Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 6 Random Variable and Mathematical Expectation Ex 6.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 1.
Value which is obtained by multiplying possible values of random variable with probability of occurrence and is equal to weighted average is called
(a) Discrete value
(b) Weighted value
(c) Expected value
(d) Cumulative value
Solution:
(c) Expected value

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 2.
Demand of products per day for three days are 21, 19, 22 units and their respective probabilities are 0.29, 0.40, 0.35. Profit per unit is 0.50 paisa then expected profits for three days are
(a) 21, 19, 22
(b) 21.5, 19.5, 22.5
(c) 0.29, 0.40, 0.35
(d) 3.045, 3.8, 3.85
Solution:
(d) 3.045, 3.8, 3,85
Hint:

x 21 19 22
P(x) 0.29 0.40 0.35

E(X) = \(\sum_{ x }\) xP(x)
For Day 1
E(X) = 21 × 0.29
= 6.09
Expected profit = 6.09 × 0,50 = 3.054
For Day 2
E(X) = 19 × 0.40
= 7.6
Expected Profit = 7.6 × 0.50 = 3.8
For Day 3
E(X) = 22 × 0.35
= 7.7
Expected Profit = 7.7 × 0.50 = 3.85

Question 3.
Probability which explains x is equal to or less than particular value is classified as
(a) discrete probability
(b) cumulative probability
(c) marginal probability
(d) continuous probability
Solution:
(b) Cumulative Probability

Question 4.
Given E(X) = 5 and E(Y) = -2, then E(X – Y) is
(a) 3
(b) 5
(c) 7
(d) -2
Solution:
(c) 7
Hint:
E(x – 7) = E(x) – E(y) = 5 – (-2) = 7

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 5.
A variable that can assume any possible value between two points is called
(a) discrete random variable
(b) continuous random variable
(c) discrete sample space
(d) random variable
Solution:
(b) Continuous random Variable

Question 6.
A formula or equation used to represent the probability distribution of a continuous random variable is called
(a) probability distribution
(b) distribution function
(c) probability density function
(d) mathematical expectation
Solution:
(c) Probability density function

Question 7.
If X is a discrete random variable and p(x) is the probability of X, then the expected value of this random variable is equal to
(a) Σf(x)
(b) Σ[x + f(x)]
(c) Σf(x) + x
(d) Σxp(x)
Solution:
(d) Σxp(x)

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 8.
Which of the following is not possible in probability distribution?
(a) Σp(x) > 0
(b) Σp(x) = 1
(c) Σxp(x) = 2
(d) p(x) = -0.5
Solution:
(d) P(x) = -0.5

Question 9.
If c is a constant, then E(c) is
(a) 0
(b) 1
(c) c f(c)
(d) c
Solution:
(d) c

Question 10.
A discrete probability distribution may be represented by
(a) table
(b) graph
(c) mathematical equation
(d) all of these
Solution:
(d) all of these

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 11.
A probability density function may be represented by:
(a) table
(b) graph
(c) mathematical equation
(d) both (b) and (c)
Solution:
(d) both (b) and(c)

Question 12.
If c is a constant in a continuous probability distribution, then p(x = c) is always equal to
(a) zero
(b) one
(c) negative
(d) does not exist
Solution:
(a) Zero

Question 13.
E[X – E(X)] is equal to
(a) E(X)
(b) V(X)
(c) 0
(d) E(X) – X
Solution:
(c) 0

Question 14.
E[X – E(X)]² is
(a) E(X)
(b) E(X)²
(c) V(X)
(d) S.D(X)
Solution:
(c) V(X)

Question 15.
If the random variable takes negative values, then the negative values will have
(a) positive probabilities
(b) negative probabilities
(c) constant probabilities
(d) difficult to tell
Solution:
(a) Positive probabilities

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 16.
If we have f(x) = 2x, 0 ≤ x ≤ 1, then f(x) is a
(a) probability distribution
(b) probability density function
(c) distribution function
(d) continuous random variable
Solution:
(b) Probability density function

Question 17.
\(\int_{ -∞ }^{∞}\) f(x) dx is always equal to
(a) zero
(b) one
(c) E(X)
(d) f(x) + 1
Solution:
(b) one

Question 18.
A listing of all the outcomes of an experiment and the probability associated with each outcome is called
(a) probability distribution
(b) probability density function
(c) attributes
(d) distribution function
Solution:
(a) Probability distribution

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 19.
Which one is not an example of a random experiment?
(a) A coin is tossed and the outcome is either a head or a tail
(b) A six-sided die is rolled
(c) Some number of persons will be admitted to a hospital emergency room during any hour.
(d) All medical insurance claims received by a company in a given year.
Solution:
(d) All medical insurance claims received by a company in a given year

Question 20.
A set of numerical values assigned to a sample space is called
(a) random sample
(b) random variable
(c) random numbers
(d) random experiment
Solution:
(b) random variable

Question 21.
A variable which can assume a finite or countably infinite number of values is known as
(a) continuous
(b) discrete
(c) qualitative
(d) none of them
Solution:
(b) Discrete

Question 22.
The probability function of a random variable is defined as
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3 1
Then k is equal to
(a) zero
(b) \(\frac { 1 }{4}\)
(c) \(\frac { 1 }{15}\)
(d) one
Solution:
(c) k = 1/15
Hint:
W.K.T Σi=1 P(xi) = 1
p(x= -1)+ p(x= -2) + p(x = 0) + p(x = 1)
+ p(x = 2) = V
k + 2k + 3k + 4k + 5k = 1
5k = 1 ⇒ k = 1/5

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 23.
If p(x) = \(\frac { 1 }{10}\), x = 10, then E(X) is
(a) zero
(b) \(\frac { 6 }{8}\)
(c) 1
(d) -1
Solution:
(c) 1
Hint:
P(x) + 1/10 and x = 10
E(x) = Xp(x) = 10(1/10) = 1

Question 24.
A discrete probability function p(x) is always
(a) non-negative
(b) negative
(c) one
(d) zero
Solution:
(a) non-negative

Question 25.
In a discrete probability distribution, the sum of all the probabilities is always equal to
(a) zero
(b) one
(c) minimum
(d) maximum
Solution:
(b) one

Question 26.
An expected value of a random variable is equal to it’s
(a) variance
(b) standard deviation
(c) mean
(d) con variance
Solution:
(c) mean

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 27.
A discrete probability function p(x) is always non-negative and always lies between
(a) 0 and ∞
(b) 0 and 1
(c) -1 and +1
(d) -∞ and ∞
Solution:
(b) 0 and 1

Question 28.
The probability density function p(x) cannot exceed
(a) zero
(b) one
(c) mean
(d) infinity
Solution:
(b) One

Question 29.
The height of persons in a country is a random variable of the type
(a) discrete random variable
(b) continuous random variable
(c) both (a) and (b)
(d) neither (a) not (b)
Solution:
(b) Continuous random variable

Question 30.
The distribution function F(x) is equal to
(a) P(X = x)
(b) P(X ≤ x)
(c) P (X ≥ x)
(d) all of these
Solution:
(b) p(x ≤ x)

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 8 Physical and Chemical Equilibrium Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium

11th Chemistry Guide Physical and Chemical Equilibrium Text Book Back Questions and Answers

Textbook Evaluation:

I. Choose the best answer:

Question 1.
If Kb and Kf for a reversible reaction are 0.8 × 10-5 and 1.6 × 10-4 respectively, the value of the equilibrium constant is,
a) 20
b) 0.2 × 10-1
c) 0.05
d) None of these
Answer:
a) 20

Question 2.
At a given temperature and pressure, the equilibrium constant values for the equilibria
3A2 + B2 + 2Cr Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 1 2A3BC and
A3BC Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 2 \(\frac{3}{2}\) [A2] + \(\frac{1}{2}\) B2 +C
The relation between K1 and K2 is
a) K1 = \(\frac{1}{\sqrt{\mathrm{K}_{2}}}\)

b)K2 = \(\mathrm{K}_{1}^{-1 / 2}\)

C) K12 = 2K2

d) \(\frac{\mathrm{K}_{1}}{2}\) = K2
Answer:
b)K2 = \(\mathrm{K}_{1}^{-1 / 2}\)

Question 3.
The equilibrium constant for a reaction at room temperature is K1 and that at 700 K is K2 If K1 > K2, then
a) The forward reaction is exothermic
b) The forward reaction is endothermic
c) The reaction does not attain equilibrium
d) The reverse reaction is exothermic
Answer:
a) The forward reaction is exothermic

Question 4.
The formation of ammonia from N2(g) and H2(g) is a reversible reaction 2NO(g) + O2(g) Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 2 2NO2(g) + Heat What is the effect of increase of temperature on this equilibrium reaction
a) equilibrium is unaltered
b) formation of ammonia is favoured
c) equilibrium is shifted to the left
d) reaction rate does not change
Answer:
c) equilibrium is shifted to the left

Question 5.
Solubility of carbon dioxide gas in cold water can be increased by
a) increase in pressure
b) decrease in pressure
c) increase in volume
d) none of these
Answer:
a) increase in pressure

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 6.
Which one of the following is incorrect statement?
a) for a system at equilibrium, Q is always less than the equilibrium constant
b) equilibrium can be attained from either side of the reaction
c) the presence of catalyst affects both the forward reaction and reverse reaction to the same extent
d) Equilibrium constant varied with temperature
Answer:
a) for a system at equilibrium, Q is always less than the equilibrium constant

Question 7.
K1 and K2 are the equilibrium constants for the reactions respectively.
N2(g) + O2(g)Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 1 2NO(g)
2NO(g) + O2(g) Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 22NO2 (g)
What is the equilibrium constant for the reaction NO2(g) ⇌ 1/2 N2(g) + O2(g)
a) \(\frac{1}{\sqrt{\mathrm{K}_{1} \mathrm{~K}_{2}}}\)

b) \(\left(\mathrm{K}_{1}=\mathrm{K}_{2}\right)^{1 / 2}\)

c) \(\frac{1}{2 \mathrm{~K}_{1} \mathrm{~K}_{2}}\)

d) \(\left(\frac{1}{\mathrm{~K}_{1} \mathrm{~K}_{2}}\right)^{3 / 2}\)
Answer:
a) \(\frac{1}{\sqrt{\mathrm{K}_{1} \mathrm{~K}_{2}}}\)

Question 8.
In the equilibrium, 2A(g) ⇌ 2B(g) + C2(g) the equilibrium concentrations of A, B and C2 at 400K are 1 × 10-4M, 2.0 × 10-3 M, 1.5 × 10-4 M respectively. The value of Kc for the equilibrium at 400 K is
a) 0.06
b) 0.09
c) 0.62
d) 3 × 10-2
Answer:
a) 0.06

Question 9.
An equilibrium constant of 3.2 × 10-6 for a reaction means, the equilibrium is
a) largely towards forward direction
b) largely towards reverse direction
c) never established
d) none of these
Answer:
b) largely towards reverse direction

Question 10.
\(\frac{\mathrm{K}_{\mathrm{C}}}{\mathrm{K}_{\mathrm{P}}}\) for the reaction,
N2(g) + 3H2(g) ⇌ 2NH3 (g) is
a) \(\frac{1}{\mathrm{RT}}\)
b) \(\sqrt{\mathrm{RT}}\)
c) RT
d) (RT)2
Answer:
d) (RT)2

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 11.
For the reaction AB (g) ⇌ A(g) + B(g), at equilibrium, AB is 20 % dissociated at a total pressure of P, the equilibrium constant Kp is related to the total pressure by the expression
a) P = 24 Kp
b) P = 8 Kp
c) 24 P = Kp
d) none of these
Answer:
a) P = 24 Kp

Question 12.
In which of the following equilibrium, Kp and Kc are not equal?
a) 2N0(g) ⇌ N2(g) + O2(g)
b) SO2(g) + NO2 ⇌ SO3(g) + NO(g)
c) H2(g) + I2(g) ⇌ 2HI(g)
d) PCl5 ⇌ PCl3(g) + Cl2(g)
Answer:
b) SO2(g) + NO2 ⇌ SO3(g) + NO(g)

Question 13.
If x is the fraction of PCl5 dissociated at equilibrium in the reaction PCl5 ⇌ PCl3 + Cl2 then starting with 0.5 mole of PCl5, the total number of moles of reactants and products at equilibrium is
a) 0.5 – x
b) x + 0.5
c) 2x + 0.5
d) x + 1
Answer:
b) x + 0.5

Question 14.
The values of Kp1 and Kp2; for the reactions, X ⇌ Y + Z, A ⇌ 2B are in the ratio 9 : 1 if degree of dissociation of X and A be equal then total pressure at equilibrium P1, and P2 are in the ratio
a) 36 : 1
b) 1 : 1
c) 3 : 1
d) 1 : 9
Answer:
a) 36 : 1

Question 15.
In the reaction
Fe(OH)3(S) ⇌ Fe3+ (aq) + 3OH (aq),
if the concentration of OH ions is decreased by 1/4 times, then the equilibrium concentration of Fe3+ will
a) not changed
b) also decreased by 1/4 times
c) increase by 4 times
d) increase by 64 times
Answer:
d) increase by 64 times

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 16.
Consider the reaction where Kp = 0.5 at a particular temperature PCl5(g) ⇌ PCl3 (g) + Cl2 (g) if the three gases are mixed in a container so that the partial pressure of each gas is initially 1 atm, then which one of the following is true.
a) more PCl3 will be produced
b) more Cl2 will be produced
c) more PCl5 will be produced
d) None of these
Answer:
c) more PCl5 will be produced

Question 17.
Equimolar concentrations of H2 and I2 are heated to equilibrium in a 1 liter flask. What percentage of the initial concentration of H2 has reacted at equilibrium if the rate constant for both forward and reverse reactions are equal
a) 33%
b) 66%
c) (33)2%
d) 16.5 %
Answer:
a) 33%

Question 18.
In a chemical equilibrium, the rate constant for the forward reaction is 2.5 × 10-2, and the equilibrium constant is 50. The rate constant for the reverse reaction is,
a) 11.5
b) 50
c) 2 × 102
d) 2 × 10-3
Answer:
b) 50

Question 19.
Which of the following is not a general characteristic of equilibrium involving physical process
a) Equilibrium is possible only in a closed system at a given temperature
b) The opposing processes occur at the same rate and there is a dynamic but stable condition
c) All the physical processes stop at equilibrium
d) All measurable properties of the system remains constant
Answer:
c) All the physical processes stop at equilibrium

Question 20.
For the formation of Two moles of SO3(g) from SO2 and O2, the equilibrium constant is K1. The equilibrium constant for the dissociation of one mole of SO3 into SO2 and O2 is
a) \(1 / \mathrm{K}_{1}\)
b) K12
c) \(\left(\frac{1}{\mathrm{~K}_{1}}\right)^{1 / 2}\)
d) \(\frac{\mathrm{K}_{1}}{2}\)
Answer:
c) \(\left(\frac{1}{\mathrm{~K}_{1}}\right)^{1 / 2}\)

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 21.
Match the equilibria with the corresponding conditions:
i) Liquid ⇌ Vapour
ii) Solid ⇌ Liquid
iii) Solid ⇌ Vapour
iv) Solute(s) ⇌ Solute (Solution)
1) Melting point
2) Saturated solution
3) Boiling point
4) Sublimation point
5) Unsaturated solution

