Category: Class 12

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 4 Chemical Thermodynamics Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 4 Chemical Thermodynamics

1. Select the most appropriate option.

Question 1.
The correct thermodynamic conditions for the spontaneous reaction at all temperatures are
(a) ΔH < 0 and ΔS > 0
(b) ΔH > 0 and ΔS < 0
(c) ΔH < 0 and ΔS < 0
(d) ΔH < 0 and ΔS = 0
Answer:
(a) ΔH < 0 and ΔS > 0

Question ii.
A gas is allowed to expand in a well-insulated container against a constant external pressure of 2.5 bar from an initial volume of 2.5 L to a final volume of 4.5 L. The change in internal energy, ΔU of the gas will be
(a) -500 J
(b) +500J
(c) -1013 J
(d) +1013 J
Answer:
(a) -500 J

Question iii.
In which of the following, entropy of the system decreases ?
(a) Crystallisation of liquid into solid
(b) Temperature of crystalline solid is increased from 0 K to 115 K
(c) H_2(g) → 2H_(g)
(d) 2NaHCO_3(s) → Na₂CO_3(s) + CO_2(g) + H₂O_(g)
Answer:
(a) Crystallisation of liquid into solid

Question iv.
The enthalpy of formation for all elements in their standard states is
(a) unity
(b) zero
(c) less than zero
(d) different elements
Answer:
(b) zero

Question v.
Which of the following reactions is exothermic ?
(a) H_2(g) → 2H_(g)
(b) C(s) → C_(g)
(c) 2Cl_(g) → Cl_2(g)
(d) H₂O_(s) → H₂O_(l)
Answer:
(c) 2Cl_(g) → Cl_2(g)

Question vi.
6.24 g of ethanol are vaporized by supplying 5.89 kJ of heat. Enthalpy of vaporization of ethanol will be
(a) 43.4 kJ mol^-1
(b) 60.2 kJ mol^-1
(c) 38.9 kJ mol^-1
(d) 20.4 kJ mol^-1
Answer:
(a) 43.4 kJ mol^-1

Question vii.
If the standard enthalpy of formation of methanol is -238.9 kJ mol^-1 then entropy change of the surroundings will be
(a) -801.7 JK^-1
(b) 801.7 JK^-1
(c) 0.8017 JK^-1
(d) -0.8017 JK^-1
Answer:
(b) 801.7 JK^-1

Question viii.
Which of the following are not state functions ?
1. Q + W 2. Q 3. W 4. H-TS
(a) 1, 2 and 3
(b) 2 and 3
(c) 1 and 4
(d) 2, 3 and 4
Answer:
(b) 2 and 3

Question ix.
For vaporization of water at 1 bar, ΔH = 40.63 kJ mol^-1 and ΔS =108.8 JK^-1 mol^-1. At what temperature, ΔG = 0?
(a) 273.4 K
(b) 393.4 K
(c) 373.4 K
(d) 293.4 K
Answer:
(c) 373.4 K

Question x.
Bond enthalpies of H – H, Cl – Cl and H – Cl bonds are 434 kJ mol^-1, 242 kJ mol^-2 and 431 kJ mol^-1, respectively. Enthalpy of formation of HCl is
(a) 245 kJ mol^-1
(b) -93 kJ mol^-1
(c) -245 kJ mol^-1
(d) 93 kJ mol^-1
Answer:
(b) -93 kJ mol^-1

2. Answer the following in one or two sentences.

Question i.
Comment on the statement: No work is involved in an expansion of a gas in vacuum.
Answer:
(1) When a gas expands against an external pressure Pex, changing the volume from V₁ to V₂, the work obtained is given by
W = -P_ex (V₂ – V₁).
(2) Hence the work is performed by the system when it experiences the opposing force or pressure.
(3) Greater the opposing force, more is the work.
(4) In free expansion, the gas expands in vaccum where it does not experience opposing force, (P = 0). Since external pressure is zero, no work is obtained.
∴ W = -P_ex (V₂ – V₁)
= -0 × (V₂ – V₁)
= 0
(5) Since during expansion in vacuum no energy is expended, it is called free expansion.

Question ii.
State the first law of thermodynamics.
Answer:
The first law of thermodynamics is based on the principle of conservation of energy and can be stated in different ways as follows :

1. Energy can neither be created nor destroyed, however, it may be converted from one form into another.
2. Whenever, a quantity of one kind of energy is consumed or disappears, an equivalent amount of another kind of energy appears.
3. The total mass and energy of an isolated system remain constant, although there may be interconservation of energy from one form to another.
4. The total energy of the universe remains constant.

Question iii.
What is enthalpy of fusion?
Answer:
Enthalpy of fusion (Δ_fusH) : The enthalpy change that accompanies the fusion of one mole of a solid into a liquid at constant temperature and pressure is called enthalpy of fusion.
For example,

This equation describes that when one mole of ice melts (fuses) at 0 °C (273 K) and 1 atmosphere, 6.1 kJ of heat will be absorbed.

Question iv.
What is standard state of a substance?
Answer:
The thermodynamic standard state of a substance (compound) is the most stable physical state of it at 298 K and 1 atmosphere (or 1 bar). The enthalpy of the substance in the standard state is represented as Δ_fH⁰.

Question v.
State whether ∆S is positive, negative or zero for the reaction 2H_(g) → H_2(g). Explain.
Answer:
(i) The given reaction, 2H_(g) → H_2(g) is the formation of H_2(g) from free atoms.
(ii) Since two H atoms form one H₂ molecule, ∆n = 1 – 2= -1 and disorder is decreased. Hence entropy change ∆S < 0 (or negative).

Question vi.
State second law of thermodynamics in terms of entropy.
Answer:
The second law of thermodynamics states that the total entropy of the system and its surroundings (universe) increases in a spontaneous process.
OR
Since all the natural processes are spontaneous, the entropy of the universe increases.
It is expressed mathematically as
∆ S_Total = ∆ S_system + ∆S_surr > 0
∆ S_Universe = ∆ S_system + ∆ S_surr > 0

Question vii.
If the enthalpy change of a reaction is ∆H how will you calculate entropy of surroundings?
Answer:
(i) For endothermic reaction, ∆H > 0. This shows the system absorbs heat from surroundings.
∴ ∆_surr H < 0.
∵ Entropy change = ∆_surr S = \(\frac{-\Delta_{\text {surr }} H}{T}\)
There is decrease in entropy of surroundings.
(ii) For exothermic reaction, ∆H < 0, hence for surroundings, ∆_surr H > 0

∴ ∆_surr > 0.

Question viii.
Comment on spontaneity of reactions for which ∆H is positive and ∆S is negative.
Answer:
Since ∆H is +ve and ∆S is -ve, ∆G will be +ve at all temperatures. Hence reactions will be non-spontaneous at all temperatures.

3. Answer in brief.

Question i.
Obtain the relationship between ∆G° of a reaction and the equilibrium constant.
Answer:
Consider following reversible reaction, aA + bB ⇌ cC + dD
The reaction quotient Q is,
Q = \(\frac{[\mathrm{C}]^{c} \times[\mathrm{D}]^{d}}{[\mathrm{~A}]^{a} \times[\mathrm{B}]^{b}}\)
The free energy change ∆G for the reaction is ∆G = ∆G° + RT in Q
Where ∆G° is the standard free energy change.
At equilibrium
Q = \(\frac{[\mathrm{C}]_{e}^{c} \times[\mathrm{D}]_{e}^{d}}{[\mathrm{~A}]_{e}^{a} \times[\mathrm{B}]_{e}^{b}}=\mathrm{K}\)
∴ ∆G = ∆G° + RT In K
∵ at equilibrium ∆G = 0
∴ 0 = AG° + RT In K
∴ ∆G° = -RT In K
∴ ∆G°= -2.303 RT log₁₀K.

Question ii.
What is entropy? Give its units.
Answer:
(i) Entropy : Being a state function and thermodynamic function it is defined as entropy change (∆S) of a system in a process which is equal to the amount of heat transferred in a reversible manner (Q_rev) divided by the absolute temperature (T), at which the heat is absorbed. Thus,

(ii) Units of entropy are JK^-1 in SI unit and cal K^-1 in c.g.s. units. It is also expressed in terms of entropy unit (e.u.). Hence 1 e.u. = 1 JK^-1.
(iii) Entropy is a measure of disorder in the system. Higher the disorder, more is entropy of the system.

Question iii.
How will you calculate reaction enthalpy from data on bond enthalpies?
Answer:
(i) In chemical reactions, bonds are broken in the reactant molecules and bonds are formed in the product molecules.
(ii) Energy is always required to break a chemical bond while energy is always released in the formation of the bond.
(iii) The enthalpy change of a gaseous reactions (Δ_fH⁰) involving substances with covalent bonds can be calculated with the help of bond enthalpies of reactants and products. (In case of solids we need lattice energy or heat of sublimation while in case of liquids we need heat of evaporation.)

If the energy required to break the bonds of reacting molecules is more than the energy released in the bond formation of the products, then the reaction will be endothermic and ∆H⁰ reaction will be positive. On the other hand if the energy released in the bond formation of the products is more than the energy required to break the bonds of reacting molecules then the reaction will be exothermic and ∆H⁰ reaction will be negative.

Question iv.
What is the standard enthalpy of combustion ? Give an example.
Answer:
Standard enthalpy of combustion or standard heat of combustion : it is defined as the enthalpy change when one mole of a substance in the standard state undergoes complete combustion in a sufficient amount of oxygen at constant temperature (298 K) and pressure (1 atmosphere or 1 bar). It is denoted by ∆_cH⁰.
E.g. CH₃OH_(l) + \(\frac {3}{2}\) O_2(g) = CO_2(g) + 2H₂O
∆_cH⁰= -726 kJ mol^-1
(∆_cH⁰ is always negative.)
[Note : Calorific value : It is the enthalpy change or amount of heat liberated when one gram of a substance undergoes combustion.

Question v.
What is the enthalpy of atomization? Give an example.
Answer:
Enthalpy of atomisation (∆_atomH) : it is the enthalpy change accompanying the dissociation of one mole of gaseous substance into its atoms at constant temperature and pressure.
For example : CH_4(g) → C_(g) + 4H_(g) ∆_atomH = 1660 kJ mol^-1

Question vi.
Obtain the expression for work done in chemical reaction.
Answer:
Consider n₁ moles of gaseous reactants A of volume V₁ change to n₂ moles of gaseous products B of volume V₂ at temperature T and pressure P.

In the initial state, PV₁ = n₁RT
In the final state, PV₂ = n₂RT
PV₂ – PV₁ = n₂RT – n₁RT = (n₂ – n₁)RT = ∆nRT
where ∆n is the change in number of moles of gaseous products and gaseous reactants.
Due to net changes in gaseous moles, there arises change in volume against constant pressure resulting in mechanical work, -P∆V.
∴ W = -P∆V = -P(V₂ – V₁) = – ∆nRT
(i) If n₁ – n₂, ∆n = 0, W = 0. No work is performed.
(ii) If n₂ > n₁, ∆n > 0, there is a work of expansion by the system and W is negative.
(iii) If n₂ < n₁, ∆n < 0, there is a work of compression, hence work is done on the system and W is positive.

Question vii.
Derive the expression for PV work.
Answer:
Consider a certain amount of an ideal gas enclosed in an ideal cylinder fitted with massless, frictionless rigid movable piston at pressure P, occupying volume V₁ at temperature T.

Fig. 4.8 : Work of expansion
As the gas expands, it pushes the piston upward through a distance d against external force F, pushing the surroundings.
The work done by the gas is,
W = opposing force × distance = -F × d
-ve sign indicates the lowering of energy of the system during expansion.
If a is the cross section area of the cylinder or piston, then,
W = \(-\frac{F}{a}\) × d × a
Now the pressure is P_ex = \(\frac{F}{a}\)
while volume change is, ΔV = d × a
∴ W = -P_ex × ΔV
If during the expansion, the volume changes from V₁ and V₂ then, ΔV = V₂ – V₁
∴ W= -P_ex(V₂ – V₁)
During compression, the work W is +ve, since the energy of the system is increased,
W = +P_ex(V₂ – V₁)

Question viii.
What are intensive properties? Explain why density is intensive property.
Answer:
(A) Intensive property : It is defined as a property of a system whose magnitude is independent of the amount of matter present in the system.
Explanation :

1. Intensive property is characteristic of the system, e.g., refractive index, density, viscosity, temperature, pressure, boiling point, melting point, freezing point of a pure liquid, surface tension, etc.
2. The intensive properties are not additive.

(B) Density is a ratio of two extensive properties namely, mass and volume. Since the ratio of two extensive properties represents an intensive property, density is an intensive property. It does not depend on the amount of a substance.

Question ix.
How much heat is evolved when 12 g of CO reacts with NO₂ ? The reaction is :
4 CO_(g) + 2 NO_2(g) → 4CO_2(g) + N_2(g), ∆H⁰ = -1200 kJ

4. Answer the following questions.

Question i.
Derive the expression for the maximum work.
Answer:

Consider n moles of an ideal gas enclosed in an ideal cylinder fitted with a massless and frictionless movable rigid piston. Let V be the volume of the gas at a pressure P and a temperature T.
If in an infinitesimal change pressure changes from P to P – dP and volume increases from V to V + dV. Then the work obtained is, dW = -(P-dP) dV
= -PdV + dPdV
Since dP.dV is negligibly small relative to PdV
dW= -PdV
Let the state of the system change from A(P₁, L₁) to B (P₂, V₂) isothermally and reversibly, at temperature T involving number of infinitesimal steps.

Then the total work or maximum work in the process is obtained by integrating above equation.

At constant temperature,

where n, P, V and T represent number of moles, pressure, volume and temperature respectively. For the process,
ΔU = 0, ΔH = 0.
The heat absorbed in reversible manner
Q_rev, is completely converted into work.
Q_rev = -W_max.
Hence work obtained is maximum.

Question ii.
Obtain the relatioship between ∆H and ∆U for gas phase reactions.
Answer:
Consider a reaction in which n₁ moles of gaseous reactant in initial state change to n₂ moles of gaseous product in the final state.
Let H₁, U₁, P₁, V₁ and H₂, U₂, P₂, V₂ represent enthalpies, internal energies, pressures and volumes in the initial and final states respectively then,

The heat of reaction is given by enthalpy change ΔH as,
ΔH = H₂ – H₁
By definition, H = U + PV
∴ H₁ = U₁ + P₁V₁ and H₂ = U₂+ P₂V₂
∴ ΔH = (U2 + P₂V₂) – (U₁ + P₁V₁)
= (U2 – U₁) + (P₂V₂ – P₁V₁)
Now, ΔU = U₂ – U₁
Since PV = nRT,
For initial state, P₁V₁= n₁RT
For final state, P₂V₂ = n₂RT
∴ P₂V₂ – P₁V₁ = n₂RT – n₁RT
= (n₂ – n₁) RT
= ΔnRT
where Δn

∴ ΔH = ΔU + ΔnRT
If Q_P and Q_V are the heats involved in the reaction at constant pressure and constant volume respectively, then since Q_P = ΔH and Q_V = ΔU.
∴ Q_P = Q_V = ΔnRT

Question iii.
State Hess’s law of constant heat summation. Illustrate with an example. State its applications.
Answer:
Statement of law of constant heat summation : It states that, the heat of a reaction or the enthalpy change in a chemical reaction depends upon initial state of reactants and final state of products and independent of the path by which the reaction is brought about (i.e. in single step or in series of steps).
OR
Heat of reaction is same whether it is carried out in one step or in several steps.
Explanation :
Consider the formation of CO_2(g).

Hess’s law treats thermochemical equations mathematically i.e., they can be added, subtracted or multiplied by numerical factors like algebraic equations.

Applications : Hess’s law is used for :

1. To calculate heat of formation, combustion, neutralisation, ionization, etc.
2. To calculate the heat of reactions which may not take place normally or directly.
3. To calculate heats of extremely slow or fast reactions.
4. To calculate enthalpies of reactants and products.

Question iv.
Although ∆S for the formation of two moles of water from H₂ and O₂ is -327 JK^-1, it is spontaneous. Explain. (Given ∆H for the reaction is -572 kJ).
Answer:
Given : ΔS= -327 JK^-1; ΔH = -572 kJ
ΔG = ΔH – TΔS, and ΔH << ΔS
∴ ΔG < 0, and hence the formation of H₂O_(l) is spontaneous.

Question v.
Obtain the relation between ∆G and ∆S_Total. Comment on spontaneity of the reaction.
Answer:
(i) Gibbs free energy, G is defined as,
G = H – TS
where H is the enthalpy, S is the entropy of the system at absolute temperature T.
Since H, T and S are state functions, G is a state function and a thermodynamic function.

(ii) At constant temperature and pressure, change in free energy ΔG for the system is represented as, ∆G = ∆H – T∆S

This is called Gibbs free energy equation for ∆G. In this ∆S is total entropy change, i.e., ∆S_Total.

(iii) The SI units of ∆G are J or kJ (or Jmol^-1 or kJmol^-1).
The c.g.s. units of ∆G are cal or kcal (or cal mol^-1 or kcal mol^-1.)

The second law explains the conditions of spontaneity as below :
(i) ∆S_total > 0 and ∆G < 0, the process is spontaneous.
(ii) ∆S_total < 0 and ∆G > 0, the process is nonspontaneous.
(iii) ∆S_total = 0 and ∆G = 0, the process is at equilibrium.

Question vi.
One mole of an ideal gas is compressed from 500 cm³ against a constant external pressure of 1.2 × 10⁵ Pa. The work involved in the process is 36.0 J. Calculate the final volume.
Answer:
Given : V₁ = 500 cm³ = 0.5 dm³;
P_ex = 1.2 × 10⁵ Pa = 1.2 bar; W= 36 J;
1 dm³ bar = 100 J; V₂ = ?
W = -P_ex (V₂ – V₁)
36 J = – 1.2 (V₂ – 0.5) dm³ bar
= -1.2 × 100 (V₂ – 0.5) J
∴ V₂ – 0.5 = \(\frac{-36}{1.2 \times 100}=-0.3\)
∴ V₂ = 0.5 -0.3 = 0.2 dm³ = 200 cm³
Ans. Final volume = 200 cm³.

Question vii.
Calculate the maximum work when 24g of O₂ are expanded isothermally and reversibly from the pressure of 1.6 bar to 1 bar at 298 K.
Answer:
Given : W₀₂ = 24 g, P₁ = 1.6 bar, P₂ = 1 bar
T = 298 K, W_max = ?

Question viii.
Calculate the work done in the decomposition of 132 g of NH₄NO₃ at 100 °C.
NH₄NO_3(s) → N₂O_(g) + 2 H₂O_(g)
State whether work is done on the system or by the system.
Answer:
NH4₄NO_3(s) → N₂O_(g) + 2 H₂O_(g)
m_NH4NO3 = 132 g; M_NH4NO3 = 80 g mol^-1
T = 273 + 100 = 373 K; Δn = ?
For the reaction,
Δn = Σn_{2 gaseous products} – Σn_{1 gaseous reactants}
= 3 – 0 = 3 mol
Since there is an increase in number of gaseous moles, the work is done by the system.
n_NH4NO3 = \(\frac{m_{\mathrm{NH}_{4} \mathrm{NO}_{3}}}{M_{\mathrm{NH}_{4} \mathrm{NO}_{3}}}\)
= \(\frac{132}{80}\)
= 1.65 mol
For 1 mol NH₄NO_3(s) Δn = 3 mol
∴ For 1.65 mol NH₄NO_3(s) Δn = 3 × 1.65 = 4.95 mol
W = -ΔnRT = -4.95 × 8.314 × 373
= – 15350 J
= – 15.35 kJ
Ans. Work is done by the system.
Work done = – 15.35 kJ

Question ix.
Calculate standard enthalpy of reaction,
Fe₂O_3(s) + 3CO_(g) → 2 Fe_(s) + 3CO_2(g),
from the following data.
∆_fH⁰(Fe₂O₃) = -824 kJ/mol,
∆_fH⁰(CO) = -110 kJ/mol,
∆_fH⁰(CO₂) = -393 kJ/mol
Answer:
Given : ∆_fH⁰Fe₂O₃ = -824 kJ/mol^-1;
∆_fH⁰(CO) = – 110 kJ mol^-1
∆_fH⁰(CO₂) = – 393 kJ/mol^-1; ∆_fH⁰ = ?
Required equation,
Fe₂O_3(s) + 3CO_(g) → 2Fe_(s) + 3CO_2(g)
∆H₁ = ? – (I)
Given equations :

= -(-824) -3 (-110) + 3(-393)
= 824 + 330 – 1179
∆_fH⁰ = -25 kJ
Ans. ∆_fH⁰ = -25 kJ

Question x.
For a certain reaction ∆H⁰ =219 kJ and ∆S⁰ = -21 J/K. Determine whether the reaction is spontaneous or nonspontaneous.
Answer:
Given : ∆H⁰ = 219 kJ; ∆S⁰ = -21 J/K, ∆G⁰ = ?
For standard state, T = 298 K
∆G⁰ = ∆H⁰ – T∆S⁰
= 219 – 298 × (-21) × 10^-3
= 219 + 6.258
= 225.3 kJ
Since ∆S < 0 and ∆G⁰ > 0, the reaction is non-spontaneous.

Question xi.
Determine whether the following reaction is spontaneous under standard state conditions.
2 H₂O_(l) + O_2(g) → 2H₂O_2(l)
if ∆H⁰ = 196 kJ, ∆S⁰ = -126 J/K
Does it have a cross-over temperature?
Answer:
Given : 2H₂O_(l) + O_2(g) → 2H₂O_2(l)
∆H⁰ = +196 kJ
∆S⁰ = -126 JK-1 =0.126 kj K-1
T= 298 K
∆G⁰ = ?
Cross over temperature = T = ?
∆G⁰ = ∆H⁰ – T∆S⁰
= 196 – 298 (-0.126)
= 196 + 37.55
= + 233.55 kJ
∵ ∆G⁰ > 0, the reaction is non-spontaneous.
∆H⁰ > 0, ∆S⁰ < 0,
Since at all temperatures, ∆G⁰ > 0, there is no cross over temperature.
Ans. The reaction is non-spontaneous.
There is no cross-over temperature for the reaction.

Question xii.
Calculate ∆U at 298 K for the reaction,
C₂H_4(g) + HCl_(g) → C₂H₅Cl_(g), ∆H = -72.3 kJ
How much PV work is done?
Answer:
Given : C₂H_4(g) + HCl_(g) → C₂H₅Cl_(g)
T = 298 K; ∆H = -72.3 kJ; PV = ?;
∆U = ?
∆n = Σn_{2gaseous products} – Σn_{1gaseous reactants}
= 1 – (1 + 1)= -1 mol
For PV work :
W = -∆nRT
= – (- 1) × 8.314 × 298
= 2478 J = 2.478 kJ
∆H = ∆U + ∆nRT
∴ ∆U = ∆H – ∆nRT
= – 72.3 – (-2.478)
= – 69.82 kJ
Ans. PV work = 2.478 kJ
∆U = -69.82 kJ

Question xiii.
Calculate the work done during synthesis of NH₃ in which volume changes from 8.0 dm3 to 4.0 dm³ at a constant external pressure of 43 bar. In what direction the work energy flows?
Answer:
Given : V₁ = 8.0 dm³; V₂ = 4.0 dm³; P_ex = 43 bar
W = ? What direction work energy flows ?
W = -P_ex(V₂ – V₁)
= -43 (4 – 8)
= 172 dm³ bar
= 172 × 100 J
= 17200 J
= 17.2 kJ
In this compression process, the work is done on the system and work energy flows into the system.

Question xiv.
Calculate the amount of work done in the
(a) oxidation of 1 mole HCl_(g) at 200 °C according to reaction.
4HCl_(g) + O_2(g) → 2 Cl_2(g) + 2 H₂O_(g)
(b) decomposition of one mole of NO at 300 °C for the reaction
2 NO_(g) → N_2(g) + O₂
Answer:
Given :
(a) 4HCl_(g) + O_2(g) → 2Cl_2(g) + 2H₂O_(g)
n_HCl = 1 mol; T = 273 + 200 = 473 K, W = ?
For 4 mol HCl ∆n = (2 + 2) – (4 + 1) = – 1 mol
∴ For 1 mol HCl ∆n = –\(\frac {1}{4}\) = -0.25 mol
W = -∆nRT = – (-0.25) × 8.314 × 473 = 983.11
(b) ∆n = (1 + 1) – 2 = 0 mol
W = -∆nRT = -(0) × 8.314 × 473 = 0
Ans. (a) W = 983.1 J
(b) W = 0.0 J

Question xv.
When 6.0 g of O₂ reacts with CIF as per
2CIF_(g) + O_2(g) → Cl₂O_(g) + OF_2(g)
The enthalpy change is 38.55 kJ. What is standard enthalpy of the reaction ?
Answer:
Given : The given reaction is for 1 mol O₂ or 32 g O₂.
∵ For 6.0 g O₂
∆ H⁰ = 38.55 kJ
∴ For 32 g O₂
∆ H⁰ = \(\frac{32 \times 38.55}{6}\)
= 205.6 kJ
Ans. ∆H⁰ = 205.6 kJ

Question xvi.
Calculate the standard enthalpy of formation of CH₃OH_(l) from the following data:
i. CH₃OH_(l) + \(\frac {3}{2}\) O_2(g) → CO_2(g) + 2H₂O_(l), ∆H⁰ = -726 kJ mol^-1
ii. C (Graphite) + O_2(g) → CO_2(g), ∆_cH⁰ = -393 kJ mol^-1
iii. H_2(g) + \(\frac {1}{2}\) O_2(g) → H₂O_(l), ∆_fH⁰ = -286 kJ mol^-1
Answer:


∴ ∆H⁰
= –\(\Delta H_{2}^{0}\) + \(\Delta H_{3}^{0}\) + 2∆\(\Delta H_{4}^{0}\)
= – (- 726) + (- 393) + 2(- 286)
= 726 – 393 – 572
= – 239 kJ mol^-1
Ans. Standard enthalpy of formation = ∆_fH⁰= -239 kJ mol^-1

Question xvii.
Calculate ∆H0 for the following reaction at 298 K
H₂B₄O_7(s) + H₂O_(l) → 4HBO_2(aq)
i. 2H₃BO_3(aq) → B₂O_3(s) + 3H₂O_(l), ∆H⁰ = 14.4 kJ mol^-1
ii. H₃BO_3(aq) → HBO_2(aq) + H₂O_(l), ∆H⁰ = -0.02 kJ mol^-1
iii. H₂B₄O_7(s) → 2B₂O_3(s) + H₂O_(l), ∆H⁰ = 17.3 kJ mol^-1
Answer:
Given equations :
i. 2H₃BO_3(aq) → B₂O_3(s) + 3H₂O_(l), ……….(i)
∆H⁰ = 14.4 kJ mol^-1
ii. H₃BO_3(aq) → HBO_2(aq) + H₂O_(l) ……….(ii)
∆H⁰ = -0.02 kJ mol^-1
iii. H₂B₄O_7(s) → 2B₂O_3(s) + H₂O_(l), ……….(iii)
∆H⁰ = 17.3 kJ mol^-1
Required equation :
(iv) H₂B₄O_7(s) + H₂O_(l) → 4HBO_2(aq) ……. (iv)
\(\Delta H_{4}^{0}=?\)
To obtain eq. (iv) add 4 times equation (ii) and eq.
(iii) and subtract 2 times equation (i).
∴ eq. (iv) = 4 eq. (ii) + eq. (iii) – 2eq. (i)
∴ \(\Delta H_{4}^{0}=4 \Delta H_{2}^{0}+\Delta H_{3}^{0}-2 \Delta H_{1}^{0}\)
= 4(-0.02) + 17.3 – 2(14.4)
= -0.08 + 17.3 – 28.8
= -11.58 kJ
∴ Enthalpy change for the reaction
= ∆_rH⁰ = -11.58 kJ
Ans. ∆_rH⁰ for the given reaction = -11.58 kJ

Question xviii.
Calculate the total heat required (a) to melt 180 g of ice at 0 °C, (b) heat it to 100 °C and then (c) vapourise it at that temperature. Given ∆_fusH_(ice) = 6.01 kJ mol^-1 at 0 °C, ∆_vapH_(H2O) = 40.7 kJ mol^-1 at 100 °C specific heat of water is 4.18 J g^-1 K^-1.
Answer:
Given : Mass of ice = m = 180 g
T₁ = 273 + 0 °C = 273 K
T₂ = 273 + 100 °C = 373 K
∆_fusH_(ice) = ∆_fusH_(H2O)(s) = 6.01 kJ mol^-1
∆_vapH_H2O(l) = 40.7 kJ mol^-1
Specific heat of water = C = 4.18 J g^-1 K^-1
For converting 180 g ice into vapour, ∆ H_Total = ?
Number of moles of H₂O = \(\frac {180}{18}\) = 10 mol
The total process can be represented as,

(i) ∆H₁ = ∆_fusH = 10 mol × 6.01 kJ mol^-1
= 60.1 kJ
(ii) When the temperature of water is raised from 0 °C to 100 °C (i.e., 273 K to 373 K), then
∆ H₂ = m × C × ∆T
= m × C × (T2 – T1)
= 180 g × 4.18 Jg^-1K^-1 × (373 – 273) × 10^-3 kJ = 75.24 kJ
∆ H₃ = ∆_vapH = 10 mol × 40.7 kJ mol^-1 = 407 kJ
Hence total enthalpy change,
∆ H_Total = ∆H₁ + ∆H₂ + ∆H₃
= 60.1 + 75.24 + 407
= 542.34 kJ
Ans. Total heat required = 542.34 kJ

Question xix.
The enthalpy change for the reaction,
C₂H_4(g) + H_2(g) → C₂H_6(g)
is -620 J when 100 ml of ethylene and 100 mL of H₂ react at 1 bar pressure. Calculate the pressure volume type of work and ∆U for the reaction.
Answer:
Given :
\(\begin{aligned}
&\mathrm{C}_{2} \mathrm{H}_{4(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6(\mathrm{~g})} \\
&100 \mathrm{~mL} \quad 100 \mathrm{ml} \quad 100 \mathrm{ml}
\end{aligned}\)
∆H = – 620 J; V_C2H4 = 100 mL; V_H2 = 100 mL
P_ex= 1 bar; W=?; ∆U = ?
∆V = 100 – (100 + 100) = -100 mL = -0.1 dm³
W = -P_ex(V₂ – V₁)
= -P_ex × ∆V
= -1 × (-0.1)
= 0.1 dm³ bar
= 0.1 × 100 J
= +10 J
∆H = ∆U + P∆V
∴ ∆U = ∆H – P∆V = -620 – (+10) = -610 J
Ans. W = +10 J; ∆U = -610 J

Question xx.
Calculate the work done and comment on whether work is done on or by the system for the decomposition of 2 moles of NH₄NO₃ at 100 °C
NH₄NO_3(s) → N₂O_(g) + 2H₂O_(g)
Answer:
Given : NH₄NO_3(s) → N₂O_(g) + 2H₂O_(g)
n_NH4NO3 = 2 mol; T = 273 + 100 = 373 K
W = ? Comment on work = ?
∆n_reaction = (1 + 2) – 0 = 3 mol
∵ For 1 mol of NH₄NO₃ ∆n_reaction = 3 mol
∴ For 2 mol of NH₄NO₃ ∆n_reaction = 6 mol
Due to 6 moles of gaseous products from 2 mol NH₄NO₃, there is work of expansion, hence work is done by the system.
W = -∆nRT
= – 6 × 8.314 × 373 = -18606 J
= -18.606 kJ
Ans. Work is done by the system.
W= -18.606 kJ

12th Chemistry Digest Chapter 4 Chemical Thermodynamics Intext Questions and Answers

(Textbook page No. 73)

Question 1.
Under what conditions ∆H = ∆U ?
Answer:
(a) ∆H = ∆U + P∆V
when ∆V = 0, ∆H = ∆U
(b) ∆H = ∆U + ∆nRT
when ∆n = 0, ∆H = ∆U

Try this… (Textbook page No. 71)

Question 1.
25 kJ of work is done on the system and it releases 10 kJ of heat. What is ∆U?
Answer:
W = 25 kJ; Q= -10 kJ
∆U = Q + W = -10 + 25
∆U = + 15 kJ

Try this… (Textbook page No. 75)

Question 1.
For KCl, ∆_LH = 699 kJ/mol^-1 and ∆_hydH = -681.8 kJ/mol^-1. What will be its enthalpy of solution?
Answer:
Enthalpy of solution :
∆_solnH = ∆_LH + ∆_hydH
= 699 + (-681.8)
∆_solnH = +17.2 kJ mol^-1

Try this… (Textbook page No. 76)

Question 1.
Given the thermochemical equation,
C₂H_2(g) + \(\frac {5}{2}\) O_2(g) → 2CO_2(g)+ H₂O_(l), ∆_rH⁰ = -1300 kJ
Write thermochemical equations when
i. Coefficients of substances are multiplied by 2.
ii. equation is reversed.
Answer:
(i) 2C₂H_2(g) + 5O_2(g) → 4CO_2(g) + 2H₂O_(l)
∆_rH⁰ = -2 × 1300 kJ
= – 2600 kJ
(ii) 2CO_2(g) + H₂O_(l) → C₂H_2(g) + \(\frac {5}{2}\)O_2(g)
∆_rH⁰ = +1300 KJ

Try this… (Textbook page No. 78)

Question 1.
(i) Write thermochemical equation for complete oxidation of one mole of H_2(g). Standard enthalpy change of the reaction is -286 kJ.
(ii) Is the value -286 kJ, enthalpy of formation or enthalpy of combustion or both? Explain.
Answer:
(i) H_2(g) + \(\frac {1}{2}\)O_2(g) → H₂O_(l) ∆_cH⁰ = -286 KJ mol^-1
(ii) The value -286 kJ is the standard enthalpy of formation of H₂O_(l) or standard enthalpy of combustion of H_2(g).

