. Xerophytic plants are those that have adapted to dry environments.
2. They have adapted to arid conditions by storing water in stems.
3. Stomata of these plants remain closed during day time to reduce the rate of transpiration to bare minimum.
4. Leaves are modified into spines or are reduced in size to check the loss of water due to transpiration.
5. The waxy surfaces of xerophytic plants prevent the loss of moisture.
6. Thus, they are able to survive in high temperature.

Can you tell? (Textbook Page No. 147)

Compare C₄ plants and CAM plants.
Answer:

C4 Plants	CAM Plants
1. These are mostly tropical and subtropical plants.	These are mostly xerophytic plants.
2. Leaves show Kranz anatomy.	Leaves does not show Kranz anatomy.
3. Stomata is open during day time.	Stomata is open during night time.
4. Photorespiration is not easily detectable.	Photorespiration is detectable in afternoon.
5. Carbon fixation takes place in mesophyll cells and Calvin Cycle takes place in bundle sheath cells.	Photosynthesis takes place in the mesophyll cells but carbon fixation takes place at night and Calvin cycle happens during day.
e.g. Sugarcane, maize, jowar, Amaranthus, etc.	e.g. Kalanchoe, Opuntia, Aloe, etc.

Balbharati Maharashtra State Board Class 11 Information Technology Solutions Practicals Skill Set 6 DBMS (PostgreSQL) Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 Information Technology Practicals Skill Set 6 DBMS (PostgreSQL)

SOP 1: Create a database, using Postgres SQL named hospital.

• In this database, create a table of patients with the following fields
  Patient_ID, Patients_Name, Address, Room_number and Doctor’s_name.
• Give appropriate data type for each field.
  

Answer:
Step 1: Open Command Terminal. Switch over to the Postgres account on your server by typing.
$ sudo -i -u Postgres

Step 2: You can now access a Postgres prompt immediately by typing.
$ psql

Step 3: To create a database hospital;
create database hospital;

Step 4: Connect to Database using \c
\c hospital;

Step 5: Create a table in the database. Create Table Command is used.
create table patients(patients_Id Integer,patients_name text,Address text,Room_number integer,Doctor_name text);

Step 6: Let’s see the result of the patient’s table.
select * from patients;
or
\d patients;

SOP 2: Create a database using PostgreSQL named Schoolmaster.

• In this database create a table of students with the following fields
  student_ID, student_name, Address, Phone_number, Date_of_Birth.
• Give appropriate data types for each field. Enter at least 5 records.

Answer:
Step 1: Create a database School-Master.
create database school_master;

Step 2: Now To connect the database use \c Command.
\c database school_master;

Step 3: Create a table of students with the following fields.Give appropriate data type for each Field.
student_ID, student_name, Address. Phone_number, Date_of_Birth.
create table students(student_ID integer, student_name text,Address_text,Phone_number integer,Date_of_Birth date);

Step 4: Enter at least 5 records.
Insert into students values(001,’ZAHRA LALANIVMAZGAON’,123456789,’20-08-2000’);
Insert into students values(002,’MUHAMMAD LALANI’/BYCULLA’, 987654210,’30-01-2000’);
Insert into students values(003,’KUNAL KAPOOR’,’WALKESHWAR’, 987224210,T5-7-2000’);
Insert into students values(004,’AKSHAY SINGH’,’CHARNI ROAD’. 937224210,’19-6-2000’);
Insert into students values(005,’RUKHSHAR BANU ’,’DIWANPARA’, 937226210,’18-8-2000’);

Step 5: Show all records using select command
seleet*from students:

SOP 3: Given the list of fields: Empld, EmpName, EmpDepartment, Salaryld, Salary Amount, Bonus in the tables Employee and Salary respectively. Define primary key, foreign key and segregate for above fields into employee and salary table. Also create a one-to-one relationship between Employee and Salary Table.
Answer:
Step 1: Create a school database
Create database school;

Step 2: connect to database \c databasename;
\c school;