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 3
Answer:
b) 3 1 4 2

Question 22.
Consider the following reversible reaction at equilibrium, A + B ⇌ C, If the concentration of the reactants A and B are doubled, then the equilibrium constant will
a) be doubled
b) become one fourth
c) be halved
d) remain the same
Answer:
d) remain the same

Question 23.
[Co(H2O)6]2+ (aq) (pink) + 4Cl (aq) ⇌ [CoCl4]2-(aq) (blue) + 6 H2O (l)
In the above reaction at equilibrium, the reaction mixture is blue in colour at room temperature. On cooling this mixture, it becomes pink in color. On the basis of this information, which one of the following is true?
a) ∆H > 0 for the forward reaction
b) ∆H = 0 for the reverse reaction
c) ∆H < 0 for the forward reaction
d) Sign of the ∆H cannot be predicted based on this information
Answer:
a) ∆H > 0 for the forward reaction

Question 24.
The equilibrium constants of the following reactions are:
N2 + 3H2 ⇌ 2NH3; K1
N2 + O2 ⇌ 2NO; K2
H2 + 1/2O2 ⇌ H2O; K3
The equilibrium constant (K) for the reaction;
2NH3 + 5/2 O2 ⇌ 2NO + 3H2O, will be
a) K23 K3/K1

b) K1 K33/K2

c) K2K33/K1

d) K2K3/K1
Answer:
c) K2K33/K1

Question 25.
A 20 litre container at 400 K contains CO2 (g) at pressure 0.4 atm and an excess (neglect the volume of solid SrO). The volume of the container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when the pressure of CO2 attains its maximum value will be: Given that: SrCO3(S) ≅ SrO + CO2 (g) [Kp = 1.6 atm]
a) 2 litre
b) 5 litre
c) 10 litre
d) 4 litre
Answer:
b) 5 litre

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

II. Write brief answer to the following questions:

Question 26.
If there is no change in concentration, why is the equilibrium state considered dynamic?
Answer:
At chemical equilibrium the rate of two opposing reactions are equal and the concentration of. reactants and products do not change with time. This condition is not static and is dynamic because both the forward and reverse reactions are still occurring at the same rate and no macroscopic change is observed. So the chemical equilibrium is in a state of dynamic equilibrium.

Question 27.
For a given reaction at a particular temperature, the equilibrium constant has a constant value. Is the value of Q also constant? Explain.
Answer:
Kc and Qc are constant at equilibrium, both are temperature dependent. When Kc is constant at a given temperature, Qc also constant.

Question 28.
What is the relation between Kp and Kc? Given one example for which Kp is equal to Kc.
Answer:
The relation between Kp and Kc is Kp = KC (RT)∆ng
Kp = equilibrium constant is terms of partial pressure.
Kc = equilibrium constant is terms of concentration.
R = gas constant
T = Temperature.
∆ng = Difference between the sum of the number of moles of products and the sum of number of moles of reactants in the gas phase. When ∆ng = 0
Kp = KC(RT)0 = KC i.e., Kp = KC
Example: H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)
∆ng = 2 – 2 = 0
∴ Kp = KC for the synthesis of HI.

Question 29.
For a gaseous homogeneous reaction at equilibrium, number of moles of products are greater than the number of moles of reactants. Is Kc is larger or smaller than Kp.
Answer:
Kp >Kc
np > nR

Question 30.
When the numerical value of the reaction quotient (Q) is greater than the equilibrium constant, in which direction does the reaction proceed to reach equilibrium?
Answer:
When Q > KC the reaction will proceed in the reverse direction, i.e, formation of reactants.

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 31.
For the reaction, A2(g) + B2(g) ⇌ 2AB(g); ∆H is -ve.
the following molecular scenes represent differenr reaction mixture.(A-green, B-blue)

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 4

i) Calculate the equilibrium constant Kp and (Kc).
ii) For the reaction mixture represented by scene (x), (y) the reaction proceed in which directions?
iii) What is the effect of increase in pressure for the mixture at equilibrium.
Answer:
Kc = \(\frac{[\mathrm{AB}]^{2}}{\left[\mathrm{~A}_{2}\right]\left[\mathrm{B}_{2}\right]}\) ; A – green, B – blue
Given that ‘V’ is constant(closed system)
At equilbrium,
Kc = \(\frac{(-)}{\left(\frac{2}{V}\right)\left(\frac{2}{V}\right)}\) = \(\frac{16}{4}\) = 4
Kp = Kc(RT)∆n = 4 (RT)0 = 4
At stage ‘x’,
Q = \(\frac{\left(\frac{6}{\mathrm{~V}}\right)^{2}}{\left(\frac{2}{\mathrm{~V}}\right)\left(\frac{1}{\mathrm{~V}}\right)}=\frac{36}{2}\) = 18
Q > Kc i.e., reverse reaction is favoured.
At stage ‘y’,
Q = \(\frac{\left(\frac{3}{V}\right)^{2}}{\left(\frac{3}{V}\right)\left(\frac{3}{V}\right)}=\frac{9}{3 \times 3}\) = 1
Kc > Q i.e., forward reaction is favoured.

Question 32.
State Le – Chateller principle.
Answer:
It states that If a system at equilibrium is disturbed, then the system shifts itself in a direction that nullifies the effect of that disturbance.

Question 33.
Consider the following reactions,
a) H2(g) + I2(g) ⇌ 2HI(g)
b) CaCO3(s) ⇌ CaO (s) + CO2(g)
c) S(s) + 3 F2(g) ⇌ SF6(g)
In each of the above reactions find out whether you have to increase (or) decrease the volume to increase the yield of the product.
Answer:
a) H2(g) + I2(g) ⇌ 2HI(g)
In this reaction, there is no effect on changing the volume. ∆ng = 0

b) CaCO3(s) ⇌ CaO (s) + CO2(g)
In this reaction, increases in volume favours forward reaction.

c) S(s) + 3 F2(g) ⇌ SF6(g)
In this reaction, decreases in volume favours forward reaction.

Question 34.
State law of mass action.
Answer:
The law states that “At any instant, the rate of chemical reaction at a given temperature is directly proportional to the product of the active masses of the reactants at that instant.”

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 35.
Explain how will you predict the direction of an equilibrium reaction.
Answer:
From the knowledge of equilibrium constant, it is possible to predict the di¬rection in which the net reaction is taking place for a given concentration or partial pressure of reactants and products.
Consider a general homogeneous reversible reaction,

xA + yB ⇌ lC + mD

For the above reaction under non-equilibrium conditions, reaction quotient ‘Q’ is defined as the ratio of the product of active masses of reaction products raised to the respective stoichiometric coefficients in the balanced chemical equation to that of the reactants.

Under non – equilibrium conditions, the reaction quotient Q can be calculated using the following expression.
Q = \(\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\)

As the reaction proceeds, there is a continuous change in the concentration of reactants and products and also the Q value until the reaction reaches equilibrium. At equilibrium, Q is equal to Kc at a particular temperature. Once the equilibrium is attained, there is no change in the Q value. By knowing the Q value, we can predict the direction of the reaction by comparing it with Kc.
If Q = Kc, the reaction is in equilibrium state.
If Q > Kc, the reaction will proceed in the reverse direction, i.e., formation of reactants.
If Q < Kc, the reaction will proceed in the forward direction i.e., formation of products.
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 5
Predicting the direction of a reaction

Question 36.
Derive a general expression for the equilibrium constant Kp and Kc for the reaction, 3H2(g) + N2(g) ⇌ 2NH3(g).
Answer:
Let us consider the formation of ammonia in which, ‘a’ moles nitrogen and h’ moles hydrogen gas are allowed to react in a container of volume V. Let ‘x’ moles of nitrogen react with 3 x moles of hydrogen to give 2x moles of ammonia.
3H2(g) + N2(g) ⇌ 2NH3(g)

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 6

Applying law of mass action,
K2 = \(\frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{~N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}}\)
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 7
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 8

Question 37.
Write the balanced chemical equation for an equilibrium reaction for which the equilibrium constant is given by expression.
Answer:
Kc = \(\frac{\left[\mathrm{NH}_{3}\right]^{4}\left[\mathrm{O}_{2}\right]^{7}}{\left.[\mathrm{NO}]^{4} \mid \mathrm{H}_{2} \mathrm{O}\right]^{6}}\)
Balanced equation is:
4NO + 6H2O ⇌ 4NH3 + 7O2

Question 38.
What is the effect of added Inert gas on the reaction at equilibrium?
Answer:
When an inert gas (i.e., a gas which does not react with any other species involved in equilibrium) is added to an equilibrium system at constant volume, the total number of moles. of gases present in the container increases, that is, the total pressure of gases increases, the partial pressure of the reactants and the products are unchanged. Hence at constant volume, the addition of inert gas has no effect on the equilibrium.

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 39.
Derive the relation between Kp and Kc.
Answer:
Let us consider the general reaction in which all reactants and products are ideal gases.
xA + yB ⇌ lC + mD
The equilibrium constant, Kc is,
Kc = \(\frac{\left[\mathrm{Cl}^{l}[\mathrm{D}]^{\mathrm{m}}\right.}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\) ……………..(1)

and Kp is,
Kp = \(\frac{p_{C}^{l} \times p_{D}^{m}}{p_{A}^{x} \times p_{B}^{y}}\) ………….(2)

The ideal gas equation is,
PV = nRT
P = \(\frac{n}{V}\) RT
since, Active mass = molar concentration = n/V
P = active mass × RT
Based on the above expression the partial pressure of the reactants and products can be expresssed as,
PXA = [A]x[RT]x
PYB = [A]y[RT]y
PlC = [A]l[RT]l
PmD = [A]m[RT]m
On substituting in Eqn.(2).,
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 9
By comparing equation (1) and (4),
we get
Kp∆ng =Kc (RT) …………..(5)
Where,
∆ng is the difference between the sum of number of moles of products and the sum of number of moles of reactants in the gas phase.
The following relations become immediately obvious.
when ∆ng = 0.
Kp = Kc (RT)0 = Kc
Example :
H2(g) + I2(g) ⇌ 2HI(g)
N2 + O2 ⇌ 2NO(g)
when ∆ng = +ve
Kp = Kc(RT)+ve
Kp > Kc
Example:
2NH3(g) ⇌ N2(g) + 3H2(g)
PCl5(g) ⇌ PCl3(g) + Cl2(g)
When ∆ng = -ve
Kp = Kc(RT)-ve
Kp < Kc
Example:
2H2(g) + O2(g) ⇌ 2H2O(g)
2SO2(g)+ O2(g) ⇌ 2SO3(g)

Question 40.
One mole of PCl5 is heated in one litre closed container. If 0.6 mole of chlorine is found at equilibrium, Calculate the value of equilibrium constant.
Solution:
Given that,
[PCl5]initial = \(\frac{1 \text { mole }}{1 \mathrm{dm}^{3}}\)
[Cl2]eq = 0.6 mole dm-3
PCl5 ⇌ PCl3 + Cl2
[PCl3]eq = 0.6 mole dm-3
[PCl5]eq = 0.4 mole dm-3
∴ Kc = \(\frac{\left[\mathrm{PCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]}=\frac{0.6 \times 0.6}{0.4}\)
= 0.9 mole dm-3

Question 41.
For the reaction SrCO3 (s) ⇌ SrO(s) + CO2(g) the value of equilibrium constant Kp = 2.2 × 10-4 at 1002 K. Calculate Kc for the reaction.
Solution :
For the reaction,
SrCO3(s) ⇌ SrO(s) + CO2(g)
∆ng = 1 – 0 = 1
∴ Kp = Kc (RT)
2.2 × 10-4 = Kc (0.0821) (1002)
Kc = \(\frac{2.2 \times 10^{-4}}{0.0821 \times 1002}\) = 2.674 × 10-6

Question 42.
To study the decomposition of hydrogen iodide, a student fills an evacuated 3 litre flask with 0.3 mol of HI gas and allows the reaction to proceed at 500°C. At equilibrium he found the concentration of HI which is equal to 0.05 M. Calculate Kc and Kp.
Solution :
V = 3 L
[HI]initial = \(\frac{0.3 \mathrm{~mol}}{3 \mathrm{~L}}\) 0.1 M
[HI]eq = 0.05 M
2 HI(g) ⇌ H2(g) + I2(g)
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 10

Kc = \(\frac{\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]}{[\mathrm{HI}]^{2}}=\frac{0.025 \times 0.025}{0.05 \times 0.05}\) = 0.25
Kp = Kc(RT)(∆ng)
∆ng = 2 – 2 = 0
Kp = 0.25(RT)0
Kp = 0.25.

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 43.
Oxidation of nitrogen monoxide was studied at 200 with initial pressures of 1 atm NO and 1 atm of O2. At equilibrium partial pressure of oxygen is found to be 0. 5 atm calculate Kp value.
Solution:
2 NO(g) + O2(g) ⇌ 2NO2(g)
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 11
Kp = \(\frac{\left(P_{N O_{2}}\right)^{2}}{\left(P_{N O}\right)^{2}\left(P_{O}\right)}\)

= \(\frac{0.96 \times 0.96}{0.04 \times 0.04 \times 0.52}\)
Kp = 1.017 × 103.