Question 2.
Write equation for bond enthalpy of Cl-Cl bond in Cl₂ molecule ∆_rH⁰ for dissociation of Cl₂ molecule is 242.7 kJ.
Answer:
Equation for bond enthalpy :
Cl_2(g) → 2Cl_(g) ∆_rH⁰ = 242.7 kJ mol^-1
∴ Bond enthalpy of Cl₂ = 242.7 kJ mol^-1

Try this… (Textbook page No. 82)

Question 1.
State whether ∆S is positive, negative or zero for the following reactions.
i. 2H_2(g) + O_2(g) → 2H₂O_(l)
ii. CaCO_3(s) → CaO_(s) + CO_2(g)
Answer:
(i) 2H_2(g) + O_2(g) → 2H₂O_(l)
Since the system is converted from gaseous state to a liquid state, the disorder is decreased, hence ∆S < O (negative).

(ii) CaCO_3(s) → CaO_(s) + CO_2(g)
Since molecules of solid CaCO₃ break giving gaseous CO₂, disorder is increased hence ∆S > O (positive).

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 8 Electrostatics Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 8 Electrostatics

1. Choose the correct option

i) A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge, potential, capacitance respectively are
(A) Constant, decreases, decreases
(B) Increases, decreases, decreases
(C) Constant, decreases, increases
(D) Constant, increases, decreases
Answer:
(A) Constant, decreases, decreases

ii) A slab of material of dielectric constant k has the same area A as the plates of a parallel plate capacitor and has thickness (3/4d), where d is the separation of the plates. The change in capacitance when the slab is inserted between the plates is

Answer:
(D) C = \(\frac{\varepsilon_{0} A}{d}\left(\frac{4 k}{k+3}\right)\)

iii) Energy stored in a capacitor and dissipated during charging a capacitor bear a ratio.
(A) 1 : 1
(B) 1 : 2
(C) 2 : 1
(D) 1 : 3
Answer:
(A) 1 : 1

iv) Charge + q and -q are placed at points A and B respectively which are distance 2L apart. C is the mid point of A and B. The work done in moving a charge +Q along the semicircle CRD as shown in the figure below is

(C) \(\frac{q Q}{6 \pi \varepsilon_{0} L}\)
(D) \(\frac{-q Q}{6 \pi \varepsilon_{0} L}\)
Answer:
(A) \(-\frac{Q q_{1}}{6 \pi \varepsilon_{0} L}\)

v) A parallel plate capacitor has circular plates of radius 8 cm and plate separation 1mm. What will be the charge on the plates if a potential difference of 100 V is applied?
(A) 1.78 × 10^-8 C
(B) 1.78 × 10^-5 C
(C) 4.3 × 10⁴ C
(D) 2 × 10^-9 C
Answer:
(A) 1.78 × 10^-8 C

2. Answer in brief.

i) A charge q is moved from a point A above a dipole of dipole moment p to a point B below the dipole in equitorial plane without acceleration. Find the work done in this process.

Answer:
The equatorial plane of an electric dipole is an equipotential with V = 0. Therefore, the no work is done in moving a charge between two points in the equatorial plane of a dipole.

ii) If the difference between the radii of the two spheres of a spherical capacitor is increased, state whether the capacitance will increase or decrease.
Answer:
The capacitance of a spherical capacitor is C = 4πε₀\(\left(\frac{a b}{b-a}\right)\) where a and b are the radii of the concentric inner and outer conducting shells. Hence, the capacitance decreases if the difference b – a is increased.

iii) A metal plate is introduced between the plates of a charged parallel plate capacitor. What is its effect on the capacitance of the capacitor?
Answer:
Suppose the parallel-plate capacitor has capacitance C₀, plates of area A and separation d. Assume the metal sheet introduced has the same area A.

Case (1) : Finite thickness t. Free electrons in the sheet will migrate towards the positive plate of the capacitor. Then, the metal sheet is attracted towards whichever capacitor plate is closest and gets stuck to it, so that its potential is the same as that of that plate. The gap between the capacitor plates is reduced to d – t, so that the capacitance increases.

Case (2) : Negligible thickness. The thin metal sheet divides the gap into two of thicknesses d₁ and d₁ of capacitances C₁ = ε₀A/d₁ and C₂ = ε₀A/d₂ in series.
Their effective capacitance is
C = \(\frac{C_{1} C_{2}}{C_{1}+C_{2}}=\frac{\varepsilon_{0} A}{d_{1}+d_{2}}=\frac{\varepsilon_{0} A}{d}\) = C₀
i.e., the capacitance remains unchanged.

iv) The safest way to protect yourself from lightening is to be inside a car. Justify.
Answer:
There is danger of lightning strikes during a thunderstorm. Because trees are taller than people and therefore closer to the clouds above, they are more likely to get hit by lightnings. Similarly, a person standing in open ground is the tallest object and more likely to get hit by a lightning. But car with a metal body is an almost ideal Faraday cage. When a car is struck by lightning, the charge flows on the outside surface of the car to the ground but the electric field inside remains zero. This leaves the passengers inside unharmed.

v) A spherical shell of radius b with charge Q is expanded to a radius a. Find the work done by the electrical forces in the process.
Answer:
Consider a spherical conducting shell of radius r placed in a medium of permittivity ε. The mechanical force per unit area on the charged conductor is
f = \(\frac{F}{d S}=\frac{\sigma^{2}}{2 \varepsilon}\)
where a is the surface charge density on the conductor. Given the charge on the spherical shell is Q, (σ = Q/πr². The force acts outward, normal to the surface.

Suppose the force displaces a charged area element adS through a small distance dx, then the work done by the force is
dW = Fdx = (\(\frac{\sigma^{2}}{2 \varepsilon}\) dS) dx
During the displacement, the area element sweeps out a volume dV = dS ∙ dx.

Therefore, the work done by the force in expanding the shell from radius r = b to r = a is

This gives the required expression for the work done.

Question 3.
A dipole with its charges, -q and + q located at the points (0, -b, 0) and (0 +b, 0) is present in a uniform electric field E whose equipotential surfaces are planes parallel to the YZ planes.
(a) What is the direction of the electric field E?
(b) How much torque would the dipole experience in this field?
Answer:
(a) Given, the equipotentials of the external uniform electric field are planes parallel to the yz plane, the electric field \(\vec{E}=\pm E \hat{\mathrm{i}}\) that is, \(\overrightarrow{E}\) is parallel to the x-axis.

(b) From above figure, the dipole moment, \(\vec{p}=q(2 b) \hat{\mathrm{j}}\)
The torque on this dipole,

So that the magnitude of the torque is τ = 2qbE.
If \(\vec{E}\) is in the direction of the + x-axis, the torque \(\vec{\tau}\) is in the direction of – z-axis, while if \(\vec{E}\) is in the direction of the -x-axis, the torque \(\vec{\tau}\) is in the direction of + z-axis.

Question 4.
Three charges – q, + Q and – q are placed at equal distance on straight line. If the potential energy of the system of the three charges is zero, then what is the ratio of Q : q?
Answer:

In the above figure, the line joining the charges is shown as the x-axis with the origin at the + Q charge. Let q₁ = +Q, and q₂ = q₃ = – q. Let the two – q charges be at (- a, 0) and (a, 0), since the charges are given to be equidistant.
∴ r₂₁ = r₃₁ = a and r₃₂ = 2a
The total potential energy of the system of three charges is

This gives the required ratio.

Question 5.
A capacitor has some dielectric between its plates and the capacitor is connected to a DC source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, the electric field, charge stored and voltage will increase, decrease or remain constant.
Answer:
Assume a parallel-plate capacitor, of plate area A and plate separation d is filled with a dielectric of relative permittivity (dielectric constant) k. Its capacitance is C = \(\frac{k \varepsilon_{0} A}{d}\) …………. (1)
If it is charged to a voltage (potential) V, the charge on its plates is Q = CV.

Since the battery is disconnected after it is charged, the charge Q on its plates, and consequently the product CV, remain unchanged.

On removing the dielectric completely, its capacitance becomes from Eq. (1),
C’ = \(\frac{\varepsilon_{0} A}{d}=\frac{1}{k} C\) ……………. (2)
that is, its capacitance decreases by the factor k. Since C’V’ = CV, its new voltage is
V’ = \(\frac{C}{C^{\prime}}\) V = kV …………… (3)

so that its voltage increases by the factor k. The stored potential energy, U = \(\frac{1}{2}\) QV, so that Q remaining constant, U increases by the factor k. The electric field, E = V/ d, so that E also increases by a factor k.

Question 6.
Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors C₁ and C₂ with their capacitances in the ratio 1 : 2, so that the energy stored in these two cases becomes the same.
Answer:

This gives the required ratio.

Question 7.
Two charges of magnitudes -4Q and +2Q are located at points (2a, 0) and (5a, 0) respectively. What is the electric flux due to these charges through a sphere of radius 4a with its centre at the origin?
Answer:
The sphere of radius 4a encloses only the negative charge Q₁ = -4Q. The positive charge Q₂ = +2Q being located at a distance of 5a from the origin is outside the sphere. Only a part of the electric flux lines originating at Q₂ enters the sphere and exits entirely at other points. Hence, the electric flux through the sphere is only due to Q₁.
Therefore, the net electric flux through the sphere = \(\frac{Q_{1}}{\varepsilon_{0}}=\frac{-4 Q}{\varepsilon_{0}}\) . The minus sign shows that the flux is directed into the sphere, but not radially since the sphere is not centred on Q₁.

Question 8.
A 6 µF capacitor is charged by a 300 V supply. It is then disconnected from the supply and is connected to another uncharged 3µF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation ?
Answer:
Data: C = 6 µF = 6 × 10^-6 F = C₁, V = 300 V
C₂ = 3 µF
The electrostatic energy in the capacitor
= \(\frac{1}{2}\)Cv² = \(\frac{1}{2}\)(6 × 10^-6)(300)²
= 3 × 10^-6 × 9 × 10⁴ = 0.27J
The charge on this capacitor,
Q = CV = (6 × 10^-6)(300) = 1.8 mC
When two capacitors of capacitances C₁ and C₂ are connected in parallel, the equivalent capacitance C
= C₁ + C₂ = 6 + 3 = 9 µF
= 9 × 10^-6F
By conservation of charge, Q = 1.8 C.
∴ The energy of the system = \(\frac{Q^{2}}{2 C}\)
= \(\frac{\left(1.8 \times 10^{-3}\right)^{2}}{2\left(9 \times 10^{-6}\right)}=\frac{18 \times 10^{-8}}{10^{-6}}\) = 0.18 J
The energy lost = 0.27 – 0.18 = 0.09 J

Question 9.
One hundred twenty five small liquid drops, each carrying a charge of 0.5 µC and each of diameter 0.1 m form a bigger drop. Calculate the potential at the surface of the bigger drop.
Answer:
Data : n = 125, q = 0.5 × 10^-6 C, d = 0.1 m
The radius of each small drop, r = d/2 = 0.05 m
The volume of the larger drop being equal to the volume of the n smaller drops, the radius of the larger drop is
R = \(\sqrt[3]{n} r\) = \(\sqrt[3]{125}\) (0.05) = 5 × 0.05 = 0.25 m
The charge on the larger drop,
Q = nq = 125 × (0.5 × 10^-6) C
∴ The electric potential of the surface of the larger drop,
V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{R}\) = (9 × 10⁹) × \(\frac{125 \times\left(0.5 \times 10^{-6}\right)}{0.25}\)
= 9 × 125 × 2 × 10³ = 2.25 × 10⁶ V

Question 10.
The dipole moment of a water molecule is 6.3 × 10^-30 Cm. A sample of water contains 1021 molecules, whose dipole moments are all oriented in an electric field of strength 2.5 × 10⁵ N /C. Calculate the work to be done to rotate the dipoles from their initial orientation θ₁ = 0 to one in which all the dipoles are perpendicular to the field, θ₂ = 90°.
Answer:
Data: p = 6.3 × 10^-30 C∙m, N = 10²¹ molecules,
E = 2.5 × 10⁵ N/C, θ₀ = θ₁ = 0°, θ = θ₂ = 90°
W = pE(cos θ₀ – cos θ)
The total work required to orient N dipoles is
W = NpE(cos θ₁ – cos θ₂)
=(10²¹)(6.3 × 10^-30)(2.5 × 10⁵)
= 15.75 × 10^-4 J = 1.575 mJ

Question 11.
A charge 6 µC is placed at the origin and another charge -5 µC is placed on the y axis at a position A (0, 6.0) m.

(a) Calculate the total electric potential at the point P whose coordinates are (8.0, 0) m
(b) Calculate the work done to bring a proton from infinity to the point P. What is the significance of the sign of the work done ?
Answer:
Data : q₁ = 6 × 10^-6 C, q₂ = -5 × 10^-6 C,
A ≡ (0, 6.0 m), P ≡ (8.0 m, 0), r₁ = OP = 8 m, q = e = 1.6 × 10^-19C, 1/4πε₀ = 9 × 10⁹ N∙m²/C²
r₂ = AP = \(\sqrt{(8-0)^{2}+(0-6)^{2}}\) = \(\sqrt{64+36}\) = 10 m

(a) The net electric potential at P due to the system of two charges is

(b) The electric potential V at the point P is the negative of the work done per unit charge, by the electric field of the system of the charges q₁ and q₂, in bringing a test charge from infinity to that point.
V = \(-\frac{W}{q_{0}}\)
∴ W = -qV= -(1.6 × 10^-19)(2.25 × 10³)
= -3.6 × 10^-16 J= -2.25 keV
That is, in bringing the positively charged proton from a point of lower potential to a point of higher potential, the work done by the electric field on it is negative, which means that an external agent must bring the proton against the electric field of the system of the two source charges.

[Note : The potential V at a point is the work done per unit charge (W_ext) by an external agent in bringing a test charge from infinity to that point. In the above case, the work done by an external agent will be positive. The question does not specify this.]

Question 12.
In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10^-3 m² and the separation between the plates is 2 mm.
i) Calculate the capacitance of the capacitor.
ii) If this capacitor is connected to 100 V supply, what would be the charge on each plate?
iii) How would charge on the plates be affected if a 2 mm thick mica sheet of k = 6 is inserted between the plates while the voltage supply remains connected ?
Answer:
Data: k = 1(air), A = 6 × 10^-3 m², d = 2
mm = 2 × 10^-3 m,V = 100V, t = 2 mm = d, k₁ = 6,
ε₀ = 8.85 × 10^-12 F/m
(i) The capacitance of the air capacitor, C₀ = \(\frac{\varepsilon_{0} A}{d}\)
= \(\frac{\left(8.85 \times 10^{-12}\right)\left(6 \times 10^{-3}\right)}{2 \times 10^{-3}}\)
= 26.55 × 10^-12 F = 26.55 pF

(ii) Q₀ = C₀V = (26.55 × 10^-12)(100)
= 26.55 × 10^-10 C = 2.655 nC

(iii) The dielectric of relative permittivity k₁ completely fills the space between the plates (∵t = d), so that the new capacitance is C = k₁C₀.
With the supply still connected, V remains the same.
∴ Q = CV = kC₀V = kQ₀ =6(2.655 nF) = 15.93 nC
Therefore, the charge on the plates increases.
[Note: Ck₁C₀ = 6(26.55 pF)= 159.3 pF.]

Question 13.
Find the equivalent capacitance between P and Q. Given, area of each plate = A and separation between plates = d.

Answer:
(i) The capacitor in figure is a series combination of three capacitors of plate separations d/3 and plate areas A, with C₁ filled with air (k₁ = 1), C₂ filled with dielectric of k₂ = 3 and C₃ filled with dielectric of k₃ = 6

(ii) In figure, a series combination of two capacitors C₂(k₂ = 3) and C₃(k₃ = 6), of plate areas A/2 and plate separations d/2, is in parallel with a capacitor C₁ (k₁ = 4) of plate area A/2 and plate separation d.

12th Physics Digest Chapter 8 Electrostatics Intext Questions and Answers

Can you recall (Textbook Page No. 188)

Question 1.
What is gravitational Potential ?
Answer:
We measure the gravitational potential energy U of a body (1) by assigning U = 0 for a reference configuration (such as the body at a reference level) (2) then equating U to the work W done by an external force to move the body up or down from that level to a point. We then define gravitational potential of the point as gravitational potential energy per unit mass of the body.

We follow the same procedure with the electric force, which is also a conservative force with the only difference that while the gravitational force is always attractive, electric force can be attractive (for unlike charges) or repulsive (for like charges).

Remember this (Textbook Page No. 191)

Question 1.
Due to a single charge at a distance r, Force (F ) α 1/r², Electric field (E ) α 1/r² but Potential (V) α 1/ r.
Answer:
At a point a distance r from an isolated point charge, the force F on a point charge and the electric field E both vary as 1/r², while the potential energy U of a point charge and the electric potential V at the point both vary as 1/r.

Use your brain power (Textbook Page No.194)

Question 1.
Is electrostatic potential necessarily zero at a point where electric field strength is zero? Justify.
Answer:
Electric potential is a scalar quantity while electric field intensity is a vector quantity. When we add potentials at a point due two or more point charges, the operation is simple scalar addition along with the sign of V, determined by the sign of the q that produces V. At a point, the net field is the vector sum of the fields due to the individual charges. Midway between the two charges of an electric dipole, the potentials due to the two charges are equal in magnitude but opposite in sign, and thus add up to zero. But the electric fields due to the charges are equal in magnitude and direction-towards the negative charge-so that the net field there is not zero. But midway between two like charges of equal magnitudes, the potentials are equal in magnitude and have the same sign, so that the net potential is nonzero. However, the fields due to the two equal like charges are equal in magnitude but opposite in direction, and thus vectorially add up to zero.

Do you know (Textbook Page No. 203)

Question 1.
If we apply a large enough electric field, we can ionize the atoms and create a condition for electric charge to flow like a conductor. The fields required for the breakdown of dielectric is called dielectric strength.
Answer:
In a sufficiently strong electric field, the molecules of a dielectric material become ionized, allowing flow of charge. The insulating properties of the dielectric breaks down, permanently or temporarily, and the phenomenon is called dielectric breakdown. During dielectric breakdown, electrical discharge through the dielectric follows random-path patterns like tree branches, called a Lichtenberg figure. Dielectric strength is the voltage that an insulating material can withstand before breakdown occurs. It usually depends on the thickness of the material. It is expressed in kV/mm. For example, the dielectric strengths of air, polystyrene and mica in kV/mm are 3, 20 and 118. Higher dielectric strength corresponds to better insulation properties.

Remember this (Textbook Page No. 205)

Question 1.
Series combination is used when a high voltage is to be divided on several capacitors. Capacitor with minimum capacitance has the maximum potential difference between the plates.
Answer:
Series combination of capacitors

1. Equivalent capacitance is less than the smallest capacitance in series. For several capacitors of given capacitances, the equivalent capacitance of their series combination is minimum.
2. All capacitors in the combination have the same charge but their potential differences are in the inverse ratio of their capacitances.
3. Series combination of capacitors is sometimes used when a high voltage, which exceeds the breakdown voltage of a single capacitor, is to be divided on more than one capacitors. Capacitive voltage dividers are only useful in AC circuits, since capacitors do not pass DC signals.

Parallel combination of capacitors

1. For several capacitors of given capacitances, the equivalent capacitance of their parallel combination is maximum.
2. The same voltage is applied to all capacitors in the combination, but the charge stored in the combination is distributed in proportion to their capacitances. The maximum rated voltage of a parallel combination is only as high as the lowest voltage rating of all the capacitors used. That is, if several capacitors rated at 500 V are connected in parallel to a capacitor rated at 100 V, the maximum voltage rating of the capacitor bank is only 100 V.
3. Parallel combination of capacitors is used when a large capacitance is required, i.e., a large charge is to be stored, at a small potential difference.

Remember this (Textbook Page No. 207)

Question 1.
(1) If there are n parallel plates then there will be (n-1) capacitors, hence
C = (n – 1) \(\frac{A \varepsilon_{0}}{d}\)
(2) For a spherical capacitor, consisting of two concentric spherical conducting shells with inner and outer radii as a and b respectively, the capacitance C is given by
C = 4πε₀\(\left(\frac{a b}{b-a}\right)\)
(3) For a cylindrical capacitor, consisting of two coaxial cylindrical shells with radii of the inner and outer cylinders as a and b, and length l, the capacitance C is given by
C = \(\frac{2 \pi \varepsilon_{0} \ell}{\log _{e} \frac{b}{a}}\)
Answer:
1. Stacking together n identical conducting plates equally spaced and with alternate plates connected to two points P and Q, forms a parallel combination of n -1 identical capacitors between P and Q. Then, the capacitance between the points is (n – 1) times the capacitance between any two adjacent plates.

2. A cylindrical capacitor consists of a solid cylindrical conductor of radius a is surrounded by coaxial cylindrical shell of inner radius b. The length of both cylinders is L, such that L is much larger than b – a, the separation of the cylinders, so that edge effects can be ignored. The capacitance of the capacitor is C = \(\frac{2 \pi \varepsilon_{0} L}{\log _{e}(b / a)}\).

The capacitance depends only on the geometrical factors, L, a, and b, as for a parallel-plate capacitor.

3. A spherical capacitor which consists of two concentric spherical shells of radii a and b. The capacitance of the capacitor is C = 4πε₀\(\left(\frac{a b}{b-a}\right)\)

Again, the capacitance depends only on the geometrical factors, a and b.

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 7 Wave Optics Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 7 Wave Optics

1. Choose the correct option.

i) Which of the following phenomenon proves that light is a transverse wave?
(A) reflection
(B) interference
(C) diffraction
(D) polarization
Answer:
(D) polarization

ii) Which property of light does not change when it travels from one medium to another?
(A) velocity
(B) wavelength
(C) amplitude
(D) frequency
Answer:
(D) frequency

iii) When unpolarized light is passed through a polarizer, its intensity
(A) increases
(B) decreases
(C) remains unchanged
(D) depends on the orientation of the polarizer
Answer:
(B) decreases

iv) In Young’s double slit experiment, the two coherent sources have different intensities. If the ratio of maximum intensity to the minimum intensity in the interference pattern produced is 25:1. What was the ratio of intensities of the two sources?
(A) 5:1
(B) 25:1
(C) 3:2
(D) 9:4
Answer:
(D) 9:4

v) In Young’s double slit experiment, a thin uniform sheet of glass is kept in front of the two slits, parallel to the screen having the slits. The resulting interference pattern will satisfy
(A) The interference pattern will remain unchanged
(B) The fringe width will decrease
(C) The fringe width will increase
(D) The fringes will shift.
Answer:
(A) The interference pattern will remain unchanged

2. Answer in brief.

i) What are primary and secondary sources of light?
Answer:
(1) Primary sources of light: The sources that emit light on their own are called primary sources. This emission of light may be due to
(a) the high temperature of the source, e.g., the Sun, the stars, objects heated to high temperature, a flame, etc.
(b) the effect of current being passed through the source, e.g., tubelight, TV, etc.
(c) chemical or nuclear reactions taking place in the source, e.g., firecrackers, nuclear energy generators, etc.

(2) Secondary sources of light: Some sources are not self luminous, i.e., they do not emit light on their own, but reflect or scatter the light incident on them. Such sources of light are called secondary sources, e.g. the moon, the planets, objects such as humans, animals, plants, etc. These objects are visible due to reflected light.
Many of the sources that we see around are secondary sources and most of them are extended sources.

ii) What is a wavefront? How is it related to rays of light? What is the shape of the wavefront at a point far away from the source of light?
Answer:
Wavefront or wave surface : The locus of all points where waves starting simultaneously from a source reach at the same instant of time and hence the particles at the points oscillate with the same phase, is called a wavefront or wave surface.

Consider a point source of light O in a homogeneous isotropic medium in which the speed of light is v. The source emits light in all directions. In time t, the disturbance (light energy) from the source, covers a distance vt in all directions, i.e., it reaches out to all points which are at a distance vt from the point source. The locus of these points which are in the same phase is the surface of a sphere with the centre O and radius vt. It is a spherical wavefront.

In a given medium, a set of straight lines can be drawn which are perpendicular to the wavefront. According to Huygens, these straight lines are the rays of light. Thus, rays are always normal to the wavefront. In the case of a spherical wavefront, the rays are radial.

If a wavefront has travelled a large distance away from the source, a small portion of this wavefront appears to be plane. This part is a plane wavefront.

iii) Why are multiple colours observed over a thin film of oil floating on water? Explain with the help of a diagram.
Answer:
Several phenomena that we come across in our day to day life are caused by interference and diffraction of light. These are the vigorous colours of soap bubbles as well as those seen in a thin oil film on the surface of water, the bright colours of butterflies and peacocks etc. Most of these colours are not due to pigments which absorb specific colours but are due to interference of light waves that are reflected by different layers.

Interference due to a thin film:

The brilliant colours of soap bubbles and thin films on the surface of water are due to the interference of light waves reflected from the upper and lower surfaces of the film. The two rays have a path difference which depends on the point on the film that is being viewed. This is shown in above figure.

The incident wave gets partially reflected from upper surface as shown by ray AE. The rest of the light gets refracted and travels along AB. At B it again gets partially reflected and travels along BC. At C it refracts into air and travels along CF. The parallel rays AE and CF have a phase difference due to their different path lengths in different media. As can be seen from the figure, the phase difference depends on the angle of incidence θ₁, i.e., the angle of incidence at the top surface which is the angle of viewing, and also on the wavelength of the light as the refractive index of the material of the thin film depends on it. The two waves propagating along AE and CF interfere producing maxima and minima for different colours at different angles of viewing. One sees different colours when the film is viewed at different angles.

As the reflection is from the denser boundary, there is an additional phase difference of π radians (or an
additional path difference λ). This should be taken into account for mathematical analysis.

iv) In Young’s double slit experiment what will we observe on the screen when white light is incident on the slits but one slit is covered with a red filter and the other with a violet filter? Give reasons for your answer.
Answer:
In Young’s double-slit experiment, when white light is incident on the slits and one of the slit is covered with a red filter, the light passing through this slit will emerge as the light having red colour. The other slit which is covered with a violet filter, will give light having violet colour as emergent light. The interference fringes will involve mixing of red and violet colours. At some points, fringes will be red if constructive interference occurs for red colour and destructive interference occurs for violet colour. At some points, fringes will be violet if constructive interference occurs for violet colour and destructive interference occurs for red colour. The central fringe will be bright with mixing of red and violet colours.

v) Explain what is optical path length. How is it different from actual path length?
Answer:
Consider, a light wave of angular frequency ω and wave vector k travelling through vacuum along the x-direction. The phase of this wave is (kx-ωt). The speed of light in vacuum is c and that in medium is v.
k = \(\frac{2 \pi}{\lambda}\) = \(\frac{2 \pi v}{v \lambda}\) = \(\frac{\omega}{v}\) as ω = 2πv and v = vλ, where v is the frequency of light.