Step 3: In this database create two tables Employee and Salary with the following fields. Define primary key, foreign key and segregate for above fields into employee and salary table. Empld, EmpName, EmpDepartment, Salaryld, SalaryAmount, Bonus.
Create table salary(salaryld Integer PRIMARY KEY,Salaryamount integer,Bonus integer);
Create table employeefEmpId integer PRIMARY KEY,EmpName text,EmpDept text,Salaryld integer,FOREIGN KEY(salary ID)”REFERENCES Salaryfsalary id));

Step 4: See both tables
select*from salary;
select*from employee;

Balbharati Maharashtra State Board Class 11 Information Technology Solutions Practicals Skill Set 5 Digital Content Creation (GIMP, Inkscape) Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 Information Technology Practicals Skill Set 5 Digital Content Creation (GIMP, Inkscape)

SOP 1: Use of Toolbox and editing an image using GIMP.

• Create an image by using Toolbox controls from GIMP.
• Insert the image in an already created image.

Answer:
Step 1: Click GIMP Image Editor

Step 2: Create a blank New image
File → New (Ctrl + N)

Step 3: Change the foreground and background colour uses the Blend Tool.

Step 4: We will create a kid’s cap using toolbox controls.

Step 5: Draw colour oval (Ellipse). Change the background colour and foreground colours with three different Ellipse(oval) . Create a layer for each and every Ellipse.

Step 6: To add text, you’ll need to access your “Text Tool”. You can find your Text Tool in your Toolbox window.

Step 7: To make this image visible to people larger, select your “Scale Tool” from your Toolbox window.

Step 8: To save your image select Export as, select proper type, and click on export.

Step 9: Insert the image in an already created image.
Click on File → Open → Image → As Layer.
Move the image to the desired location and merge to the original image.

SOP 2: Use GIMP for the following.

• Create a new image
• Put your name using the text tool.
• Use various filters to make a logo of your name.
• Auto crop image to text size.

Answer:
Use of Toolbox and editing an image using GIMP.
Create an image by using Toolbox controls from GIMP.

Step 1: click on the GIMP image Editor icon

Step 2: Create blank image File → Text document (Ctrl+N)

Step 3: Creates balloon image having very attractive colours using toolbox controls.

Step 4: Now change the foreground and background colour using the Blend tool (shot cut L). Drag the mouse from left to right on your canvas after release the mouse canvas will be filled by the gradient of foreground and background colour.

Step 5: Next click on the Text Tool (Short cut T)

Step 6: Using the Move tool, we can move the text roughly to the center of the image but instead we prefer to put it precisely in the center using the Align tool. Click on align tool (Short cut Q) and then select the text Layer and text.

Step 7: Drawing coloured oval (Ellipse). Change the background and foreground colours again with two different colours. We have to choose red and yellow colour respectively.

Step 8: Using the move tool place these ovals near the text. Select the layers one by one and merge them.
Layer → Merge Down. Now we have colourful single layer. Let us create a ballon from the canvas.

Step 9: Do Filters → Map → Map Object. Check ‘Update preview live’ and uncheck the ‘Transparent Background’.

Step 10: We will tweak the balloon further and give it a more realistic shape. Apply the Distort Filter with curve Bend option, i.e, Filter → Distort → Curve Bend. In the resulting dialog choose automatic preview and Lower curve border. Then drag the Mid Point of the Curve Indicator line using the mouse.

SOP 3: Use Inkscape for the following.

• Draw a simple landscape using basic geometric shapes.
• Use gradient tool for the same.

Answer:
Step 1: Select the tool button from the left toolbar showing the circle icon.

Step 2: Drag it with the mouse to any place in the client area. You will get an ellipse. To get the exact shape, press the CTRL key while dragging.

Step 3: After the ellipse is complete choose the selection tool from the left toolbar, then click on the ellipse, it will show 8 size handles. You can resize it by dragging those handles.

Step 4: To fill the ellipse with color, just click on any color given in the bottom color palette. Also, you can click on Menu object → Fill and stroke. It will give RGB color options, you can use any combinations from it.

SOP 4: Use Inkscape for the following.

• Load an Id size image,
• Make 12 copies of it.
• Arrange in 4 rows × 3 columns on an A4 size page.