Question 44.
1 mol of CH4, 1 mole of CS2 and 2 mol of H2S are 2 mol of H2 are mixed in a 500 ml flask. The equilibrium constant for the reaction Kc = 4 x 10-2 mol2 lit-2. In which direction will the reaction proceed to reach equilibrium ?
Solution :
CH4(g) + 2 H2S (g) ⇌ CS2(g) + 4H2(g)
Kc = 4 x 10-2 mol2 lit-2
Volume = 500 ml = \(\frac{1}{2}\) L
[CH4]in = \(\frac{1 \mathrm{~mol}}{\frac{1}{2} L}\) = 2 mol L-1;

[CH2]in = \(\frac{1 \mathrm{~mol}}{\frac{1}{2} L}\) = 2 mol L-1
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 12

Question 45.
At particular temperature Kc = 4 × 10-2 for the reaction, H2S ⇌ 2H2(g) + S2(g). Calculate the Kc for each of the following reaction.
i) 2H2S(g) ⇌ 2H2 + S2(g)
ii) 3H2S(g) ⇌ 3H2(g) + 3/2 S2(g)
Solution :
Kc = 4 × 10-2 for the reaction,
H2S ⇌ 2H2(g) + S2(g)
Kc = \(\frac{\left[\mathrm{H}_{2}\right]\left[\mathrm{S}_{2}\right]^{1 / 2}}{\left[\mathrm{H}_{2} \mathrm{~S}\right]}\)

⇒ 4 × 10-2 = \(\frac{\left[\mathrm{H}_{2}\right]\left[\mathrm{S}_{2}\right]^{1 / 2}}{\left[\mathrm{H}_{2} \mathrm{~S}\right]}\)

(i) 2H2S(g) ⇌ 2H2 + S2(g)
For the reaction,
2H2S(g) ⇌ 2H2 + S2(g)
Kc = \(\frac{\left[\mathrm{H}_{2}\right]^{2}\left[\mathrm{~S}_{2}\right]}{\left[\mathrm{H}_{2} \mathrm{~S}\right]^{2}}\)
= (4 × 10-2)2 = 16 × 10-4

(ii) 3H2S(g) ⇌ 3H2(g) + 3/2 S2(g)
For the reaction,
3H2S(g) ⇌ 3H2(g) + 3/2 S2(g)
Kc = \(\frac{\left[\mathrm{H}_{2}\right]^{3}\left[\mathrm{~S}_{2}\right]^{3 / 2}}{\left[\mathrm{H}_{2} \mathrm{~S}\right]^{3}}\)
= (4 × 10-2)3 = 64 × 10-6

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 46.
28 g of Nitrogen and 6 g of hydrogen were mIed In a 1 litre closed container. At equIlibrium 17 g NH3 was produced. Calculate the weight of nitrogen, hydrogen at equilibrium.
Solution:
Given mN2 =28g;
mH2 = 6g;
V = 1 L.
(nN2)initial = \(\frac{28}{28}\) = 1

(nH2)initial = \(\frac{6}{2}\) = 3
N2(g) + 3 H2(g) ⇌ 2NH3(g)
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 13
[NH3] = \(\frac{17}{17}\) = 1 mol
Weight of N2 = (no. of moles of N2) × (molar mass of N2)
= 0.5 × 28 = 14g
Weight of H2 = (no. of moles of H2) × (molar mass of H2)
= 1.5 × 2 = 3g

Question 47.
The equilibrium for the dissociation of XY2 is given as,
2 XY2(g) ⇌ 2 XY(g) + Y2(g)
if the degree of dissociation x is so small compared to one. Show that 2 Kp = PX3 where P is the total pressure and Kp is the dissociation equilibrium constant of XY2.
Answer:
2 XY2(g) ⇌ 2 XY(g) + Y2(g)
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 14

Question 48.
A sealed container was filled with 0.3 mol H2(g), 0.4 mol I2(g) and 0.2 mol HI(g) at 800 K and total pressure 1.00 bar. Calculate the amounts of the components in the mixture at equilibrium given that K = 870 for the reaction, H2(g) + I2(g) ⇌ 2 HI (g).
Solution:
A2(g) + B2(g) ⇌ 2 AB (g)
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 15

Given that, Kp = 1;
\(\frac{4 x^{2}}{(1-x)^{2}}\) = 1
⇒ 4x2 = (1 – x)2 = 1
⇒ 4x2 = 1 + x2 – 2x
3x2 + 2x – 1 = 0
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 16
x = 0.33 – 1(not possible)
∴ [A2]eq = 1 – x = 1 – 0.33 = 0.67
[B2]eq = 1 – x = 1 – 0.33 = 0.67
[AB2]eq = 2x × 0.33 = 0.66

Question 49.
Deduce the Vant Hoff equation.
Answer:
This equation gives that quantitative temperature dependence of equilibrium constant (K). The relation between standard free energy change (∆G°) and equilibrium constant is
∆G° = – RT In K ………………..(1)
We know that, ∆G° = ∆H° – T∆S° …………(2)
Substituting (1) in equation (2)
– RT In K = ∆H° – T∆S°
Rearranging, In K = \(\frac{\Delta \mathrm{H}^{0}}{\mathrm{RT}}+\frac{\Delta \mathrm{S}^{0}}{\mathrm{RT}}\) ……………(3)
Differentiating equation (3) with respect to temperature,
\(\frac{\mathrm{d}(\ln \mathrm{K})}{\mathrm{dT}}=\frac{\Delta \mathrm{H}^{0}}{\mathrm{RT}^{2}}\) ……………….(4)
Equation (4) is known as differential form of van,t Hoff equation.
On integrating the equation (4), between T1 and T2 with their respective equilbrium consatnts K1 and K2.
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 17
Equation 5 is known as integrated form of Van’t Hoff equation.

Question 50.
The equilibrium constant Kp for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g) is 8.19 × 102 at 298 K and 4.6 × 10-1 at 498 K. Calculate ∆H° for the reaction.
Solution :
Kp1 = 8.19 × 102;
T1 = 298 K
Kp1 = 8.19 × 102;
T1 = 298 K
Kp2 = 4.16 × 10-1;
T2 = 498 K
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 18

Question 51.
The partial pressure of carbon dioxide in the reaction CaCO3(s) ⇌ CaO(s) + CO2(g) is 1.017 × 10-3 atm at 500°C. Calculate Kp at 600°C for the reaction. H for the reaction is 181 KJ mol-1 and does not change in the given range of temperature.
Solution:
PCO2 = 1.017 × 10-3 atm
T = 500°C;
Kp = PCO2
∴ Kp1 = 1.017 × 10-3;
T = 500 + 273 = 773 K
Kp2 = ?
T = 600 + 273 = 873 K
∆H° = 181 KJ mol-1
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 19

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

11th Chemistry Guide Physical and Chemical Equilibrium Additional Questions and Answers

I.Choose the best answer:

Question 1.
Which of the following represents physical equilibrium?
(a) PCl5(g) \(\rightleftharpoons\) PCl3(g) + Cl2(g)
(b) H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)
(c) H2O(l) \(\rightleftharpoons\) H2O(g)
(d) N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g)
Answer:
(c) H2O(l) \(\rightleftharpoons\) H2O(g)
Solution:
Physical states arc in equilibrium i.e., liquid – vapour equilibrium.

Question 2.
In which of the following reaction, the value of Kp will be equal to Kc?
a) H2 + I2 ⇌ 2HI
b) PCl5 ⇌ PCl3 + Cl2
c) 2NH3 ⇌ N2 + 3H2
d) 2SO2 +O2 ⇌ 2SO3
Answer:
a) H2 + I2 ⇌ 2HI

Question 3.
Which one of the following does not undergo sublimation?
(a) Iodine
(b) water
(c) Camphor
(d) Ammonium chloride
Answer:
(b) Water

Question 4.
The reaction, 2SO2(g) + O2(g) ⇌ 2SO3(g) is carried out in a 1 dm3 vessel and 2 dm3 vessels separately. The ratio of the reaction velocities will be
a) 1 : 8
b) 1 : 4
c) 4 : 1
d) 8 : 1
Answer:
d) 8 : 1

Question 5.
Which of the following is an example of homogeneous equilibrium?
(a) H2O(1) \(\rightleftharpoons\) H2O(g)
(b) CaCO3(s) \(\rightleftharpoons\) CaO(s) + CO(g)
(c) H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)
(d) 2CO(g) \(\rightleftharpoons\) CO2(g) + C(s)
Answer:
(c) H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)
Solution:
Here all reactants and products are in the same phase i.e., gaseous phase.

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 6.
The equilibrium constant expression for the equilibrium :
2NH2(g) + 2O2(g) ⇌ N2O(g) + 3H2O(g) is
a) Kc = \(\frac{\left[\mathrm{N}_{2} \mathrm{O}\right]\left[\mathrm{H}_{2} \mathrm{O}\right]^{3}}{\left[\mathrm{NH}_{3}\right]\left[\mathrm{O}_{2}\right]}\)

b) Kc = \(\frac{\left[\mathrm{H}_{2} \mathrm{O}\right]^{3}\left[\mathrm{~N}_{2} \mathrm{O}\right]}{\left[\mathrm{NH}_{3}\right]^{2}\left[\mathrm{O}_{2}\right]^{2}}\)

c) Kc = \(\frac{\left[\mathrm{NH}_{3}\right]^{2}\left[\mathrm{O}_{2}\right]^{2}}{\left[\mathrm{~N}_{2} \mathrm{O}\right]\left[\mathrm{H}_{2} \mathrm{O}\right]^{3}}\)

d) Kc = \(\frac{\left[\mathrm{NH}_{3}\right]\left[\mathrm{O}_{2}\right]}{\left[\mathrm{N}_{2} \mathrm{O}\right]\left[\mathrm{H}_{2} \mathrm{O}\right]}\)
Answer:
b) Kc = \(\frac{\left[\mathrm{H}_{2} \mathrm{O}\right]^{3}\left[\mathrm{~N}_{2} \mathrm{O}\right]}{\left[\mathrm{NH}_{3}\right]^{2}\left[\mathrm{O}_{2}\right]^{2}}\)

Question 7.
Statement I: In dissociation of PCI5 to PCI3 and CI2, Kp > KC
Statement II: In dissociation of PCI5, Δng = – ve and so Kp > KC.
(a) Statement I & II are correct and statement II is the correct explanation of statement I.
(b) Statement I & II are correct but statement II is not the correct explanation of statement I.
(c) Statement I is correct but statement II is wrong.
(d) Statement I is wrong but statement II is correct.
Answer:
(c) Statement I is correct but statement II is wrong.
Solution:
Δng = 2 – 1 = 1 = +ve

Question 8.
For the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) at 250°C, the value of Kc is 26, then the value of Kp at the same temperature will be
a) 0.61
b) 0.57
c) 0.83
d) 0.46
Answer:
a) 0.61

Question 9.
In which of the following reaction, Kp is equal to KC ?
(a) N2(g) ± O2(g) \(\rightleftharpoons\) 2NO(g)
(b) 2NH3(g) \(\rightleftharpoons\) N2(g) + 3H2(g)
(c) 2H2(g) + O2(g) \(\rightleftharpoons\) 2H2O(g)
(d) PCI5(g) \(\rightleftharpoons\) PCI3(g) + CI2
Answer:
(a) N2(g) ± O2(g) \(\rightleftharpoons\) 2NO(g)
Solution:
KC (RT)Δng when Δng = 0 then K = K for option (a), Δng = 2 – 2 = 0

Question 10.
In which of the following equilibria, the value of Kp is less than Kc?
a) H2 + I2 ⇌ 2HI
b) N2 + 3 H2 ⇌ 2NH3
c) N2 + O2 ⇌ 2NO
d) CO + H2O ⇌ CO2 + H2
Answer:
b) N2 + 3 H2 ⇌ 2NH3

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 11.
In the reversible reaction A + B ⇌ C + D. The concentration of each C and D at equilibrium was 0.8-mole lit, then the equilibrium constant will be
a) 6.4
b) 0.64
c) 1.6
d) 16.0
Answer:
d) 16.0

Question 12.
Statement I: A pure solid in an equilibrium reaction has the same concentration at a given temperature.
Statement II: The solid does not expand to fill its container and it has the same number of moles of its volume.
(a) Statement I and II are correct and statement II is the correct explanation of the statement of I.
(b) Statement I and II are correct but II is not the correct explanation of!.
(c) Statement I and II are not correct.
(d) Statement I is wrong but 11 is correct.
Answer:
(a) Statement I and II are correct and statement II is the correct explanation of the statement of I.

Question 13.
For the reaction C(s) + CO2(g) ⇌ 2CO (g), the partial pressure of CO2 and CO are 2.0 and 4.0 atm respectively at equilibrium. The Kp for the reaction is
a) 0.5
b) 4.0
c) 8.0
d) 32.0
Answer:
c) 8.0

Question 14.
When the rate of forward reaction becomes equal to backward reaction, this state is termed as
a) chemical equilibrium
b) Reversible state
c) Equilibrium
d) All of these
Answer:
d) All of these

Question 15.
Which of the following does not alter the equilibrium?
(a) catalyst
(b) concentration
(c) temperature
(d) pressure
Answer:
(a) catalyst

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 16.
For the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g), the position of equilibrium can be
shifted to the right by _______.
a) increasing the temperature
b) Doubling the volume
c) Addition of Cl2 at constant volume
d) Addition of equimolar quantities of PCl3 and PCl5
Answer:
b) Doubling the volume

Question 17.
In which of the following reaction, pressure has no effect?
(a) N2 + 3N2 \(\rightleftharpoons\) 2NH3(g)
(b) 2SO2(g) + O2(g) \(\rightleftharpoons\) 2SO3(g)
(c) N2O4(g) \(\rightleftharpoons\) 2NO2(g)
(d) H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)
Answer:
(d) H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)
Solution:
In the reaction H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g) the volumes are equal on both sides and so pressure has no effect.

Question 18.
Rate of reaction curve for equilibrium can be like [rf = forward rate, rb = backward rate]
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 20
Answer:
a) Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 21

Question 19.
For a reaction, N2 + 3H2 ⇌ 2NH3(g), the value of Kc does not depend upon
A) initial concentration of the reactants
B) pressure
C) temperature
D) Catalyst
a) only C
b) A, B, C
c) A, B, C
d) A, B, C, D
Answer:
c) A, B, C

Question 20.
Which one of the following equations is not correct?
(a) ΔG° = – RTInK
(b) ΔG° = ΔH° – TΔS°
(c) – RTInK = ΔH° – TΔS°
(d) In k = \(\frac { ΔH° }{ T }\) – \(\frac { ΔS° }{ R }\)
Answer:
(d) In k = \(\frac { ΔH° }{ T }\) – \(\frac { ΔS° }{ R }\)

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 21.
At 1000 K, the value of Kp for the reaction, A(g) + 2B(g) ⇌ 3C (g) + D(g) is 0.05 atm. The value of Kc in terms of R would be
a) 20000 R
b) 0.02 R
c) 5 × 10-5 R
d) 5 × 10-5 × R-1
Answer:
d) 5 × 10-5 × R-1

Question 22.
In a chemical equilibrium, the rate constant for the backward reaction is 7.5 x 10-4 and the equilibrium constant is 1.5. The rate constant for the forward reaction
a) 2 × 10-3
b) 5 × 10-4
c) 1.12 × 10-3
d) 9.0 × 10-4
Answer:
c) 1.12 × 10-3

Question 23.
A reversible reaction is one which
a) Proceeds in one direction
b) proceeds in both direction
c) proceeds spontaneously
d) All the statements are wrong
Answer:
b) proceeds in both direction

Question 24.
What is the correct expression for the representation of the solubility product constant of Ag2CrO4?
(a) [Ag+]2 [CrO42-]
(b) [2Ag+] [CrO42-]
(c) [Ag+] [CrO42-]
(d) [2Ag+]2 [CrO42-]
Answer:
(a) [Ag+]2 [CrO42-]

Question 25.
For the system A(g) + 2B(g) ⇌C (g), the equilibrium concentrations are (A) 0.06 mole / lit (B) 0.12 mole / lit and (C) 0.216 mole / lit. The Keq for the reaction is
a) 250
b) 416
c) 4 × 10-3
d) 125
Answer:
a) 250

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 26.
What effect does a catalyst have on the equilibrium position of a reaction?
(a) a catalyst favours the formation of products
(b) a catalyst favours the formation of reactants
(c) a catalyst does not change the equilibrium position of a reaction
(d) a catalyst may favour reactants or product formation, depending upon the direction in which the reaction is written.
Answer:
(c) a catalyst does not change the equilibrium position of a reaction

Question 27.
Partial pressures of A, B, C and D on the basis of gaseous system A + 2B ⇌ C + 3D are A = 0.20, B = 0.10, C = 0.30 and D = 0.50 atm. The numerical value of equilibrium constant is _______.
a) 11.25
b) 18.75
c) 5
d) 3.75
Answer:
b) 18.75