If the wave travels a distance ∆x, its phase changes by ∆φ = k∆x = ω∆x/v.
Similarly, if the wave is travelling in vacuum, k = ω/c and ∆φ = ω∆x/c
Now, consider a wave travelling a distance ∆x in the medium, the phase difference generated is,
∆φ’ = k’∆x = ωn∆x/c = ω∆x’/c … (1)
where ∆x’ = n∆x … (2)
The distance n∆x is called the optical path length of the light in the medium; it is the distance the light would have travelled in the same time t in vacuum (with the speed c).

The optical path length in a medium is the corresonding path in vacuum that the light travels in the same time as it takes in the given medium.

Thus, a distance d travelled in a medium of refractive index n introduces a path difference = nd – d = d(n – 1) over a ray travelling equal distance through vacuum.

Question 3.
Derive the laws of reflection of light using Huygens’ principle.
Answer:
Consider a plane wavefront AB of monochromatic light propagating in the direction A’A incident obliquely at an angle i on a plane refracting surface MN. This plane refracting surface MN separates two uniform and optically transparent mediums.
Let v₁ and v₂ be the speeds of light in medium 1 (say, a rarer medium) and medium 2 (a denser medium) respectively.

When the wavefront reaches MN at point A at t = 0, A becomes a secondary source and emits secondary waves in the second medium, while ray B’B reaches the surface MN at C at time t = T. Thus, BC = v₁T. During the time T, the secondary wavelet originating at A covers a distance AE in the denser medium with radius v₂T.

As all the points on CE are in the same phase of wave motion, CE represents the refracted wavefront in the denser medium. CE is the tangent to the secondary wavelet starting from A. It is also a common tangent to all the secondary wavelets emitted by points between A and C. PP’ is the normal to the boundary at A.
∠A’AP = ∠BAC = the angle of incidence (i) and ∠P’AE = ∠ACE = the angle of refraction (r).
From ∆ABC and ∆AEC,

By definition, the refractive index of medium 2 with respect to medium 1,

Here, n₁ and n₂ are the absolute refractive indices of medium 1 and medium 2 respectively. Eq. (1) is Snell’s law of refraction. Also, it can be seen from the figure, that the incident ray and the refracted ray lie on the opposite sides of the normal and all three of them lie in the same plane.

Thus, the laws of refraction of light can be deduced by Huygens’ construction of a plane wavefront.
If v₁ > v₂, i.e. n₁ < n₂, then r < i (bending of the refracted ray towards the normal).
[Notes :
(1) Quite often, the terms vacuum and free space are used in the same sense. Absolute vacuum or a perfect vacuum-a region of space devoid of material, particles – does not exist. The term vacuum is also used to mean a region of space occupied by a gas at very low pressure. Free space means a region of space devoid of matter and fields. Its refractive index is 1 (by definition). Its temperature is 0 K. ε₀ and μ₀ are defined for free space. The refractive index of a medium with respect to air is very close to the absolute refractive index of the medium as the speed of light in air is very close to that in free space.
(2) There is no lateral inversion in refraction.
(3) There is no bending of light when the angle of incidence is zero (normal incidence), r = 0 for i = 0.]

Question 4.
Derive the laws of refraction of light using Huygens’ principle.
Answer:
Consider a plane wavefront AB of monochromatic light propagating in the direction A’A incident obliquely at an angle i on a plane refracting surface MN. This plane refracting surface MN separates two uniform and optically transparent mediums.
Let v₁ and v₂ be the speeds of light in medium 1 (say, a rarer medium) and medium 2 (a denser
medium) respectively.

When the wavefront reaches MN at point A at t = 0, A becomes a secondary source and emits secondary waves in the second medium, while ray B’B reaches the surface MN at C at time t = T. Thus, BC = v₁T. During the time T, the secondary wavelet originating at A covers a distance AE in the denser medium with radius v₂T.

As all the points on CE are in the same phase of wave motion, CE represents the refracted wavefront in the denser medium. CE is the tangent to the secondary wavelet starting from A. It is also a common tangent to all the secondary wavelets emitted by points between A and C. PP’ is the normal to the boundary at A.
∠A’AP = ∠BAC = the angle of incidence (i) and ∠P’AE = ∠ACE = the angle of refraction (r).
From ∆ABC and ∆AEC,

By definition, the refractive index of medium 2 with respect to medium 1,

Here, n₁ and n₂ are the absolute refractive indices of medium 1 and medium 2 respectively. Eq. (1) is Snell’s law of refraction. Also, it can be seen from the figure, that the incident ray and the refracted ray lie on the opposite sides of the normal and all three of them lie in the same plane.

Thus, the laws of refraction of light can be deduced by Huygens’ construction of a plane wavefront.
If v₁ > v₂, i.e. n₁ < n₂, then r < i (bending of the refracted ray towards the normal).
[Notes :

1. Quite often, the terms vacuum and free space are used in the same sense. Absolute vacuum or a perfect vacuum-a region of space devoid of material, particles – does not exist. The term vacuum is also used to mean a region of space occupied by a gas at very low pressure. Free space means a region of space devoid of matter and fields. Its refractive index is 1 (by definition). Its temperature is 0 K. ε₀ and μ₀ are defined for free space. The refractive index of a medium with respect to air is very close to the absolute refractive index of the medium as the speed of light in air is very close to that in free space.
2. There is no lateral inversion in refraction.
3. There is no bending of light when the angle of incidence is zero (normal incidence), r = 0 for i = 0.]

Question 5.
Explain what is meant by polarization and derive Malus’ law.
Answer:
According to the electromagnetic theory of light, a light wave consists of electric and magnetic fields vibrating at right angles to each other and to the direction of propagation of the wave. If the vibrations of \(\vec{E}\) in a light wave are in all directions perpendicular to the direction of propagation of light, the wave is said to be unpolarized.

If the vibrations of the electric field \(\vec{E}\) in a light wave are confined to a single plane containing the direction of propagation of the wave so that its electric field is restricted along one particular direction at right angles to the direction of propagation of the wave, the wave is said to be plane-polarized or linearly polarized.

This phenomenon of restricting the vibrations of light, i.e., of the electric field vector in a particular direction, which is perpendicular to the direction of the propagation of the wave is called polarization of light.

Consider an unpolarized light wave travelling along the x-direction. Let c, v and λ be the speed, frequency and wavelength, respectively, of the wave. The magnitude of its electric field (\(\vec{E}\)) is,
E = E₀ sin (kx – ωt), where E₀ = E_max = amplitude of the wave, ω = 2πv = angular frequency of the wave and k = \(\frac{2 \pi}{\lambda}\) = magnitude of the wave vector or propagation vector.

The intensity of the wave is proportional to \(\left|E_{0}\right|^{2}\). The direction of the electric field can be anywhere in the y-z plane. This wave is passed through two identical polarizers as shown in below figure.

When a wave with its electric field inclined at an angle φ to the axis of the first polarizer is passed through the polarizer, the component E₀ cos φ will pass through it. The other component E₀ sin φ which is perpendicular to it will be blocked.

Now, after passing through this polarizer, the intensity of this wave will be proportional to the square of its amplitude, i.e., proportional to |E₀ cos φ|².

The intensity of the plane-polarized wave emerging from the first polarizer can be obtained by averaging | E₀ cos φ|² over all values of φ between 0 and 180°. The intensity of the wave will be
proportional to \(\frac{1}{2}\)|E₀|² as the average value of cos² φ over this range is \(\frac{1}{2}\). Thus the intensity of an unpolarized wave reduces by half after passing through a polarizer.

When the plane-polarized wave emerges from the first polarizer, let us assume that its electric field (\(\overrightarrow{E_{1}}\)) is along the y-direction. Thus, this electric field is,
\(\overrightarrow{E_{1}}\) = \(\hat{\mathrm{j}}\) E₁₀ sin (kx – ωt) …. (1)
where, E₁₀ is the amplitude of this polarized wave. The intensity of the polarized wave,
I₁ ∝ |E₁₀|² …(2)

Now this wave passes through the second polarizer whose polarization axis (transmission axis) makes an angle θ with the y-direction. This allows only the component E₁₀ cos θ to pass through it. Thus, the amplitude of the wave which passes through the second polarizer is E₂₀ = E₁₀ cos θ and its intensity,
I₂ ∝ | E₂₀|²
∴ I₂ ∝ | E₁₀|² cos²θ
∴ I₂ = I₁ cos₂θ … (3)

Thus, when plane-polarized light of intensity I₁ is incident on the second identical polarizer, the intensity of light transmitted by the second polarizer varies as cos²θ, i.e., I₂ = I₁ cos²θ, where θ is the angle between the transmission axes of the two polarizers. This is known as Malus’ law

[Note : Etienne Louis Malus (1775-1812), French military engineer and physicist, discovered in 1809 that light can be polarized by reflection. He was the first to use the word polarization, but his arguments were based on Newton’s corpuscular theory.]

Question 6.
What is Brewster’s law? Derive the formula for Brewster angle.
Answer:
Brewster’s law : The tangent of the polarizing angle is equal to the refractive index of the reflecting medium with respect to the surrounding (₁n₂). If θ_B is the polarizing angle,
tan θ_B = ₁n₂ = \(\frac{n_{2}}{n_{1}}\)
Here n₁ is the absolute refractive index of the surrounding and n₂ is that of the reflecting medium.
The angle θ_B is called the Brewster angle.

Consider a ray of unpolarized monochromatic light incident at an angle θ_B on a boundary between two transparent media as shown in below figure. Medium 1 is a rarer medium with refractive index n₁ and medium 2 is a denser medium with refractive index n₂. Part of incident light gets refracted and the rest

gets reflected. The degree of polarization of the reflected ray varies with the angle of incidence.
The electric field of the incident wave is in the plane perpendicular to the direction of propagation of incident light. This electric field can be resolved into a component parallel to the plane of the paper, shown by double arrows, and a component perpendicular to the plane of the paper shown by dots, both having equal magnitude. Generally, the reflected and refracted rays are partially polarized, i.e., the two components do not have equal magnitude.

In 1812, Sir David Brewster discovered that for a particular angle of incidence. θ_B, the reflected wave is completely plane-polarized with its electric field perpendicular to the plane of the paper while the refracted wave is partially polarized. This particular angle of incidence (θ_B) is called the Brewster angle.

For this angle of incidence, the refracted and reflected rays are perpendicular to each other.
For angle of refraction θ_r,
θ_B + θ_r = 90° …… (1)
From Snell’s law of refraction,
∴ n₁ sin θ_B = n₂ sin θ_r … (2)
From Eqs. (1) and (2), we have,
n₁ sin θ_B = n₂ sin (90° – θ_B) = n₂ cos θ_B

This is called Brewster’s law.

Question 7.
Describe Young’s double slit interference experiment and derive conditions for occurrence of dark and bright fringes on the screen. Define fringe width and derive a formula for it.
Answer:
Description of Young’s double-slit interference experiment:

1. A plane wavefront is obtained by placing a linear source S of monochromatic light at the focus of a convex lens. It is then made to pass through an opaque screen AB having two narrow and similar slits S₁ and S₂. S₁ and S₂ are equidistant from S so that the wavefronts starting simultaneously from S and reaching S₁ and S₂ at the same time are in phase. A screen PQ is placed at some distance from screen AB as shown in below figure
   
2. S₁ and S₂ act as secondary sources. The crests/-troughs of the secondary wavelets superpose and interfere constructively along straight lines joining the black dots shown in above figure. The point where these lines meet the screen have high intensity and are bright.
3. Similarly, there are points shown with red dots where the crest of one wave coincides with the trough of the other. The corresponding points on the screen are dark due to destructive interference.
4. These dark and bright regions are called fringes or bands and the whole pattern is called interference pattern.

Conditions for occurence of dark and bright fringes on the screen :

Consider Young’s double-slit experimental set up. Two narrow coherent light sources are obtained by wavefront splitting as monochromatic light of wavelength λ emerges out of two narrow and closely spaced, parallel slits S₁ and S₂ of equal widths. The separation S₁S₂ = d is very small. The interference pattern is observed on a screen placed parallel to the plane of and at considerable distance D (D » d) from the slits. OO’ is the perpendicular bisector of segment S₁S₂.

Consider, a point P on the screen at a distance y from O’ (y « D). The two light waves from S₁ and S₂ reach P along paths S₁P and S₂P, respectively. If the path difference (∆l) between S₁P and S₂P is an integral multiple of λ, the two waves arriving there will interfere constructively producing a bright fringe at P. On the contrary, if the path difference between S₁P and S₂P is half integral multiple of λ, there will be destructive interference and a dark fringe will be produced at P.
From above figure,
(S₂P)² = (S₂S₂)² + (PS₂‘)²
= (S₂S₂‘)² + (PO’ + O’S₂‘)²

Expression for the fringe width (or band width) : The distance between consecutive bright (or dark) fringes is called the fringe width (or band width) W. Point P will be bright (maximum intensity), if the path difference, ∆l = y_n\(\frac{d}{D}\) = nλ where n = 0,1, 2, 3…, Point P will be dark (minimum intensity equal to zero), if y_m\(\frac{d}{D}\) = (2m – 1)\(\frac{\lambda}{2}\), where, m = 1, 2, 3…,
Thus, for bright fringes (or bands),

These conditions show that the bright and dark fringes (or bands) occur alternately and are equally – spaced. For Point O’, the path difference (S₂O’ – S₁O’) = 0. Hence, point O’ will be bright. It corresponds to the centre of the central bright fringe (or band). On both sides of O’, the interference pattern consists of alternate dark and bright fringes (or band) parallel to the slit.

Let y_n and y_{n + 1}, be the distances of the nth and (n + 1)^th bright fringes from the central bright fringe.

Alternately, let y_m and y_{m + 1} be the distances of the m th and (m + 1)^th dark fringes respectively from the central bright fringe.

Eqs. (7) and (11) show that the fringe width is the same for bright and dark fringes.

[Note : In the first approximation, the path difference is d sin θ.]

Question 8.
What are the conditions for obtaining good interference pattern? Give reasons.
Answer:
The conditions necessary for obtaining well defined and steady interference pattern :

1. The two sources of light should be coherent:
   The two sources must maintain their phase relation during the time required for observation. If the phases and phase difference vary with time, the positions of maxima and minima will also change with time and consequently the interference pattern will change randomly and rapidly, and steady interference pattern would not be observed. For coherence, the two secondary sources must be derived from a single original source.
2. The light should be monochromatic :
   Otherwise, interference will result in complex coloured bands (fringes) because the separation of successive bright bands (fringes) is different for different colours. It also may produce overlapping bands.
3. The two light sources should be of equal brightness, i.e., the waves must have the same amplitude.
   The interfering light waves should have the same amplitude. Then, the points where the waves meet in opposite phase will be completely dark (zero intensity). This will increase the contrast of the interference pattern and make it more distinct.
4. The two light sources should be narrow :
   If the source apertures are wide in comparison with the light wavelength, each source will be equivalent to multiple narrow sources and the superimposed pattern will consist of bright and less bright fringes. That is, the interference pattern will not be well defined.
5. The interfering light waves should be in the same state of polarization :
   Otherwise, the points where the waves meet in opposite phase will not be completely dark and the interference pattern will not be distinct.
6. The two light sources should be closely spaced and the distance between the screen and the sources should be large : Both these conditions are desirable for appreciable fringe separation. The separation of successive bright or dark fringes is inversely proportional to the closeness of the slits and directly proportional to the screen distance.

Question 9.
What is meant by coherent sources? What are the two methods for obtaining coherent sources in the laboratory?
Answer:
Coherent sources : Two sources of light are said to be coherent if the phase difference between the emitted waves remains constant.

It is not possible to observe interference pattern with light from any two different sources. This is because, no observable interference phenomenon occurs by superposing light from two different sources. This happens due to the fact that different sources emit waves of different frequencies. Even if the two sources emit light of the same frequency, the phase difference between the wave trains from them fluctuates randomly and rapidly, i.e., they are not coherent.
Consequently, the interference pattern will change randomly and rapidly, and steady interference pattern would not be observed.

In the laboratory, coherent sources can be obtained by using
(1) Lloyd’s mirror and
(2) Fresnel’s biprism.

(1) Lloyd’s mirror : A plane polished mirror is kept at some distance from the source of monochromatic light and light is made incident on the mirror at a grazing angle.

Some light falls directly on the screen as shown by the black lines in above figure, while some light falls on the screen after reflection from the mirror as shown by red lines. The reflected light appears to come from a virtual source and thus two sources can be obtained. These two sources are coherent as they are derived from a single source. Superposition of the waves coming from these coherent sources, under appropriate conditions, gives rise to interference pattern consisting of alternate bright and dark bands on the screen as shown in the figure.

(2) Fresnel’s biprism : It is a single prism having an obtuse angle of about 178° and the other two angles of about 1° each. The biprism can be considered as made of two thin prisms of very small refracting angle of about 1°. The source, in the form of an illuminated narrow slit, is aligned parallel to the refracting edge of the biprism. Monochromatic light from the source is made to pass through that narrow slit and fall on the biprism.

Two virtual images S₁ and S₂ are formed by the two halves of the biprism. These are coherent sources which are obtained from a single secondary source S. The two waves coming from S₁ and S₂ interfere under appropriate conditions and form interference fringes, like those obtained in Young’s double-slit experiment, as shown in the figure in the shaded region. The formula for y is the same as in Young’s experiment.

Question 10.
What is diffraction of light? How does it differ from interference? What are Fraunhoffer and Fresnel diffractions?
Answer:
1. Phenomenon of diffraction of light: When light passes by the edge of an obstacle or through a small opening or a narrow slit and falls on a screen, the principle of rectilinear propagation of light from geometrical optics predicts a sharp shadow. However, it is found that some of the light deviates from its rectilinear path and penetrates into the region of the geometrical shadow. This is a general characteristic of wave phenomena, which occurs whenever a portion of the wavefront is obstructed in some way. This bending of light waves at an edge into the region of geometrical shadow is called diffraction of light.

2. Differences between interference and diffraction :

1. The term interference is used to characterise the superposition of a few coherent waves (say, two). But when the superposition at a point involves a large number of waves coming from different parts of the same wavefront, the effect is referred to as diffraction.
2. Double-slit interference fringes are all of equal width. In single-slit diffraction pattern, only the non-central maxima are of equal width which is half of that of the central maximum.
3. In double-slit interference, the bright and dark fringes are equally spaced. In diffraction, only the non-central maxima lie approximately halfway between the minima.
4. In double-slit interference, bright fringes are of equal intensity. In diffraction, successive non-central maxima decrease rapidly in intensity.

[Note : Interference and diffraction both have their origin in the principle of superposition of waves. There is no physical difference between them. It is just a question of usage. When there are only a few sources, say two, the phenomenon is usually called interference. But, if there is a large number of sources the word diffraction is used.]

3. Diffraction can be classified into two types depending on the distances involved in the experimental setup :

(A) Fraunhofer diffraction : In this class of diffraction, both the source and the screen are at infinite distances from the aperture. This is achieved by placing the source at the focus of a convex lens and the screen at the focal plane of another convex lens.

(B) Fresnel diffraction : In this class of diffraction, either the source of light or the screen or both are at finite distances from the diffracting aperture. The incident wavefront is either cylindrical or spherical depending on the source. A lens is not needed to observe the diffraction pattern on the screen.

Question 11.
Derive the conditions for bright and dark fringes produced due to diffraction by a single slit.
Answer:
When a parallel beam of monochromatic light of wavelength λ illuminates a single slit of finite width a, we observe on a screen some distance from the slit, a broad pattern of alternate dark and bright fringes. The pattern consists of a central bright fringe, with successive dark and bright fringes of diminishing intensity on both sides. This is called ‘ the diffraction pattern of a single slit.

Consider a single slit illuminated with a parallel beam of monochromatic light perpendicular to the plane of the slit. The diffraction pattern is obtained on a screen at a distance D (» a) from the slit and at the focal plane of the convex lens, Fig. 7.33.

We can imagine the single slit as being made up of a large number of Huygens’ sources evenly distributed over the width of the slit. Then the maxima and minima of the pattern arise from the interference of the various Huygens’ wavelets.

Now, imagine the single slit as made up of two adjacent slits, each of width a/2. Since, the incident plane wavefronts are parallel to the plane of the slit, all the Huygens sources at the slit will be in phase. They will therefore also in phase at the point P₀ on the screen, where P₀ is equidistant from all the Huygens sources. At P₀, then, we get the central maximum.

For the first minimum of intensity on the screen, the path difference between the waves from the Huygens sources A and O (or O and B) is λ/2, which is the condition for destructive interference. Suppose, the nodal line OP for the first minimum subtends an angle θ at the slit; θ is very small. With P as the centre and PA as radius, strike an arc intersecting PB at C. Since, D » a, the arc AC can be considered a straight line at right angles to PB. Then, ∆ABC is a right-angled triangle similar to ∆OP₀P.
This means that, ∠BAC = θ
∴ BC = a sinθ
∴ Difference in path length,
BC = PB – PA = (PB – PO) + (PO – PA)

(∵ θ is very small and in radian)
The other nodal lines of intensity minima can be understood in a similar way. In general, then, for the with minimum (m = ±1, ±2, ±3, …).
θ_m = \(\frac{m \lambda}{a}\) (mth minimum) … (2)
as θ_m is very small and in radian.
Between the successive minima, the intensity rises to secondary maxima when the path difference is an odd-integral multiple of \(\frac{\lambda}{2}\) :
a sin θ_m = (2m + 1)\(\frac{\lambda}{2}\) = (m + \(\frac{1}{2}\))λ
i.e., at angles given by,
θ_m \(\simeq\) sin θ_m = (m + \(\frac{1}{2}\))\(\frac{\lambda}{a}\)
(with secondary maximum) … (3)

Question 12.
Describe what is Rayleigh’s criterion for resolution. Explain it for a telescope and a microscope.
Answer:
Rayleigh’s criterion for minimum resolution : Two overlapping diffraction patterns due to two point sources are acceptably or just resolved if the centre of the central peak of one diffraction pattern is as far as the first minimum of the other pattern.

The ‘sharpness’ of the central maximum of a diffraction pattern is measured by the angular separation between the centre of the peak and the first minimum. It gives the limit of resolution.

Two overlapping diffraction patterns due to two point sources are not resolved if the angular separation between the central peaks is less than the limit of resolution. They are said to be just separate, or resolved, if the angular separation between the central peaks is equal to the limit of resolution. They are said to be well resolved if the angular separation between the central peaks is more than the limit of resolution.

Resolving power of an optical instrument:
The primary aim of using an optical instrument is to see fine details, whether observing a star system through a telescope or a living cell through a microscope. After passing through an optical system, light from two adjacent parts of the object should produce sharp, distinct (separate) images of those parts. The objective lens or mirror of a telescope or microscope acts like a circular aperture. The diffraction pattern of a circular aperture consists of a central bright spot (called the Airy disc and corresponds to the central maximum) and concentric dark and bright rings.

Light from two close objects or parts of an object after passing through the aperture of an optical system produces overlapping diffraction patterns that tend to obscure the image. If these diffraction patterns are so broad that their central maxima overlap substantially, it is difficult to decide if the intensity distribution is produced by two separate objects or by one.

The resolving power of an optical instrument, e.g., a telescope or microscope, is a measure of its ability to produce detectably separate images of objects that are close together.

Definition : The smallest linear or angular separation between two point objects which appear just resolved when viewed through an optical instrument is called the limit of resolution of the instrument and its reciprocal is called the resolving power of the instrument.

Rayleigh’s criterion for minimum resolution : Two overlapping diffraction patterns due to two point sources are acceptably or just resolved if the centre of the central peak of one diffraction pattern is as far as the first minimum of the other pattern.

The ‘sharpness’ of the central maximum of a diffraction pattern is measured by the angular separation between the centre of the peak and the first minimum. It gives the limit of resolution.

Two overlapping diffraction patterns due to two point sources are not resolved if the angular separation between the central peaks is less than the limit of resolution. They are said to be just separate, or resolved, if the angular separation between the central peaks is equal to the limit of resolution. They are said to be well resolved if the angular separation between the central peaks is more than the limit of resolution.

The resolving power of a telescope is defined as the reciprocal of the angular limit of resolution between two closely-spaced distant objects so that they are just resolved when seen through the telescope.

Consider two stars seen through a telescope. The diameter (D) of the objective lens or mirror corresponds to the diffracting aperture. For a distant point source, the first diffraction minimum is at an angle θ away from the centre such that

Dsinθ = 1.22 λ
where λ is the wavelength of light. The angle θ is usually so small that we can substitute sin θ \(\approx\) θ (θ in radian). Thus, the Airy disc for each star will be spread out over an angular half-width θ = 1.22 λ/D about its geometrical image point. The radius of the Airy disc at the focal plane of the objective lens is r = fθ = 1.22fλ/D, where / is the focal length of the objective.

When observing two closely-spaced stars, the Rayleigh criterion for just resolving the images as that of two point sources (instead of one) is met when the centre of one Airy disc falls on the first minimum of the other pattern. Thus, the angular limit (or angular separation) of resolution is
θ = \(\frac{1.22 \lambda}{D}\) …. (1)
and the linear separation between the images at the focal plane of the objective lens is
y = fθ …(2)
∴ Resolving power of a telescope,
R = \(\frac{1}{\theta}\) = \(\frac{D}{1.22 \lambda}\) … (3)
It depends

1. directly on the diameter of the objective lens or mirror,
2. inversely on the wavelength of the radiation.

Question 13.
Whitelight consists of wavelengths from 400 nm to 700 nm. What will be the wavelength range seen when white light is passed through glass of refractive index 1.55? [Ans: 258.1 – 451.6 nm]
Answer:
Let λ₁ and λ₂ be the wavelengths of light in water for 400 nm and 700 nm (wavelengths in vacuum) respectively. Let λ_a be the wavelength of light in vacuum.

The wavelength range seen when white light is passed through the glass would be 258.06 nm to 451.61 nm.

Question 14.
The optical path of a ray of light of a given wavelength travelling a distance of 3 cm in flint glass having refractive index 1.6 is same as that on travelling a distance x cm through a medium having refractive index 1.25. Determine the value of x.
[Ans: 3.84 cm]
Answer:
Let d_fg and d_m be the distances by the ray of light in the flint glass and the medium respectively. Also, let n_fg and n_m be the refractive indices of the flint glass and the medium respectively.
Data : d_fg = 3 cm, n_fg = 1.6, nm = 1.25,
Optical path = n_m × d_m = n_fg × d_fg

Thus, x cm = 3.84 cm
∴ x = 3.84
This is the value of x.

Question 15.
A double-slit arrangement produces interference fringes for sodium light (λ = 589 nm) that are 0.20° apart. What
is the angular fringe separation if the entire arrangement is immersed in water (n = 1.33)? [Ans: 0.15°]
Answer:
Data : θ₁ = 0.20°, n_w = 1.33
In the first approximation,
D sin θ₁ = y₁ and D sin θ₂ = y₂

∴ θ₂ = sin^-10.026
= 9′ = 0.15°
This is the required angular fringe separation.
OR
In the first approximation,

Question 16.
In a double-slit arrangement the slits are separated by a distance equal to 100 times the wavelength of the light passing through the slits.
(a) What is the angular separation in radians between the central maximum and an adjacent maximum?
(b) What is the distance between these maxima on a screen 50.0 cm from the slits?
[Ans: 0.01 rad, 0.5 cm]
Answer:
Data : d = 100λ, D = 50.0 cm
(a) The condition for maximum intensity in Young’s experiment is, d sin θ = nλ, n = 0, 1, 2 …,
The angle between the central maximum and its adjacent maximum can be determined by setting n equal to 1,
∴ d sin θ = λ

(b) The distance between these maxima on the screen is D sin θ = D\(\left(\frac{\lambda}{d}\right)\)
= (50.0 cm)\(\left(\frac{\lambda}{100 \lambda}\right)\)
= 0.50 cm

Question 17.
Unpolarized light with intensity I₀ is incident on two polaroids. The axis of the first polaroid makes an angle of 50° with the vertical, and the axis of the second polaroid is horizontal. What is the intensity of the light after it has passed through the second polaroid? [Ans: I₀/2 × (cos 40°)²]
Answer:
According to Malus’ law, when the unpolarized light with intensity I₀ is incident on the first polarizer, the polarizer polarizes this incident light. The intensity of light becomes I₁ = I₀/2.
Now, I₂ = I₁ cos²θ
∴ I₂ = \(\left(\frac{I_{0}}{2}\right)\) cos²θ
Also, the angle θ between the axes of the two polarizers is θ₂ – θ₁.

The intensity of light after it has passed through the second polaroid = \(\left(\frac{\boldsymbol{I}_{0}}{\mathbf{2}}\right)\)cos²40° = \(\frac{I_{0}}{2}\)(0.7660)²
= 0.2934 I₀

Question 18.
In a biprism experiment, the fringes are observed in the focal plane of the eyepiece at a distance of 1.2 m from the slits. The distance between the central bright band and the 20^th bright band is 0.4 cm. When a convex lens is placed between the biprism and the eyepiece, 90 cm from the eyepiece, the distance between the two virtual magnified images is found to be 0.9 cm. Determine the wavelength of light used. [Ans: 5000 Å]
Answer:
Data : D = 1.2 m
The distance between the central bright band and the 20^th bright band is 0.4 cm.
∴ y₂₀ = 0.4 cm = 0.4 × 10^-2 m
W = \(\frac{y_{20}}{20}\) = \(\frac{0.4}{20}\) × 10^-2 m = 2 × 10^-4 m,
d₁ = 0.9 cm = 0.9 × 10^-2m, v₁ = 90 cm = 0.9 m
∴ u₁ = D – v₁ = 1.2m – 0.9m = 0.3 m

Question 19.
In Fraunhoffer diffraction by a narrow slit, a screen is placed at a distance of 2 m from the lens to obtain the diffraction pattern. If the slit width is 0.2 mm and the first minimum is 5 mm on either side of the central maximum, find the wavelength of light. [Ans: 5000 Å]
Answer:
Data : D = 2 m, y_1d = 5 mm = 5 × 10^-3 m, a = 0.2 mm = 0.2 × 10^-3 m = 2 × 10^-4 m

This is the wavelength of light.