Answer:
Step 1: Open Inkscape Vector Graphics Editor.

Step 2: File → New (Ctrl +N). Now Click on Document properties and select paper size A4.

Step 3: Import a bitmap or Svg Image into this document (Ctrl+l)

Step 4: Create 12 copies of it (4 rows and 3 columns) Create duplicate (Ctrl + D) images and set them in the proper grid.

Step 5. Arrange in 4 rows and 3 columns on an A4 size page.

SOP 5: Use Inkscape for the following.

• You are starting a new business.
• Create an advertisement to be published in a local newspaper promoting your product or services.
• Size should be 210 × 210 mm.
• Create your own visiting card using Inkscape.

Answer:
Steps to create advertisement:

Step 1: Choose a distinctive font and type some articles.

Step 2: Resize the text and press the long CTRL key to preserve the aspect ratio.

Step 3: Then outset it (Path → Outset)

Step 4: Change the color and resize down to the size.

Step 5: Now duplicate the text for the border.

Step 6: Color the duplicate temporarily in any random, non-white color (we need to see it over the white background) and move it under the initial text.

Step 7: Apply the outset effect (no need to resize this time, we are working on a rough border) and make it white.

Step 8: Duplicate the white border and make the duplicate black.

Step 9: Move the black duplicate at the bottom of the stack and shift it one or two pixels down and to the left for a drop shadow effect.

Step 10: Open the “Fill and Stroke” dialog and add a bit of Gaussian Blur and decrease the opacity.

Step 11: Post your finished work.

Steps to Create Visiting Card:

Step 1: To create a new empty document, use File → New → Default or press Ctrl+N.
To use File → New → Templates… or press Ctrl+Alt+N

Step 2: To click on document properties (Shift+Ctrl=D) and select Business card paper size.

Step 3: Click on the text menu and select Align and Distribute to align the card.

Step 4: Use the snap to the page border tool and Drag the Rectangles on the card.

Step 5: Specify the Height and width in inches. Width=3.75 Height=2.25

Step 6: Select the rectangle → Open Fill and stroke Toolbar → select linear gradient tool to show a border. Align the content.

Step 7: Set the ruler position.

Step 8: Create Layers. Now Draw the Circle on the card and set the fill and stroke properties.

Step 9: Change the object colour and create a duplicate object (ctrl+D)

Step 10: Select both objects → to go to the path menu → select difference. Now Select the Second object → to create duplicated object → change the object colour.

Step 11: Select Second duplicate Object → path → select difference → select flip selected object.
Select Edit Text Object Control → Type your Name, your website, Email ID, Phone in one, and Insert Logo.

SOP 6: Using Inkscape make the following picture.

Answer:
Step 1: Open Inkscape Vector Graphics Editor File → New

Step 2: Select Circles, Ellipses to create circles.

Step 3: Change the colours of the object. Use Draw freehand line tool.

Step 4: For the middle body, again click on the ellipse, select white color and click and drag on the place you want.

Step 5: For Head again choose ellipse tool ALT + CTRL to create a perfect circle, resize it and drag it on the proper place.

Step 6: For eyes, zoom in and choose the circle tool, select the proper color, press ALT+CTRL to resize it, and choose another circle, set black color, and draw a circle inside it.

Step 7: Press CTRL+D to make a duplicate and place it on another side.

Step 8: Create Hands. Create duplicate images → path → differences.

Step 9: Create legs using star tools → merge all starts → path → union.

Balbharati Maharashtra State Board Class 11 Information Technology Solutions Practicals Skill Set 4 Accounting Package (GNUKhata) Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 Information Technology Practicals Skill Set 4 Accounting Package (GNUKhata)

SOP 1: Use of Accounting Package to create a company.
Create a company with the following particulars.
Company Name: B.B Enterprises
Case: Upper Case
Company Type: Profit Making
Financial Year: 01-04-2019 to 31-03-2020
Use GNUKhata for: Accounting Only
Create a profile with relevant data for any company. Create an Admin account for the company.
Answer:
Step 1: Open GNUKhata Application. Click on ‘company Setup Wizard’ or (press shift + control + C)

Step 2: While creating a company the following details are to be given.
Company Name: B.B Enterprises
Case: Upper Case
Company Type: Profit Making
Financial Year: 01/04/2019 to 31/03/2020
Use GNUKhata for: Accounting Only
Fill in all the required details and click on proceed button on the right button corner.