Question 28.
The molar concentration of 96 g of O2 contained in a 2 L vessel is ______.
a) 16 mol / L
b) 1.5 mol / L
c) 4 mol / L
d) 24 mol / L
Answer:
b) 1.5 mol / L

Question 29.
According to the law of mass action rate of a chemical reaction is proportional to
a) concentration of reactants
b) molar concentration of reactants
c) concentration of products
d) molar concentration of products
Answer:
b) molar concentration of reactants

Question 30.
The equilibrium constant of the reaction SO2(g) + \(\frac{1}{2}\) O2(g) ⇌ SO3 (g) is 4 × 10-3 atm-1/2. The equilibrium constant of the reaction 2SO3(g) ⇌ 2SO2(g) + O2(g) would be
a) 250 atm
b) 4 × 103 atm
c) 0.25 × 104 atm
d) 6.25 × 104 atm
Answer:
d) 6.25 × 104 atm

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 31.
Which of the following is not a general characteristic of equilibria involving physical processes?
(a) Equilibrium is possible only in a close system at a given temperature
(b) All measurable properties of the system remain constant
(c) All the physical processes stop at equilibrium
(d) The opposing processes occur at the same rate and there is a dynamic but stable condition
Answer:
(c) All the physical processes stop at equilibrium

Question 32.
The active mass of 64 g of HI is a two-liter flask would be
a) 2
b) 1
c) 5
d) 0.25
Answer:
d) 0.25

Question 33.
At a certain temperature, 2HI ⇌ H2 + I2. Only 50 % HI is dissociated at equilibrium. The equilibrium constant is
a) 0.25
b) 1.0
c) 3.0
d) 0.50
Answer:
a) 0.25

Question 34.
Unit of equilibrium constant for the reversible reaction, H2 + I2 ⇌ 2HI is ________.
a) mol-1 litre
b) mol-2 litre
c) mol litre-1
d) None of these
Answer:
d) None of these

Question 35.
Consider the reaction CaCO3(s) \(\rightleftharpoons\) CaO(s) + CO2(g) is a closed container at equilibrium. What would be the effect of the addition of CaCO3 on the equilibrium?
(a) increases
(b) remains unaffected
(c) decreases
(d) unpredictable
Answer:
(b) remains unaffected

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 36.
The following equilibrium are given
N2 +3H2 ⇌ 2NH2 ………K1
N2 + O2 ⇌ 2N0 ………….K2
H2 + 1/2 O2 ⇌ H2O …….. K3
The equilibrium constant of the reaction.
2 NH3 + 5/2 O2 ⇌ 2NO + 3H2O, in terms of K1, K2 and K3 is _________.
a) \(\frac{\mathbf{K}_{1} \mathbf{K}_{2}}{\mathbf{K}_{3}}\)

b) \(\frac{\mathrm{K}_{1} \mathrm{~K}_{3}^{2}}{\mathrm{~K}_{2}}\)

c) \(\frac{\mathrm{K}_{2} \mathrm{~K}_{3}^{2}}{\mathrm{~K}_{1}}\)

d) K1 K2 K3
Answer:
c) \(\frac{\mathrm{K}_{2} \mathrm{~K}_{3}^{2}}{\mathrm{~K}_{1}}\)

Question 37.
For the reaction
2NO2(g) ⇌ 2NO(g) + O2(g) Kc = 1.8 × 10-6 at 184°C and R = 0.083 Jk-1 mol-1. When Kp and are compared at 184C, it
is found that
a) Kp > Kc
b) Kp < Kc
c) Kp – Kc
d) Kp × Kc( depends upon pressure of gases)
Answer:
a) Kp > Kc

Question 38.
The rate of forwarding reaction is two times that of reverse reaction at a given temperature and identical concentration. K equilibrium is
a) 2.5
b) 2.0
c) 0.5
d) 1.5
Answer:
b) 2.0

Question 39.
Hemoglobin (Hb) forms a bond with oxygen and given oxyhemoglobin (HbO2). This process is partially regulated by the concentration of H3O+ and dissolved CO2 in blood as HbO2 +H3O+ + CO2 \(\rightleftharpoons\) H+ – Hb – CO2 + O2 + H2O. If there is the production of lactic acid and CO2 during a muscular exercise, then
(a) more HbO2 is formed
(b) more O2 is released
(c) CO2 is released
(d) both (b) and (c)
Answer:
(b) more O2 is released

Question 40.
For the following thre. reactions 1, 2 and 3 equilibrium constants are given:
1. CO(g) + H2O(g) ⇌ CO(g) + H2(g); K1
2.CH4(g) + H2O(g) ⇌ CO + 3H(g); K2
3. CH4(g) + 2H2O(g) ⇌ CO(g) + 4H2(g); K3
Which of the following relations is correct?
a) K1√K2 = K3
b) K2K3 = K1
c) K3 = K1K2
d) K3 = K23 K12
Answer:
c) K3 = K1K2

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 41.
4 moles of A is mixed with 4 moles of B. At equilibrium for the reaction A + B ⇌ C + D. 2 moles of C and D is formed. The equilibrium constant for the reaction will be
a) \(\frac{1}{4}\)
b) \(\frac{1}{2}\)
c) 1
d) 4
Answer:
c) 1

Question 42.
In a reaction PCl5 ⇌ PCl3 + Cl2 degree of dissociation is 30%. If the Initial moles of PCl5 is one then total moles at equilibrium is
a) 1.3
b) 0.7
c) 1.6
d) 1.0
Answer:
a) 1.3

Question 43.
When 3 moles of A and 1 mole of B are mixed in a 1 lit vessel, the following reaction takes place A(g) + B(g) ⇌ 2C(g), 1.5 moles of ‘C’ are formed. The equilibrium constant for the reaction is
a) 0.12
b) 0.25
c) 0.50
d) 4.0
Answer:
d) 4.0

Question 44.
In which of the following, the reaction proceeds towards completion
a) K = 103
b) K = 10-2
c) K = 10
d) K = 1
Answer:
a) K = 103

Question 45.
Two moles of NH3 when putting into a previously evacuated vessel (1L) partially dissociate into N2 and H2. If at equilibrium one mole of NH3 is present, the equilibrium constant is ________.
a) \(\frac{3}{4}\) mol2 lit-2

b) \(\frac{27}{64}\) mol2 lit-2

c) \(\frac{27}{32}\) mol2 lit-2

d) \(\frac{27}{16}\) mol2 lit-2
Answer:
d) \(\frac{27}{16}\) mol2 lit-2

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 46.
For the reaction N2(g) + O2(g) ⇌ 2NO(g) the value of Kc at 800°C is 0.1. When the equilibrium concentration of both reactions is 0.5 mol, what is the value of Kp at the same temperature?
a) 0.5
b) 0.1
c) 0.01
d) 0.025
Answer:
b) 0.1

Question 47.
28 g of N2(g) and 6g of H2(g) were mixed at equilibrium 17 g NH3 was produced. The weight of N2 and H2 at equilibrium are respectively.
a) 11g, 0g
b) 1g, 3g
c) 14g, 3g
d) 11g, 3g
Answer:
c) 14g, 3g

Question 48.
2SO3 ⇌ 2SO2 + O2 If Kc = 100, α = 1, half of the reaction is completed, the concentration of SO3 and SO2 are equal, the concentration of O2 is _________.
a) 0.001 M
b) \(\frac{1}{2}\) SO2
c) 2 times of SO2
d) Data in complete
Answer:
d) Data incomplete

Question 49.
For reaction HI ⇌ \(\frac{1}{2}\)H2 + \(\frac{1}{2}\)I2 value of Kc is \(\frac{1}{8}\), then value of K is \(\frac{1}{8}\) for H2 + I2 ⇌ 2HI.
a) \(\frac{1}{64}\)
b) 64
c) \(\frac{1}{8}\)
d) 8
Answer:
b) 64

Question 50.
2 moles of PCl5 were heated in a closed vessel of 2-liter capacity. At equilibrium, 40% of PCl5 is dissociated into PCl3 and Cl2. The value of the equilibrium constant is
a) 0.266
b) 0.53
c) 2.66
d) 5.3
Answer:
a) 0.266

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 51.
For a reaction H2 + I2 ⇌ 2HI at 721 k, the value of the equilibrium constant is 50. If 0.5 mol each of H2 and I2 is added to the system the value of the equilibrium constant will be
a) 40
b) 60
c) 50
d) 30
Answer:
c) 50

Question 52.
In a reaction the rate of reaction is proportional to its active mass, this statement is known as
a) Law of mass action
b) Le – Chatelier principle
c) Faradays’s law of electrolysis
d) law of constant proportion
Answer:
b) Le – Chatelier principle

Question 53.
In a chemical equilibrium A + B ⇌ C + D, when one mole each of the two reactants are mixed, 0.6 mole each of the products are formed. The equilibrium constant calculated is
a) 1
b) 0.36
c) 2.25
d) \(\frac{4}{9}\)
Answer:
c) 2.25

Question 54.
Under a given set of experimental conditions, with an increase in the concentration of the reactants, the rate of a chemical reaction.
a) Decreases
b) Increases
c) Remains unaltered
d) First decreases and then increases
Answer:
c) Remains unaltered

Question 55.
A + B ⇌ C + D, If finally the concentrations of A and B are both equal but at equilibrium concentration of D will be twice of that of A then what will be the equilibrium constant of reaction.
a) \(\frac{4}{9}\)
b) \(\frac{9}{4}\)
c) \(\frac{1}{9}\)
d) 4
Answer:
d) 4

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 56.
Theory of ‘active mass’ indicates that the rate of chemical reaction is directly proportional to the ___________.
a) Equilibrium constant
b) Properties of reactants
c) Volume of apparatus
d) Concentration of reactants
Answer:
d) Concentration of reactants

Question 57.
In the reaction, A + B ⇌ 2C, at equilibrium the concentration of A and B is 0.20 mol L-1 each and that of C was found to be 0.60 mol L-1. The equilibrium constant of the reaction is
a) 2.4
b) 18
c) 4.8
d) 9
Answer:
d) 9

Question 58.
The rate at which substances react depends on their _______.
a) Atomic weight
b) Molecular weight
c) Equivalent weight
d) Active mass
Answer:
c) Equivalent weight

Question 59.
If in the reaction N2O4 ⇌ 2NO2, α is that part of N2O4 which dissociates, then the number of moles at equilibrium will be
a) 3
b) 1
c) (1- α)2
d) (1 + α)
Answer:
d) (1 + α)

Question 60.
On decomposition of NH4HS, the following equilibrium is established NH4HS(s) ⇌ NH3(g) + H2S(g). If the total pressure is P atm, then the equilibrium constant Kp is equal to _______.
a) P atm
b) P2 atm2
c) \(\frac{\mathrm{P}^{2}}{4}\) atm2
d) 2P atm
Answer:
c) \(\frac{\mathrm{P}^{2}}{4}\) atm2

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 61.
Some gaseous equilibrium are given below:
i) CO + H2O ⇌ CO2 + H2
ii) 2CO + O2 ⇌ 2CO
iii) 2H2 + O2 ⇌ 2H3O
find out the relation between equilibrium constants.
a) K = K1K2
b) K = (K1K2)2
c) K = (K1K2)–\(\frac{1}{2}\)
d) K = \(\left(\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}\right)^{\frac{1}{2}}\)
Answer:
d) K = \(\left(\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}\right)^{\frac{1}{2}}\)

Question 62.
Kc = 9 for the reaction A + B ⇌ C + D. If A and B are taken In equal amounts, the the amount of ‘C’ in equilibrium is
a) 1
b) 0.25
c) 0.75
d) None of these
Answer:
c) 0.75

Question 63.
In the reaction C(S) + CO2(g) ⇌ 2CO(g) the equilibrium pressure is 12 atm. If 50 % of CO2 reacts, then Kp will be ________.
a) 12 atm
b) 16 atm
c) 20 atm
d) 24 atm
Answer:
b) 16 atm

Question 64.
For the following gas equilibrium, N2O4(g) ⇌ 2NO2(g) Kp is found to be equal to Kc. This is attained when,
a) 0°C
b) 273 k
c) 1 k
d) 12.19 k
Answer:
d) 12.19 k

Question 65.
Consider the following reversible gaseous reactions {At 298 K)
a) N2O4 ⇌ 2NO2
b) 2SO2 + O2 ⇌ 2SO3
c) 2HI ⇌ H2 + I2
d) X + Y ⇌ 4Z
Answer:
a) N2O4 ⇌ 2NO2

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 66.
For the reaction A + 2B ⇌ 2C at equilibrium [C] = 1.4 M, [A]0 = 1 M, [B]0 = 2M, [C]0 = 3 M. The value of Kc is .
a) 0.084
b) 8.4
c) 84
d) 840
Answer:
a) 0.084

Question 67.
For the reaction H2(g) + I2(g) ⇌ 2HI(g). Kc = 66.9 at 350°C and Kc = 50.0 at 448°C. The reaction has
a) ∆H = +ve
b) ∆H = -ve
c) ∆H = zero
d) ∆H = not found the signs.
Answer:
b) ∆H = -ve

Question 68.
In an equilibrium reaction H2(g) + I2(g) ⇌ 2HI(g), ∆H = -3000 calories, which factor favours dissociation of HI
a) Low temperature
b) High pressure
c) High temperature
d) Low pressure
Answer:
c) High temperature

Question 69.
In an equilibrium reaction for which ∆G°= 0, the equilibrium constant K Is
a) 0
b) 1
c) 2
d) 10
Answer:
b) 1

Question 70.
Consider the following reversible reaction at equilibrium, 2H2O(g) ⇌ 2H2(g) + O2(g). Which one of the following ∆H = 241. 7 kJ changes in a conditions will lead to maximum decomposition of H2O(g)?
a) Increasing both temperature and pressure
b) Decreasing temperature and increasing pressure
c) Increasing temperature and decreasing pressure
d) Increasing temperature at constant pressure.
Answer:
c) Increasing temperature and decreasing pressure

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 71.
In which of the following system doubling the volume of the container causes a shift to the right.
a) H2(g) + Cl2(g) ⇌ 2HCl(g)
b) 2CO(g) + O2(g) ⇌ 2CO2(g)
c) N2(g) + 3H2(g) ⇌ 2NH3(g)
d) PCl5(g) ⇌ PCl3(g) + Cl2(g)
Answer:
d) PCl5(g) ⇌ PCl3(g) + Cl2(g)

Question 72.
∆G° (HI, g) ≈ 1.7 kJ. What is the equilibrium constant at 25° C for 2HI(g) ⇌ H2(g) + I2(g)
a) 24.0
b) 3.9
c) 2.0
d) 0.5
Answer:
d) 0.5

Question 73.
In the reaction A2(g) + 4B2(g) ⇌ 2AB4 ∆H < 0 the formation of AB4 is will, be favoured at
a) Low temperature, high pressure
b) High temperature, low pressure
c) Low temperature, low pressure
d) High temperature, high pressure
Answer:
a) Low temperature, high pressure

Question 74.
N2 + 3H2 ⇌ 2NH3. If the temperature of the following equilibrium reaction increases then the reaction.
a) Shifts Right side
b) Shifts left side
C) Remains unchanged
d) No change
Answer:
b) Shifts left side

Question 75.
Consider the equilibrium N2(g) + 3H2(g) ⇌ 2NH3 ∆H = -93.6 kJ. The maximum yield of ammonia is obtained by
a) Decrease of temperature and increase of pressure.
b) Increase of temperature and decreases of pressure.
c) Decrease of both the temperature and pressure
d) Increase of both the temperature and pressure
Answer:
a) Decrease of temperature and increase of pressure.