Question 20.
The intensity of the light coming from one of the slits in Young’s experiment is twice the intensity of the light coming from the other slit. What will be the approximate ratio of the intensities of the bright and dark fringes in the resulting interference pattern? [Ans: 34]
Answer:
Data : I₁ : I₂ = 2 : 1
If E₁₀ and E₂₀ are the amplitudes of the interfering waves, the ratio of the maximum intensity to the minimum intensity in the fringe system is


The ratio of the intensities of the bright and dark fringes in the resulting interference pattern is 34 : 1.

Question 21.
A parallel beam of green light of wavelength 550 nm passes through a slit of width 0.4 mm. The intensity pattern of the transmitted light is seen on a screen which is 40 cm away. What is the distance between the two first order minima? [Ans: 1.1 mm]
Answer:
Data : λ = 550 nm = 546 × 10^-9 m, a = 0.4 mm = 4 × 10^-4 m, D = 40 cm = 40 × 10^-2 m
y_md = m\(\frac{\lambda D}{a}\)
∴ y_1d = 1\(\frac{\lambda D}{a}\) and
2y_1d = \(\frac{2 \lambda D}{a}\)
= \(\frac{2 \times 550 \times 10^{-9} \times 40 \times 10^{-2}}{4 \times 10^{-4}}\)
= 2 × 550 × 10^-6 = 1092 × 10^-6
= 1.100 × 10^-3m = 1.100 mm
This is the distance between the two first order minima.

Question 22.
What must be the ratio of the slit width to the wavelength for a single slit to have the first diffraction minimum at 45.0°? [Ans: 1.274]
Answer:
Data : θ = 45°, m = 1
a sin θ = mλ for (m = 1, 2, 3… minima)
Here, m = 1 (First minimum)
∴ a sin 45° = (1) λ
∴ \(\frac{a}{\lambda}\) = \(\frac{1}{\sin 45^{\circ}}\) = 1.414
This is the required ratio.

Question 23.
Monochromatic electromagnetic radiation from a distant source passes through a slit. The diffraction pattern is observed on a screen 2.50 m from the slit. If the width of the central maximum is 6.00 mm, what is the slit width if the wavelength is
(a) 500 nm (visible light);
(b) 50 µm (infrared radiation);
(c) 0.500 nm (X-rays)?
[Ans: 0.4167 mm, 41.67 mm, 4.167 × 10^-4 mm]
Answer:
Data:2W = 6mm ∴ W= 3 mm = 3 × 10^-3 m, y = 2.5 m,
(a) λ₁ = 500 nm = 5 × 10^-7 m
(b) λ₂ = 50 μm = 5 × 10^-5 m
(c) λ₃ = 0.500 nm = 5 × 10^-10 m
Let a be the slit width.

Question 24.
A star is emitting light at the wavelength of 5000 Å. Determine the limit of resolution of a telescope having an objective of diameter of 200 inch. [Ans: 1.2 × 10^-7 rad]
Answer:
Data : λ = 5000 Å = 5 × 10^-7 m,
D = 200 × 2.54 cm = 5.08 m
θ = \(\frac{1.22 \lambda}{D}\)
= \(\frac{1.22 \times 5 \times 10^{-7}}{5.08}\)
= 1.2 × 10^-7 rad
This is the required quantity.

Question 25.
The distance between two consecutive bright fringes in a biprism experiment using light of wavelength 6000 Å is 0.32 mm by how much will the distance change if light of wavelength 4800 Å is used? [Ans: 0.064 mm]
Answer:
Data : λ₁ = 6000 Å = 6 × 10^-7m, λ₂ = 4800 Å = 4.8 × 10^-7m, W₁ = 0.32 mm = 3.2 × 10^-4m
Distance between consecutive bright fringes,

∴ ∆W = W₁ – W₂
= 3.2 × 10^-4m – 2.56 × 10^-4m
= 0.64 × 10^-4m
= 6.4 × 10^-5
= 0.064 mm
This is the required change in distance.

12th Physics Digest Chapter 7 Wave Optics Intext Questions and Answers

Use your brain power (Textbook Page No. 167)

What will you observe if

Question 1.
you look at a source of unpolarized light through a polarizer ?
Answer:
When a source of unpolarized light is viewed through a polarizer, the intensity of the light transmitted by the polarizer is reduced and hence the source appears less bright.

Question 2.
you look at the source through two polarizers and rotate one of them around the path of light for one full rotation?
Answer:
When a source of unpolarized light is viewed through two polarizers and the second polarizer is rotated gradually, the intensity of the light transmitted by the second polarizer goes on decreasing. When the axes of polarization of the two polarizers are at 90° to each other, light almost disappears depending on the quality of the polarizers. (Ideally the intensity of the transmitted light should be zero.) The light reappears, i.e., its intensity increases, when the second polarizer is rotated further, and the intensity of the light becomes maximum when the axes of polarization are parallel again.

Question 3.
instead of rotating only one of the polaroids, you rotate both polaroids simultaneously in the same direction?
Answer:
If both the polaroids are rotated simultaneously in the same direction with the same angular velocity, then there would be no change in the intensity of the transmitted light observed.

Can you tell? (Textbook Page No. 168)

Question 1.
If you look at the sky in a particular direction through a polaroid and rotate the polaroid around that direction what will you see ?
Answer:
As the scattered light is polarized, the sky appears bright and dim alternately.

Question 2.
Why does the sky appear to be blue while the clouds appear white ?
Answer:
The blue colour of the sky is because of the scattering of light by air molecules and dust particles in the atmosphere. As the wavelength of blue light is less than that of red light, the blue light is preferentially scattered than the light corresponding to other colours in the visible region. Clouds are seen due to scattering of light from lower parts of the atmosphere. The clouds contain the dust particles and water droplets which are sometimes large enough to scatter light of all the wavelengths such that the combined effect makes the clouds appear white.

Remember this (Textbook Page No. 171)

Question 1.
For the interference pattern to be clearly visible on the screen, the distance (D) between the slits and the screen should be much larger than the distance (d) between the two slits (S₁ and S₂), i.e., D » d.
Answer:
The condition for constructive interference at P is,
∆l = y_n\(\frac{d}{D}\) = nλ …. (1)
y_n being the position (y-coordinate) of nth bright fringe (n = 0, ±1, ±2, …).
∴ y_n = nλ\(\frac{D}{d}\) ….. (2)
Similarly, the position of mth (m = +1, ±2,…) dark fringe (destructive interference) is given by,
∆l = y_m\(\frac{d}{D}\) = (2m – 1)λ giving
y_m = (2m – 1)λ\(\frac{D}{d}\) …(3)
The distance between any two consecutive bright or dark fringes, i.e., the fringe width
= W = ∆y = y_{n + 1} – y_n = λ\(\frac{D}{d}\) …(4)
Conditions given by Eqs. (1) to (4) and hence the location of the fringes are derived assuming that the two sources S₁ and S₂ are in phase. If there is a non-zero phase difference between them it should be added appropriately. This will shift the entire fringe pattern but will not change the fringe width.

Do you know (Textbook Page No. 172 & 173)

Several phenomena that we come across in our day to day life are caused by interference and diffraction of light. These are the vigorous colours of soap bubbles as well as those seen in a thin oil film on the surface of water, the bright colours of butterflies and peacocks etc. Most of these colours are not due to pigments which absorb specific colours but are due to interference of light waves that are reflected by different layers.

Interference due to a thin film :
The brilliant colours of soap bubbles and thin films on the surface of water are due to the interference of light waves reflected from the upper and lower surfaces of the film. The two rays have a path difference which depends on the point on the film that is being viewed. This is shown in above figure.

The incident wave gets partially reflected from upper surface as shown by ray AE. The rest of the light gets refracted and travels along AB. At B it again gets partially reflected and travels along BC. At C it refracts into air and travels along CF. The parallel rays AE and CF have a phase difference due to their different path lengths in different media. As can be seen from the figure, the phase difference depends on the angle of incidence θ₁ i.e., the angle of incidence at the top surface which is the angle of viewing, and also on the wavelength of the light as the refractive index of the material of the thin film depends on it. The two waves propagating along AE and CF interfere producing maxima and minima for different colours at different angles of viewing. One sees different colours when the film is viewed at different angles.

As the reflection is from the denser boundary, there is an additional phase difference of π radians (or an additional path difference λ). This should be taken into account for mathematical analysis.

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 8 Respiration and Circulation Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 8 Respiration and Circulation

1. Multiple choice questions

Question 1.
The muscular structure that separates the thoracic and abdominal cavity is …………………..
(a) pleura
(b) diaphragm
(c) trachea
(d) epithelium
Answer:
(b) diaphragm

Question 2.
What is the minimum number of plasma membrane that oxygen has to diffuse across to pass from air in the alveolus to haemoglobin inside a R.B.C.?
(a) two
(b) three
(c) four
(d) five
Answer:
(a) two

Question 3.
…………………. is a sound producing organ.
(a) Larynx
(b) Pharynx
(c) Tonsils
(d) Trachea
Answer:
(a) Larynx

Question 4.
The maximum volume of gas that is inhaled during breathing in addition to T.V. is …………………..
(a) residual volume
(b) IRV
(c) GRV.
(d) vital capacity
Answer:
(b) IRV

Question 5.
………………….. muscles contract when the external intercostals muscles contract.
(a) Internal abdominal
(b) Jaw
(c) Muscles in bronchial walls
(d) Diaphragm
Answer:
(d) Diaphragm

Question 6.
Movement of cytoplasm in unicellular organisms is called …………………..
(a) diffusion
(b) cyclosis
(c) circulation
(d) thrombosis
Answer:
(b) cyclosis

Question 7.
Which of the following animals do not have closed circulation?
(a) Earthworm
(b) Rabbit
(c) Butterfly
(d) Shark
Answer:
(c) Butterfly

Question 8.
Diapedesis is performed by …………………..
(a) erythrocytes
(b) thrombocytes
(c) adipocytes
(d) leucocytes
Answer:
(d) leucocytes

Question 9.
Pacemaker of heart is …………………..
(a) SA node
(b) AV node
(c) His bundle
(d) Purkinje fibers
Answer:
(a) SA node

Question 10.
Which of the following is without nucleus?
(a) Red blood corpuscle
(b) Neutrophil
(c) Basophil
(d) Lymphocyte
Answer:
(a) Red blood corpuscle

Question 11.
Cockroach shows which kind of circulatory system?
(a) Open
(b) Closed
(c) Lymphatic
(d) Double
Answer:
(a) Open

Question 12.
Diapedesis can be seen in …………………..
(a) RBC
(b) WBC
(c) Platelet
(d) neuron
Answer:
(b) WBC

Question 13.
Opening of inferior vena cava is guarded by …………………..
(a) bicuspid valve
(b) tricuspid valve
(c) Eustachian valve
(d) Thebesian valve
Answer:
(c) Eustachian valve

Question 14.
…………………. wave in ECG represent atrial depolarization.
(a) P
(b) QRS complex
(c) Q
(d) T
Answer:
(a) P

Question 15.
The fluid seen in the intercellular spaces in Human is …………………..
(a) blood
(b) lymph
(c) interstitial fluid
(d) water
Answer:
(b) lymph

2. Match the columns

Question 1.
Respiratory surface Organism

Respiratory surface	Organism
(1) Plasma membrane	(a) Insect
(2) Lungs	(b) Salamander
(3) External gills	(c) Bird
(4) Internal gills	(d) Amoeba
(5) Trachea	(e) Fish

Answer:

Respiratory surface	Organism
(1) Plasma membrane	(d) Amoeba
(2) Lungs	(c) Bird
(3) External gills	(b) Salamander
(4) Internal gills	(e) Fish
(5) Trachea	(a) Insect

3. Very Short Answer Questions

Question 1.
Why does trachea have ‘C’-shaped rings of cartilage?
Answer:
Trachea is supported by ‘C’-shaped rings of J cartilage which prevent it from collapsing and always keep it open.

Question 2.
Why is respiration in insect called direct respiration?
Answer:
Respiration in insect is called direct because tracheal tubes exchange O₂ and CO₂ directly with the haemocoel which then exchange them with tissues.

Question 3.
Why is gas exchange very rapid at alveolar level?
OR
Why does gas exchange in the alveolar region very rapid?
Answer:
Gas exchange is very rapid at alveolar level because numerous alveoli (about 700 millions) in the lungs provide large surface area for gaseous exchange.

Question 4.
Name the organ which prevents the entry of food into the trachea while eating.
Answer:
Epiglottis prevents the entry of food into trachea while eating.

4. Short Answer Questions

Question 1.
Why is it advantageous to breathe through the nose than through the mouth?
Answer:
Breathing through nose is better than breathing through the mouth because of the following reasons:

1. The nostrils are smaller than the mouth so air exhaled through the nose creates a backflow of air into the lungs.
2. As we exhale more slowly through the nose than we do through the mouth, the lungs have more time to extract oxygen from the air that we have already taken in.
3. The hairs inside nostrils filter any dust particles and microbes in the air and it only lets the clean air pass through.
4. The air gets warm and humidified in nostrils as it passes into our bodies.
5. Moreover breathing through the mouth can dry the oral cavity and lead to bad breath, gum disease and tooth decay.

Question 2.
Identity the incorrect statement and correct it.
(a) A respiratory surface area should have a. large surface area.
(b) A respiratory surface area should be kept dry.
(c) A respiratory surface area should be thin, may be 1 mm or less.
Answer:
Statement (a) and statement (c) are correct whereas statement (b) is incorrect. A respiratory surface area should be kept moist, is the correct statement.

Question 3.
Given below are the characteristics of some modified respiratory movement. Identify them.
a. Spasmodic contraction of muscles of expiration and forceful expulsion of air through nose and mouth.
Answer:
Sneezing

b. An inspiration followed by many short convulsive expiration accompanied by facial expression.
Answer:
Laughing, Crying.

Question 4.
Blood plasma.
Answer:

1. Plasma is a straw coloured, slightly alkaline viscous fluid part of the blood, having 90-92% water and 8-10% soluble proteins.
2. Serum albumin, serum globulin, heparin, fibrinogen and prothrombin are the plasma proteins which form 7% of the plasma.
3. Glucose, amino acids, fatty acids and glycerol are the nutrients dissolved in plasma.
4. Nitrogenous wastes (urea, uric acid, . ammonia and creatinine) and respiratory gases (oxygen and carbon dioxide) is present in plasma.
5. Enzymes and hormones too are transported Ada plasma.
6. Inorganic minerals are also present in plasma such as bicarbonates, chlorides, phosphates and sulphates of sodium, potassium, calcium and magnesium.

Question 5.
Blood clotting/Coagulation of blood.
OR
Explain blood clotting in short.
Answer:

1. The process of converting the liquid blood into a semisolid form is called blood clotting or coagulation.
2. The process of clotting may be initiated by contact of blood with any foreign surface (intrinsic process) or with damaged tissue (extrinsic process).
3. Intrinsic and extrinsic processes involve interaction of various substances called clotting factors by a step wise or cascade mechanism.
4. There are in all twelve clotting factors numbered as I to XII (factor VI is not in active use).
5. Interaction of these factors in a cascade manner leads to formation of enzyme, Thromboplastin which helps in the formation of enzyme prothrombinase.
6. Prothrombinase inactivates heparin and also converts inactive prothrombin into active thrombin.
7. Thrombin converts soluble blood protein- fibrinogen into insoluble fibrin. Fibrin forms a mesh in which platelets and other blood cells are trapped to form the clot.
8. These reactions occur in 2 to 8 minutes. Therefore, clotting time is said to be 2 to 8 minutes.

Question 6.
Describe pericardium.
Answer:

1. Pericardium is the double layered peritoneum that encloses the heart. It consists of two layers, viz. fibrous pericardium and serous pericardium.
2. Fibrous pericardium is the outer layer having tough, inelastic fibrous connective tissue whereas serous pericardium is the v inner double layered membrane. It has in turn an outer parietal layer and inner visceral layer.
3. Parietal layer of serous pericardium lies on the inner side of fibrous pericardium.
4. Visceral layer also known as epicardium adheres to heart and thus forms outer covering over the heart.
5. There is a pericardial fluid in the pericardial space which is present in between the parietal and visceral layers of serous pericardium.

Question 7.
Describe valves in the human heart.
Answer:
Human heart has following main valves:

1. Tricuspid valve : Tricuspid valve is present between the right atrium and right ventricle. It has three cusps or flaps. It prevents the backflow of blood into right atrium.
2. Bicuspid valve : Bicuspid valve, also called mitral valve is present between the left atrium and left ventricle. It has two flaps. It prevents the backflow of blood in left atrium. Both tricuspid and bicuspid valves are attached to papillary muscles with tendinous chords or chordate tendinae to prevent valves from turning back into atria at the time of systole.
3. Semilunar valve : These are present at the opening of pulmonary artery and systemic aorta. They prevent the back flow of blood when ventricles undergo systole.
4. Thebesian valve : Thebesian valve is present at the opening of coronary sinus.
5. Eustachian valve : Eustachian valve is present at the opening of inferior vena cava.

Question 8.
What is the role of papillary muscles and chordae tendinae in human heart?
Answer:

1. Papillary muscles are large and well- developed muscular ridges present along the inner surface of the ventricles.
2. Bicuspid and tricuspid valves are attached to papillary muscles of ventricles by chordae tendinae.
3. Chordae tendinae are inelastic fibres present in the lumen of ventricles.
4. The chordae tendinae prevent the valves from turning back into the atria during the contraction of ventricles and regulate the opening and closing of bicuspid and tricuspid valves.

Question 9.
Explain in brief the factors affecting blood pressure.
Answer:

1. Cardiac output : Normal cardiac output is 5 lit/min. Increase in cardiac output increases systolic pressure.
2. Peripheral resistance : Peripheral resistance depends upon the diameter of blood vessels. Decrease in diameter of arterioles and capillaries under the effect of vasopressin cause increase in peripheral resistance and thereby increase in blood pressure.
3. Blood volume : Loss of blood in accidents decreases blood volume and thus cause decrease in blood pressure.
4. Viscosity of blood : Blood pressure is directly proportional to viscosity of blood.
5. Age : Blood pressure increases with age due to increase in inelasticity of blood vessels.
6. Venous return : Amount of blood brought to the heart via the veins per unit time is called the venous return and it is directly proportional to blood pressure.
7. Length and diameter of blood vessels : Blood pressure is directly proportional to the total length of the blood vessel. Blood pressure can also be affected by vasoconstriction or vasodilation.
8. Gender : Females have slightly lower BP than males of her age before menopause. However, the risk of high B. P increases in the females after menopause sets in.

5. Give Scientific Reason

Question 1.
Closed circulation is more efficient than open circulation.
Answer:

1. Closed circulation considerably enhances the speed, precision and efficiency of circulation.
2. The blood flows more rapidly, it takes less time to circulate through the closed system and return to the heart.
3. This fastens the supply and removed of materials to and from the tissues by the blood as compared to open circulation.
4. In open circulation, there are no blood vessels such as arteries or veins, to pump the blood.
5. Therefore, the blood pressure is very low.
6. Organisms with an open circulatory system typically have a relatively high volume of hemolymph and low blood pressure. Closed circulation is thus more efficient than open circulation.

Question 2.
Human heart is called as myogenic and autorhythmic?
Answer:

1. The heart shows auto rhythmicity because the impulse for its rhythmic movement develops inside the heart. Such heart is called myogenic.
2. Some of the cardiac muscle fibres become auto rhythmic (self-excitable) and start generating impulse during development.
3. These autorhythmic fibres perform two important function, viz. acting as a pacemaker and setting the rhythm for heart.
4. They also form conducting system for conduction of nerve impulses throughout the heart muscles.

Question 3.
In human heart, the blood flows only in one direction.
Answer:

1. In veins there are valves, which prevent the back flow of the blood.
2. In arteries, blood flows with unidirectional pressure.
3. Hence the circulation takes place only in one direction.

Question 4.
Arteries are thicker than veins.
Answer:

1. Arteries have relatively thick walls to enable them to withstand the high pressure of blood ejected from the heart.
2. Arteries expand when the pressure increases as the heart pushes blood out but then recoil (shrink) Wn the pressure decreases when the heart relaxes between heartbeats.
3. This expansion and recoiling occurs to maintain a smooth blood flow.
4. Veins, on the other hand, have thinner walls and larger lumen veins have no need for thick walls as then need not have to withstand high pressure like arteries.
5. Moreover, as veins transport relatively low pressure blood, they are commonly equipped with valves to promote the unidirectional flow of blood towards the heart.

Question 5.
Left ventricle is thick than all other chambers of heart.
OR
Left ventricle has thicker wall than the right ventricle.
Answer:

1. Left ventricle pumps oxygenated blood to all parts of the body. Therefore, there is greater pressure from the blood in left ventricle.
2. Right ventricle sends deoxygenated blood to lungs for oxygenation. This does not put more pressure and lungs are in vicinity of the heart.
3. Due to these functional differences between the two ventricles, left ventricle has thicker wall than that of the right ventricle.

6. Distinguish Between

Question 1.
Open circulation and Closed circulation
Answer:

Open circulation	Closed circulation
1. In open circulation, blood flows through large open spaces and channels called lacunae and haemocoels among the tissues.	1. In closed circulation, blood flows through a network of blood vessels all over the body.
2. Tissues are in direct contact with the blood.	2. Blood does not come in direct contact with tissue.
3. Blood flows with low pressure and usually does not contain any respiratory pigment like haemoglobin.	3. Blood flows with high pressure and contains respiratory pigment like haemoglobin.
4. Exchange of material takes place directly between blood and cells or tissues of the body.	4. Exchange of material takes place between blood and body tissues through an intermediate fluid called lymph.
5. Volume of blood flowing through a tissue cannot be controlled as blood flows out in open space.	5. Volume of blood can be regulated by the contraction and relaxation of the smooth muscles of the blood vessels.
6. Open circulatory system is found in arthropods and some molluscs.	6. Closed circulatory system is found in annelids, echinoderms and all vertebrates.

Question 2.
Arteries and veins.
Answer:

Arteries	Veins
1. The blood vessels that arise from the heart and carry blood away from heart are called arteries.	1. The blood vessels that bring blood to the heart are called veins.
2. Arteries are thick walled blood vessels, situated in deep layers in the body.	2. Veins cure thin walled blood vessels, situated superficially in the body.
3. Arteries do not have valves.	3. Veins have valves.
4. Tunica adventitia, the outermost layer of arteries is thick and elastic.	4. Tunica externa, the outermost layer of veins is thin.
5. Tunica media is very thick and contain elastic fibres.	5. Tunica media is thin layer and contain involuntary muscle fibres.
6. The lumen of arteries is small.	6. The lumen of the veins is very spacious.
7. With the exception of pulmonary arteries, all other arteries carry oxygenated blood.	7. With the exception of pulmonary veins, all other veins carry deoxygenated blood.
8. Blood in the arteries show high blood pressure.	8. Blood in the veins show lesser blood pressure.

Question 3.
Blood and Lymph.
Answer:

Blood	Lymph
1. Contains blood plasma with proteins and all three types of blood cells namely RBCs, WBCs and blood platelets.	1. Contains blood plasma without blood proteins, RBCs and platelets and contains lymphocytes.
2. Red in colour due to presence of RBCs.	2. Light yellow in colour and does not contain RBCs.
3. Carries oxygen in the body.	3. Does not carry oxygen.
4. The flow of blood in blood vessels is fast.	4. The flow of lymph in lymph capillaries is slow.
5. Lymphocytes are present.	5. Lymphocytes are present, more in number than those present in the blood.

Question 4.
Blood capillary and Lymph capillary.
Answer:

Blood capillary	Lymph capillary
1. Reddish, easy to observe.	1. Colourless, difficult to observe.
2. Joined to arterioles at one end and to venules at another end.	2. Blind (closed at the tip).
3. Narrower than lymph capillaries.	3. Wider than blood capillaries.
4. Wall consists of normal endothelium and basement membrane.	4. Wall consists of thin endothelium and poorly developed basement membrane.
5. Contains red blood.	5. Contains colourless lymph.
6. Have relatively high pressure.	6. Have relatively low pressure.

Question 5.
Intrinsic and Extrinsic process of clotting.
Answer:

Intrinsic process	Extrinsic process
1. The intrinsic pathway requires only clotting factors found within the blood itself – in particular, clotting factor XII (Hageman factor) from the platelets.	1. The extrinsic pathway is initiated by factors external to the blood, in the tissues adjacent to damaged blood vessel – in particular, it is initiated by clotting factor III, thromoboplastin from the damaged tissues.
2. It is a longer, multistep process and it takes a little longer for the blood to clot by this mechanism.	2. It involves fewer chemical reaction steps and produce a clot a little more quickly than the intrinsic pathway.

7. Long Answer Questions

Question 1.
Smita was working in a garage with the doors closed and automobiles engine running. After some time she felt breathless and fainted. What would be the reason? How can she be treated
OR
While working with the car engine in a closed garage, John suddenly felt dizzy and fainted what is the possible reason?
Answer:

1. As Smita and John were working with the car engine running in a closed garage, they must be suffering from carbon monoxide poisoning.
2. Carbon monoxide (CO) is a highly toxic gas produced when fuels burn incompletely from automobile engines.
3. Because of strong affinity of haemoglobin with carbon monoxide, it readily combines with carbon monoxide to from a stable compound, carboxyhaemoglobin. Thus, less haemoglobin is available for oxygen transport depriving the cells of oxygen.
4. Exposure to carbon monoxide can usually leads to throbbing headache, drowsiness, breathlessness and often person gets fainted. In extreme cases carbon monoxide poisoning usually leads to unconsciousness, convulsions, cardiovascular failure, coma and eventually death.

The breathless persons can be treated by following method:

1. Oxygen treatment : The best way to treat carbon monoxide poisoning is to breathe in pure oxygen (high-dose oxygen treatment)
2. Oxygen chamber : Doctor may temporarily place her in a pressurized oxygen chamber (also known as a hyperbaric oxygen chamber)

Question 2.
Shreyas went to a garden on a wintry morning. When he came back, he found it difficult to breath and stated wheezing. What could be the possible condition and how can he be treated?
Answer:
(1) It indicates that Shreyas might be suffering from allergic reactions. He may have come in contact with allergens such as pollen, dust, pet dander or other environmental substances on his way in the garden. Or Shreyas may be already a patient of Asthma and his symptoms may have aggrevated due to wintry climate.

(2) If a person is allergic to a substance, such as pollen, his immune system reacts to the substance as if it was foreign and harmful, and tries to destroy it.

(3) The body reacts to these allergens by making and releasing substances known as IgE antibodies. These IgE antibodies attach to most cells in the body which release histamine. Histamine is the main substance responsible for pollen allergy symptoms such as difficulty in breathing, wheezing, sneezing, itchy throat, etc.

(4) Treatment : There are several drugs to treat the allergic reactions:

• Antihistamines such as cetirizine or diphenhydramine.
• Decongestants, such as pseudoephedrine or oxymetazoline.
• Medications that combine an antihistamine and decongestant such as Actifed and Claritin-D.

Question 3.
Why can you feel a pulse when you keep a finger on the wrist or neck but not when you keep them on a vein?
Answer:
(1) When the heart contracts, it creates pressure that pushes blood out of heart. This pressure acts like a wave. This “wave” of pressure is the pulse you feel. But this pressure is not constant.

(2) When the heart pumps the blood out of it at the time of systole, there is maximum pressure in the arteries. This pressure weakens considerably when it reaches capillaries, and so the veins which are away from the heart are under less pressure. Due to low pressure veins have valves to prevent backflow of blood.

(3) The pressure in the arteries can be felt every time the heart beats, especially in arteries which come to surface of the body like that of the wrist and neck but not in veins.

(4) The pressure in veins is always weaker than in arteries, resulting in a weaker pulse to the point that it is undetectable by touch
alone.

(5) Owing to this, when we keep finger on the arteries of wrist or neck, we feel a pulse but not when we keep it on a vein.

Question 4.
A man’s pulse rate is 68 and cardiac output is 5500 cm3. Find the stroke volume.
Answer:
Cardiac output is the volume of blood pumped out per min for a normal adult human being it is calculated as follows:
Cardiac output = Heart rate × Stroke volume
Given : Cardiac output = 5500 cm³
Pulse rate = Heart rate = 68
By using these values stroke volume of is calculated as follows:
∴ Cardiac output = Heart rate × Stroke volume
∴ Stroke volume = Cardiac output/Heart rate
= 5500/68
= Approx. 80. ∴ Stroke volume is 80 ml.

Question 5.
Which blood vessel leaving from the heart will have the maximum content of oxygen and why?
Answer:

1. The Aorta leaving the heart from left ventricle carry the maximum content of oxygen.
2. Deoxygenated blood becomes oxygenated in the pulmonary capillaries surrounding the alveoli of lungs. The oxygenated blood from lungs is collected by the four pulmonary veins.
3. These pulmonary veins carry that oxygenated blood to left atrium of heart. During atrial systole that blood is carried to left ventricle.
4. Left ventricle then pumps that oxygenated blood to Aorta during ventricular systole. Therefore, aorta has the maximum content of oxygen.

Question 6.
If the duration of the atrial ‘systole is 0.1 second and that of complete diastole is 0.4 second, then how does one cardiac cycle complete in 0.8 second?
Answer:

1. The time duration required to complete one cardiac cycle is 0.8 second.
2. Cardiac cycle is divided into three important phases, viz, atrial systole, ventricular systole and joint diastole.
3. Atrial systole in normal condition lasts for 0.1 second, ventricular systole follows atrial systole and lasts for 0.3 second whereas joint diastole or complete diastole lasts for about 0.4 second.
4. In this way one cardiac cycle is completed in 0.8 second.

Question 7.
How is blood kept moving in the large veins of the legs?
Answer:
1. When heart undergoes systole, it pushes the blood with pressure in aorta. This pressure moves the entire circulation of the blood throughout the body. Aorta gives rise to dorsal aorta after supplying to upper parts of body. Then it divides into two arteries which enter two legs. The blood is forced to move in the legs due to blood pressure and also aided by gravity.