Step 3: Enter Appropriate Company Information in the Given Field.
Company Profile window appears, in that window fill in the details and click on proceed button on the right bottom corner.

Step 4: Create Admin Next step is the ‘create Admin’ which is mandatory with proper user name and password. Fill in the fields and click on ‘Create & Login button, the company will be created.
Now your Company is ready.

Step 5: Admin Dashboard After login, the following admin dashboard, appears.

SOP 2: Create ledger accounts using the accounting Package.
Create ledger accounts for the following and allocate proper groups.

1. Import duty
2. Insurance
3. Machinery
4. Audit Fee
5. Purchase
6. Sales
7. Telephone charges
8. Interest Received
9. Salary
10. Professional fees

Answer:
Step 1: You Can select an already created company using the ‘Select Existing Company ’ Option on Opening Screen.

Step 2: Log in with User Name and Password

Step 3: To create an account (ledger account) click on the Hamburger Menu available at the left top corner of the dashboard.
Click on the Master Account → It allows you to create an account.
The following table shows the group name, sub-group name, and account name which are to be created.

Group Name	Sub Group Name	Account / Ledger Name
Direct Expenses	None	Import Duty
Indirect Expenses	None	Insurance
Fixed Assets	Plant & Machinery	Machinery
Indirect Expenses	None	Audit fee
Direct Expense	None	Purchase
Direct Income	None	Sales
Indirect Expense	None	Telephone charges
Indirect Income	None	Interest received
Indirect Expense	None	Salary
Direct Income	None	Professional Fees

Balbharati Maharashtra State Board Class 11 Information Technology Solutions Practicals Skill Set 3 Client Side Scripting (JavaScript) Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 Information Technology Practicals Skill Set 3 Client Side Scripting (JavaScript)

SOP 1: Create a JavaScript program for the following using appropriate variables, JavaScript inbuilt functions, and control structures.

• To accept an integer and display the result by multiplying it with 3.
• To accept two integers and display a larger number of them.
• To check whether the user entered number is positive or negative.

Answer:
To accept integer and display the result by multiplying it with 3.
<!DOCTYPE html>
<head>
<title>To accept integer and display the result by multiplying it with 3.</title>
</head>
<body>
<script language=”javascript”> var a,no,ans;
a=prompt(Enter any value’);
no=parseInt(a);
ans=no*3;
document.
write(“The answer is :”+ans);
</script>
</body>
</html>

To accept two integers and display larger number of them.
<!DOCTYPE html>
<head>
<title>To accept two integers and display larger number of them.</title>
</head>
<body>
<script language=”javascript”>
var a,b;
a=prompt(‘Enter first value’);
b=prompt(‘Enter second value’);
if(a>b)
document.write(“a is large number than b “);
else
document. write(“b is large number than a”);
</script>
</body>
</html>

To check whether, user entered number is positive or negative.
<!DOCTYPE html>
<head>
<title>To check whether, user entered number is positive or negative</title>
</head>
<body>
<script language=”javascript”>
var a,no;
a=prompt(‘Enter any number’);
no=parseInt(a);
if(no>0)
document.write(“Number is Positive”);
else
document.write(“Number is Negative”);
</script>
</body>
</html>

SOP 2: Create a JavaScript program for the following using appropriate variables, JavaScript inbuilt functions, and control structures.

• To accept two positive or negative numbers and check whether they are equal or not.
• To accept a number and display the square of it.
• To check whether the accepted integer is multiple of 3 or multiple of 7.