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 76.
H2(g) + I2(g) ⇌ 2HI(g) ∆H = +q cal, then formation of HI
a) Is favoured by lowering the temperature
b) Is favoured by increasing the pressure
c) Is unaffected by change in pressure
d) Is unaffected by change in temperature
Answer:
c) Is unaffected by change in pressure

Question 77.
The formation of SO3 takes place according to the following reaction, 2SO2 + O2 ⇌ 2SO3; ∆H = 45.2 kcal. The formation of SO3 is favoured by
a) Increasing in temperature
b) Removal of oxygen
c) Increase of volume
d) Increase of pressure
Answer:
d) Increase of pressure

Question 78.
Which of the following equilibrium is not shifted by an increase in the pressure?
a) H2(g) + 3H2(g) ⇌ 2HI(g)
b) N2(g) + 3H2(g) ⇌ 2NH3(g)
c) 2CO(g) + O2(g) ⇌ 2CO2(g)
d) 2C(g) + O2(g) ⇌ 2CO(g)
Answer:
a) H2(g) + 3H2(g) ⇌ 2HI(g)

Question 79.
Consider the heterogeneous equilibrium in a closed container. NH4HS(s) ⇌ NH3(g) + H2S(g) if more NH4HS is added to the equilibrium
a) Partial pressure of NH3 increases
b) Partial pressure of H2S increases
c) Total pressure in the container increases
d) No effect on partial pressure of NH3 and H2S
Answer:
d) No effect on partial pressure of NH3 and H2S

Question 80.
According to Le – Chatelier’s principle adding heat to a solid and liquid in equilibrium with endothermic nature will cause the ____________.
a) Temperature to rise
b) Temperature to fall
c) Amount of solid to decrease
d) Amount of liquid to decrease
Answer:
c) Amount of solid to decrease

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 81.
According to Le-chatelier’s principle, adding heat to a solid to liquid in equilibrium will cause the
a) temperature to increase
b) temperature to decrease
c) amount of liquid to increase
d) amount of solid to increase
Answer:
c) amount of liquid to increase

Question 82.
For the reaction A + B + Q ⇌ C + D, if the temperature is increased, then the concentration of the products will
a) increase
b) decrease
c) Remain same
d) become zero
Answer:
a) increase

Question 83.
H2(g) + I2(g) ⇌ 2HI(g). In this reaction when pressure increases, the reaction direction
a) does not change
b) is forward
c) is backward
d) decreases
Answer:
a) does not change

Question 84.
Which of the following factor is shifted the reaction PCl3 + Cl2 ⇌ PCl5 at the left side?
a) Adding PCl5
b) increase pressure
c) constant temperature
d) catalyst
Answer:
a) Adding PCl5

Question 85.
The rate of reaction of which of the following is not affected by pressure?
a) PCl3 + Cl2 ⇌ PCl5
b) N2 + 3 H2 ⇌ 2NH3
c) N2 + O2 ⇌ 2NO
d) 2 SO2 + O2 ⇌ 2 SO3
Answer:
c) N2 + O2 ⇌ 2NO

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 86.
Which reaction is not affected by change in pressure?
a) H2 + I2 ⇌ 2HI
b) 2 C + O2 ⇌ 2CO
c) N2 + 3 H2 ⇌ 2NH3
d) PCl5 ⇌ PCl3 + Cl2
Answer:
a) H2 + I2 ⇌ 2HI

Question 87.
Which of the following reactions proceeds at low pressure?
a) N2 + 3H2 ⇌ 2NH3
b) H2 + I2 ⇌ 2HI
c) PCl5 ⇌ PCl3 + Cl2
d) N2 + O2 ⇌ 2NO
Answer:
c) PCl5 ⇌ PCl3 + Cl2

Question 88.
In the following reaction PCl5 ⇌ PCl3(g) + Cl2(g) at constant temperature, rate of backward reaction, is increased by
a) inert gas mixed at constant volume
b) Cl2 gas mixed at constant volume
c) inert gas mixed at constant pressure
d) PCl5 mixed in constant volume
Answer:
b) Cl2 gas mixed at constant volume

Question 89.
On cooling of following system at equilibrium CO2(s) ⇌ CO2(g)
a) There is no effect on the equilibrium state
b) more gas is formed
c) more gas solidifies
d) None
Answer:
c) more gas solidifies

Question 90.
On the velocity in a reversible reaction the correct explanation of the effect of catalyst is _______.
a) it provides a new reaction path of low activation energy
b It increases a kinetic energy of the reacting molecules
c) it displaces the equilibrium state on right side
d) it decreases the velocity of backward reaction
Answer:
a) it provides a new reaction path of low activation energy

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 91.
According to Le – Chatelier’s principle, if heat is given to solid – liquid system, then
a) Quantity of solid will reduce
b) Quantity of liquid will reduce
c) increase in temperature
d) decrease in temperature
Answer:
a) Quantity of solid will reduce

Question 92.
In a gives system, water and ice are in equilibrium, if pressure is applied to the system then .
a) more of ice is formed
b) amount of ice and water will remain same
c) more of ice is melted
d) either (a) or (c)
Answer:
c) more of ice is melted

Question 93.
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 22
Answer:
b) Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 23

Question 94.
The value of ∆G° for a reaction in aqueous phase having Kc = 1, would be
a) – RT
b) -1
c) 0
d) + RT
Answer:
c) 0

Question 95.
The equilibrium constants for the reaction, Br2 ⇌ 2 Br at 500 K and 700 K are 1 × 10-10 and 1 × 10-5 respectively. The reaction is
a) endothermic
b) exothermic
c) fast
d) slow
Answer:
a) endothermic

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 96.
In the reaction A2(g) + 4 B2(g) ⇌ 2AB4(g), ∆H < 0, the decomposition of AB4(g) will be favoured at
a) low temperature and high pressure
b) high temperature and low pressure
c) low temperature and low pressure
d) high temperature and high pressure
Answer:
c) low temperature and low pressure

Question 97.
The equilibrium constant for the reaction N2(g) + O2(g) ⇌ 2NO (g) is 4 × 10-4 at 200 K. In the presence of a catalyst the equilibrium is attained 10 times faster. Therefore the equilibrium constant in pressure of the catalyst at 200 K is
a) 4 × 10-3
b) 4 × 10-4
c) 4 × 10-5
d) None
Answer:
b) 4 × 10-4

Question 98.
Change in volume of the system does not alter the number of moles in which of the following equilibrium.
a) N2(g) + O2(g) ⇌ 2 NO (g)
b) PCl5(g) ⇌ PCl3(g) + Cl2(g)
c) N2(g) + 3H2(g) ⇌ 2 NH3(g)
d) SO2Cl2(g) ⇌ SO2(g) + Cl2(g)
Answer:
a) N2(g) + O2(g) ⇌ 2NO2 (g)

Question 99.
In which of the following equilibrium reactions the equilibrium would shift to the right, if total pressure is increased
a) N2 + 3H2 ⇌ 2 NH3
b) H2 + I2 ⇌ 2HI
c) H2 + Cl2 ⇌ 2 HCl
d) N2O4 ⇌ 2NO
Answer:
a) N2 + 3H2 ⇌ 2 NH3

Question 100.
The graph in K (eq) vs \(\frac{1}{T}\) relates for a reaction must be
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 24
a) Exothermic
b) Endothermic
c) ∆H is negligible
d) High spontaneous at ordinary temperature
Answer:
a) Exothermic

Question 101.
Which of the following statements is cor¬rect for a reversible process in state of equilibrium?
a) ∆G° = -2.30 RT log K
b) ∆G° = 2.30 RT log K
c) ∆G = -2.30 RT log K
d) ∆G = 2.30 RT log K
Answer:
a) ∆G° = -2.30 RT log K

Question 102.
Which of the following is a characteristic of a reversible reaction?
a) Number of moles of reactants and products are equal
b) it can be influenced by a catalyst
c) it can never proceed to completion
d) none of the above
Answer:
c) it can never proceed to completion

Question 103.
The equilibrium constant in a reversible reaction at a given temperature
a) depends on the initial concentration of the reactants
b) depends on the concentration of the products at equilibrium
c) does not depend on the initial concentration
d) it is not characteristic of the reaction
Answer:
c) does not depend on the initial concentration

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

II. Very short question and answers(2 Marks):

Question 1.
Define the state of equilibrium.
Answer:
At a particular stage, the rate of the reverse reaction is equal to that of the forward reaction indicating a state of equilibrium.

Question 2.
What are irreversible reactions?
Answer:
Reactions in which the products formed do not react to give back reactants under normal conditions are called irreversible reactions.

Question 3.
Why the chemical equilibrium is referred to as Dynamic Equilibrium?
Answer:
At equilibrium, the forward and the backward reactions are proceeding at the same rate and no macroscopic change is observed. So the chemical equilibrium is often called dynamic equilibrium.

Question 4.
How is a gas – solution equilibrium exist?
Answer:
When a gas dissolves in a liquid under a given pressure, there will be an equilibrium between gas molecuLes in the gaseous state and those dissolved in the liquid. Example – In carbonate beverages, the following equilibrium exists.
CO2(g) \(\rightleftharpoons\)  CO2 (Solution)

Question 5.
Write the Applications of the equilibrium constant?
Answer:
The knowledge of the equilibrium constant helps us in

  • predicting the extent of a reaction, predicting the direction in which the net reaction is taking place, and
  • calculation equilibrium constants and concentration of the reactants and products.

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 6.
What is Le – Chatelier’s principle?
Answer:
It states that “If a system at equilibrium is subjected to a change of concentration, pressure or temperature, then the equilibrium is shifted in such a way to nullify the effect of that disturbance.

Question 7.
For the H2(g) + I2(g) ⇌ 2HI(g) reaction the value of Kc at 717 K is 48. If at a particular instant, the concentration of H2, I2, and HI is 0.2 mol lit-1 and 0.6 mol lit-1 respectively, then calculate Q and predict the direction of the reaction.
Solution:
Q = \(\frac{\left[\mathrm{H}_{1}\right]^{2}}{\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]}=\frac{0.6 \times 0.6}{0.2 \times 0.2}\) = 9
Q < Kc, the reaction will proceed in the forward direction.

Question 8.
Define Degree of dissociation.
Answer:
The fraction of the total number of moles of the reactants are dissociated.
Degree of dissociation ‘x’
‘x’ = \(\frac{\text { no.of moles dissociated }}{\text { Total no. of moles present initially }}\)

Question 9.
Dissociation of PCl5 decreases in presence of increase in Cl2, Why?
Answer:

  1. The dissociation equilibrium of PCl5 in gaseous state is written as PCl5(g) ⇌ PCl3(g) + Cl2(g)
  2. In a chemical equilibrium increasing the concentration of the products results in shifting the equilibrium in favour of the reactants. So in the presence of increased Cl2 reversed reaction only favoured.
  3. Hence dissociation of PCl5 decreases in presence of increase in Cl2.

Question 10.
The equilibrium constant for Kc for A(g) ⇌ B(g) is 2.5 × 10-2 . The rate constant of the forward reaction is 0.05 sec-1. Calculate the rate constant of the reverse reaction.
Answer:
Kc = 2.5 × 10-2;
Kf = 0.05 sec-1
Kc = \(\frac{\mathrm{K}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{r}}}\)
= \(\frac{0.05}{2.5 \times 10^{-2}}\) = 2
Rate constant of the reverse reaction,
Kr = 2 sec-1.

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 11.
The equilibrium constant for the reaction 2SO3(g) ⇌ 2SO2(g) + O2(g) is 0.15 mol dm-3 at 900K. Calculate the equilibrium constant for the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g) at the same temperature. 2SO3(g) ⇌ 2SO2(g) + O2(g).
Answer:
Kc = \(\frac{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}{\left[\mathrm{SO}_{3}\right]^{2}}\) = 0.15
K’ = \(\frac{1}{\mathrm{~K}_{\mathrm{c}}}=\frac{1}{0.15}\)
= 6.6 mole dm-3
K’ = 6.6 mole dm-3

III. Short question and answers (3 Marks):

Question 1.
Explain about the extent of reaction of dissociation of bromine mono chloride at 1000 K.
Answer:
2BrCl(g) \(\rightleftharpoons\) Br2(g) + Cl2(g) KC = 5
10-3 <  KC <  103
both forward and backward reaction make significant progress. Neither forward nor reverse reaction predominates.

Question 2.
What is liquid – vapour equilibrium?
Answer:
The equilibrium of a liquid with its gas at its boiling point at 1 atm pressure is called liquid-vapour equilibrium.
Rate of evaporation = Rate of condensation
H2O(l) ⇌ H2O(g).

Question 3.
What is Solid – vapour equilibrium?
Answer:

  • When solid iodine is placed in a closed transparent vessel, after some time the vessel gets filled up with violet vapour.
  • The intensity of the colour increases and finally becomes constant, i.e., equilibrium is attained.
    I2(s) ⇌ I2(g)

Question 4.
What are Homogeneous and Heterogeneous equilibriums?
Answer:
Homogeneous equilibrium:
In a homogeneous equilibrium, all the reactants and products are in the same phase.
Example:
H2(g) + I2(g) ⇌ 2HI (g)
In the above equilibrium, H2, I2, and HI are in the gaseous state.
Similarly, for the following reactions,
CH3COOCH3(aq) + H2O(aq) ⇌ CH3COOH (aq) + CH3OH (aq) all the reactants and products are in homogenous solution phase.

Heterogeneous equilibrium:
If the reactants and products are present in different phases, it is known as heterogeneous equilibrium.
Example:
H2O(l) ⇌ H2O (g)
CaCO3(s) ⇌ CaO(s) + CO2(g).

Question 5.
Define Q value for a chemical equilibrium reaction.
Answer:
Consider a homogeneous reversible reaction xA + yB \(\rightleftharpoons\) lC + mD For the above reaction under non-equilibrium conditions, reaction quotient Q is defined as the ratio of the product of active masses of reaction products raised to the respective stoichiometric coefficients in the balanced chemical equation to that of the reactants. Under non-equilibrium conditions,
Q = \(\frac { d(Ink) }{ dt }\) ΔH° = \(\frac { { \triangle H }^{ 0 } }{ { RT }^{ 2 } }\)

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 6.
What is the reaction quotient? Write the relation of K & Q.
Answer:
Reaction quotient ‘Q’ is defined as the ratio of the product of active mass of reaction products raised to the respective stoichiometric coefficients in the balanced chemical equation to that of the reactants under non equilibrium.
Reaction quotient Q.
Q = \(\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\)
If Q = Kc, the reaction is in equilibrium
If Q > Kc, the reaction will proceed in the reverse direction i.e., formation of reactants.
If Q < Kc, the reaction will proceed in the forward direction i.e., formation of products.