2. In addition, the muscles in legs help transport blood back to our heart. As the muscles of our body contract and relax to move our limbs, they squeeze the blood in veins and the blood is then pushed towards the heart.

3. The veins in legs also have valves to keep this process going and prevent blood from flowing back down towards the feet.

4. In this way blood is kept moving in the large veins of the legs.

Question 8.
Describe histological structure of artery, vein and capillary.

Answer:
Histological structure of artery and vein.

1. Artery is a thick walled blood vessel that carries oxygenated blood. (Exception is pulmonary artery which carries deoxygenated blood from heart to lungs for oxygenation.)
2. All the arteries arise from heart and carry blood away from the heart.
3. Each artery is made up of three layers, viz. tunica externa, tunica media and tunica interna.
4. Tunica externa or adventitia is the thickest layer of all. It is the outermost coat made up of connective tissue with elastic and collagen fibres.
5. Tunica media is the middle coat made up of smooth muscle fibres and elastic fibres. It withstands high blood pressure during ventricular systole. It is also thick.
6. Tunica interna or intima is the innermost coat made of endothelium and elastic layer.

Histology of Capillaries:

1. Capillaries are the smallest and thinnest blood vessels. Capillaries are formed by the division and re-division of the arterioles.
2. The wall of the capillary is made up of endothelium or squamous epithelium.
3. The capillary wall is permeable to water and dissolved substances.
4. Exchange of respiratory gases, nutrients, excretory products, etc. takes place through the capillary wall.
5. Capillaries unite to form venules.

Question 9.
What is blood pressure? How is it measured? Explain factors affecting blood pressure.
Answer:
1. Blood pressure:

1. The pressure exerted by blood on the wall of the blood vessels is called blood pressure. Pressure exerted by blood on the wall of arterial wall is arterial blood pressure. Blood pressure is described in two terms viz. systolic blood pressure and diastolic blood pressure.
2. Systolic blood pressure is the pressure exerted on arterial wall during ventricular contraction (systole). For a normal healthy adult the average value is 120 mmHg.
3. Diastolic blood pressure is the pressure on arterial wall during ventricular relaxation (diastole). For a normal healthy adult it is 80 mmHg.
4. B. E = SP/DP = 120/80 mmHg. Blood pressure is normally written as 120/80 mmHg. Difference between systolic and diastolic pressure is called pulse pressure normally, it is 40 mmHg.

2. Measurement of blood pressure:

1. Blood pressure is measured with the help of an instrument called sphygmomanometer.
2. The instrument consists of inflatable rubber bag cuff covered by a cotton cloth. It is connected with the help of tubes to a mercury manometer on one side and a rubber bulb on the other side.
3. During measurement, the person is asked to lie in a sleeping position. The instrument is placed at the level of heart and the cuff is tightly wrapped around upper arm.
4. The cuff is inflated till the brachial artery is blocked due to external pressure. Then pressure in the cuff is slowly lowered till the first pulsatile sound is produced. At this moment, pressure indicated in manometer is systolic pressure. Sounds heard during this measurement of blood pressure are called as Korotkoff sounds.
5. Pressure in the cuff is further lowered till any pulsatile sound cannot be heard due to smooth blood flow. At this moment, pressure indicated in manometer is diastolic pressure an optimal blood pressure (normal) level reads 120/80 mmHg.

3. Factors affecting blood pressure:

1. Cardiac output : Normal cardiac output is 5 lit/min. Increase in cardiac output increases systolic pressure.
2. Peripheral resistance : Peripheral resistance depends upon the diameter of blood vessels. Decrease in diameter of arterioles and capillaries under the effect of vasopressin cause increase in peripheral resistance and thereby increase in blood pressure.
3. Blood volume : Loss of blood in accidents decreases blood volume and thus cause decrease in blood pressure.
4. Viscosity of blood : Blood pressure is directly proportional to viscosity of blood.
5. Age : Blood pressure increases with age due to increase in inelasticity of blood vessels.
6. Venous return : Amount of blood brought to the heart via the veins per unit time is called the venous return and it is directly proportional to blood pressure.
7. Length and diameter of blood vessels : Blood pressure is directly proportional to the total length of the blood vessel. Blood pressure can also be affected by vasoconstriction or vasodilation.
8. Gender : Females have slightly lower BP than males of her age before menopause. However, the risk of high B. P increases in the females after menopause sets in.

Question 10.
Describe human blood and give its functions.
Answer:
Blood Composition:

1. Blood is a red coloured fluid connective tissue derived from embryonic mesoderm.
2. It has two components – the fluid plasma (55%) and the formed elements i.e. blood cells (44%).
3. Plasma is a straw coloured, slightly alkaline and viscous fluid having 90% water and 10% solutes such as proteins, nutrients, nitrogenous wastes, salts, hormones, etc.
4. Blood corpuscles are of three types, viz. erythrocytes (RBCs), white blood corpuscles (WBCs) and thrombocytes (platelets).
   

(5) Red blood corpuscles or Erythrocytes:

1. Erythrocytes or red blood corpuscles. They are circular, biconcave, enucleated cells.
2. The RBC size : 7 pm in diameter and 2.5 pm in thickness.
3. The RBC count : 5.1 to 5.8 million RBCs/ cu mm of blood in an adult male and 4.3 to 5.2 million/cu mm in an adult female.
4. The average life span of RBC : 120 days.
5. RBCs are formed by the process of erythropoiesis. In foetus, RBC formation takes place in liver and spleen whereas in adults it occurs in red bone marrow.
6. The old and worn out RBCs are destroyed in liver and spleen.
7. Polycythemia is an increase in number of RBCs while erythrocytopenia is decrease in their (RBCs) number.

Functions of RBCs:

1. Transport of oxygen from lungs to tissues and carbon dioxide from tissues to lungs with the help of haemoglobin.
2. Maintenance of blood pH as haemoglobin acts as a buffer.
3. Maintenance of the viscosity of blood.

(6) White blood corpuscles / Leucocytes:

1. Leucocytes or White Blood Corpuscles (WBCs) are colourless, nucleated, amoeboid and phagocytic cells.

2. Their size ranges between 8 to 15 pm. Total WBC count is 5000 to 9000 WBCs/cu mm of blood. The average life span of a WBC is about 3 to 4 days.

3. They are formed by leucopoiesis in red bone marrow, spleen, lymph nodes, tonsils, thymus and Payer’s patches, whereas the dead WBCs are destroyed by phagocytosis in blood, liver and lymph nodes.

4. Leucocytes are mainly divided into two types, viz., granulocytes and agranulocytes.

5. Granulocytes : Granulocytes are cells with granular cytoplasm and lobed nucleus. Based on their staining properties and shape of nucleus, they are of three types, viz. neutrophils, eosinophils and basophils.

(I) Neutrophils:

1. In neutrophils, the cytoplasmic granules take up neutral stains.
2. Their nucleus is three to five lobed.
3. It may undergo changes in structure hence they are called polymorphonuclear leucocytes or polymorphs.
4. Neutrophils are about 70% of total WBCs.
5. They are phagocytic in function and engulf microorganisms.

(II) Eosinophils or acidophils:

1. Cytoplasmic granules of eosinophils take up acidic dyes such as eosin. They have bilobed nucleus.
2. Eosinophils are about 3% of total WBCs.
3. They are non-phagocytic in nature.
4. Their number increases (i.e. eosinophilia) during allergic conditions.
5. They have antihistamine property.

(III) Basophils:

1. The cytoplasmic granules of basophils take up basic stains such as methylene blue.
2. They have twisted nucleus.
3. In size, they are smallest and constitute about 0.5% of total WBCs.
4. They too are non-phagocytic.
5. Their function is to release heparin which acts as an anticoagulant and histamine that is involved in inflammatory and allergic reaction.

6. Agranulocytes : There are two types of agranulocytes, viz. monocytes and lymphocytes. Agranulocytes do not show cytoplasmic granules and their nucleus is not lobed. They are of two types, viz. lymphocytes and monocytes.
(I) Lymphocytes:

1. Agranulocytes with a large round nucleus are called lymphocyte.
2. They are about 30% of total WBCs.
3. Agranulocytes are responsible for immune response of the body by producing antibodies.

(II) Monocytes:

1. Largest of all WBCs having large kidney shaped nucleus are monocytes. They are about 5% of total WBCs.
2. They are phagocytic in function.
3. They can differentiate into macrophages for engulfing microorganisms and removing cell debris. Hence they are also called scavengers.
4. At the site of infections they are seen in more enlarged form.

(7) Thrombocytes/Platelets:

1. Thrombocytes or platelets are non- nucleated, round and biconvex blood corpuscles.
2. They are smallest corpuscles measuring about 2.5 to 5 mm in diameter with a count of about 2.5 lakhs/cu mm of blood.
3. Their life span is about 5 to 10 days.
4. Thrombocytes are formed from megakaryocytes of bone marrow. They break from these cells as fragments during the process of thrombopoiesis.
5. Thrombocytosis is the increase in platelet count while thrombocytopenia is decrease in platelet count.
6. Thrombocytes possess thromboplastin which helps in clotting of blood.
7. Therefore, at the site of injury platelets aggregate and form a platelet plug. Here they release thromboplastin due to which further blood clotting reactions take place.

(8) Functions of blood:

1. Transport of oxygen and carbon dioxide
2. Transport of food
3. Transport of waste product
4. Transport of hormones
5. Maintenance of pH
6. Water balance
7. Transport of heat
8. Defence against infection
9. Temperature regulation
10. Blood clotting/coagulation
11. Helps in healing

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 7 Plant Growth and Mineral Nutrition Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 7 Plant Growth and Mineral Nutrition

1. Multiple choice questions

Question 1.
Which of the hormone can replace vernalization?
(a) Auxin
(b) Cytokinin
(c) Gibberellins
(d) Ethylene
Answer:
(c) Gibberellins

Question 2.
The principle pathway of water translocation in angiosperms is ………………..
(a) Sieve cells
(b) Sieve tube elements
(c) Xylem
(d) Xylem and phloem
Answer:
(c) Xylem

Question 3.
Abscissic acid controls ………………..
(a) cell division
(b) leaf fall and dormancy
(c) shoot elongation
(d) cell elongation and wall formation
Answer:
(b) leaf fall and dormancy

Question 4.
Which is employed for artificial ripening of banana fruits?
(a) Auxin
(b) Ethylene
(c) Cytokinin
(d) Gibberellin
Answer:
(b) Ethylene

Question 5.
Which of the following is required for stimulation of flowering in plants?
(a) Adequate oxygen
(b) Definite photoperiod
(c) Adequate water
(d) Water and minerals
Answer:
(b) Definite photoperiod

Question 6.
For short day plants, the critical period is ………………..
(a) light
(b) dark/night
(c) UV rays
(d) Both (a) and (c)
Answer:
(b) dark/night

Question 7.
Which of the following is NOT day neutral plant?
(a) Tomato
(b) Cotton
(c) Sunflower
(d) Soybean
Answer:
(d) Soybean

Question 8.
Essential macro elements are ………………..
(a) manufactured during photosynthesis
(b) produced by enzymes
(c) obtained from soil
(d) produced by growth hormones
Answer:
(c) obtained from soil

Question 9.
Function of Zinc is ………………..
(a) closing of stomata
(b) biosynthesis of 3-IAA
(c) synthesis of chlorophyll
(d) oxidation of carbohydrates
Answer:
(b) biosynthesis of 3-LAA

Question 10.
Necrosis means ………………..
(a) yellow spot on the leaves
(b) death of tissue
(c) darkening of green colour in leaves
(d) wilting of leaves
Answer:
(b) death of tissue

Question 11.
Conversion of nitrates to nitrogen is called ………………..
(a) ammonification
(b) nitrification
(c) nitrogen fixation
(d) denitrification
Answer:
(d) denitrification

Question 12.
How many molecules of ATP are required to fix one molecule of nitrogen?
(a) 12
(b) 20
(c) 6
(d) 16
Answer:
(d) 16

2. Very short answer questions

Question 1.
Enlist the phases of growth in plants.
Answer:
The three phases of growth are phase of cell division, phase of cell enlargement and phase of cell maturation.

Question 2.
Give full form of IAA.
Answer:
Full form is Indole Acetic Acid.

Question 3.
What does it mean by ‘open growth’?
Answer:
In plants the growth is indeterminate and takes place throughout the life at specific regions having meristems.

Question 4.
Plant stress hormone.
Answer:
Abscissic acid.

Question 5.
What is denitrification?
Answer:
Anaerobic bacteria can convert nitrates of soil back into nitrogen gas. That process performed by denitrifying bacteria is denitrification.

Question 6.
Bacteria responsible for conversion of nitrite to nitrate.
Answer:
Nitrobacter.

Question 7.
What is the role of gibberellins in rosette plants?
Answer:
In rosette plants like beet and cabbage, bolting, i.e. elongation of internodes before flowering is observed due to effect of gibberellins.

Question 8.
Vernalization
Answer:
The response of plant to the influence of low temperature on flowering in plants is called vernalization.

Question 9.
Photoperiodism
Answer:
The response of plant to the influence of light for initiation of flowering is known as photoperiodism.

Question 10.
What is grand period of growth?
Answer:
There are three phases of growth and the total time required for all phases to occur is called grand period of growth.

3. Short Answer Questions

Question 1.
(i) Differentiation
Answer:

1. It is a process of maturation of cells derived from apical meristems.
2. Differentiation is a permanent change in structure and function of cells that leads to its maturation.
3. Cell undergoes major anatomical and physiological change during differentiation process.
4. In hydrophytic plants parenchyma cells develop large schizogenous cavities which help them in aeration, buoyancy and mechanical support.

(ii) Redifferentiation
Answer:

1. It is a process in which cells produced by de-differentiation lose their capacity of division and become mature.
2. The cells mature to perform specific function.
3. Interfascicular cambium is formed by process of dedifferentiation loses its capacity to divide.
4. Secondary xylem and secondary phloem is formed form this cambium in vascular cylinder.

Question 2.
Arithmetic growth and Geometric growth
Answer:

Arithmetic growth	Geometric growth
1. In arithmetic growth only one daughter cell continues to divide, while the other undergoes differentiation and maturation.	1. In geometric growth both the daughter cells continue to divide and redivide again and again.
2. Rate of growth is constant.	2 Rate growth is initially slow but later on rapid rate.
3. Linear curve is obtained.	3. Exponential curve is obtained.
4. Mathematical expression is Lt = Lo + rt whereLt = length of time ‘t’ Lo = Length at time zero rt = growth rate, t = time of growth	4. Mathematical expression is Wt = Woe rt where, Wt = final size, Wo = initial size, r = growth rate, t = time of growth E = base of natural logarithm
5. e.g. Elongation of root	5. e.g. Divisions of zygote during embryo development.

Question 3.
Enlist the role and deficiency symptoms of: (a) nitrogen (b) phosphorus (c) potassium.
Answer:
(a) Nitrogen:
Role : Constituent of proteins as amino acids, nucleic acids, vitamins, hormones, coenzymes, ATP and chlorophyll molecule.
Deficiency symptoms : stunted growth and chlorosis.

(b) Phosphorus:
Role : Constituent of cell membrane, certain proteins, nucleic acids and nucleotides, required for all phosphorylation reactions.
Deficiency symptoms : Poor growth, leaves dull green

(c) Potassium :
Role : Determination of anion – cation balance in cell, necessary for protein synthesis, involved in formation of cell membrane, opening and closing of stomata, activates enzymes, helps in maintenance of turgidity of cells.
Deficiency symptom : Yellow edges in leaves, premature death.

Question 4.
What is short day plant? Give any two examples.
Answer:
The plants which flower when the day length or light period is shorter than the critical photoperiod are called short day plants or SDP
SDPs usually flower during winter and late summer.
Examples – Dahlia, Aster, Tobacco, Chrysanthemum, Soybean (Glycine max) and Cocklebur (Xanthium).

Question 5.
What is vernalization? Give its significance.
Answer:
A low temperature or chilling treatment that induces early flowering in plants is known as vernalization.

Significance:

1. Due to chilling treatment crops can be produced earlier.
2. Crops can be grown in areas where they do not grow naturally.

4. Long answer questions

Question 1.
Explain sigmoid growth curve with the help of diagram.
Answer:

1. When growth occurs in plants three distinct phases of growth are noticed.
2. Phase of cell formation is first phase where meristematic cells divide and new cells added.
3. In phase of cell enlargement newly formed cells elongate and with turgidity there is cell enlargement.
4. In phase of cell maturation cells get differentiated.
5. When we compare the growth rate it differs in these three phases.
6. In first phase or lag phase it is slow, while in log phase or exponential phase, growth rate accelerates and it reaches maximum.
7. In stationary phase of maturation growth rate slows down and comes to steady state.
8. When this changing rate of growth is plotted against time duration in a graph a sigmoid or S-shaped growth curve is obtained.

Question 2.
Describe the types of plants on the basis of photoperiod required, with the help of suitable examples.
Answer:

1. Effect of light duration on flowering of plants is known as photoperiodism.
2. Depending on photoperiodic response, plants are categorised into three types – Short day plants, long day plants and day neutral plants.

1. Short day plants : Plants that flower under short day length conditions are called short day plants. Plants such as Dahlia, Xanthium, Soybean, Aster, Tobacco and Chrysanthemum are short day plants or SDR. Short day plants require a long uninterrupted dark period for flowering. Therefore, they are also called long night plants.

2. Long day plants : Plants that flower only when they are exposed to light period longer than their critical photoperiod are called long day plants or LDP Long day plants require a short dark or night period for flowering. Hence, they are also called short night plants. Plants such as radish, spinach, wheat, poppy, cabbage, pea, sugar beet, etc. are long day plants.

3. Day neutral plants : Plants in which the flowering is not affected by the day length period are called day neutral plants or DNP or photoneutral plants. Plants such as cucumber, sunflower, cotton, balsam, maize, tomato, etc. are day neutral plants.

Question 3.
Explain biological nitrogen fixation with example.
Answer:

1. Conversion of atmospheric nitrogen into nitrogenous salts to make it available to plants for its update is described as nitrogen fixation.
2. When living organisms are involved in nitrogen fixation process it is known as biological nitrogen fixation.
3. The process is mainly carried out by prokaryotic organisms, i.e. different kinds of bacteria present in soil.
4. The nitrogen fixing organisms are known as diazotrophs or nitrogen fixers and about 70% nitrogen is fixed by them.
5. The nitrogen fixers are either free living bacteria or symbiotic associated with other higher plants e.g. Rhizobium.
6. The cyanobacteria have specialized cells heterocysts which help in process of nitrogen fixation.
7. Nitrogen fixation is high energy requiring process and 16 ATP molecules are needed for fixation of one molecule of nitrogen to ammonia.
8. Soil bacteria like Nitrosomonas, Nitrosocyccus convert ammonia to nitrate and the Nitrobacter convert nitrite to nitrate. This is known as nitrification, biological oxidation.
9. These bacteria are chemoautotrophic and utilize these processes for their metabolism.
10. Fabaceae plants like pea, bean have root nodules which harbour symbiotic bacterium Rhizobium which fixes nitrogen. It is host specific, soil bacterium, Nitrogen is made available to host plant.

Question 4.
Write on macro and micro nutrients required for plant growth.
Answer:

1. Plants absorb mineral nutrients from their surroundings.
2. For a proper growth of plants about 35 to 40 different elements are required.
3. Plants absorb these nutrients in ionic or dissolved form from soil with their root system e.g. Phosphorus as PO₄^–, Sulphur as SO₄^2- etc.
4. Based on their requirement in quantity, they are classified as major nutrients or macronutrients and those needed in small amounts Eire minor or micronutrients.
5. Macroelements are required in large amounts, as they play nutritive and structural roles e.g. C, H, O, R Mg, N, K, S and Ca. – Ca pectate cell wall component, Mg component of chlorophyll.
6. C, H, O are non-mineral major elements obtained from air and water e.g. CO₂ is source of carbon, Hydrogen from water.
7. Microelements are required in traces as they mainly have catalytic role as co-factors or activators of enzymes.
8. Microelements may be needed for certain activity in life cycle of plant e.g. B for pollen germination, Si has protective role during stress conditions and fungal attacks, Al enhances availability of phosphorus.
9. The important micronutrients for plant growth are Mn, B, Cu, Zn, Cl.

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 6 Plant Water Relation Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 6 Plant Water Relation

1. Multiple Choice Questions

Question 1.
In soil, water available for absorption by root is ……………..
(a) gravitational water
(b) capillary water
(c) hygroscopic water
(d) combined water
Answer:
(b) capillary water

Question 2.
The most widely accepted theory for ascent of sap is ……………..
(a) capillarity theory
(b) root pressure theory
(c) diffusion
(d) transpiration pull theory
Answer:
(d) transpiration pull theory

Question 3.
Water movement between the cells is due to ……………..
(a) T.E
(b) W.P
(c) D.P.D.
(d) incipient plasmolysis
Answer:
(c) D.P.D.

Question 4.
In guard cells, when sugar is converted into starch, the stomata pore ……………..
(a) closes almost completely
(b) opens partially
(c) opens fully
(d) remains unchanged
Answer:
(a) closes almost completely

Question 5.
Surface tension is due to ……………..
(a) diffusion
(b) osmosis
(c) gravitational force
(d) cohesion
Answer:
(d) cohesion

Question 6.
Which of the following type of solution has lower level of solutes than the solution?
(a) Isotonic
(b) Hypotonic
(c) Hypertonic
(d) Anisotonic
Answer:
(b) Hypotonie

Question 7.
During rainy season wooden doors warp and become difficult to open or to close because of ……………..
(a) plasmolysis
(b) imbibition
(c) osmosis
(d) diffusion
Answer:
(b) imbibition

Question 8.
Water absorption takes place through ……………..
(a) lateral root
(b) root cap
(c) root hair
(d) primary root
Answer:
(c) root hair

Question 9.
Due to low atmospheric pressure the rate of transpiration will ……………..
(a) increase
(b) decrease rapidly
(c) decrease slowly
(d) remain unaffected
Answer:
(a) increase

Question 10.
Osmosis is a property of ……………..
(a) solute
(b) solvent
(c) solution
(d) membrane
Answer:
(c) solution

2. Very short answer question

Question 1.
What is osmotic pressure?
Answer:
The pressure exerted due to osmosis is osmotic pressure.

Question 2.
Name the condition in which protoplasm of the plant cell shrinks.
Answer:
Plasmolysis

Question 3.
What happens when a pressure greater than the atmospheric pressure is applied to pure water or a solution?
Answer:
When a pressure greater than the atmospheric pressure is applied to pure water or a solution then water potential of pure water or solution increases.

Question 4.
Which type of solution will bring about deplasmolysis ?
Answer:
Placing a plasmolysed cell in hypotonic solution will bring about deplasmolysis.

Question 5.
Which type of plants have negative root pressure?
Answer:
Plants showing excessive transpiration have negative root pressure.

Question 6.
In which conditions transpiration pull will be affected?
Answer:
Due to temperature fluctuations during day and night gas bubbles may be formed which affects transpiration pull.

Question 7.
Mention the shape of guard cells in Cyperus.
Answer:
Kidney shaped and dumbbell shaped guard cells are seen.

Question 8.
Why do diurnal changes occur in osmotic potential of guard cells?
Answer:
Enzyme activity of phosphorylase converts starch into sugar during daytime and sugar is converted to starch during night. This causes changes in osmotic potential of guard cells.

Question 9.
What is symplast pathway?
Answer:
When water is absorbed by root hair it passes across from one living cell to other living cell through the plasmodesmatal connections between them, then it is called symplast pathway across the root.

3. Answer the Following Questions

Question 1.
Describe mechanism of absorption of water.
Answer:

1. The absorption of water takes place by two modes, i.e. active absorption and passive absorption.
2. Passive absorption is the chief method of absorption (98%).
3. There is no expenditure of energy in passive absorption.
4. Transpiration pull is a driving force and water moves depending upon concentration gradient. Water is pulled upwards.
5. It occurs during daytime when there is active transpiration.
6. Active absorption occurs usually during night time as due to closure of stomata transpiration stops.
7. Water absorption is against D.ED. gradient, A.T.R energy is required which is available from respiration.
8. Active absorption may be osmotic or non- osmotic type.
9. For osmotic absorption root pressure has a role.

Question 2.
Discuss theories of water translocation.
Answer:

1. Translocation of water is transport of water along with dissolved minerals from roots to aerial parts.
2. The movement is against the gravity and described as ascent of sap.
3. The translocation occurs through lumen of water conducting tissue xylem mainly vessels and tracheids.
4. Different theories have been discussed for translocation mechanism like vital force theory (Root pressure), relay pump, physical force (capillary), etc.
5. Cohesion tension theory or transpiration pull theory is most widely accepted theory.

Question 3.
What is transpiration? Describe mechanism of opening and closing of stomata.
Answer:

1. The loss of water in the form of vapour is called transpiration.
2. Stomatal transpiration is a main type of transpiration where minute pores are concerned with it.
3. Stomata are bounded by two guard cells which in turn are surrounded by accessory cells.
4. Opening and closing of stomata is controlled by turgidity of guard cells.
5. When guard cells become turgid due to endosmosis their lateral thin and elastic wall bulges or stretch out.
6. The inner thick and inelastic wall is pulled apart, thus the stoma opens during daytime.
7. At night when guard cells become flaccid due to exosmosis the wall relaxes and stoma closes.
8. Endosmosis and exosmosis takes place due to changes in osmotic potential of guard cells.

Question 4.
What is transpiration? Explain role of transpiration.
Answer:
Transpiration : The loss of water from plant body in the form of vapour is called transpiration.

Role of transpiration:

1. Removal of excess of water
2. Helps in passive absorption of water and minerals
3. Helps in ascent of sap – transpiration pull
4. Maintains turgor of cells
5. Imparts cooling effect by reducing temperature 90% – 93% is stomatal transpiration and hence when stomata are open gaseous exchange takes place.

Question 5.
Explain root pressure theory and its limitations.
Answer:

1. Root pressure theory is proposed by J. Pristley.
2. For translocation of water, activity of living cells of root is responsible.
3. Absorption of water by root hair is a constant and continuous process and due to this a hydrostatic pressure is developed in cortical cells.
4. Owing to this hydrostatic pressure i.e. root pressure, water is forced into xylem and further conducted upwards.
5. Root pressure is an osmotic phenomenon.

Limitation of this theory:

1. Not applicable to tall plants above 20 metres.
2. Even in absence of root pressure ascent of sap is noticed.
3. In actively transpiring plants, root pressure is not developed.
4. In taller gymnosperms, root pressure is zero.
5. Xylem sap is under tension and shows negative hydrostatic pressure.

Question 6.
Explain capillarity theory of water translocation.
Answer:

1. Capillarity theory of water translocation is proposed by Bohem.
2. Capillarity is because of surface tension and cohesive forces and adhesive forces of water molecules.
3. Xylem vessels and tracheids are tubular elements having their lumen.
4. In these elements water column exists due to combined action of cohesive and adhesive forces of water and lignified wall.
5. As a result of this capillarity water is raised upwards.

Question 7.
Why is transpiration called ‘a necessary evil’?
Answer:

1. The loss of water in the form of water vapour is called transpiration.
2. About 90 – 93% of transpiration occurs through stomata, small apertures located in the epidermis of leaves.
3. For this process stomata must remain open and then only gaseous exchange by diffusion takes places.
4. Gaseous exchange is necessary for respiration and photosynthesis. If stomata remain closed then it will affect productivity of plant.
5. The process is necessary evil because water which is important for plant is lost in the process.
6. At the same time it helps in absorption of water and its translocation. Hence it cannot be avoided.
   So Curtis has rightly called it as necessary evil.

Question 8.
Explain movement of water in the root.
Answer:

1. Root hairs absorb water by imbibition then diffusion which is followed by osmosis.
2. As water is taken inside the root hair cell it becomes turgid i.e. increase in turgor pressure (T.E)
3. Root hair cell has less D.ED. but adjacent cortical cell has more D.PD.
4. The inner cortical cell has more osmotic potential so it will suck water from root hair cell.
5. Root hair cell becomes flaccid and ready to absorb soil water.
6. Water is passed on similarly in inner cortical cells.
7. Water moves rapidly through loose cortical cells up to endodermis and through passage cells in pericycle.
8. From pericycle due to hydrostatic pressure developed it is forced into protoxylem.

Question 9.
(i) Osmosis
Answer:
It is a special type of diffusion of solvent through a semipermeable membrane.

(ii) Diffusion
Answer:
It is the movement of ions/ atoms/molecules of a substance from the region of higher concentration to the region of their lower concentration.

(iii) Plasmolysis
Answer:
Exo-osmosis in a living cell when placed in hypertonic solution is called plasmolysis.

(iv) Imbibition
Answer:
It is swelling up of hydrophilic colloids due to adsorption of water.

(v) Guttation
Answer:
The loss of water in the form of liquid is called guttation.

(vi) Transpiration
Answer:
The loss of water from plant body in the form of vapour is called transpiration.

(vii) Ascent of sap
Answer:
The transport of water with dissolved minerals in it from root to other aerial parts of plant against the gravity is called ascent of sap.

(viii) Active absorption
Answer:
Water absorption by activity of root which is against the D.PD. gradient along with expenditure of A.T.E energy generated by respiration is the process of active absorption.

(ix) Diffusion Pressure Deficit (D.P.D.)
Answer:
The difference in the diffusion pressures of pure solvent and the solvent in a solution is called diffusion pressure deficit.

(x) Turgor pressure
Answer:
It is the pressure exerted by turgid cell sap on to the cell membrane and cell wall.

(xi) Water potential
Answer:
Chemical potential of water is called water potential.

(xii) Wall pressure
Answer:
Thick and rigid cell wall exerts a counter pressure to turgor pressure developed on the cell sap is called wall pressure that operates in opposite direction.

(xiii) Root pressure
Answer:
As absorption of water by root hair being a continuous process, a sort of hydrostatic pressure is developed in living cells of root, this is called root pressure.