Answer:
To accept two positive or negative numbers and check whether they are equal or not.
<!DOCTYPE html>
<head>
<title>program3</title>
</head>
<body>
<script language=”javascript”> var no1,no2;
no1=prompt(‘Enter first number’);
no2=prompt(‘Enter Second number’);
if(no1==no2)
document.write(“Both are equal”);
else
document.write(“Given numbers are not equal”);
</script>
</body>
</html>

To accept number and display square of it.
<!DOCTYPE html>
<head>
<title>To accept number and display square of it</title>
</head>
<body>
<script language-’j avascript”>
var no,sqr;
no=prompt(‘Enter Any number’);
sqr=no*no;
document, write (“The Square is=”+sqr);
</script>
</body>
</html>

To check whether the accepted integer is multiple of 3 or multiple of 7.
<!DOCTYPE html>
<html>
<head>
<title>To check whether the accepted integer is multiple of 3 or multiple of 7.</title>
</head>
<body>
<script language=JavaScript>
var a;
a=prompt(“Enter your first interger / number”);
if(a%3==0 | | a%7==0)
alert(“multiple of 3 or 7”);
else
alert(“not a multiple of 3 or 7”);
</script>
</body>
</html>

SOP 3: Create a JavaScript program for the following using appropriate variables, JavaScript inbuilt string functions, and control structures.

• To accept a string and calculate its length.
• To accept a string and display it in lowercase and uppercase.
• To check whether the length of the string is 4 or greater.

Answer:
To accept a string and calculate its length.
<!DOCTYPE html>
<head>
<title>program8</title>
</head>
<body>
<script language=”javascript”>
var a;
a=prompt(‘Enter string’);
document.write(“The length is=”+a.length);
</script>
</body>
</html>

To accept string and display it into lowercase and uppercase.
<!DOCTYPE html>
<head>
<title> To accept string and display it into lowercase and uppercase</title>
</head>
<body>
<script language=”javascript”>
var a;
a=prompt(‘Enter any string’);
document.write(“<br>Entering Strings “+a);
document.write(“<br>Lowercase=”+a.toLowerCase());
document.writeln(“<br>Uppercase=”+a.toUpperCase());
</script>
</body>
</html>

To check whether the length of string is 4 or greater.
<!DOCTYPE html>
<html>
<head>
<title> JavaScript</title>
</head>
<body>
<script language=JavaScript>
var a,b;
a=prompt(“Enter your text”);
b=a.length;
if(b=4 || b>=4)
alert(‘‘your length is 4 or greater”);
else
alert(“your text length is below 4”);
</script>
</body>
</html>

SOP 4: Create event-driven JavaScript programs for the following using appropriate variables, JavaScript inbuilt functions, and control structures.
To accept numbers and validate if the given value is a number or not by clicking on the button.

To calculate the addition and division of two numbers.

Answer:
To accept number and validate if the given value is a number or not by click
<!DOCTYPE html>
<html>
<head>
<title> To accept number and validate if the given value is a number or not by clicking on the button.
</title>
</head>
<script language-’JavaScript”>
function display()
{
var a,b;
a=form1.t1;value;
if(a>=0)
alert(“Value is a number”+” “+a);
else
alert(“Value is a string”+” “+a);
}
</script>
</head>
</body>
<form name=form1>
Enter Value: <Input type=text name=t1><br><br>
<Input type=button value=Check onClick=”display()”>
</form>
</body>
</html>

To calculate addition and division of two numbers.
<!DOCTYPE html>
<head>
<title>To calculate addition and division of two numbers.</title>
<script langauge=”javascript”>
function add()
{
var a,b,result;
a=f1.t1.value;
b=f1.t2.value;
result=parselnt(a)+parselnt(b);
document.write(“The Addition is =”+result);
}
function div()
{
var a,b,d;
a=f1.t1.value;
b=f1.t2.value;
d=parselnt(a)/parselnt(b);
document.write(“The Divide is =”+d);
}
</script>
</head>
<body>
<form name=”f1”>
1st Number : <input type=”text” name=”t1”><br>
2nd Number : <input type=”text” name=”t2”><br>
<input type=”button” value=”Addition” name=”b1” onClick=”add()”>
<input type=”button” value=”Division” name=”b2” onClick=”div()”>
</form>
</body>
</html>

Balbharati Maharashtra State Board Class 11 Information Technology Solutions Practicals Skill Set 2 Web Designing (HTML – 5) Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 Information Technology Practicals Skill Set 2 Web Designing (HTML – 5)

SOP 1: Write a program using HTML with the following specifications.