Question 7.
Explain effect of pressure of formation of HI.
Answer:
Pressure has no effect on the equilibrium when the total number of the moles of the gaseous reactants and the gaseous products are equal. ( ∆ng = 0).
Therefore, pressure has effect only on the equilibrium with ∆ng = 0.
Let us consider the following reactions
H2(g) + I2(g) ⇌ 2HI(g)
N2(g) + O2(g) ⇌ 2NO(g)
Here,
The number of moles on reactants and products are equal. So pressure has no effect on them.

Question 8.
The equilibrium concentrations of NH3, N2 and H2 are 1.8 × 10-2 M, 1.2 × 10-2 M and 3 × 10-2 M respectively. Calculate the equilibrium constant for the formation of NH3 from N2 and H2. [Hint: M= mol lit-1]
Solution:
Given data:
[NH3] = 1.8 × 10-2 M ;
[N2] = 1.2 × 10-2 M ;
[H2] = 3 × 10-2 M ;
Kc = ?
N2(g) + 3H2(g) ⇌ 2NH3(g)
Kc = \(\frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{~N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}}\)
Kc = \(\frac{1.8 \times 10^{-2} \times 1.8 \times 10^{-2}}{1.2 \times 10^{-2} \times 3 \times 10^{-2} \times 3 \times 10^{-2} \times 3 \times 10^{-2}}\)

Kc = 1 × 10-2 l2 mol-2

Question 9.
One mole of H2 and one mole of I2 are allowed to attain equilibrium. If the equilibrium mixture contains 0.4 mole of HI. Calculate the equilibrium constant.
Answer:
Given data:
[H2] = 1 mole;
[I2] = 1 mole
At equilibrium, [HI] = 0.4 mole; Kc = ?
H2(g) + I2 ⇌ 2HI(g)
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 25

Question 10.
Relate the three equilibrium constants.
i) N2 + O2 ⇌ 2NO; K1
ii) 2NO + O2 ⇌ 2NO2; K2
iii) N2 + 2O2 ⇌ 2NO2; K3
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 26

Question 11.
Write the equilibrium constant for the following:
i)H2O2(g) ⇌ H2O + 1/2O2(g)
ii) CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
iii) N2O4(g) ⇌ 2NO2(g)
Answer:
i)H2O2(g) ⇌ H2O + 1/2O2(g)
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 27

ii) CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 28

iii) N2O4(g) ⇌ 2NO2(g)
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 29

Question 12.
Calculate the ∆ng, for the following reactions:
i) H2(g) + I2(g) ⇌ 2HI
ii) 2H2O(g) + 2Cl2(g) ⇌ 4HCl(g) + O2(g)
∆ng = (Total number of moles of gaseous products) – (Total number of moles of gaseous reactants)
Answer:
i) H2(g) + I2(g) ⇌ 2HI
∆ng = 2 – 2 = 0
∆ng = 0

ii) 2H2O(g) + 2Cl2(g) ⇌ 4HCl(g) + O2(g)
∆ng = 5 – 4 = 1
∆ng = 1

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 13.
For the reaction A+B ⇌ 3C at 25°C, a 3 litre volume reaction vessel contains 1, 2, and 4 moles of A, B, and C respectively at equilibrium, calculate the equilibrium constant Kc of the reaction at 25°C.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 30

Question 14.
At temperature T1 the equilibrium constant of reaction is K1 At a higher temperature T2, K2 is 10% of K1. Predict whether the equilibrium is endothermic or exothermic.
Answer:
At temperature, T1, the equilibrium constant is K1
At a high temperature, T2, the equilibrium constant K2 = 10%KX
K2 = \(\frac{10}{100}\) K1
K2 = 0.1 K1
K1 > K2
Hence, the equilibrium is exothermic.

Question 15.
Calculate the Kc when a mixture containing 8.07 moles of H2 and 9.08 moles of I2 are reacted at 448°C until 13.38 moles of HI was formed at the equilibrIum.
Answer:
The equilibrium reaction is,
H2(g) + I2(g) ⇌ 2HI(g)
Kc = \(\frac{(2 x)^{2}}{(a-x)(b-x)}\)
a = 8.07 moles;
b = 9.08 moles
2x = 13.38 moles
x = 6.69 moles
Kc = \(\frac{4(6.69)^{2}}{(8.07-6.69)(9.08-6.69)}\)

= \(\frac{4 \times 44.75}{1.38 \times 2.39}\)
Kc = 54.29

Question 16.
For the equilibrium 2NOCl(g) ⇌ 2NO(g) + Cl2(g) the value of the equilibrium constant Kc is 3.75 × 10-6 at 790°C. Calculate Kp for this equilibrium at the same temperature. [Hint: Kp = Kc(RT)∆ng]
Answer:
Kc = 3.75 × 10-6;
T = 790 + 273 = 1063 K
Kp = Kc × (RT)∆ng
For the reaction,
2NOCl(g) ⇌ 2NO(g) + Cl2(g)
∆ng = 3 – 2 = 1
Kp = 3.75 × 10-6 × (8.314 × 1063)
= 33141.6 × 10-6
= 3.31 × 104 × 10-6
Kp = 3.31 × 10-2

Question 17.
How much PCl5 must be added to one litre volume reaction vessel at 250°C in order to obtain a concentration of 0.1 mole of Cl2, Kc for PCl5 ⇌ PCl3 + Cl2 is 0.0414 mol dm-3 at 250°C.
Answer:
PCl5 ⇌ PCl3 + Cl2
Number of moles of PCl5 at equilibrium = a – 0.1
Number of moles PCl3 at equilibrium = 0.1
Number of moles Cl2 at equilibrium = 0.1
Kc = \(\frac{\left[\mathrm{PCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]}\)
= 0.0414
= \(\frac{0.1 \times 0.1}{a-0.1}\)
0.01 = (a – 0.1) 0.0414
0.01 = 0.0414a – 0.00414
0.0414a = 0.01414
a = 0.3415 mole

Question 18.
At 540 K, the equlllbrium constant Kp for PCl5 dissociation equilibrium at 1.0 atm is 1.77 atm. Calculate equilibrium constant in molar concentration (Kc) at same temperature and pressure.
Answer:
Kp = 1.77 atm;
T = 540 K;
a = 0.082
PCl5 ⇌ PCl3 + Cl2
∆ng = 2 – 1 = 1
Kp = Kc(RT)∆ng
1.77 = Kc × (0.0821 × 450)1
Kc = \(\frac{1.77}{(0.0821 \times 540)}\)
= 4 × 10-2 moles/litres.

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

III. Long question and answers (5 Marks):

Question 1.
Explain the formation of solid-liquid equilibrium with a suitable example.
Answer:
1. Consider melting of ice in a closed container at 273 K. This system reaches a state of physical equilibrium in which the amount of water in the solid phase and liquid phase does not change with time.

2. In this process. the total number of water molecules leaving from and returning to the solid phase at any instant are equal.

3. If some ice cubes and water are placed in a thermos flask (at 273K and I atm) then there will be no change in the mass of ice and water.

4. At equilibrium: Rate of melting of ice Rate of freezing of water
H2O(l) \(\rightleftharpoons\) H2O(g)

Question 2.
Derive Kp & Kc relation for the formation of HI.
Answer:
Synthesis of HI:
Consider that ‘a’ moles and ‘b’ moles are the initial concentration of H2 and I2 respectively, taken in a container of volume V. Let ‘x’ moles of each of H2 and I2 react together to form 2x moles of HI.
H2(g) + I2(g) ⇌ 2HI(g)
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 31
The equilibrium constant Kp can also be calculated as follows:
Total number of moles at equilibrium = a – x + b – x + 2x = a + b.
If the total pressure of the system at equilibrium be P, then
Partial pressure of H2,
pH2 = \(\left(\frac{a-x}{a+b}\right)\)P

Partial pressure of I2,
pI2 = \(\left(\frac{b-x}{a+b}\right)\)P

Partial pressure of HI,
pHI = \(\left(\frac{2x}{a+b}\right)\)P

Kp = \(\frac{\left(\frac{2 x}{a+b}\right)^{2} \mathrm{P}^{2}}{\left(\frac{a-x}{a+b}\right) \mathrm{P}\left(\frac{b-x}{a+b}\right) \mathrm{P}}=\frac{4 x^{2}}{(a-x)(b-x)}\)
In the above equilibrium, Kc and Kp are thus shown to be equal.

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 3.
Derive the equilibrium constant Kp & Kc for the Dissociation of PCl5.
Answer:
Dissociation of PCl5:
Consider that ‘a’ moles is the initial concentration of PCl5 taken in a container of volume V. Let ‘x’ moles of PCl5 be dissociated into x moles each of PCl3 and Cl2.
PCl5(g) ⇌ PCl3(g) + PCl2(g)
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 32
The equilibrium constant Kp can also be calculated as follows:
Total number of moles at equilibrium = a – x + x + x = a + x
If the total pressure of the system at equilibrium be P, then
Partial pressure of PCl5,
pPCl5 = \(\left(\frac{a-x}{a+x}\right)\)P;

Partial pressure of PCl3,
pPCl3 = \(\left(\frac{x}{a+x}\right)\)P;

Partial pressure of Cl2
pCl2 = \(\left(\frac{x}{a+x}\right)\)P

Kp = \(\frac{\left(\frac{x}{a+x}\right) \mathrm{P}\left(\frac{x}{a+x}\right) \mathrm{P}}{\left(\frac{a-x}{a+x}\right) \mathrm{P}}=\frac{x^{2} \mathrm{P}}{(a-x)(a+x)}\)

Question 4.
Explain the effect of Concentration of formation of HI?
Answer:
Effect of concentration:
At equilibrium, the concentration of the reactants and the products does not change. The addition of more reactants or products to the reacting system at equilibrium causes an increase in their respective concentrations.

According to Le Chatelier’s principle, the effect of increase in concentration of a substance is to shift the equilibrium in a direction that consumes the added substance. Suppose at equilibrium if one of the reactants is added, then the equilibrium will shift in the forward direction and if one of the products is added, then the equilibrium will shift in backward direction.
Let us consider the reaction,
H2(g) + I2(g) ⇌ 2HI(g)

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 33

On addition of H2 or I2 to the equilibrium mixture product HI will tend to increase. In order to minimize the stress, the system shifts the reaction in a direction where H2 and I2 are consumed i.e., the formation of HI would relieve the stress. Hence, the equilibrium shifts to the right (forward direction).

Similarly, removal of HI (product) also boosts the forward reaction and increases the concentration of products.
If HI is added to the equilibrium mixture, the concentration HI is increased, then the system proceeds in the reverse direction to nullify the stress.

Question 5.
The equilibrium constant at 298 K for a reaction is 100. A + B ⇌ C + D. If the initial concentration of all the four species were 1 M each what will be equilibrium concentration of D (in mol lit-1)
Solution:
Given data:
[A] = [B] = [C] = 1 M;
Kc = 100; [D] = ?
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 34
10(1 – x) = 1 + x
10 – 10x – 1 – x = 0
9 – 11x = 0
11x = 9
x = 9/11 = 0.818
So, [D] at equilibrium = 1 + 0.818 = 1.818 M

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 6.
What happens when ∆ng = 0, ∆ng = -ve, ∆ng= +ve In a gaseous reaction?
Answer:
(i) When ∆ng = 0, the total number of moles of gaseous products are equal to the total number of moles of gaseous reactants.
Eg:
H2(g) + I2(g) ⇌ 2HI(g)
∆ng = 2 – 2 = 0
Kp = Kc(RT)∆ng
If ∆ng = 0,
Kp = Kc(RT)0
∴ Kp = Kc

(ii) When ∆ng = +ve, the total number of moles of gaseous products are greater than the total number of moles of gaseous reactants.
Eg:
2H2O + 2Cl2(g) ⇌ 4HCl(g) + O2(g)
∆ng = 5 – 4 = 1
Kp = Kc(RT)1
Kp = Kc(RT)
∴ Kp > Kc

(iii) When ∆ng = -ve, the total number of moles of gaseous products are greater than the total number of moles of gaseous reactants.
N2(g) + 3H2(g) ⇌ 2NH3(g)
∆ng = 2 – 4 = -2
Kp = Kc(RT)-2

Kp = \(\frac{K_{c}}{(R T)^{2}}\)

∴ Kp < Kc

Question 7.
What is the relationship between the dissociation and formation equilibrium constant?
Answer:
In a chemical equilibrium reaction, the equilibrium constant for the dissociation equilibrium reaction which is also known as the dissociation constant is found to be the reciprocal value of the equilibrium constant for the formation equilibrium reaction.
Formation of equilibrium reaction,
2SO2(g) + O2 ⇌ 2SO3
Kc = \(\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}\) dm3 /mole.

Dissociation equilibrium reaction of SO3 the reactants became products and vice versa.
2SO3(g) ⇌ 2SO2(g) + O2(g)
Kc‘ = \(\frac{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}{\left[\mathrm{SO}_{3}\right]^{2}}\) mole/dm3
= \(\frac{1}{\mathrm{~K}_{\mathrm{c}}}\)

Kc‘ is considered as the dissociation constant of SO3 gas. Usually, the equilibrium constant of the dissociation equilibrium is the reciprocal of the equilibrium constant of the formation equilibrium reaction.

Question 8.
In the equilibrium H2 + I2 ⇌ 2HI, the number of moles of H2, I2, and HI are 1, 2, 3 moles respectively. Total pressure of the reaction mixture is 60 atm. Calculate the partial pressures of H2, I2, and HI in the mixture.
Solution:
H2 + I2 ⇌ 2HI
Number of moles of H2 at equilibrium = 1 mole
Number of moles of I2 at equilibrium = 2 mole
Number of moles of HI at equilibrium = 3 mole
Total number of moles of equilibrium = 1 + 2 + 3 = 6 moles
Mole fraction of H2 = 1/6;
Mole fraction of I2 = 2/6;
Mole fraction of HI = 3/6
Partial pressure = Mole fraction × Total pressure
PH2 = \(\frac{1}{6}\) × 60 = 10 atm;
PI2 = \(\frac{2}{6}\) × 60 = 20 atm;
PHI = \(\frac{3}{6}\) × 60 = 30 atm.