Question 10.
Osmotic Pressure (O.P) and Turgor Pressure (T.P)
Answer:

Osmotic Pressure (O.R)	Turgor Pressure (T.P.)
1. The pressure exerted due to osmosis is called osmotic pressure.	1. The pressure exerted by turgid cell sap on cell membrane and cell wall, is called turgor pressure.
2. It is pressure caused by water when it moves by osmosis.	2. It is pressure caused by content of cell (cell sap).
3. It is generated by the osmotic flow of water through a semipermeable membrane.	3. It is maintained by osmosis.

Question 11.
How are the minerals absorbed by the plants ?
Answer:

1. Soil is the chief source of minerals for the plants.
2. Minerals get dissolved in the soil water.
3. Minerals are absorbed by the plants in the ionic form mainly through roots.
4. Absorption of minerals is independent of water.
5. Absorbed minerals are pulled upwards along with xylem sap.
6. Mineral ions can be remobilized in the plant body form older parts to young plants E.g. Ions of S, P and N.

4. Long answer questions

Question 1.
Describe structure of root hair.
Answer:

1. Water from soil is absorbed by plants with the help of root hairs.
2. Root hairs are present in zone of absorption.
3. Epidermal cells form unicellular extensions which are short lived (ephemeral) structures i.e. root hairs.
4. Root hairs are nothing but cytoplasmic extensions of epiblema cell.
5. Root hairs are long tube like structures of about 1 to 10 mm.
6. They are colourless, unbranched and very delicate structures.
7. A large central vacuole is surrounded by thin layer of cytoplasm, plasma membrane and outer cell wall.
8. The cell wall of root hair is thin and double layered with outer layer of pectin and inner layer of cellulose which is freely permeable.

Question 2.
Write on journey of water from soil to xylem in roots.
Answer:

1. Unicellular root hairs which are tubular extensions of epiblema cells absorb readily available capillary water from soil.
2. The three physical processes imbibition, diffusion and osmosis are concerned with absorption of water.
3. Water molecules get adsorbed on cell wall of root hair (imbibition).
4. They enter the root hair cell by diffusion through cell wall which is freely permeable.
5. By process of osmosis they enter through plasma membrane which is semipermeable.
6. The root hair cell becomes turgid and hence its turgor pressure increases and D.ED. value decreases.
7. The adjacent cell of cortex has more D.ED. value as its osmotic potential is more.
8. The cortical cell thus takes water from epidermal cell which is turgid. This process goes on due to gradient of suction pressure developed from cell to cell till thin walled passage cells of endodermis.
9. From endodermis it will enter pericycle and then due to hydrostatic pressure it is forced in protoxylem cell.
10. The pathway of water is by apoplast and symplast.
11. When water passes through cell wall and intercellular spaces of cortex it is apoplast pathway.
12. When water passes across living cells through their plasmodesmatal connections it is symplast pathway.

Question 3.
Explain cohesion theory of translocation of water.
Answer:

1. This is very widely accepted theory of ascent of sap proposed by Dixon and Joly.
2. It is based on principles of adhesion and cohesion of water molecules and transpiration by plants.
3. A strong force of attraction existing between water molecules is cohesion and the force of attraction between water molecules and lignified walls of xylem elements is adhesion.
4. Ascent of sap occurs through lumen of xylem elements.
5. Owing to cohesive and adhesive forces a continuous water column is maintained in xylem from root to aerial parts i.e. leaves.
6. Transpiration occurs through stomata and transpiration pull is developed in leaf vessels.
7. This tension or pull is transmitted downwards through vein to roots which triggers ascent of sap.
8. In transpiration, water is lost in vapour form and this increases D.PD. of mesophyll cells that are near guard cells.
9. Mesophyll cells absorb water from xylem in leaf and a gradient of D.PD. or suction pressure (S. E) is set.
10. Owing to this gradient from guard cell to xylem in leaf, a transpiration pull or tension is created in xylem.
11. Hence water column is pulled upward passively against gravity.

Question 4.
Write on mechanism of opening and closing of stomata.
Answer:

1. Transpiration takes place through stomata. Turgidity of guard cells controls opening and closing of stomata
2. Turgor pressure exerted on unevenly thickened wall of guard cell is responsible for the movement.
3. The outer thin wall which is elastic is stretched out which pulls inner thick inelastic wall and thus stomata open.
4. When guard cells are flaccid that results in closure of stomata.
5. According to starch-sugar in ter conversion theory enzyme phosphorylase converts starch to sugar during daytime.
6. Sugar being osmotically active, the O.E of guard cells is increased. The water is absorbed from subsidiary cells. Due to turgidity walls are stretched and stoma opens.
7. During night-time sugar is converted to starch and hence guard cells loose water and become flaccid. Hence there is closure of stomata.
8. According to proton transport theory, the movement is due to transport of H⁺ and K⁺ ions.
9. Subsidiary cells are reservoirs of K⁺ ions. Starch is converted to malic acid which dissociate into malate and proton (H⁺) during day.
10. Proton transported to subsidiary cells and K⁺ ions are taken from it. This forms potassium malate in guard cells.
11. Potassium malate increases osmotic potential and endo osmosis occurs hence turgidity of guard cells. → stomata opens,
12. The uptake of K⁺ and Cl^– ions is stopped by abscissic acid formed during night. This changes permeability. Guard cells become hypotonic and loose water as they become flaccid stomata close.

Question 5.
What is hydroponics? How is it useful in identifying the role of nutrients?
Answer:
(1) Growing plants in aqueous (soilless) medium is known as hydroponics. [Greek word hudor = water and ponos = work]

(2) It is technique of growing plants by supplying all necessary nutrients in the water supply given to plant.

(3) A nutrient medium is prepared by dissolving necessary salts of micronutrients and macronutricnts In desired quantity and roots of plants are suspended in this liquid with appropriate support.

(4) Hydroponics is of great use in studying the deficiency symptoms of different mineral nutrients.

(5) The plants uptake mineral nutrients in the form of dissolved ions with the help of root hairs from the surrounding medium or nutrient solution supplied.

(6) While preparing the required nutricnt medium particular nutrient can be totally avoided and then the effect of lack of that nutrient can be studied in variation of plant growth.

(7) Any visible change noticed from normal structure and function of the plant is the symptom or hunger sign considered.

(8) For e.g. Yellowing of leaf is observed due to loss of chlorophyll pigments or Chiorosis is noticed if Magnesium is lacking as it is a structural componen of chlorophyll pigment.

Question 6.
Explain the active absorption of minerals.
Answer:

1. Plants absorb minerals from the soil with their root system.
2. MInerals are absorbed from the soil In the form of charged particles, positively charged cations and negatively charged anions.
3. The absorption of minerals against the concentration gradient which requires expenditure of metabolic energy is called active absorption.
4. The ATP energy derived from resp’ration in root cells Is utilized for active absrption.
5. Ions get accumulated in the root hair against the concentration gradient.
6. These ions pass into cortical cells and finally reach xylem of roots.
7. Along with the water these minerals are carried to other parts of plant.

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 5 Origin and Evolution of Life Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 5 Origin and Evolution of Life

1. Multiple Choice Questions

Question 1.
Who proposed that the first form of life could have come from pre-existing non-living organic molecules?
(a) Alfred Wallace
(b) Oparin and Haldane
(c) Charles Darwin
(d) Louis Pasteur
Answer:
(b) Oparin and Haldane

Question 2.
The sequence of origin of life may be
(a) Organic materials – inorganic materials – Eobiont – colloidal aggregates – cell.
(b) Inorganic materials – organic materials – colloidal aggregates – Eobiont – cell.
(c) Organic materials – inorganic materials – colloidal aggregates – cell.
(d) Inorganic materials – organic materials – Eobiont – colloidal aggregates – cell.
Answer:
(b) Inorganic materials – organic materials- colloidal aggregates – Eobiont – cell.

Question 3.
In Hardy-Weinberg equation, the frequency of homozygous recessive individual is represented by-
(a) p²
(b) pq
(c) q²
(d) 2pq
Answer:
(c) q²

Question 4.
Select the analogous organs.
(a) Forelimbs of whale and bat
(b) Flippers of dolphins and penguin
(c) Thorn and tendrils of bougainvillea and Cucurbita
(d) Vertebrate hearts or brains
Answer:
(b) Flippers of dolphins and penguin

Question 5.
Archaeopteryx is known as missing link because it is a fossil and share characters of both
(a) Fishes and amphibians
(b) Annelida and Arthropoda
(c) Reptiles and birds
(d) Chordates and non-chordates
Answer:
(c) Reptiles and birds

Question 6.
Identify the wrong statement regarding evolution.
(a) Darwin’s variations are small and directional.
(b) Mutations are random and non- directional.
(c) Adaptive radiations leads to divergent evolution.
(d) Mutations are non-random and directional.
Answer:
(d) Mutations are non-random and directional

Question 7.
Gene frequency in a population remain constant due to ……………….
(a) Mutation
(b) Migration
(c) Random mating
(d) Non-random mating
Answer:
(c) Random mating

Question 8.
Which of the following characteristic is not : shown by the ape?
(a) Prognathous face
(b) Tail is present
(c) Chin is absent
(d) Forelimbs are longer than hind limbs
Answer:
(b) Tail is present

Question 9.
………………. can be considered as connecting link between ape and man.
(a) Australopithecus
(b) Homo habilis
(c) Homo erectus
(d) Neanderthal man
Answer:
(a) Australopithecus

Question 10.
The Cranial capacity of Neanderthal man was ……………….
(a) 600 cc
(b) 940 cc
(c) 1400 cc
(d) 1600 cc
Answer:
(c) 1400 cc

2. Very short answer questions

Question 1.
Define
(i) Gene pool
Answer:
The sum total of genes of all individuals of interbreeding population or Mendelian population is called gene pool.

(ii) Gene frequency
Answer:
The proportion of an allele in the gene pool as compared with other alleles at the same locus is termed as gene frequency.

(iii) Organic evolution
Answer:
Organic evolution can be defined as slow, gradual, continuous and irreversible changes through which the present-day complex forms of the life developed (or evolved) from their simple pre-existing forms.

(iv) Population
Answer:
All individuals of the same species form a group which is called a population.

(v) Speciation
Answer:
Formation of new species from the pre-existing single group of organisms is called speciation.

Question 2.
What is adaptive radiation?
Answer:
The process of evolution which results in transformation of original species to many different varieties is called adaptive radiation.

Question 3.
If the variation occurs in population by chance alone and not by natural selection and bring change in frequencies of an allele, what is it called?
Answer:
If the variation occurs in population by chance alone and not by natural selection to bring change in frequencies of an allele, it is called genetic drift.

Question 4.
State the Hardy-Weinberg equilibrium law.
Answer:
The Hardy-Weinberg equilibrium law states that at equilibrium point both the allelic frequency and genotypic frequency remain constant from generation to generation, in the diploid, sexually reproducing, large, free interbreeding population in which mating is random and there is absence of any other factors that change the allele frequency.

Question 5.
What is homologous organs?
Answer:
Homologous organs are those organs, which are structurally similar but perform different functions.

Question 6.
What is vestigial organ?
Answer:
Vestigial organs are imperfectly developed and non-functional organs which are in degenerate form, they may be functional in some related and other animals or in ancestor.

Question 7.
What is the scientific name of modern man?
Answer:
Homo sapiens sapiens is the scientific name of modern man.

Question 8.
What is coacervate?
Answer:
Coacervates are colloidal aggregations of hydrophobic proteins and lipids which grew in size by taking up material from surrounding aqueous medium.

Question 9.
Which period is known as ‘age of Reptilia’?
Answer:
Jurassic period from Mesozoic era is known as age of Reptilia.

Question 10.
Name the ancestor of human which is described as man with ape brain.
Answer:
Australopithecus, the ancestor of human which is described as man with ape brain.

Short Answer Questions

Question 1.
Genetic drift.
Answer:

1. Genetic drift is random, directionless fluctuation that takes place in allele frequency.
2. It occurs by pure chance, in small sized population.
3. Genetic drift becomes an evolutional factor as it can change the gene frequency.
4. Sewall wright has given this concept and hence it is also known as Sewall wright effect.
5. Due to genetic drift, some alleles of a population are lost or reduced by chance and some others may be increased.
6. Some time, a few individuals become isolated from the large population and they produce new population in new geographical area.
7. Genetic drift is also called founders’ effect because original drifted population becomes ‘founders’ in the new area.
   E.g. Non-adaptive character of huge horns in Antelope is fixed due to genetic drift.

Question 2.
Enlist the different factors that are responsible for changing gene frequency.
Answer:
Gene flow, genetic drift, gene mutations, chromosomal aberrations such as deletion, duplication, inversion and translocation, genetic recombinations, natural selection, isolation are some of the factors which are responsible for changing the gene frequency.

Question 3.
Draw a graph to show that natural selection leads to disruptive change.
Answer:

Question 4.
Significance of fossils
Answer:

1. Fossils are studied under palaeontology. They are used in reconstruction of phylogeny.
2. Fossil study helps in studying various forms and structures of extinct animals.
3. By understanding the structure of fossil, record of missing link between two groups of organisms can be deduced.
4. By studying fossils various body forms and their evolution can be understood. They also help to understand the habit and habitat.
5. Some fossils provide the evolutionary evidences such a connecting links.

Question 5.
Write the objections to Mutation theory of Hugo de Vries.
Answer:
Objections to Mutation Theory:

1. Hugo de Vries observed the large and discontinuous variation. But these were chromosomal aberrations. Only gene mutations usually bring about minor changes.
2. Rate by which mutations take place is very slow as compared to the requirement of evolution.
3. Chromosomal aberrations are very unstable.
4. The organisms with chromosomal aberration are usually sterile and thus chromosomal aberrations have little significance in evolution.

Question 6.
What is disruptive selection? Give example.
Answer:
Disruptive selection:

1. The natural selection that disrupts the mean characters of the population, is called disruptive selection.
2. Greater number of individuals acquire peripheral character value at both ends of the distribution curve. E.g. Finches with large size or small size, both will be selected.
3. Extreme phenotypes are selected in evolutionary process and intermediate forms are eliminated.
4. When distribution curve is plotted it shows two peaks for two extremes.
5. Disruptive selection is rare because, nature always tries to balance the characters.
6. It ensures the effect on the entire gene pool of a population, considering all mating types or systems.

Example of disruptive selection:
African seed cracker finches are types of seed-feeder birds which have different sizes of beak. The seeds available to them were of small and large sized. Large beak sized birds feeds on large seeds while small beak sized birds feed on small seeds.

Such large and small birds thus thrive well. However, intermediate beak sized birds are unable to feed on either type of seeds so they starve and their population was decreased gradually. Natural selection eliminated them and thus the population of finches appear disrupted.

4. Match the columns

Question 1.

Column I	Column II
(1) August Weismann	(a) Mutation theory
(2) Hugo de Vries	(b) Germplasm theory
(3) Charles Darwin	(c) Theory of acquired characters
(4) Lamarck	(d) Theory of natural selection

Answer:

Column I	Column II
(1) August Weismann	(b) Germplasm theory
(2) Hugo de Vries	(a) Mutation theory
(3) Charles Darwin	(d) Theory of natural selection
(4) Lamarck	(c) Theory of acquired characters

5. Long Answer Questions

Question 1.
Would you consider wings of butterfly and bat as homologous or analogous and why?
Answer:
Wings of butterfly are made up of chitin. They neither have bones, nor muscles in the wings. The bat’s wings are actually patagium. They have muscles and bones just as those seen in all vertebrate limb series. Therefore, these two examples cannot be homologous. However, both the animals use the wings for flight. This is an indication that their function is similar but structure is different, hence they are analogous organs.

Question 2.
What is adaptive radiation? Explain with suitable example.
Answer:

1. Adaptive radiation is the process of evolution which results in transformation of original species to many different varieties.
2. The well-known example of adaptive radiation is Darwin’s Finches. When Charles Darwin went on his voyage to Galapagos islands, he noticed finches which is a variety of small birds.
3. According to Darwin’s observations, the American main land species of finches was the original one which must have migrated to the different islands of Galapagos.
4. Since environmental conditions here were different, they adapted in various ways to the differing environmental conditions of these islands.
5. Original bird had a beak suited for eating seeds, but the changed feeding pattern has changed the shape of beaks too. Some birds also show altered beaks for insectivorous mode. Thus, this demonstrated adaptive radiation.
6. Adaptive radiation in Australian Marsupials is also well studied. In Australia, there are many marsupial mammals who evolved from common ancestor.
7. Adaptive radiation leads to divergent evolution.

Question 3.
By talking industrial melanism as one example, explain the concept of natural selection.
OR
Explain natural selection in action by quoting the example of industrial melanism.
Answer:
1. Industrial melanism is the best example of natural selection which was studied by Kettlewell. In U.K. there are two varieties of peppered moths, Biston betularia and Biston carbonaria.

2. Before industrialisation, in Great Britain, Biston betularia were more in number than Biston carbonaria. B. Betularia is greyish white while B.carbonaria is melanic form.

3. These nocturnal moths rest on tree trunk during day. White-winged moth can camouflage well with the lichen covered whitish barks of trees. They thus escaped the attention of the predatory birds. But at the same time melanic forms were visible due to white barks of the trees. Their number was thus reduced as they were preyed upon by birds.

4. Later there was an industrial revolution, which ultimately resulted in air pollution causing dark soot to settle on the barks of the trees. Lichens too were destroyed and the melanic forms were now at advantage. Melanic forms could camouflage with black tree trunks and their number increased. White-winged moth become clearly seen in changed colours of the trees and thus they were easily caught by predatory birds. This caused decrease in their number.

5. Natural selection thus acted in changed environmental conditions and helped in the establishment of a phenotypic traits. The changed traits were more adaptive and hence were selected. Natural selection encourages those genes or traits that assure highest degree of adaptive efficiency between population and its environment.

Question 4.
Describe the Urey and Miller’s experiment.
Answer:

1. Urey and Miller performed an experiment to prove Oparin’s theory of chemical evolution.

2. They selected a spark discharge apparatus that consisted of closed system of glass having tungsten electrodes, flask for water boiling, a side tube connected to a vacuum pump, a cooling jacket and U-shaped trap.

3. The entire apparatus was first evacuated and made sterile and pre-biotic atmosphere was created in it.

4. The flask was filled with some water and mixture of methane, ammonia and hydrogen in the ratio of 1 : 2 : 2 were slowly passed through the stopcock, without allowing air.

5. Heat was supplied to the flask at very low temperature causing water to boil. The flask simulated the ocean present on primitive earth. Process of evaporation and precipitation was simulated by using heating mantle and condenser respectively.

6. Water vapours along with other gases were circulated continuously through continuous electric sparks. These sparks were given to the mixture for several days causing the gases to interact. This too simulated lightning.

7. Mixture of CH₄, NH₃ and H₂ gases passed through a condenser and was condensed to liquid.

8. The liquefied mixture was collected in the U-shaped trap, present at the bottom of the apparatus. It was found that variety of simple organic compounds (urea, amino acids, lactic acid and sugars) were formed in the apparatus.

This experiment provides the evidence in support to the fact that simple molecules present in the earth’s early atmosphere combined to form the organic building blocks of life.

Question 5.
What is Isolation? Describe the different types of reproductive isolations.
Answer:
1. Isolation means separation of the population of a particular species into smaller units. The organisms belonging to these subunits are prevented from interbreeding due to some barrier. These barriers are called isolating mechanisms.

2. They prevent the genetic exchange and gene flow.

3. Due to isolating mechanisms in nature the divergence among organisms takes place gradually leading to speciation. The isolating mechanisms are of two types namely, geographical isolation and reproductive isolation.

I. Geographical Isolation : The barrier in the form of physical distance or geographical barrier is called geographical isolation. The original population gets divided into two or more groups by geographical barriers such as river, ocean, mountain, glacier, etc. Organisms cannot cross the barriers on their own and hence interbreeding is prevented between isolated groups.

The separated groups experience different environmental factors and they acquire new traits by mutations. The separated populations develop distinct gene pool and they do not interbreed. Each subgroup then evolves differently which results into formation of new species. E.g. Darwin’s Finches, African elephant, Loxodonta and Indian elephant, Elephas.

II. Reproductive Isolation : Two populations may be occupying the same area, they may not be separated by geographical barrier, but then also they are reproductively isolated. Such reproductive isolation occurs due to change in genetic material, gene pool and structure of genital organs. Such differences prevent interbreeding between population. Such isolation later leads to speciation.

III. Different types of reproductive isolations : Reproductive isolation is of two types, viz. pre-zygotic and post-zygotic isolating mechanisms.

1. Pre-zygotic or pre-mating isolating mechanisms do not allow individuals to mate with each other at all.
2. By various mechanisms the two groups remain isolated.
3. In post-zygotic or post-mating isolating mechanisms, the two individuals can mate but the result of mating is not favourable.
4. Thus the populations remain isolated without the actual genetic exchange.

Question 6.
What is Genetic variations? Explain the different factors responsible for genetic variations.
Answer:
Genetic variations : The change in gene and gene frequencies is known as genetic variation. Genetic variations are caused by following factors:
(i) Mutations : Sudden permanent heritable change is called mutation. Mutation can occur in the gene, in the chromosome structure and in chromosome number. Mutation that occurs within the single gene is called point mutation or gene mutation. This leads to the change in the phenotype of the organism, causing variations.

(ii) Genetic recombination : In sexually reproducing organisms, during gamete formation, exchange of genetic material occurs between non-sister chromatids of homologous chromosomes. This is called crossing over. It produces new genetic combinations which result in variation. Fertilization between opposite mating gametes leads to various recombinations resulting into the phenotypic variations. These result in change in the frequencies of alleles.

(iii) Gene flow : Gene flow is movement of genes into or out of a population. Gene movement may be in the form of migration of organism, or gametes (dispersal of pollens) or segments of DNA (transformation). Gene flow also alters gene frequency causing evolutionary changes.

(iv) Genetic drift : Any random fluctuation (alteration) in allele frequency, occurring in the natural population by pure chance, is called genetic drift. For example, when the size of a population is severely reduced due to natural disasters like earthquakes, floods, fires, etc. elimination of particular alleles from a population becomes possible. Smaller populations have greater chances for genetic drift. It results in the change in the gene frequency. Genetic drift is also an important factor for evolutionary change.

(v) Chromosomal aberrations : The structural, morphological change in chromosome due to rearrangement of genes is called chromosomal aberrations. Due to changes in the gene arrangement or gene sequence variations are caused.

6. Complete the chart

Era	Dominating group of animals
1. Coenozoic	————–
2. ————-	Reptiles
3. Palaeozoic	————-
4. ————	Lower Invertebrates

Answer:

Era	Dominating group of animals
1. Coenozoic	Mammals
2. Mesozoic	Reptiles
3. Palaeozoic	Insects, Fishes, Amphibians
4. Proterozoic	Lower Invertebrates

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 4 Molecular Basis of Inheritance Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 4 Molecular Basis of Inheritance

1. Multiple Choice Questions

Question 1.
Griffith worked on ………………..
(a) Bacteriophage
(b) Drosophila
(c) Frog eggs
(d) Streptococci
Answer:
(d) Streptococci

Question 2.
The molecular knives of DNA are ………………..
(a) Ligases
(b) Polymerases
(c) Endonucleases
(d) Transcriptase
Answer:
(c) Endonucleases

Question 3.
Translation occurs in the ………………..
(a) Nucleus
(b) Cytoplasm
(c) Nucleolus
(d) Lysosomes
Answer:
(b) Cytoplasm

Question 4.
The enzyme required for transcription is ………………..
(a) DNA polymerase
(b) RNApolymerase
(c) Restriction enzyme
(d) RNase
Answer:
(b) RNA polymerase

Question 5.
Transcription is the transfer of genetic information from ………………..
(a) DNA to RNA
(b) t-RNA to m-RNA
(c) DNA to m-RNA
(d) m-RNA to t-RNA
Answer:
(a) DNA to RNA

Question 6.
Which of the following is NOT part of protein synthesis?
(a) Replication
(b) Translation
(c) Transcription
(d) All of these
Answer:
(a) Replication

Question 7.
In the RNA molecule, which nitrogen base is found in place of thymine?
(a) Guanine
(b) Cytosine
(c) Thymine
(d) Uracil
Answer:
(d) Uracil

Question 8.
How many codons are needed to specify three amino acids?
(a) 3
(b) 6
(c) 9
(d) 12
Answer:
(a) 3

Question 9.
Which out of the following is NOT an example of inducible operon?
(a) Lactose operon
(b) Histidine operon
(c) Arabinose operon
(d) Tryptophan operon
Answer:
(d) Tryptophan operon

Question 10.
Place the following event of translation in the correct sequence ………………..
i. Binding of met-t-RNA to the start codon.
ii. Covalent bonding between two amino acids.
iii. Binding of second t-RNA.
iv. Joining of small and large ribosome subunits.
(a) iii, iv, i, ii
(b) i, iv, iii, ii
(c) iv, iii, ii, i
(d) ii, iii, iv, i
Answer:
(b) i, iv, iii, ii

2. Very Short Answer Questions

Question 1.
What is the function of an RNA primer during protein synthesis?
Answer:
During DNA replication, RNA primer provides 3’ OH to which DNA polymerase enzyme can add nucleotides to synthesize new strand using parental strand of DNA as template.
[Note : RNA primer has no direct role in protein synthesis.]

Question 2.
Why is the genetic code considered as commaless?
Answer:
The triplet codon are arranged one after the other on m-RNA molecule without any gap or space and therefore genetic code is considered as commaless.

Question 3
Genome
Answer:
Genome is the total genetic constitution of an organism or a complete copy of genetic information (DNA) or one complete set of chromosomes (monoploid or haploid) of an organism.

Question 4.
Which enzyme does remove supercoils from replicating DNA?
Answer:
Super-helix relaxing enzyme (Topoisomerase) removes supercoils from replicating DNA.

Question 5.
Why are Okazaki fragments formed on lagging strand only?
Answer:
Okazaki fragments are formed only on lagging template as only short stretch of lagging template becomes available for replication at one time.

Question 6.
When does DNA replication take place?
Answer:
In eukaryotes DNA-replication takes place during S-phase of interphase of cell cycle and in prokaryotes. DNA replicates prior to cell division.

Question 7.
Define term Codogen and Codon
Answer:
Codogen is a triplet of nucleotides present on the DNA which specifies one particular amino acid.
Codon is a triplet of nucleotides present on the m-RNA which specifies one particular amino acid.

Question 8.
What is degeneracy of genetic code?
Answer:
Genetic code is degenerate as 61 codons code for 20 amino acids, that is two or more codons can specify the same amino acid. E.g. Cysteine has two codons, while isoleucine has three codons.

Question 9.
Which are the nucleosomal ‘core’ histones?
Answer:
Two molecules each of histone proteins, viz. H₂A. H₂B, H₃ and H₄ are the nucleosomal ‘core’ histones.

3. Short Answer Questions

Question 1.
DNA packaging in eukaryotic cell.
Answer:

1. In eukaryotic cells, DNA (2.2 metres) is condensed, coiled and supercoiled to be packaged efficiently in the nucleus (10^-16 m).
2. DNA is associated with histone and non-histone proteins.
3. Histones are a set of positively charged, basic proteins, rich in basic amino acid residues lysine and arginine.
4. Nucleosome consists of nucleosome core (two molecules of each of histone proteins viz. H₂A, H₂B, H₃ and H₄ forming histone octamer) and negatively charged DNA (146 bps) that wraps around the histone octamer by 1 3/4 turns.
5. H₁ protein binds the DNA thread where it enters and leaves the nucleosome.
6. Adjacent nucleosomes are linked with linker DNA (varies in length from 8 to 114 bp, average length of linker DNA is about 54 bp).
7. Each nucleosome contains 200 bp of DNA.
8. Packaging involves formation of – Beads on string (10 nm diameter), Solenoid fibre (looks like coiled telephone wire, 30 nm diameter/300Å), Chromatin fibre and Chromosome.
9. Non-Histone Chromosomal Proteins (NHC) contribute to the packaging of chromatin at a higher level.

Question 2.
Enlist the characteristics of genetic code.
Answer:
The characteristics of genetic code are

1. Genetic code is triplet, commaless and non-overlapping.
2. It is degenerate and non-ambiguous.
3. It is universal
4. It has polarity.

Question 3.
Applications of DNA fingerprinting.
Answer:
Applications of DNA fingerprinting are as follows:

1. In forensic science to solve rape and murder cases.
2. Finds out the biological father or mother or both, of the child, in case of disputed parentage.
3. Used in pedigree analysis in cats, dogs, horses and humans.

Question 4.
Explain the role of lactose in ‘Lac Operon’.
Answer:

1. A small amount of beta-galactoside permease enzyme is present in cell even when Lac operon is switched off and it allows a few molecules of lactose to enter into the cell.
2. Lactose binds to repressor and inactivates it.
3. Repressor – lactose complex cannot bind with the operator gene, which is then turned on.
4. RNA polymerase transcribes all the structural genes to produce lac m-RNA which is then translated to produce all enzymes.
5. Thus, lactose acts as an inducer.
6. When the inducer level falls, the operator is blocked again by repressor and structural genes are repressed again. This is negative feedback.

4. Short Answer Questions

Question 1.
Human genome project.
Answer:
1. Human Genome Project (HGP) was initiated in 1990 under the International administration of the Human Genome Organization (HUGO) and it was completed r in 2003.

2. The main aims:

• To sequence 3 billion base pairs of DNA in human genome and to map an estimated 33000 genes.
• To store the information collected from the project in databases.
• To develop tools and techniques for analysis of the data.
• Transfer of the related technologies to the private sectors, such as industries.
• Taking care of the legal, ethical and social issues which may arise from project.
• To sequence the genomes of several other organisms such as bacteria e.g. E.coli, Caenorhabditis elegans, Saccharomyces cerevisiae, Drosophil, rice, Arabidopsis), Mus musculus, etc.

3. Significance:

1. HGP has a major impact in the fields like Medicine, Biotechnology, Bioinformatics and the Life sciences.
2. More understanding of functions of genes, proteins and human evolution.

Question 2.
Describe the structure of operon.
Answer:

1. An operon is a unit of gene expression and regulation.
2. It includes the structural genes and their control elements. Control elements are promoters and operators.
3. The structural genes code for proteins, r-RNA and t-RNA that are necessary for all the cells.
4. Promoters are signal sequences in DNA. They start the RNA synthesis. They also act as sites where the RNA polymerases are bound during transcription.
5. Operators are present between the promoters and structural genes.
6. There is repressor protein that binds to the operator region of the operon.
7. There are regulatory genes which are responsible for the formation of repressors which interact with operators.