• The background colour should be green.
• The text colour should be red.
• The heading should be as large in size as ‘My First Web Page’.
• Display a horizontal line after the heading.
• Display your name in Bold, address in Italics, and standard as 11th.

Answer:
<!DOCTYPE html>
<html>
<head>
<title>sop1</title>
</head>
<body bgcolor=green text= red>
<h1> My First Web Page </h1>
<hr>
<b> Reliable Publications</b>
<br>
<i> Chira bazar,Charni road,Mumbai</i>
<br>
Standard 11th.
</body>
</html>

SOP 2: Create a web page with, following specifications.

• Image of any scientist with an alternate text as his name.
• Create a paragraph related to the information of that scientist.
• Create a table of his/her inventions.

Answer:
<!DOCTYPE html>
<html>
<head>
<title>sop2</title>
</head>
<body>
<img src=”Albert Einstein.jpg” alt=”Albert Einstein”>
<br>
<p>
Albert Einstein was a German-born theoretical physicist who developed the theory of relativity,<br> one of the two pillars of modern physics.His work is also known for its influence on the philosophy of science.
</p>
<table border=5 bordercolor=pink>
<tr>
<th> Sr no.</th>
<th> Invention </th>
<th> Year</th>
</tr>
<tr>
<td>1</td>
<td>Quantum Theory of Light </td>
<td>1905</td>
</tr>
<tr>
<td>2</td>
<td>Theory of Relativity</td>
<td>1907</td>
</tr>
</table>
</body>
</html>

SOP 3: Create a webpage with the following specifications.

• Display heading ‘Application Form’ in the highest heading with center alignment.
• Accept name, standard 11th or 12th with only one selection choice.
• Submit the form.

Answer:
<! DOCTYPE html>
<html>
<head> <title> sop3</title>
</head>
<body>
<h1 align=center> Application Form </h1>
<form>
Enter Name:<input type=text name=t1>
<br><br>
Standard:<br>
<input type=”radio” name=r1>11th<br>
<input type=”radio” name=r1>12th<br>
<br><br>
<input type=”submit” value=”Submit”>
</form>
</body>
</html>

SOP 4: Write a program using HTML with the following specification.
A webpage with details about a class with a total number of students-100, (Boys – 50), Girls – 50 in tabular form.
e.g.

Link this page to another page as follows.

Demo.html
Answer:
<!DOCTYPE html>
<html>
<head>
<title>New Page 1</title>
</head>
<body>
<center>
<table border=”1” width=”69%” >
<tr bgcolor=pink>
<td>Number of Students</td>
<td>Boy</td>
<td>Girls</td>
</tr>
<tr bgcolor=”lightgreen”>
<td>100</td>
<td>50</td>
<td>50</td>
</tr>
</table>
</body>
</html>

Demo
<!DOCTYPE html>
<html>
<head>
<title>the heading</title>
</head>
<BODY>
<header>
<center>
<table border=”1” align=center>
<tr align=center>
<td><a href=”sop4.html”>STD – XI</a>
<br>Stream – Science<br>
Div – A<br></td>
</tr>
</table>
</body>
</html>

Balbharati Maharashtra State Board Class 11 Information Technology Solutions Practicals Skill Set 1 Daily Computing (Libre Office) Textbook Exercise Questions and Answers.

Maharashtra State Board Class 11 Information Technology Practicals Skill Set 1 Daily Computing (Libre Office)

SOP 1: Create a Resume

• The resume should contain the following:
• Title at the center with applicable font and size.
• It should contain points such as Name, Address, Mobile Number, Date of Birth, Nationality, Caste, Category, Hobbies, etc. Add some extra points.
• For educational qualifications, a student should insert a table.
• In the end, students should write a few lines about their aim.

Answer:
Step 1: Click on the LibreOffice Writer icon.

Step 2: Create a New File.
File Menu → New → Text Document

Step 3: Write Title at the center with applicable font size. (Resume).