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 9.
In 1 litre volume reaction vessel, the equilibrium constant K of the reaction PCl5 ⇌ PCl3 + Cl2 is 2 × 10-4 What will be the degree of dissociation assuming only a small extent of 1 mole of PCl5 has dissociated?
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 35
since ‘x’ is very small compared to 1,
then K2 = x2
2 × 10-4 = x2
Degree of dissociation x = \(\sqrt{2 \times 10^{-4}}\)
= 1.414 × 10-2

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 10.
At 35°C, the value of K for the equilibrium reaction N2O4 ⇌ 2NO2 is 0.3174, Calculate the degree of dissociation when P is 0.2382 atm.
Answer:

N2O4 ⇌ 2NO2 N2O4 NO2
The initial number of moles 1 0
Number of moles dissociated x 2x
Number of moles  equilibrium 1 – x 2x

Total number of moles at equilibrium = 1 – x + 2x = 1 + x
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 36

Question 11.
Degree of dissociation of PCl5 at 1 atm and 25°C is 0.2. Calculate the pressure at which PCl5 is half dissociated at 25°C.
Answer:
For PCl5 dissociation equilibrium,
Kp = \(\frac{x^{2} p}{1-x^{2}}\) [ P = total pressure = 1 atm]
x = 0.2
Kp = \(\frac{(0.2)^{2}(1.0)}{1-(0.2)^{2}}\)
= \(\frac{0.04}{1-0.04}\)

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 37

Question 12.
Initially, 0.1 moles each of H2 and I2 gases and 0.02 moles of HI gas are mixed in a reaction vessel of constant volume at 300K. Predict the direction towards which the reaction proceeds [K = 3.5 x 10-2].
Answer:
The formation of equilibrium is
H2(g) + I2(g) ⇌ 2HI(g)
Kc = \(\frac{[\mathrm{HI}]_{\mathrm{e}}^{2}}{\left[\mathrm{H}_{2}\right]_{\mathrm{e}}\left[\mathrm{I}_{2}\right]_{\mathrm{e}}}\)
= 3.5 × 10-2 at 300 K.

Under non equilibrium conditions;
Q = \(\frac{[\mathrm{HI}]^{2}}{\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]}\)

= \(\frac{\left(2 \times 10^{-2}\right)^{2}}{0.1 \times 0.1}\)

= 4 × 10-2
Thus Q > Kc.
Therefore, the reaction will proceed initially before attaining the equilibrium in a direction such that the Q value is reduced. That is, concentrations of H2 and I2 should be increases, which is a reverse reaction. The Traction proceeds in the left side of fomiation equilibrium of HI and HI decomposes initially to H2 and I2 until Q = Kc.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 8 Biomolecules Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 8 Biomolecules

11th Bio Botany Guide Biomolecules Text Book Back Questions and Answers

Part I

Question 1.
The most basic amino acid is
a. Arginine
b. Histidene
c. Glycine
d. Glutamine
Answer:
c. GIycine

Question 2.
An example of feed back inhibition is
a. cyanide action on cytochrome
b. Sulpha drug on folic acid
c. Allosteric inhibition of hexokinase by glucose- 6- phosphate
d. The inhibition of succinic dehydrogenase by malonate
Answer:
c. Allosteric inhibition of hexokinase by glucose-6-phosphate

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 3.
Enzymes the catalyse interconversion of optical, geometrical or positional isomers are
a. Ligases
b. Lyases
c. Hydrolases
d. Isomerases
Answer:
d. Isomerases

Question 4.
Proteins perform many physiological functions, for example, some functions as enzymes one of the following represents an additional function that some proteins discharge
a. Antibiotics
b. Pigment conferring colour to skin
c. Pigments making colours of flowers
d. Hormones
Answer:
d.Hormones

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 5.
Given below is the diagrammatic representation of one of the categories of small molecular weight organic compounds in the living tissues. Identify the category shown &one Blank component ‘X’ in it.
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 1

Category   Compound
I. Cholesterol A. Guanine
II. Amino acid B. IVH2
III. Nucleotide C. Adenine
IV. Nucleoside D. Uracil

Answer:
IV

Question 6.
Distinguish  between Nitrogen base and a base found in inorganic chemistry
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 2
Question 7.
What are the factors affecting the rate of enzyme reactions?
Answer:
Enzymes being bio-molecules sensitive to environmental condition

(i) Temperature

  • Heating increases molecular motion-quicken enzyme reaction
  • Optimum temperature is the temperature that promotes maximum activity

(ii) pH

  • Change in the pH – leads to an alteration of enzyme shape (active site)
  • Extremes of pH’ denatures enzymes
  • Optimum pH’ is that at which the maximum rate of reaction occurs

(iii) Substrate concentration
For a given enzyme concentration, the rate of reaction increase with increasing substrate concentration

(iv) Enzyme concentration
The rate of reaction is directly proportional to enzyme concentration.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 8.
Briefly outline the classification of enzymes?
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 3

Question 9.
Write down the characteristic features of DNA?
Answer:
The characteristic feature of DNA.

  1. If one strand runs in the 5′ – 3′ direction, the other runs in 3′ – 5′ direction and thus are antiparallel (they run in the opposite direction). The 5′ end has the phosphate group and 3’end has the OH group.
  2. The angle at which the two sugars protrude from the base pairs is about 120°, for the narrow-angle and 240° for the wide-angle. The narrow-angle between the sugars generates a minor groove and the large angle on the other edge generates major groove.
  3. Each base is 0.34 nm apart and a complete turn of the helix comprises 3.4 nm or 10 base pairs per turn in the predominant B form of DNA.
  4. DNA helical structure has a diameter of 20 Å and a pitch of about 3 Å. X-ray crystal study of DNA takes a stack of about 10 bp to go completely around the helix (360°).
  5. Thermodynamic stability of the helix and specificity of base pairing includes
    • The hydrogen bonds between the complementary bases of the double helix
    • stacking interaction between bases tend to stack about each other perpendicular to the direction of the helical axis.
  6. Electron cloud interactions (\({ \Pi -{ \Pi } }\)) between the bases in the helical stacks contribute to the stability of the double helix.
  7. The phosphodiester linkages give an inherent polarity to the DNA helix. They form strong covalent bonds, gives strength and stability to the polynucleotide chain.
  8. Plectonemic coiling – the two strands of the DNA are wrapped around each other in a helix, making it impossible to simply move them apart without breaking the entire structure. Whereas in paranemic coiling the two strands simply lie alongside one another, making them easier to pull apart.
  9. Based on the helix and the distance between each turn, the DNA is of three forms – A DNA, B DNA and Z DNA.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 10.
Explain the structure and function of different types of RNA?
Answer:
I. mRNA (messenger RNA)

  • single-stranded
  • carries a copy of instructions to carry out amino acid assembling &protein synthesis
  • unstable
  • 5% of total RNA
  • In Prokaryotes – it is (polycistronic carrying coding sequence for many polypeptides
  • Eukaryotes – (monocistronic) contain information for only one polypeptide

II. tRNA (transfer RNA)

  • single-stranded clover-shaped with 4 arms highly folded -3 D structure
  • translates the code from mRNA and transfers amino acid to ribosomes (to built proteins)
  • unstable (also known as soluble RNA)
  • 15% of total RNA

III. rRNA (ribosomal RNA)

  • single-stranded
  • make up the 2 subunits of ribosomes
  • metabolically stable
  • 80% total RNA
  • A polymer with varied length from 120 – 3000 nucleotides & give ribosomes their shape
  • Genes of rRNA employed for phylogenetic studies

Part II

11th Bio Botany Guide Biomolecules Additional Important Questions and Answers

I Choose the right answer.

Question 1.
Who invented the electron microscope? (2010 AIIMS, 2008 JIPMER)
(a) Janssen
(b) Edison
(c) Knoll and Ruska
(d) Landsteiner
Answer:
(c) Knoll and Ruska

Question 2.
Polysaccharides also called
a. Polymers
b. Glycans
c. Glycosidic compounds
d. Glycones
Answer:
b. Glycans

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 3.
Omnis – cellula – e – cellula was given by ……………. (2007 AIIMS)
(a) Virchow
(b) Hooke
(c) Leeuwenhoek
(d) Robert Brown
Answer:
(a) Virchow

Question 4.
Nitrocellulose is used in making
a. cellophane
b. drapers
c. explosives
d. pain balms
Answer:
c. explosives

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 5.
Genes present in the cytoplasm of eukaryotic cells are found in ……………. (2006 AIIMS)
(a) mitochondria and inherited via egg cytoplasm
(b) lysosomes and peroxisomes
(c) Golgi bodies and smooth endoplasmic reticulum
(d) Plastids inherited via male gametes
Answer:
(a) mitochondria and inherited via egg cytoplasm

Question 6.
Chitin when added with amino acid becomes
a.myeopolysaccharide
b. amylopolysaceharide
c .mucopolysaccharide
d. peptidopolysaccharide
Answer:
c. mucopolysaccharide

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 7.
A quantosome is present in …………… . (JIPMER 2012)
(a) Mitochondria
(b) Chloroplast
(c) Golgi bodies
(d) ER
Answer:
(b) Chloroplast

Question 8.
Among the following one is not a non-polar solvent
a. benzene
b. sulphuric acid
c. ether
d. chloroform
Answer:
b. sulphuric acid

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 9.
One of the given below is a complex found in the cell membrane of animal cell
a. cholesterol
b. myelin
c. proline
d. Ieeithin
Answer:
a. cholesterol

Question 10.
A major site for the synthesis of lipids ……………. (2013 NEET)
(a) Rough ER
(b) smooth ER
(c) Centriole
(d) Lysosome
Answer:
(b) smooth ER

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 11.
Principle information molecules of the cell are known as
a. Nucleus
b. DNA
c. RNA
d. Nucleic acids
Answer:
d. Nucleic acids

Question 12.
(I) Cellulose – A most abundant organic compound
(II) Morphine – Pain relieving alkaloid
(III) Aldose – reducing sugar & Ketose
(IV) Glycogen – mucopolysaccharide
Answer:
(IV) Glycogen – mucopolysaccharide

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 13.
The following is a general formula of
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 4
a. Amino acid
b. Fatty acid
c. Nucleotide
d. Monosaccharide
Answer:
a. Amino acid

Question 14.
Lactose is a disaccharide of
a. Glucose – Glucose
b. Fructose – Fructose
c. Glucose – Galactose
d. Fructose – Galactose
Answer:
c. Glucose – Galactose

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 15.
Number of fatty acids in triglyceride is …………… .
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 16.
Heparin the anti-coagulant is got from
a. D – glucuronic acid
b. Polymer of fructose
c. Mucopolysaccharide from red algae
d. Glucosaminoglycan
Answer:
d. Glycosaminoglycan

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 17.
The pH at which Zwitterion is formed is known as
a. Iso ionic balance
b. Isoelectric potential
c. Isoelectric point
d. Iso ionic point
Answer:
c. Isoelectric point

Question 18.
Aspartate and Glutamate are amino acids of
a. Negatively charged ‘R’ groups
b. Positively charged ‘R’ groups
c. Non-polar aliphatic ‘W groups
d. Non-polar aromatic ‘R’ groups
Answer:
a. Negatively charged ‘R’ groups

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 19.
The test for protein ¡s
a. iodine test
b. Biuret test
c. Benedict’s test
d. Hydrolysis test
Answer:
b. Biuret test

Question 20.
The competitive inhibitor is …………… for succinic dehydrogenase.
(a) malonate
(b) succinate
(c) oxalate
(d) citrate
Answer:
(a) malonate

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 21.
Formation of new chemical bonds using ATP as a source of energy
a. Lyase
b. Hydrolase
c. Telomerase
d. Ligase
Answer:
d. Ligase

Question 22.
Uridylic acid is an
a. Dinucleotide
b. Nucleoside
c. Nucleotide
d. Ribo nucleotide
Answer:
d. Ribonucleotide

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 23.
Phosphate forming linkage with sugar is known as
a. diester linkage
b. peptide linkage
c. phosphodiester linkage
d. Ionic linkage
Answer:
c. phosphodiester linkage

Question 24.
…………… is a catalytic RNA.
(a) mRNA
(b) Ribozyme
(c) Ribonuclease
(d) rRNA
Answer:
(b) Ribozyme

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 25.
A class of lipid that serves as a major component of the cell membrane is
a. triglyceride
b. glycerol
c. phospho lipid
d. lipoprotein
Answer:
c. phospholipid

Question 26.
One molecule of sucrose on hydrolysis give
a. 2 molecules of glucose
b. 1 molecule glucose & 1 molecule fructose
c. 2 molecules of glucose & 1 molecule of fructose
d. 2 molecules of fructose
Answer:
b. 1 molecule glucose & 1 molecule fructose

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 27.
In fibrous proteins polypeptide chai are held together by
a. Vander Waals forces
b. disulphide linkage
c. electrostatic forces
d. hydrogen bonds
Answer:
a. Vander Waals forces

Question 28.
According to Chargaff’s rule, the hydrogen bonding between Adenine and Thymine is …………….
(a) 2
(b) 3
(c) 4
(d) Nil
Answer:
(a) 2

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 29.
Which polymer is stored in liver
a. Amylose
b. Amylo pectin
c. Cellulose
d. Glycogen
Answer:
d. Glycogen

Question 30.
The bond that is not needed for protein formation is
a. Hydrogen bond
b. Peptide bond
c. Ionic bond
d. glucosidic bond
Answer:
d. glucosidic bond

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 31.
A complete turn of the helix comprises …………….
(a) 34 nm
(b) 3.4 nm
(c) 20 nm
(d) 2nm
Answer:
(b) 3.4 nm

Question 32.
The acid is also known as vitamin C
a. Aspartic acid
b. Tartaric acid
c. Ascorbic acid
d. Adipic acid
Answer:
c. Ascorbic acid

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 33.
Which is the left-handed DNA?
(a) B – DNA
(b) A – DNA
(c) Z – DNA
(d) dsDNA
Answer:
(c) Z – DNA

II. Choose the wrong answer.

Question 1.
(a) Hydrolase – Amylase
(b) Oxidoreductase – Dehydrogenase
(c) Transferase – Transaminase
(d) Isomerase – Hexokinase
Answer:
(d) Isomerase – Hexokinase

Question 2.
This has nothing to do with the structure of
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 5
a. Cytosine
b. Pyrimidine
c. Adenine
d. Thyamine
Answer:
c. Adenine

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 3.
Choose the right answer
a) Amylose – linear unbranched polymer of with 20% starch
b) Amylopectin – a polymer with some 1,6 linkages that give it a linear structure
c) Inulin – Polymer of galactose
d) Amino acid – Here a basic structure of carbon linked to a basic amino group
Answer:
d)  Amino acid – Here a basic structure of carbon linked to a basic amino group

III. Match The Following And Find The Correct Answer.

Question 1.
(I) Morphine – A. Hectins
(II) Concanavalin A – B. Drug
(III) Vinblastin – C. Pigment
(IV) Anthocyanin – D. Toxin
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 6
Answer:
(a) B – A – D – C

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 2.
(I) Lactose – A. Penta saccharide
(II) Ramnose – B. Tetra saccharide
(III) Stachyose – C. Disaccharide
(IV) Verbascose – D. Tri saccharide
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 7
Answer:
(b) C-D-B-A

Question 3.
(I) Fred Sanger 1st sequenced
(II) Di sulphide bridges formed between sulphur & amino acids
(III) non-protein enzyme
(IV) homo polysaccharide with amino acid
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 8
Answer:
(b) D-A-B-C

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 4.
(I) Amino acid chain is twisted into coiled configuration call a helix – A. Tertiary Protein
(II) Protein fold into a globular structure called domains – B. Quaternary protein
(III) Linear arrangement of amino acids in a Polypeptide chain – C. Secondary protein
(IV) more than one polypeptide forms a large multiunit multimer – D. Primary Protein
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 9
Answer:
(c) C – A – D – B

Question 5.
(I)) Esters formed between long-chain alcohol another negative. – A. a molecule with two or more & saturated fatty acid function group one +ve and
(II) lipids have both hydrophobic & hydrophilic end known for permeability – B. fluid nature & selective
(III) The amino acids are both acidic & basic exoskeleton of insects – C. waxy substance coating
(IV) Zwitter is also called dipolar – D. amophoteric in nature
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 10
Answer:
(c) C-B-D-A

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

IV. Find Out The True And False Statements From The Following And That Basis Find Out The Right Answer.

Question 1.
(I) Esters are formed between long-chain alcohol & saturated fatty acid.
(II) Lecithin is a food additive & dietary supplement. ‘
(III) Lipids in their structure have two hydrophilic ends
(IV) Solid fats are usually unsaturated
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 11
Answer:
(a) True – True – False – False

Question 2.
(I) In saturated fatty acids, the hydrocarbon chain is single-bonded
(II) Triglycerides are composed of a single molecule of glycerol bound to 2 fatty acids
(III) Palmitic acid is an example of saturated fatty acid.
(IV) Oleic acid is an example of unsaturated fatty acid.
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 12
Answer:
(d) Deoxyribose sugar – Phosphate – Nitrogen base – Nucleotide

V.