Question 3.
In the figure below A, B and C are three types of

Answer:
Answer: A, B and C are A : m-RNA, B : r-RNA, C : t-RNA

Question 4.
Identify the labelled structures on the following diagram of translation.

Part A is the ………………
Part B is the ………………
Part C is the ………………
Answer:
Part A is the anti-codon.
Part B is the amino acid.
Part C is the larger subunit of ribosome.

Question 5.
Match the entries in Column I with those of Column II and choose the correct answer.

Column I	Column II
A. Alkali treatment	i. Separation of DNA fragments on gel slab
B. Southern blotting	ii. Splits DNA fragments into single strands
C. Electrophoresis	iii. DNA transferred to nitrocellulose sheet
D. PCR	iv. X-ray photography
E. Autoradiography	v. Produce fragments different sizes
F. DNA treated with REN	vi. DNA amplification

Answer:

Column I	Column II
A. Alkali treatment	ii. Splits DNA fragments into single strands
B. Southern blotting	iii. DNA transferred to nitrocellulose sheet
C. Electrophoresis	i. Separation of DNA fragments on gel slab
D. PCR	vi. DNA amplification
E. Autoradiography	iv. X-ray photography
F. DNA treated with REN	v. Produce fragments different sizes

5. Long Answer Questions

Question 1.
Explain the process of DNA replication.
Answer:

DNA replication is semi-conservative replication. It involves following steps:
Activation of Nucleotides:

1. Nucleotides (dAMP dGMR dCMP and dTMP) present in the nucleoplasm, are activated by ATP in presence of an enzyme phosphorylase.
2. This phosphorylation results in the formation of deoxyribonucleotide triphosphates i.e. dATE dGTR dCTP and dTTE

Point of Origin or Initiation point:

1. Replication begins at specific point ‘O- Origin and terminates at point ‘T’.
2. At the point ‘O’, enzyme endonuclease nicks (breaks the sugar-phosphate backbone or the phosphodiester bond) one of the strands of DNA, temporarily.

Unwinding of DNA molecule:

1. Enzyme DNA helices breaks weak hydrogen bonds in the vicinity of ‘O’.
2. The strands of DNA separate and unwind. This unwinding is bidirectional.
3. SSBP (Single strand binding proteins) remains attached to both the separated strands and prevent them from recoiling (rejoining).

Replicating fork:

1. Y-shape replication fork is formed due to unwinding and separation of two strands.
2. The unwinding of strands results in strain which is released by super-helix relaxing enzyme.

Synthesis of new strands:

1. Each separated strand acts as a template for the synthesis of new complementary strand.
2. A small RNA primer (synthesized by activity of enzyme RNA primase) get attached to the 3′ end of template strand and attracts complementary nucleotides from surrounding nucleoplasm.
3. These nucleotides bind to the complementary nucleotides on the template strand by hydrogen bonds (i.e. A = T or T = A; G = C or C = G, CEG).
4. The phosphodiester bonds are formed between nucleotides of new strand to form a polynucleotide strand.
5. The enzyme DNA polymerase catalyses synthesis of new complementary strand always in 5′ – 3′ direction.

Leading and Lagging strand:

1. The template strand with free 3′ is called the leading template.
2. The template strand with free 5′ end is called the lagging template.
3. The replication always starts at C-3 end of template strand and proceeds towards C-5 end.
4. New strands are always formed in 5′ → 3′ direction.
5. The new strand which develops continuously towards replicating fork is called the leading strand.
6. The new strand which develops discontinuously away from the replicating fork is called the lagging strand.
7. Maturation of Okazaki fragments : The lagging strand is synthesized in the form of small Okazaki fragments which are joined by enzyme DNA ligase.
8. Later RNA primers are removed by the combined action of RNase H, an enzyme that degrades the RNA strand of RNA-DNA hybrids, and polymerase I.
9. Gaps formed are filled by complementary DNA sequence with the help of DNA polymerase-I in prokaryotes and DNA polymerase-a in eukaryotes.
10. Finally, DNA gyrase (topoisomerase) enzyme forms double helix to form daughter DNA molecules.

Formation of two daughter DNA molecules:

1. In each daughter DNA molecule, one strand is parental and the other one is newly synthesized.
2. Thus, 50% part (i.e. one strand of the helix) is contributed by mother DNA. Hence, it is described as semiconservative replication.

Question 2.
Describe the process of transcription in protein synthesis.
Answer:

Transcription involves three stages, viz. Initiation, Elongation and Termination.
(1) Initiation:

1. RNA polymerase binds to promoter site.
2. It then moves along the DNA and causes local unwinding of DNA duplex into two strands in the region of the gene.
3. Only antisense strand functions as template.

(2) Elongation:

• The complementary ribonucleoside tri-phosphates get attached to exposed bases of DNA template chain.
• As transcription proceeds, the hybrid DNA-RNA molecule dissociates and makes m-RNA molecule free.
• As the m-RNA grows, the transcribed region of DNA molecule becomes spirally coiled and regains double helical form.

(3) Termination:
When RNA polymerase reaches the terminator site on the DNA, both enzyme and newly formed m-RNA (primary transcript) gets released.

Question 3.
Describe the process of translation in protein synthesis.
Answer:

Translation involves the following steps:
1. Activation of amino acids and formation of charged t-RNA (t-RNA – amino acid complex):
i. In the presence of an enzyme amino acyl t-RNA synthetase, the amino acid is activated and then attached to the specific t-RNA molecule at 3’ end to form charged t-RNA (t-RNA – amino acid complex).

ii. ATP is essential for the reaction.

2. Initiation of Polypeptide chain:

• Small subunit of ribosome binds to the m-RNA at 5’ end.
• Start codon is positioned properly at P-site.
• Initiator t-RNA, (carrying amino acid methionine in eukaryotes or formyl methionine in prokaryotes) binds with initiation codon (AUG) of m-RNA, by its anticodon (UAC) through hydrogen bonds.
• The large subunit of ribosome joins with the smaller subunit in the presence of Mg⁺⁺.
• Thus, initiator charged t-RNA occupies the P-site and A – site is vacant.

3. Elongations of polypeptide chain:
Addition of amino acid occurs in 3 Step cycle-
i. Codon recognition.
Anticodon of second (and subsequent) amino acyl t-RNA molecule recognizes and binds with codon at A-site by hydrogen bonds.

ii. Peptide bond formation.

1. Ribozyme catalyzes the peptide bond formation between amino acids on the initiator t-RNA at P-site and t-RNA at A-site.
2. It takes less than 0.1 second for formation of peptide bond.
3. Initiator t-RNA at ‘P’ site is then released from E-site.

iii. Translocation.

1. Translocation is the process in which sequence of codons on m-RNA is decoded and accordingly amino acids are added in specific sequence to form a polypeptide on ribosomes.
2. Due to this A’-site becomes vacant to receive next charged t-RNA molecule.
3. The events like arrival of t-RNA – amino acid complex, formation of peptide bond, ribosomal translocation and release of previous t-RNA, are repeated.
4. As ribosome move over the m-RNA, all the codons on m-RNA are exposed oiie by one for translation.

4. Termination and release of polypeptide:
When stop codon (UAA, UAG, UGA) gets exposed at the A-site, the release factor binds to the stop codon, thereby terminating the translation process
The polypeptide gets released in the cytoplasm.
Two subunits of ribosome dissociate and last t-RNA and m-RNA are released in the cytoplasm.
m-RNA gets denatured by nucleases immediately.

Question 4.
Describe Lac ‘Operon’.
Answer:

Lac operon consists of the following components:
(1) Regulator gene:

• Regulator gene precedes the promoter gene.
• It may not be present immediately adjacent to operator gene.
• Regulator gene codes for a repressor protein which binds with operator gene and represses (stops) its action.

(2) Promoter gene:

• It precedes the operator gene.
• It is present adjacent to operator gene.
• RNA Polymerase enzyme binds at promoter site.
• Promoter gene base sequence determines which strand of DNA acts a template.

(3) Operator gene:

• It precedes the structural genes.
• When operator gene is turned on by an inducer, the structural genes get transcribed to form m-RNA.

(4) Structural gene:

• There are 3 structural genes in the sequence lac-Z, lac-Y and lac-A.
• Enzymes produced are β-galactosidase, β-galactoside permease and transacetylase respectively.
  Inducer Allolactose acts as an inducer. It inactivates the repressor by binding with it.

Question 5.
Justify the statements. If the answer is false, change the underlined word(s) to make the statement true.
(i) The DNA molecule is double stranded and the RNA molecule is single stranded.
Answer:

1. DNA as the genetic material has to be chemically and structurally stable.
2. It should be able to generate its replica.
3. Sugar-phosphate backbone and complementary base pairing between the two strands, give stability to DNA.
4. Both the strands of DNA act as template for synthesis of their complementary strands. This allows accurate replication of DNA.
5. Single stranded RNA can be folded to form complex structures and perform specific functions such as synthesis of proteins.

(ii) The process of translation occurs at the ribosome.
Answer:

1. Translation is the process in which sequence of codons of m-RNA is decoded and accordingly amino acids are added in specific sequence to form a polypeptide on ribosomes.
2. Ribosome has one binding site for m-RNA. It orients m-RNA molecule in such a way that all the codons are properly read.
3. Ribosome has three binding sites for t-RNA : P-site (peptidyl t-RNA-site), A-site (aminoacyl t-RNA-site) and E-site (exit site).
4. t-RNAs place the required amino acids in correct sequence and translate the coded message of RNA.
5. In eukaryotes, a groove which is present between two subunits of ribosomes, protects the polypeptide chain from the action of cellular enzymes and also protects m-RNA from the action of nucleases.
6. Thus ribosome plays an important role in translation.

(iii) The job of m-RNA is to pick up amino acids and transport them to the ribosomes.
Answer:
The job of t-RNA is to pick up amino acids and transport them to ribosomes. t-RNA is an adapter molecule. It reads the codons of m-RNA and also simultaneously transfer specific amino acid to m-RNA Ribosome complex. It binds with amino acid at its 3′ end.

(iv) Transcription must occur before translation may occur.
Answer:
In prokaryotes, translation can start before transcription is complete, as both these processes occur in the same compartment, i.e. cytoplasm. But in eukaryotes, transcription and processing of hnRNA occurs in nucleus. hnRNA then comes out of the nucleus through nuclear pores and then it is translated at ribosomes in the cytoplasm.

Question 6.
Guess
(i) the possible locations of DNA on the collected evidence from a crime scene and
(ii) the possible sources of DNA.

Evidence	Possible location of DNA on the evidence	Sources of DNA
e.g. Eyeglasses	e.g. Earpieces	e.g. Sweat, Skin
Bottle, Can, Glass	Sides, mouthpiece	—————-
————–	Handle	Sweat, skin, blood
Used cigarette	Cigarette butt	—————–
Bite mark	—————–	Saliva
————-	Surface area	Hair, semen, sweat, urine

Answer:

Evidence	Possible location of DNA on the evidence	Sources of DNA
e.g. Eyeglasses	e.g. Earpieces	e.g. Sweat, Skin
Bottle, Can, Glass	Sides, mouthpiece	Saliva
Door	Handle	Sweat, skin, blood
Used cigarette	Cigarette butt	Saliva
Bite mark	Teeth impression	Saliva
Clothes	Surface area	Hair, semen, sweat, urine

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 3 Inheritance and Variation Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 3 Inheritance and Variation

1. Multiple Choice Questions

Question 1.
Phenotypic ratio of incomplete dominance in Mirabilis jalapa.
(a) 2 : 1 : 1
(b) 1 : 2 : 1
(c) 3 : 1
(d) 2 : 2
Answer:
(b) 1 : 2 : 1

Question 2.
In dihybrid cross, F₂ generation offspring show four different phenotypes while the genotypes are ……………….
(a) six
(b) nine
(c) eight
(d) sixteen
Answer:
(b) nine

Question 3.
A cross between an individual with unknown genotype for a trait with recessive plant for that trait is ……………….
(a) back cross
(b) reciprocal cross
(C) monohybrid cross
(d) test cross
Answer:
(d) test cross

Question 4.
When phenotypic and genotypic ratios are the same, then it is an example of ……………….
(a) incomplete dominance
(b) complete dominance
(c) multiple alleles
(d) cytoplasmic inheritance
Answer:
(a) incomplete dominance

Question 5.
If the centromere is situated near the end of the chromosome, the chromosome is called ……………….
(a) Metacentric
(b) Acrocentric
(c) Sub-Metacentric
(d) Telocentric
Answer:
(d) Telocentric

Question 6.
Chromosomal theory of inheritance was proposed by ……………….
(a) Sutton and Boveri
(b) Watson and Crick
(c) Miller and Urey
(d) Oparin and Halden
Answer:
(a) Sutton and Boveri

Question 7.
If the genes are located in a chromosome as p-q-r-s-t, which of the following gene pairs will have least probability of being inherited together ?
(a) p and q
(b) r and s
(c) s and t
(d) p and s
Answer:
(d) p and s

Question 8.
Find the mismatched pair:
(a) Down’s syndrome = 44 + XY
(b) Turner’s syndrome = 44 + XO
(c) Klinefelter’s syndrome = 44 + XXY
(d) Super female = 44 + XXX
Answer:
(a) Down’s syndrome = 44 + XY

Question 9.
A colourblind man marries a woman, who is homozygous for normal colour vision, the probability of their son being colour blind is ……………….
(a) 0%
(b) 25%
(c) 50%
(d) 100%
Answer:
(a) 0%

2. Very Short Answer Questions

Question 1.
Explain the statements
a. Test cross is back cross but back cross is not necessarily a test cross.
b. Law of dominance is not universal.
Answer:
a. (1) Test cross is the cross between F₁ hybrid and its homozygous recessive parent.
(2) Back cross is the cross of offspring with any one of the parents, either dominant or recessive.
(3) Therefore, test cross can be a back cross – but back cross cannot be a test cross.

b. (1) There are many traits in many organisms which show dominance. For example, widow’s peak in human beings is a dominant trait. Yellow seed colour and round seed shape are dominant traits in pea plant.
(2) However, there are characters which are either co-dominant, such as genes for human blood group A and B or incompletely dominant as in flower colour of Mirabilis jalapa.
(3) Therefore the law of dominance is not universally applicable.

Question 2.
Define the following terms:
a. Dihybrid cross
b. Homozygous
c. Heterozygous
d. Test cross
Answer:
a. A cross between parents differing in two heritable traits is called dihybrid cross.
b. An individual possessing identical alleles for a particular trait is called homozygous or pure for that trait. E.g. TT for tallness and tt for dwarfness.
c. An individual possessing contrasting allele for a particular trait is called heterozygous. E.g. Tt showing tallness.
d. The cross of F₁ progeny with homozygous recessive parent is called a test cross.

Question 3.
What are allosomes?
Answer:
Allosomes are the chromosomes which decide the sex of an organism.

Question 4.
What is crossing over?
Answer:
Crossing over is the process of forming new recombinations by interchanging and exchanging non-sister chromatid arms of the homologous chromosomes.

Question 5.
Give one example of autosomal recessive disorder.
Answer:
Thalassemia is an example of autosomal recessive disorder.

Question 6.
What are X-linked genes?
Answer:
Genes located on the non-homologous region of X chromosome are called X-linked genes.

Question 7.
What are holandric traits?
Answer:
Genes located on the non-homologous region of Y chromosome are called Y-linked genes. The traits due to such genes are called holandric traits which are seen only in male sex.

Question 8.
Give an example of chromosomal disorder caused due to non-disjunction of autosomes.
Answer:
Down’s syndrome is an example of chromosomal disorder caused due to non-disjunction of autosomes.

Question 9.
Give one example of complete sex linkage.
Answer:
Sex linkage can be complete X linkage and complete Y linkage. X linkage is haemophilia and Y linkage is hypertrichosis.

3. Short Answer Questions

Question 1.
Enlist seven traits of pea plant selected / studied by Mendel.
Answer:
Seven traits in pea selected by Mendel:

1. Tall habit versus dwarf habit (Height of the plant).
2. Purple flowers versus white flowers. (Colour of flowers)
3. Yellow seeds versus green seeds. (Colour of seeds)
4. Round seeds versus wrinkled seeds. (Shape of seeds)
5. Green pods versus yellow pods. (Colour of pods)
6. Inflated pods versus constricted pods. (Shape of pods)
7. Axial flower versus terminal flower. (Position of a flower)

Question 2.
Why law of segregation is also called the law of purity of gametes?
Answer:
(1) Mendel’s law of segregation is also called Law of purity of gametes because, during formation of gametes, the alleles separate/ segregate from each other and only one allele enters a gamete.

(2) The separation of one allele does not affect other. Since single allele enters a gamete means gametes will be pure for a trait.
E.g. The contrasting characters such as tall (T) and dwarf (t) present in F₁ hybrid (Tt) segregate during the formation of gametes.

(3) Owing to this, two types of gametes i.e. T and t are formed which are pure for the characters which they carry.
(4) Thus for example:

Question 3.
Pleiotropy.
Answer:

1. When a single gene controls two or more different traits, it is called a pleiotropic gene and the phenomenon is known as pleiotropy or pleiotropism.
2. The pleiotropic ratio is always 1 : 2 instead . of normal 3 : 1.
3. Sickle-cell anaemia is caused by the gene HbS. The healthy or normal gene which is dominant is HbA. The heterozygotes or carriers i.e., HbA/Hbs show anaemia as there is deficiency of haemoglobin due to sickling of RBCs. Abnormally low concentration of oxygen can cause sickling of RBCs.
4. The homozygotes possessing the recessive gene HbS die because of fatal anaemia because the gene for sickle-cell anaemia is lethal in homozygous condition and causes sickle-cell trait in heterozygous carrier.
5. Thus a single gene produces two different expressions.
6. When two carriers are married they will produce normal carriers and Sickle-cell anaemic children in the ratio of 1 : 2 : 1. Out of these three children sickle-cell anaemic child will die leaving the ratio 1 : 2 instead of 3 : 1.

Question 4.
What are the reasons of Mendel’s success?
Answer:
Reasons for Mendel’s success:

1. Mendel planned his experiments carefully and these experiments consisted of large sample.
2. He always recorded the results of number of plants of each type and their ratios.
3. The contrasting characters that he chose were easily recognizable.
4. The seven pairs of contrasting characters that he selected were under control of a single factor each. They were present on separate chromosomes and were transmitted from one generation to the next.
5. Mendel studied and introduced concept of dominance and recessiveness.

Question 5.
“Father is responsible for determination of sex of child and not the mother”. Justify.
Answer:

1. Human made is heterogame tic, i.e. he produces two different types of sperms. One is bearing X chromosome along with 22 autosomes and the other is Y bearing sperm with 22 autosomes.
2. Mother, on the other hand, is homogametic, producing all similar types of ova, i.e 22 + X chromosomal combination.
3. If 22+X bearing sperm fertilise an egg, female child is formed while if Y bearing sperm fertilizes an egg, male child is formed.
4. Thus the sex of the child is dependent upon type of sperm that father gives, therefore, it is said that father is responsible for determination of sex of a child and not the mother.

Question 6.
What is linkage? How many linkage groups do occur in human being and maize?
Answer:

1. Linkage is defined as the tendency of the genes to be inherited together because they are present in the same chromosome. Linkage group is group of genes situated on a chromosome.
2. Humans have 23 linkage groups because they have 23 pairs of chromosomes.
3. Maize plant has 10 linkage groups because they have 10 pairs of chromosomes.

Question 7.
PKU.
Answer:

1. PKU means phenylketonuria which is an autosomal recessive inborn error.
2. In this disorder the metabolism of phenylalanine does not occur due to deficiency of phenylalanine hydroxylase (PAH) enzyme.
3. This enzyme is necessary to metabolize the amino acid phenylalanine to the amino acid tyrosine.
4. When PAH activity is reduced, phenylalanine accumulates in blood and cerebrospinal fluid and is converted into phenylpyruvate or phenyl-ketone which is a toxic compound. This may cause mental retardation. Excess phenylalanine is excreted in urine, hence this disease is called phenylketonuria.
5. PKU is caused by mutations in the PAH gene on chromosome no. 12.
6. Untreated PKU causes abnormal phenotype which includes growth failure, poor skin pigmentation, microcephaly, seizures, global developmental delay and severe intellectual impairment. However, at birth if an infant is checked for PKU, the further abnormalities can be avoided.

Question 8.
Compare X-chromosome and Y-chromosome.
Answer:

X-chromosome	Y-chromosome
1. X-chromosome is straight, rod like and longer 1. than Y chromosome. It is metacentric.	1. Y-chromosome is shorter chromosome which is acrocentric.
2. X-chromosome has large amount of euchromatin and small amount of heterochromatin.	2. Y-chromosome has small amount of euchromatin and large amount of heterochromatin.
3. X-chromosome has large amount of DNA, hence it is genetically active due to more genes.	3. Y-chromosome has less amount of DNA, hence it is genetically less active or inert due to lesser genes.
4. Non-homologous region of X-chromosome is longer and contains more genes.	4. Non-homologous region of Y-chromosome is shorter and contains lesser genes.
5. Contains X-linked genes on non-homologous region.	5. Contains Y-linked genes on non-homologous region.
6. X-chromosome is present in men as well as women.	6. Y-chromosome is present only in men.

Question 9.
Explain the chromosomal theory of inheritance.
Answer:
Chromosomal theory of inheritance was put forth by Sutton and Boveri after studying paraillel behaviour of genes and chromosomes during meiotic division. This theory states following points:

1. Chromosomal theory identifies chromosomes as the carrier of genetic material.
2. All the hereditary characters are transmitted by gametes. Nucleus of gametes, i.e. sperms and ova of the parents contain chromosomes which transmit the heredity to offspring.
3. Chromosomes are found in pairs in somatic or diploid cells.
4. During gamete formation, homologous chromosomes pair and segregate independently at meiosis. The diploid condition is converted into haploid condition. Thus each gamete contains only one chromosome of a pair.
5. During fertilization, the union of sperm and egg restores the diploid number of chromosomes.

Question 10.
Observe the given pedigree chart and answer the following questions

(a) Identify whether the trait is sex-linked or autosomal.
(b) Give an example of a trait in human beings which shows such a pattern of inheritance.
Answer:
Pedigree given above shows:

1. First Generation : Carrier woman marrying a sufferer man. Their three children are in following birth order.
2. Second generation : First son is normal, second daughter is carrier and third daughter is sufferer.
3. Third generation : The sufferer daughter marries a normal man. Her children are normal daughter and sufferer son.

(a) The above pedigree show sex-linked (X-linked) trait. Since criss-cross inheritance is seen in the trait, it must be sex-linked inheritance.
(b) Such trait and its inheritance can be seen in colour blindness.

4. Match the Columns

rewrite the matching pairs.

Column I	Column II
(1) 21 trisomy	(a) Turner’s syndrome
(2) X-monosomy	(b) Klinefelter’s syndrome
(3) Holandric traits	(c) Down’s syndrome
(4) Feminized male	(d) Hypertrichosis

Answer:

Column I	Column II
(1) 21 trisomy	(c) Down’s syndrome
(2) X-monosomy	(a) Turner’s syndrome
(3) Holandric traits	(d) Hypertrichosis
(4) Feminized male	(b) Klinefelter’s syndrome

5. Long Answer Questions

Question 1.
What is dihybrid cross? Explain with suitable example and checker board method.
Answer:
1. A cross which involves two pairs of alleles is called a dihybrid cross. A phenotypic ratio of 9 : 3 : 3 : 1 obtained in the F₂ generation of a dihybrid cross is called a dihybrid ratio.

(2) Thus for example, when we cross a true breeding pea plant bearing round and yellow seeds with a true breeding pea plant bearing wrinkled and green seeds we get pea plants bearing round and yellow seeds in the F₁ generation.

(3) When F₁ plants are selfed, we get a ratio of 9 : 3 : 3 : 1 in the F₂ generation, where 9 plants bear yellow round seeds, 3 plants bear yellow wrinkled seeds, 3 plants bear green round seeds and 1 plant bears green wrinkled seeds.

(4) Parents (P₁) : RRYY × rryy
Gametes of P₁ RY and ry
F₁ generation : RrYy(Yellow round)
On selfing F₁ : RrYy × RrYy
Gametes of F₁ : RY, Ry, rY, ry

P₂ generation:

Round Yellow : 9 Round green : 3 Wrinkled yellow : 3 Wrinkled green : 1
Phenotypic ratio : 9 : 3 : 3 : 1
Genotypic ratio : 1 : 2 : 1 : 2 : 4 : 2 : 1 : 2 : 1

Question 2.
Explain with suitable example an independent assortment.
Answer:
(1) The law of independent assortment states that when hybrid possessing two or more pairs of contrasting characters bearing alleles form gametes, the alleles in each pair segregate independently of the other pair. Therefore, the inheritance of one pair of characters is independent of that of the other pair of characters.
(2) For example, when we cross a pea plant which is tall and having purple flowers with dwarf plant having white flowers we obtain all tall plants with purple flowers in F₁ generation. When F₁ generation are selfed, 9 : 3 : 3 : 1 ratio was obtained in F₂ generation with 9 tall and purple flower, 3 tall with white flowers, 3 dwarf with purple flowers and 1 which was dwarf and white. Tallness and purple colour are dominant traits while dwarfness and white colour are recessive traits.

(i) Homozygous tall purple – TTPP
(ii) Homozygous dwarf white – ttpp

Tall purple = 9. Tall white = 3
Dwarf purple = 3, Dwarf white = 1,
Phenotypic ratio = 9 : 3 : 3 : 1
Results : The offspring of F₁ generation will be in the proportion of 9 : 3 : 3 : 1, where 9 are tall purple, 3 are tall white, 3 are dwarf purple and 1 is dwarf white.

Question 3.
Define test cross and explain its significance.
Answer:
1. Definition of test cross : A cross between F₁ offspring and its homozygous recessive parent is called a test cross.
2. Significance of test cross:

• Test cross can be used to find out the genotype of any plant which shows dominant characters.
• Whether the plant is homozygous or heterozygous can be understood by performing test cross.
• Test cross is used to introduce useful recessive traits in the hybrids of self- pollinated plants.
• Test cross is quicker method to improve the variety of crop plants and thus it is useful for breeders and geneticists.
• Test cross can be used for verifying the laws of inheritance.

Question 4.
What is parthenogenesis? Explain the haplodiploid method of sex determination in honey bee.
Answer:
I. Parthenogenesis is a natural form of asexual reproduction in which growth and development of embryos occur without fertilization by sperm. In some insects like honey bees, parthenogenesis means development of an embryo from an unfertilized egg cell.

II. In honey bee:

1. Sex determination is by haplodiploid system.
2. Sex is determined by the number of sets of chromosomes received by an individual.
3. The egg which is fertilized by sperm, becomes diploid and develops into female.
4. The egg which is not fertilized develops by parthenogenesis and develops into a male.
5. The queen and worker bee therefore contain 32 chromosomes. The drone, i.e. male bears 16 chromosomes.
6. The sperms are produced by mitosis while eggs are produced by meiosis.

Question 5.
In the answer for inheritance of X-linked. genes, Madhav had shown carrier male. His answer was marked incorrect. Madhav was wondering why his marks were cut. Explain the reason.
Answer:
Males can never be carriers. They have single X and other Y chromosome. In X linked inheritance, the genes are present on the non-homologous region of X chromosome. Males do not have other X and hence if the genes are present on his X chromosome, they will not be suppressed in them. The Y chromosome does not have dominant gene to hide this expression as there is no homolorous region too. But in case of females, there are double X chromosomes and hence if X-linked gene is recessive, the other X can hide the expression of such X-linked gene.

Thus she becomes a carrier without showing any physical characters. She is physically normal and does not suffer from such X-linked recessive disorder. Thus, Madhav will get his answer wrong due to incorrect concept.

Question 6.
With the help of neat labelled diagram, describe the structure of chromosome.
Answer:

(1) A chromosome is best visible during metaphase, when it is highly condensed.

(2) Chromosome shows two identical halves, called sister chromatids. Chromatids are held together at centromere which is also called primary constriction.

(3) Primary constriction has disc shaped plate called kinetochore. This plate is useful for attachment of spindle fibres at the time of cell division.

(4) Additional narrow areas called secondary constrictions are seen in some chromosomes which are known as nucleolar organizers. They help in the formation of nucleolus. At secondary constriction (i) there is nucleolar organising region. Secondary constriction (ii) shows attachment of satellite body or SAT body.

(5) Each chromatid is made up of sub¬chromatids called chromonemata. Each chromonema consists of a long, unbranched, slender, highly coiled DNA thread. This double stranded DNA molecule extends throughout the length of the chromosome.

(6) The ends of the chromatid arms are called telomeres.

Question 7.
What is criss-cross inheritance? Explain with suitable example.
Answer:
Criss-cross inheritance is the type of inheritance in which the genes are passed on from father to daughter and then to her son, i.e. from male to female and from female to male (grandson). In other words, it is also said that the transmission is from the grandfather to his grandson through his daughter.

I. Inheritance of Colour blindness show criss-cross pattern.
(1) Colour blindness is a sex-linked disorder in which the person concerned cannot distinguish between red and green colours.

(2) It is recessively X-linked disorder, which is expressed in males. It is rarely seen in females.

(3) The genes for normal vision are dominant whereas those for colour blindness are recessive.

(4)

• Gene for normal vision : X^C
• Gene for colour blindness : X^c
• Normal female : X^CX^C
• Normal male : X^CY
• Colour blind female : X^cX^c
• Carrier female : X^CX^c
• Colour blind male : X^c Y

II. Crosses showing the inheritance of colour blindness:
(i) A cross between normal female and colour-blind male.

(ii) A cross of carrier female with normal male.

(1) Normal female with Colour blind male. Such cross produces 50% carrier daughters and 50% normal sons.

(2) Carrier female with normal male. Such a cross produces 25% normal daughters, 25% normal sons, 25% carrier daughters and 25% colour blind sons.