Step 4: Type Name, Address, Mobile Number, Date of Birth, Nationality, Caste, Category, Hobbies, etc.

Step 5: Write Student educational qualifications (Insert Table).
Choose Table Menu → Insert Table → Select number of rows and columns → Insert.

Step 6: Write some lines about the aim, select the text and make it bold by clicking on the bold option from the formatting toolbar.

After Completing this practical students will learn how to create resumes using LibreOffice, also tab setting and table creation.

SOP 2: By using Mail Merge send an invitation for your birthday party.

• Use the mail merge feature.
• Send an invitation to at least 5 friends.

Answer:
Choose Tools – Mail Merge Wizard.
Step 1: Select Start from a template, and click the Browse button. You see the New dialog.

Step 2: Select five friends Details in the left list, and then the Invitation letter in the right list.
Click OK to close the Templates dialog, and click Next in the wizard.

Step 3: Select Letter and click Next.

Step 4: On the next step of the wizard, click the Select Address List button to check that you are using the correct address list. If you want to use an address block, select an address block type, match the data fields if necessary, and click Next.

Step 5: Next follows the Create a salutation step. Deselect the Insert personalized salutation box. Under General salutation, select the salutation that you want on top of all letters.

Step 6: If you want to place mail merge fields anywhere else in the Invitation document select the corresponding column in your address data source and then drag and drop the column header into the invitation document where you would like the field to be. Be sure to select the entire column.

Step 7: Click Next and finally Finish creating the mail merge.

SOP 3: Create a mark list. The mark list should display:

• Fields as Name, Math, Physics, Chemistry, Biology, Total, Percentage.
• Below each subject find out the lowest marks and highest marks.
• Enter a minimum of 10 records.
• Declare the first three ranker students.
• Create a chart based on the above data.

Answer:
Step 1: Open LibreOffice Calc and add fields like Name, Math, Physics, Chemistry, Biology, Total, Percentage.

Step 2: Enter 10 records in it.

Step 3: To Calculate Total
Click inside the cell where the total has to be calculated.
(Use =SUM(A2:D2) formula) – Type the range of cell.

Step 4: Now click inside the cell where the percentage has to be calculated.
Calculate Percentage using formula = E2* 100/400, drag the formula for remaining cells.

Step 5: Enter a minimum of 10 records. And calculate the result.

Step 6: Find out the Lowest Marks
Calculate Lowest marks using formula = min(A2:A11), drag formula for remaining cells.

Step 7: Find out the Highest Marks
Calculate Highest marks using formula =max(A2:A11), drag formula for remaining cells.

Step 8: Now Show the first three ranker students. Use Sort Option for sorting and auto filter.

Step 9: To find the first ranker use formula =large(F2:F11,1)

Step 10: To find the second ranker use formula =large(F2:F11,2)

Step 11: To find the third ranker use formula =large(F2:F11,3)

Step 12: To create a chart Select Complete Table → Click on Insert Menu → Click on Chart Option.
A window will appear, which starts with the chart wizard → Select Chart type.

SOP 4: Create an Informative presentation on your college.

• The presentation should contain a minimum of 8 slides.
• One slide should contain a chart.
• One slide with an image.
• Each slide should contain custom animation & slide transition effect.

Answer:
Step 1: Preparing an Eight slide for an Informative presentation of your college.
Press the Ctrl + N Keys. OR choose Slide → New Slide from Menu bar.
OR
Click on the New Slide icon on the Standard Toolbar.

Step 2: Specify the background image. OR background colour.
Now First Choose Slide Menu → Click on Insert Menu → Click on Image Submenu (inserting college Image with information )

Step 3: Click on Insert Menu → Click on Chart Option.

Step 4: Click on object presentation → from slide bar select Custom Animation each object → Choose category, effect, duration, direction, etc.

Step 5: Click on Press F5 function key for slide show
OR
Select slide menu → Start from the first Slide from the menu bar.
OR
Click on the start from the first slide Icon on the standard Conclusion: After Completing this practical students will be known how to create an informative presentation, custom animation & slide transition effect using Libre Office Impress.