(3) Colour blind father transmits the disorder to his grandson through his carrier daughter. The inheritance of characters from the father to his grandson through his daughter is called criss-cross inheritance.

Question 8.
Describe the different types of chromosomes.
Answer:
I. Chromosomes are classified into the following four types according to the position of the centromere in them:
(1) Metacentric : In metacentric chromosome, the centromere is situated in the middle of the chromosome. The two arms of the chromosome are nearly equal. It appears ‘V’-shaped during anaphase.

(2) Sub-metacentric : In sub-metacentric chromosome, the centromere is situated some distance away from the middle. Due to this, one arm of the chromosome is shorter than the other. It appears T-shaped during anaphase.

(3) Acrocentric : In acrocentric chromosome, the centromere is situated near the end of the chromosome. One arm of the acrocentric chromosome is very short while the other is long making it appear like ‘J’-shaped during anaphase.

(4) Telocentric : In telocentric chromosome, the centromere is situated at the tip of the chromosome. Telocentric chromosome has only one arm thus it appears rod-shaped.

II. Based on the functions, chromosomes are divided into autosomes and allosomes. Autosomes are somatic chromosomes which decide the body characters. Allosomes are sex chromosomes which decide the sex of the individual.

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 2 Reproduction in Lower and Higher Animals Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 2 Reproduction in Lower and Higher Animals

1. Multiple choice questions

Question 1.
The number of nuclei present in a zygote is ……………….
(a) two
(b) one
(c) four
(d) eight
Answer:
(b) one

Question 2.
Which of these is the male reproductive organ in human?
(a) Sperm
(b) Seminal fluid
(c) Testes
(d) Ovary
Answer:
(c) Testes

Question 3.
Attachment of embryo to the wall of the uterus is known as ……………….
(a) fertilization
(b) gestation
(c) cleavage
(d) implantation
Answer:
(d) implantation

Question 4.
Rupturing of follicles and discharge of ova is known as ……………….
(a) capacitation
(b) gestation
(c) ovulation
(d) copulation
Answer:
(c) ovulation

Question 5.
In human females, the fertilized egg gets implanted in uterus ……………….
(a) after about 7 days of fertilization
(b) after about 30 days of fertilization
(c) after about two months of fertilization
(d) after about 3 weeks of fertilization
Answer:
(a) after about 7 days of fertilization

Question 6.
Test tube baby technique is called ……………….
(a) In vivo fertilization
(b) In situ fertilization
(c) In Vitro Fertilization
(d) Artificial Insemination
Answer:
(c) In Vitro Fertilization

Question 7.

The given figure shows a human sperm. Various parts of it are labelled as A, B, C, and D. Which labelled part represents acrosome?
(a) B.
(b) C
(c) D
(d) A
Answer:
(d) A

Question 8.
Presence of beard in boys is a ……………….
(a) primary sex organ
(b) secondary sexual character
(c) secondary sex organ
(d) primary sexual character
Answer:
(b) secondary sexual character

2. Very short answer questions

Question 1.
What is the difference between a foetus and an embryo?
Answer:
Embryo is a growing egg after fertilization until the main parts of the body and the internal organs have started to take shape while foetus is a stage which has the appearance of a fully developed offspring.

Question 2.
Outline the path of sperm up to the urethra.
Answer:
The path of sperm up to the urethra in male is as follows :
Seminiferous tubules → Rete testis → Vasa efferentia → Epididymis → Vas deferens → Ejaculatory ducts Urethra.

Question 3.
Which glands contribute fluids to the semen?
Answer:
The glands which contribute fluids to the semen are seminal vesicles, prostate gland.

Question 4.
Name the endocrine glands involved in maintaining the sexual characteristics of males.
Answer:
Interstitial cells of Leydig which lie in between the seminiferous tubules are involved in maintaining the sexual characteristics of male by secreting the male hormone androgen or testosterone. Adenohypophysis also regulates this secretion from the testis.

Question 5.
Where does fertilization and implantation occur?
Answer:
Fertilization of ovum takes place in the ampulla region of fallopian tube whereas implantation occur in the endometrium of uterus.

Question 6.
Enlist the external genital organs in female.
Answer:
The external genital organs in female include the following parts such as vestibule, labia minora, clitoris, labia majora and mons Veneris.

Question 8.
What is the difference between embryo and zygote?
Answer:
Zygote is the unicellular diploid structure formed as a result of fusion of sperm and ovum whereas embryo is a multicellular structure formed from zygote in the uterus 3 weeks after fertilization.

3. Fill in the blanks

Question 1.
The primary sex organ in human male is ……………….
Answer:
testis

Question 2.
The ……………… is also called the womb.
Answer:
uterus

Question 3.
Sperm fertilizes ovum in the ……………….. of fallopian tube.
Answer:
ampulla

Question 4.
The disc like structure which helps in the transfer of substances to and from the foetus’s body is called ………………..
Answer:
placenta

Question 5.
Gonorrhoea is caused by ……………….. bacteria.
Answer:
Neisseria gonorrhoeae

Question 6.
The hormone produced by the testis is ……………………
Answer:
testosterone / androgen.

4. Short Answer Questions

Question 1.
Budding in Hydra.
Answer:

1. Budding is a type of asexual reproduction method seen in Hydra.
2. Budding takes place during favourable period.
3. Towards the basal end of the body, small outgrowth is produced which is called a bud.
4. It grows and forms tentacles and gradually forms a new individual.
5. The young Hydra after complete development detaches from the parent and becomes an independent new organism.

Question 2.
Explain the different methods of reproduction occurring in sponges.
Answer:

1. Sponges reproduce both asexually and sexually and they also possess the power of regeneration. Their sexual reproduction is similar to higher animals even though their body organization is primitive type.
2. Asexual reproduction in sponges takes place by regeneration, budding and gemmule formation.
3. In sponges, during unfavourable period, gemmule is produced. It is an internal bud.
4. Archaeocytes which are dormant cells are seen in the aggregation in gemmule. These cells are capable of developing into a new organism.
5. Amoebocytes are other cells which secrete thick resistant layer of secretion which is coated around archaeocytes.
6. When favourable conditions of water and temperature return back, the gemmules can develop into new individuals by hatching, e.g. Spongilla.

Question 3.
IVF.
Answer:

1. In laboratory under sterile conditions, oocyte and sperms are placed in a test tube or glass plate to form a zygote. This process is called In Vitro Fertilization.
2. The zygote with 8 blastomeres is then transferred into the fallopian tube for further development.
3. IVF technique is used when childless couple wants to have a baby, but there are issues of sterility.
4. IVF is also called test tube baby technique.

Question 4.
Comment on any two mechanical contraceptive methods.
Answer:
Two mechanical contraceptive methods are as follows:
A. Condom or Nirodh:

1. It is a protective barrier in the form of thin rubber sheath which is used by male partner during the sexual coitus. It covers the penis and does not allow semen to flow during copulation.
2. Thus the entry of ejaculated semen into the female reproductive tract is obstructed. This can prevent conception. It is a simple and effective method and has no side effects.
3. “Nirodh” is a condom, most widely used in India as a contraceptive by males.
4. Condom also protects both the partners against sexually transmitted diseases such as AIDS and others.

B. Diaphragm, cervical caps and vaults:

1. Diaphragm and cervical caps are to be used by females as mechanical contraceptive measures.
2. They are made up of rubber. They are fitted on the cervix in vagina so that they prevent the entry of sperms into the uterus.
3. They are kept at least six hours after sexual intercourse in order to inhibit sperms from entering female genital tract.

Question 5.
Tubectomy.
Answer:

1. The permanent birth control method in women, is called tubectomy.
2. It is a surgical method, also called sterilization.
3. In tubectomy, a small part of the fallopian tube is tied and cut.
4. Tubectomy blocks transport of oocytes and also blocks sperms, thus preventing fertilization from reaching the oocyte.

Question 6.
Give the name of causal organism of Syphilis and write on its symptoms.
Answer:
1. Syphilis is a sexually transmitted veneral disease caused by a Spirochaete bacterium Treponema pallidum.

2. The site of infection is the mucous membrane in genital, rectal and oral region.

3. Symptoms of syphilis:

• Primary lesion known as chancre at the site of infection.
• They are seen on the external genitalia in males and inside the vagina in females.
• Skin rashes accompanied by fever, inflammed joints and loss of hair.
• Paralysis
• Degenerative changes in the heart and brain.

Question 7.
What is colostrum?
Answer:

1. The fluid secreted by the mammary glands soon after childbirth is called colostrum.
2. Colostrum is the sticky and yellow fluid. It contains proteins, lactose and mother’s antibodies, e.g. IgA.
3. The fat content in colostrum is low.
4. The antibodies present in colostrum helps in developing resistance for the newborn baby at a time when its own immune response is not fully developed.

5. Answer the Following Questions

Question 1.
Describe the phases of menstrual cycle and their hormonal control.
Answer:
Menstrual cycle (Ovarian cycle):
i. Menstrual cycle involves a series of cyclic, changes in the ovary and uterus. The cyclic events are regulated by gonadotropins from pituitary and the hormones from ovary.
ii. The cyclic events in woman are repeated within approximately 28 days.
iii. Menstrual cycle is divided into following phases, viz.

1. Menstrual phase (Day 1-5)
2. Follicular phase in ovary that coincides with proliferative phase in uterus. Post menstrual phase (Day 5-14)
3. Ovulatory phase (Day 14-15)
4. Luteal phase in ovary which coincides with secretory phase in uterus (Day 16 to 28).

1. Menstrual Phase:

• Menstrual phase occurs in the absence of fertilization.
• During menstruation, uterine endometrium is sloughed off. Level of progesterone and estrogen decrease during this phase resulting into release of prostaglandins which cause this rupture.
• Blood about 45-100 ml, tissue fluid, mucus, endometrial lining and unfertilized oocyte and other cellular debris is discharged through vagina as a menstrual flow. The endometrial lining becomes about 1 mm thin.
• Fibrinolysin does not allow blood to clot during this period.
• Pituitary starts secreting FSH, which further makes many primordial follicles to develop into primary and few of them into secondary follicles.

2. Proliferative phase/Follicular phase/Post menstrual phase:

• During this phase in the ovary the follicles develop while in uterus the endometrium starts proliferating. 6 to 12 secondary follicles start developing but usually only one of them becomes Graafian follicle due to action of FSH.
• Developing secondary follicles secrete the hormone estrogen.
• Estrogen brings about regeneration of endometrium. Further proliferation of endometrium causes formation of endothelial cells, endometrial or uterine glands and network of blood vessels. Endometrium’s thickness becomes 3-5 mm.

(3) Ovulatory phase:

• Ovulation occurs in this phase. Mature Graafian follicle ruptures and secondary oocyte is released into the pelvic region of abdomen.
• Ovulation occurs due to surging quantity of LH from pituitary.

(4) Luteal phase/Secretary phase :
(i) Since the empty Graafian follicle converts itself into corpus luteum under the influence of LH, this phase is called luteal phase in ovary. At the same time, the uterine endometrium thickens and becomes more secretory and hence it is called secretory phase in uterus.

(ii) Corpus luteum secretes progesterone, some amount of estrogens and inhibin. These hormones stimulate the growth of endometrial glands which later start uterine secretions.

(iii) Endometrium becomes more vascularized becomes 8-10 mm. in thickness. These changes are the preparation for the implantation of the ovum if fertilization occurs.

(iv) In absence of fertilization, corpus luteum can survive for only two weeks and then degenerate into a non-secretory white scar called corpus albicans.

(v) If ovum is fertilized, woman becomes pregnant and hormone hCG (human Chorionic Gonadotropin) is secreted by chorionic membrane of embryo which keeps corpus luteum active till the formation of placenta.

Question 2.
Explain the steps of parturition.
Answer:
Parturition involves the following three steps:
1. Dilation stage:

• Dilation stage means dilating the birth canal or passage though which baby is pushed out. In the beginning uterine contractions start from top and baby is moved to cervix. Due to compression of blood vessels and movements of flexible joints in pelvic girdle, mother experiences labour pains.
• Oxytocin is secreted later in more amount causing severe uterine contractions. This pushes baby in a head down position and closer to cervix.
• Cervix and vagina both are dilated.
• This stage lasts for about 12 hours.
• At the end, amniotic sac ruptures and amniotic fluid is passed out.

2. Expulsion stage:

• During second stage of about 20 to 60 minutes, the uterine and abdominal contractions become stronger.
• Foetus moves out with head down position through cervix and vagina.
• The umbilical cord which connects the baby to placenta is tied and cut off close to the baby’s navel.

3. After birth or placental stage : In the last stage of 10 to 45 minutes, once the baby is out then the placenta is also separated from uterine wall and is expelled out as “after birth”. This is accompanied by severe contractions of the uterus.

Question 3.
Explain the histological structure of testis.
Answer:
Histological structure of testis:

1. The external covering of testis is a fibrous connective tissue called tunica albuginea.
2. Then there is an incomplete peritoneal covering called tunica vaginalis.
3. Interior to this there is a covering called tunica vascularis formed by capillaries.
4. The testis is composed of many seminiferous tubules that are lined by cuboidal germinal epithelial cells.
5. In the seminiferous tubules various stages of developing sperms are seen as spermatogenesis takes place here. These stages are spermatogonia, primary and secondary spermatocytes, spermatids and sperms.
6. Large, pyramidal sub tentacular cells, nurse cells or Sertoli cells are present between germinal epithelium. Sperm bundles remain attached to Sertoli cells with their heads.
7. Seminiferous tubules form sperms whereas Sertoli cells provide nourishment to the sperms till maturation.
8. In between the seminiferous tubules there are interstitial cells of Leydig which are endocrine in nature. They secrete testosterone.

Question 4.
Describe the structure of blastocyst or blastulation
Answer:

1. The outer layer of cells of the morula is called trophoblast or trophoectoderm. This layer absorbs the nutritive fluid secreted by uterine endometrial membrane.
2. As more and more fluid is absorbed by trophoblast cells, the cells become flat and a cavity called blastocyst cavity or blastocoel or segmentation cavity is formed.
3. This causes trophoblast cells to get separated from inner cell mass except at one side.
4. The trophoblast cells in contact with embryonal knob are known as cells of Rauber. As the quantity of fluid increases, the morula enlarges rapidly and assumes the shape of a cyst. This stage is called blastocyst.
5. The side of the blastocyst to which embryonal knob is attached is known as the embryonic or animal pole and the opposite side as abembryonic pole.
6. The trophoblast produces extra embryonic membranes and does not participate in the formation of embryo proper.
7. Zona pellucida disappears allowing the blastula to increase in size and volume. The blastocyst stage is reached in about five days after fertilization.
8. Blastocyst depends on mother for nutrition which it obtained through placenta.

Question 5.
Explain the histological structure of ovary in human.
Answer:
Histological structure of ovary:
(1) Each ovary is a compact structure differentiated into a central part called medulla and the outer part called cortex.

(2) The cortex is covered externally by a layer of germinal epithelium while the medulla contains the stroma or loose connective tissue with blood vessels, lymph vessels and nerve fibres.

(3) Different stages of developing ovarian follicles are seen in the cortex. Each primordial follicle has at its centre a large primary oocyte (2n) surrounded by a single layer of flat follicular cells, then gradually it matures.

(4) In the ovary during each menstrual cycle there is a maturation of primordial follicles into multilayered primary, secondary and Graafian follicles.

(5) Every Graafian follicle has three layers, viz. theca externa, theca interna and membrana granulosa which are from outer to inner side. A space called antrum filled with liquor folliculi is present inside the follicle. There is a small hillock of cells called cumulus oophours or discus proligerus over which the ovum is lodged. The ovum in turn is covered by vitelline membrane, zona pellucida and corona radiata from inner side to outer surface.

(6) Ovarian cortex also shows corpus luteum, or yellow body formed from empty Graafian follicle after ovulation. Corpus luteum is converted into corpus albicans or white body in case of absence of conception.

Question 6.
Describe the various methods of birth control to avoid pregnancy.
Answer:
Birth control/Contraceptive methods are of two main types, viz. temporary and permanent.
A. Temporary methods:
(1) Natural method/Safe period/Rhythm method : A week before and a week after menstrual bleeding is considered the safe period for sexual intercourse. It is based on the fact that ovulation occurs on the 14th day of menstrual cycle.

(2) Coitus Interruptus or withdrawal : In this method, the male partner withdraws his penis from the vagina before ejaculation, so as to avoid insemination. This method also has some drawbacks, as the pre-ejaculation fluid may contain sperms and this can cause fertilization.

(3) Lactational amenorrhoea (absence of menstruation) : This method is based on the fact that ovulation does not occur during the period of intense lactation following parturition so chances of conception are almost negligible. However, this method also has high chances of failure.

(4) Chemical means (spermicides) : In this method chemicals like foam, tablets, jellies and creams are introduced into the vagina before sexual intercourse, they adhere to the mucous membrane, immobilize and kill the sperms.

(5) Mechanical means/Barrier methods:
(i) Condom : It is a thin rubber sheath that is used to cover the penis of the male. Condom should be used before starting coital activity. It also prevents STDs and AIDS.

(ii) Diaphragm, cervical caps and vaults : These devices made of rubber are inserted into the female reproductive tract to cover the cervix diming copulation. They prevent conception by blocking the entry of sperms through the cervix.

(iii) Intra-uterine devices (IUDs) : These are plastic or metal objects placed in the uterus by a doctor. These include Lippes loop, copper releasing IUDs (Cu-T, Cu 7, multiload 375) and hormone releasing IUDs (LNG-20, progestasert). They prevent fertilization of the egg or implantation of the embryo.

(6) Physiological (Oral) Devices : Birth control pills (oral contraceptive pills) check ovulation as they inhibit the secretion of follicle stimulating hormone (FSH) and luteinizing hormone (LH) that are necessary for ovulation. The pill ‘Saheli’ is taken weekly.

(7) Other contraceptives : The birth control implant is similar to that of pills in their mode of action. It is implanted under the skin of the upper arm of the female.

B. Permanent methods surgical operations : In men surgical operation is called vasectomy and in women it is called tubectomy. This method blocks gamete transport and prevent pregnancy.

Question 7.
What are the goals of RCH programmes?
Answer:
Goals of RCH programmes are as follows:

1. Various aspects related to reproduction are made aware to general public.
2. Facilities are provided to people to understand and build up reproductive health.
3. Support is given for building up a reproductively healthy society.
4. Three critical health indicators, i.e. reducing total infertility rate, infant mortality rate and maternal mortality rate are well looked after.

Question 8.
What is parturition? Which hormones are involved in parturition?
Answer:

1. Parturition is the act of expelling out the mature foetus from the uterus of mother via the vagina.
2. When the foetus is fully mature, it starts secreting ACTH (Adreno Cortico Trophic Hormone) from its pituitary.
3. ACTH stimulates adrenal glands of foetus to produce corticosteroids.
4. These corticosteroids diffuse from foetal blood to mother’s blood across the placenta. Corticosteroids accumulate in mother’s blood that results in decreased amount of progesterone. Corticosteroids also increase secretion of prostaglandins.
5. Simultaneously estrogen levels rise bringing about initation of contractions of uterine muscular wall.
6. Reduced progesterone level and increased estrogen level cause secretion of oxytocin from mother’s pituitary. This causes greater stimulation of myometrium of uterus.
7. Prostaglandins cause increased forceful contraction of uterus which expels the foetus out of the uterus.
8. Hormone relaxin secreted by the placenta makes the pubic ligaments and sacroiliac joints of the mother loosen. This causes widening of birth canal which facilitates the normal birth of the baby.

Question 9.
What are the functions of male accessory glands?
OR
Write a brief account of accessory sex glands associated with human male reproductive system.
Answer:
Seminal Vesicles, prostate gland and Cowper’s glands are associated with human male reproductive system.
(i) Seminal Vesicles:

1. Seminal vesicles occur in pair present on the posterior side of urinary bladder. Its secretion consists about 60% of the total volume of the semen. The secretion is an alkaline seminal fluid containing fructose, fibrinogen and prostaglandins.
2. Fructose helps in the movement of sperms by providing energy to them.
3. Semen is coagulated in bolus by fibrinogen. This helps in faster movements of sperms in vagina after insemination.
4. Reverse peristalsis in vagina and uterus for faster movement of sperms towards the egg in the female body is elided by prostaglandins.

(ii) Prostate gland:

1. It is a single gland located under the urinary bladder. It has about 20 to 30 separate lobes which open separately into the urethra.
2. Prostatic fluid secreted by this gland is milky white and slightly acidic. It forms 30 % of the semen and is secreted in urethra.
3. Its contents are citric acid, acid phosphatase and various other enzymes.
4. The sperms are protected from the acidic environment of vagina by acid phosphatase.

(iii) Cowper’s glands (Bulbo-urethral glands):

1. Cowper’s glands occur in pair on either side of urethra. They are small and pea shaped.
2. Cowper’s glands secrete an alkaline, viscous, mucous-like fluid. It helps as lubricant during copulation.

Question 10.
What is capacitation? Give its importance.
Answer:

1. Capacitation is the process by which the sperms are made capable to swim up to the fallopian tubes. This process takes place in 5-6 hours.
2. 50% of ejaculated sperms die due to unfavourable vaginal and uterine conditions.
3. The remaining sperms are capacitated with the help of prostaglandin and vestibular secretions of female tract. It involves the changes in the membrane covering the acrosome.
4. Due to capacitation, acrosome membrane becomes thin, Calcium ions enters the sperm and their tail begin to show rapid whiplash movements.
5. Sperms become extra active and then they ascend upwards to reach fallopian tubes.
6. After capacitation the sperms swim through the vagina and uterus and reach ampulla of fallopian tube within 5 minutes.

Long answer questions

Question 1.
Explain the following parts of male reproductive system along with labelled diagram showing these parts – Testis, vasa deferentia, epididymis, seminal vesicle, prostate gland and penis.
OR
With the help of a neat, labelled diagram, describe the human male reproductive system.
Answer:

(i) Testis, the male gonad, the accessory ducts and glands along with external genitalia form the male reproductive system.

(ii) Testes:

1. Testes are male gonads with dimensions of about 4.5. cm length, 2.5 cm width and 3 cm thickness.
2. There are about 200 to 300 lobules in each testis in which there are seminiferous tubules that form rete testis.
3. Testes produce sperms and secrete male sex hormone, androgen or testosterone.

(iii) Accessory ducts : Rete testis, vasa efferentia, epididymis, vas deferens, ejaculatory duct and urethra together form the accessory ducts of male reproductive system.
1. Vasa efferentia : Vasa efferentia are 12-20 fine tubules. They arise from rete testis and end into the epididymis. The sperms from the testis are carried by these ducts to the epididymis.

2. Epididymis : Epididymis are long and coiled tubes having three parts, viz. caput, corpus and cauda epididymis. They are located on the posterior border of each testis. The sperms undergo maturation in epididymis.

3. Vasa deferentia:

• Vasa deferentia are a pair of 40 cm long tubular structures that arise from cauda epididymis.
• Each vas deferens enters the abdominal cavity through the inguinal canal and then ascends in the form of spermatic cord.
• Vas deferens of each side is joined by the duct from seminal vesicle to form ejaculatory duct.

4. Ejaculatory duct : About 2 cm long pair of ducts formed by joining of vas deferens and a duct of seminal vesicle are the ejaculatory ducts. Both ejaculatory ducts open into urethra near the prostate gland. Seminal fluid containing spermatozoa are carried by ejaculatory duct to the urethra.

5. Urethra : The male urethra provides a common passage for the urine and semen hence is also called urinogenital duct.

(iv) Accessory glands : Associated with male reproductive system are : (a) Seminal vesicles (b) Prostate gland and (c) Cowper’s or Bulbourethral glands. Every accessory gland has secretion which helps in functions of reproductive system.

(v) External genitalia : External genitalia consists of penis and scrotum.
1. Penis:

• Penis is the copulatory organ used for insemination or deposition of sperms in female genital tract.
• It is cylindrical, erectile and pendulous organ through which passes the urethra.
• It contains three columns of erectile tissues which has abundant blood sinuses.
• The tip is called glans penis while the retractible fold of skin on penis is called prepuce.

2. Scrotum : The scrotum is a pouch of pigmented skin arising from lower abdominal wall. It protects testes within it. Scrotum acts as thermoregulator. Testis are suspended in scrotum by spermatic cord.

Question 2.
Describe female reproductive system of human
Answer:

The female reproductive system consists of internal organs and external genitalia.

Internal organs are pair of ovaries and pair of fallopian ducts or oviducts, single median uterus and vagina. External genitalia is called vulva. There are a pair of vestibular glands in external genitalia. Mammary glands or breasts are also associated with reproductive system of female.

(1) Ovaries:

• Ovaries are situated in the abdomen in upper lateral part of the pelvis near the kidneys. Their dimensions are about 3 cm in length, 1.5 cm in breadth and 1.0 cm thick. They are solid, oval or almond shaped organs.
• Ovaries produce ova and they are also endocrine in nature as they produce estrogen, progesterone, relaxin, activin and inhibin.
• Ovarian hormones bring about secondary sexual characters. They also control menstrual cycle, pregnancy and parturition.

(2) Fallopian tubes/oviducts:
(i) Fallopian tubes lie horizontally over peritoneal cavity. These are about 10 to 12 cm long, narrow, muscular structure lined by ciliated epithelium.
(ii) They transport the ovum after ovulation from the ovary to the uterus.
(iii) Fallopian tube can be subdivided into the following three parts:

• The infundibulum which bears a number of finger-like processes called fimbriae at its free border.
• Infundibulum is funnel-shaped having ostium which receives ova released from the ovary.
• The second part is the ampulla where the fertilization takes place.
• The last part is short cornua or isthmus which opens into the uterus.

(3) Uterus/Womb:
(i) Uterus is a pear-shaped, highly muscular, thick walled, hollow organ measuring about 8 cm in length, 5 cm in width and 2 cm in thickness.

(ii) Uterus has the following three parts : Fundus, Body or corpus and Cervix.

(iii) The cervix communicates above with the body of the uterus by an aperture, the internal os and with vagina below by an opening the external os.

(iv) Uterus has three-layered wall. These layers are:

• Perimetrium : An outer serous layer.
• Myometrium : The middle thick muscular layer of smooth muscles.
• Endometrium : The inner highly vascular mucosa that has many uterine glands.

(v) Uterus receives the ovum from fallopian tube. It develops placenta during pregnancy for the nourishment of foetus. At the time of parturition, it expels the young one at birth.

(4) Vagina:

• Vagina is a highly distensible fibro-muscular tube that lies between the cervix and the vestibule.
• It is about 7 to 9 cm in length and is internally lined by stratified and non- keratinised epithelium. The vaginal wall has inner mucous lining.
• Vagina acts as a birth canal as well as copulatory passage. It also allows passage of menstrual flow.
• Vagina opens into vestibule by vaginal orifice which may be covered with hymen which is also a mucous membrane.

(5) External genitalia or vulva or pudendum : The external genitalia consists of five parts; viz. labia majora, labia minora, mons veneris, clitoris and vestibule.

(6) A pair of vestibular glands / Bartholin’s glands : These glands open into the vestibule and release a lubricating fluid.

(7) A pair of mammary glands/breasts : These are the accessory organs of female reproductive system for production and release of milk after parturition.

Question 3.
Describe the process of fertilization.
Answer:
(1) Fertilization is the process of fusion of the haploid male and female gametes which results in the formation of a diploid zygote (2n).

(2) In human beings fertilization is internal. Sperms deposited in vagina, swim across the uterus and fertilize the ovum in ampulla of the fallopian tube.

(3) Fertilization involves the following events:
(i) Insemination : Discharge of semen into the vagina at the time of copulation is called insemination.

(ii) Movement of sperm towards egg : Sperms reaching the vagina undergo capacitation process for 5-6 hours. During capacitation acrosomal membrane of sperm becomes thin and Ca⁺⁺ enters the sperm making it extra active. Sperms reach up to the ampulla by swimming aided with contraction of uterus and fallopian tubes. These contractions are stimulated by oxytocin of female. By capacitation sperm can reach ampulla within 5 minutes, they remain 5 viable for 24 to 48 hours, whereas ovum remains viable for 24 hours.

(iii) Entry of sperm into the egg : Though many sperms reach the ampulla, only a single sperm fertilizes the ovum. The acrosome of sperm after coming in contact with the ovum, releases lysins; hyaluronidase and corona penetrating enzymes. Due to these enzymes cells of corona radiata are separated and dissolved. The sperm head then passes through zona pellucida of egg. The zona pellucida has glycoprotein fertilizin receptor proteins. These bind to specific acid protein-antifertilizin of sperm. This makes sperm and ovum to come together. Fertilizin-Antifertilizin interaction is species- specific.

(iv) Acrosome reaction : When the sperm head comes in contact with the zona pellucida, its acrosome covering ruptures to release lytic enzymes, acrosin or zona lysin. These enzymes dissolve plasma membrane of egg so that the sperm nucleus and the centrioles enter the egg, while other parts remain outside. Now the vitelline membrane of egg changes into fertilization membrane which prevents any further entry of other sperms into the egg, thus polyspermy is prevented.

(v) Activation of ovum : After the entry of sperm head into ovum, it gets activated to resume and complete its meiosis-II. With this it gives out the second polar body. The germinal vesicle organises into female pronucleus. At this stage, it is true ovum.

(vi) Fusion of egg and sperm : The coverings of male and female pronuclei degenerate results in the formation of a synkaryon by a process called syngamy or karyogamy. The zygote is thus formed.

Question 4.
Explain the process by which zygote divides and redivides to form the morula.
Answer:
(1) Cleavage is a rapid mitotic division to form a blastula. These divisions takes place immediately after fertilization. The cells formed by cleavage are called blastomeres.

(2) The type of cleavage in human is holoblastic, i.e. the whole zygote gets divided, radial and indeterminate, i.e. fate of each blastomere is not predetermined.

(3) Cleavage show faster synthesis of DNA and high consumption of oxygen.

(4) Since there is no growth phase between the cleavages, the size of blastomeres will be reduced with every successive cleavage.

(5) The cleavages occur as follows:

(6) Successive divisions produce a solid ball of cells called morula of 16 cells. It consists of an outer layer of smaller clearer cells and an inner mass of larger cells.

(7) Morula reaches the uterus about 4-6 days after fertilization.