Question 1.
Label the diagram parts correctly by choosing the right option.
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 13

A

B C

D

a Deoxyribose sugar Nitrogen base Nucleotide Phosphate
b Deoxyribose sugar Phosphate Nucleotide Nitrogen base
c Deoxyribose sugar Nitrogen base Nucleotide Phosphate
d Deoxyribose sugar Phosphate Nitrogen base Nucleotide

Answer:
(d) Deoxyribose sugar – Phosphate Nitrogen base – Nucleotide

Question 2.
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 14

A

B C

D

a Q arm Centromere Sister Chromatids Q arm
b P arm Centromere Sister Chromatids Q arm
c Sister Chromatids Centromere Q arm P arm
d Q arm Centromere P arm Sister Chromatids

Answer:
(b) P arm – Centromere – Sister chromatids – Q arm

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

VI. Assertion & Reason – Find Out The Correct Answer.

Question 1.
Assertion (A): Adhesion refers to the tendency of water molecules to cling together
Reason (R): Because of hydrogen bonding, water molecules interact with one another continuous column of water is raised in xylem vessels.
(a) Assertion & Reason correct Reason Explaining Assertion
(b) Assertion & Reason correct- Reason not explaining Assertion
(c) Assertion is true but Reason is wrong
(d) Assertion is true but Reason is not explaining Assertion.
Answer:
(a) Assertion & Reason correct Reason Explaining Assertion.

Question 2.
Assertion (A): Glycine is a non-essential amino acid
Reason (R): It must be taken through diet
(a) Assertion & Reason correct Reason Explaining Assertion
(b) Assertion & Reason correct- Reason not explaining Assertion
(c) Assertion is true but Reason is wrong
(d) Assertion is true but Reason is not explaining Assertion.
Answer:
(c) Assertion is true but Reason is wrong.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 3.
Assertion (A): In the presence of enzyme substance molecules can be attached by the reagent.
Reason (R): Active sites of enzymes hold the substance in a suitable position.
(a) Assertion & Reason correct Reason Explaining Assertion
(b) Assertion & Reason correct- Reason not explaining Assertion
(c) Assertion is true but Reason is wrong
(d) Assertion is true but Reason is not explaining Assertion
Answer:
(a) Assertion & Reason correct Reason Explaining Assertion.

Question 4.
Assertion (A): Aminoacids behave like salt rather than simple amines or carboxylic acid.
Reason (R): In aqueous solution, the COOH group of amino acid loses a protein and the NH2 group accepts a proton to form zwitterion (salt).
(a) Assertion & Reason correct Reason Explaining Assertion
(b) Assertion & Reason correct- Reason not explaining Assertion
(c) Assertion is true but Reason is wrong
(d) Assertion is true but Reason is not explaining Assertion
Answer:
(a) Assertion & Reason correct Reason Explaining Assertion.

2 Marks

Question 1.
Define Micronutrients.
Answer:
Nutrients which are required in trace amounts is called micronutrients.
Cobalt, zinc, boron, copper, molybdenum, and manganese – essential for enzyme action.
Eg – Molybdenum is necessary for the fixation of nitrogen by enzyme nitrogenase.

Question 2.
Write down the properties of Water.
Answer:

  • It has Adhesion & cohesion property
  • High latent heat of vaporisation
  • High melting and boiling point
  • Universal solvent
  • Has specific heat capacity.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 3.
Differentiate between Primary and Secondary Metabolites.
Answer:

Primary metabolites

Secondary Metabolites

Required for the basic metabolic processes, like Photosynthesis, Respiration, Protein & lipid metabolism. No direct function in the growth and development of organisms.

Question 4.
Define Polymerisation.
Answer:
A process in which repeating subunits termed monomers is bound into chains of different lengths called polymers.
Eg – Starch – Polynucleotide.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 5.
Distinguish between Glycogen and Cellulose.
Answer:

Glycogen Cellulose
Storage polysaccharide Structural polysaccharide
Animal starch Plant starch
made up of glucose with ( α- 1-6) linked branches Made up of 1000s of glucose units held by β  glucose units held by 1,4 glucosidic linkage
Seen in liver cells skeletal muscle fibre throughout the human body except brain. Occur in cotton. In the form of nitrocellulose used as explosives.

Question 6.
Distinguish between Dinucleotide & Polynucleotide.
Answer:

Dinucleotide

Polynucleotide

2 nucleotide joined to form Dinucleotide
They are linked through 3′- 5′- Phospho – diester linkage by condensation between phosphate group of one with the sugar of other.
Like dinucleotide,when many nucleotides then it leads to the formation of polynucleotides Eg-DNA-RNA

Question 7.
Differentiate between Nucleoside & Nucleotide.
Answer:

Nucleoside Nucleotide
Nitrogen + Sugar → Nucleoside
Eg – Adenine + Ribose → Adenosine
Nucleoside +→ Nucleotide Phosphoric acid (N + S) + P
Adenine + Ribose → Adenosine
Adenosine + Phosphoric  Acid → Adenylic acid

Question 8.
State Chargaff’s Law.
Answer:
Chargaffs Law in 1949
I. A = T & G = C
Between A & T double bond Between G & C triple bond
II. A + G number equal to T + C
III. But A : T, need not be equal to G : C

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 9.
Differentiate Between DNA & RNA.
Answer:

DNA

RNA

Double-stranded Single-stranded
The genetic material in almost all living
organism except for RNA virus
Not genetic material except RNA virus
2 types
Prokaryotic DNA is circular
Eukaryotic DNA is linear
3 types
m RNA
t RNA
r RNA
Controls all aspects of a cell plays important role in protein synthesis

Question 10.
Distinguish between Cation, Zwitterion & Anion.
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 15

Question 11.
How will you test reducing sugar?
Answer:

Substrate

Reagent

Result

1. Glucose is taken in a test tube (Aldehyde) An alkaline solution of copper di sulfate (Benedicts’ reagent) added & heated So brick-red precipitate of copper oxide is formed (i.e.) (Cu + is reduced to Cu+) Aldehyde is oxidised to COOH group.

Question 12.
Draw the structure of a basic amino acid.
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 16

3 Marks

Question 1.
Classify Polysaccharides.
Answer:
Polysaccharides have more than 10 monosaccharides
They can be divided in to

  • Homopolysaccharides
  • Hetero polysaccharides
1. Homopolysaccharides 2. Heteropolysaccharides
a. Starch a. Peptidoglycan
b. Glycogen b. Hyaluronic acid
c. Cellulose c. chondroitin
d. Chitin d. keratan sulphate
e. Inulin e. Agar Agar

Question 2.
Why do we call Glucose and Fructose Isomers -Discuss.
Answer:
Glucose and Fructose have a same molecular formula, C6H1206 but different structural formulas- so they are known as Isomers.
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 17

Question 3.
Explain the formation of Disaccharide Lactose.
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 18

Question 4.
Draw the structure of Fatty acid.
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 19

Question 5.
Distinguish between Waxes & Steroids.
Answer:

Waxes

Steroids

Esters formed between long-chain alcohol and saturated fatty acids
Fur feathers, fruits, leaves, skin, and insect exoskeleton are waterproofed with a coating of wax.
Complex compounds found in cell membrane & animal hormones. Eg – Cholesterol
It reinforces the structure of the life cell membrane in animal cells also in Mycoplasma.

Question 6.
Draw the structural formula of 3 simple amino acids – Glycine, Alanine & Valine.
Answer:
The non-polar aliphatic R group has 6 amino acids Glycine, Alanine, Valine, Leucine, Methionine & Isoleucine.
Structure of Glvcine Alanine Valine
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 20

Question 7.
Distinguish between Macronutrients & Micronutrients.
Answer:
Macronutrients:

  • Nutrients required in larger quantities for plant growth are called Macronutrients.
  • e.g. Potassium and Calcium

Micronutrients:

  • Nutrients required in trace amount for plant growth are called Micronutrients
  • e.g. Zinc and Bora

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 8.
Define Activation energy?
Answer:

  • The minimum quantity of energy the reactants must possess in order to undergo a specified reaction is known as Activation energy.
  • Energy being the biocatalysts reduce the activation energy, thereby help the reaction occurs.
  • The rate of reaction increases if activation energy decreases.

Question 9.
Distinguish between Primary metabolite & Secondary metabolite.
Answer:
Between Primary metabolite & Secondary metabolite:

  • Primary metabolites are those that are required for the basic metabolic processes like photosynthesis, respiration, etc Example: Lipase, a protein.
  • Secondary metabolites do not show any direct function in growth and development of organisms. Example: Ricin, gums.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 10.
Give examples for Secondary metabolites.
Answer:
Secondary metabolites :

  • Pigments – Carotenoids/Anthocyanins
  • Alkaloids – Morphine, codeine
  • Essential oil – Lemongrass oil, Rose oil
  • Toxins – Abrin & ricin
  • Lectins – Concanavalin. A
  • Drugs – Vinbiastine, curcumin
  • Polymeric substances – Rubber, gums, cellulose

Question 11.
Draw the various structures of Protein.
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 21

Question 12.
Draw the structure of Purine – Adenine & Guanine Nucleotides.
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 22

Question 13.
Why do some people have curly hair?
Answer:
Human hair is made of protein. The more the distance between the sulphur atoms, the more the proteins bend; the more the hair curls.

Question 14.
Deoxyribose (C5H10O4) is not a carbohydrate – Discuss.
Answer:
Carbohydrates are hydrates of carbon Deoxy ribose is a carbohydrate, but its formula C5H10O4 -does not apply the general formula of Carbohydrate (C2H20)X formula has Carbohydrates formula.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

5 Marks Questions

Question 1.
How will you identify the presence of glucose in a given food sample?
Answer:
Aldoses and ketoses are reducing sugars. This means that, when heated with an alkaline solution of copper (II) sulphate (a blue solution called Benedict’s solution), the aldehyde or ketone group reduces Cu2+ ions to Cu+ ions forming brick red precipitate of copper (I) oxide. In the process, the aldehyde or ketone group is oxidised to a carboxyl group (-COOH).

This reaction is used as test for reducing sugar and is known as Benedict’s test. The results of Benedict’s test depends on concentration of the sugar. If there is no reducing sugar it remains blue. Sucrose is not a reducing sugar The greater the concentration of reducing sugar, the more is the precipitate formed and greater is the colour change.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 2.
What is protein denaturation.
Answer:
Exposure to heat causes atoms to vibrate violently distrust hydrogen and ionic bonds. There is the loss of 3D structure protein become elongated, disorganised strands. Soaps, detergents, acid, alcohol and some disinfectants disrupt the interchain bond cause the molecule to be non-functional.

Question 3.
Tabulate other sugar compounds
Answer:

Other Polysaccharides

Structure

Functions

Inulin Polymer of fructose It is not metabolised in the human body and is readily filtered through the kidney
Hyaluronic acid Hcteropolymcr of d glucuronic acid and D-N acetyl glucosamine 11 accounts for the toughness and flexibility of cartilage and tendon
Agar Mucopolysaccharide from red algae Used as solidifying agent in culture medium in laboratory
Heparin Glycosamino glycan contains variably sulphated disaccharide unit Drcscnt in liver Used as an anticoagulant
Chondroitin sulphate Sulphated glycosaminoglycan composed of altering sugars (N-acetylglucosamine and glucuronic acid Dietary supplement for treatment of osteoarthritis
Kcratan sulphate Sulphated glycosaminoglycan and is a structural carbohydrate Acts as cushion to absorb mechanical shock

Question 4.
Classify enzyme reactions:
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 23
II. a. Extracellular enzymes
Enzymes secreted outside & work externally Eg digestive enzymes. b. Intracellular Enzymes
Remain within cells & work there Eg Insulin.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 5.
Explain the three types of Co-Factors.
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 24

II. Prosthetic groups:
Eg. Vit B2(Riboflavin) – & Flavin adenine dinucleotide (FAD) Kreb’s cycle
III. Co- Enzymes:
These are co-factors but don’t remain attached to enzymes Eg. NAD, NADP, Co-enzyme A, ATP etc.

Question 6.
Tabulate the uses of enzymes
Answer:

Enzyme Source Application
Bacterial protease Bacillus Biological detergents
Bacterial glucose isomerase Bacillus Fructose- Syrup manufacture
Fungal lactose Kluyveromyces Breaking down of lactose
glucose + glactose
Amylases Aspergillus Removal of starch in woven cloth production

Question 7.
Enumerate the properties of Enzyme.
Answer:
The properties of Enzyme:

  • Enzymes are globular proteins.
  • They act as catalysts and effective even in small quantities.
  • They remain unchanged at the end of the reaction.
  • They are highly specific.
  • They have an active site where the reaction takes place.
  • Enzymes lower the activation energy of the reaction they catalyse.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 8.
Explain lock and key mechanism of enzymes.
Answer:
The substrate Enzyme product
The substrate binds to a specific pocket in an enzyme known as the Active site.
Active site = Lock
Substrate = Key
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 25

  • The substrate binds to the active site of the enzyme
  • As enzyme and substrate form ES- Complex, the substrate is raised in Energy —
  • This was explained by Fischer
  • transition stage break down in to, products, Enzyme remain unchanged.

Question 9.
What are the various types of inhibitors of enzymes.
Answer:
Definition:
Substances present in the cells may react with enzyme and lower the rate of reactions Inhibitors
I. Competitive Inhibitors:
Substances resemble the shape of substrate & compete to occupy active sites
Eg. 1. Enzyme RUBISCO – is competitively inhibited by oxygen/carbon dioxide in the chloroplast
2. Succinic dehydrogenase – Inhibited by malonate.

II. Non-Competitive Inhibitors
Unlike substrates, blocks by binding on active sites, change its shape so enzyme unable to accept substrate.
Enzyme – pyruvate kinase- inhibited by amino acid Alanine.

III. Non-reversible/ Irreversible Inhibitors
They bind to an enzyme tightly & permanently destroying their catalytic nature Enzyme cytochrome oxidase inhibited by cyanide ions Neurotransmitter – blocked by nerve gas sarin.

Question 10.
Distinguish between feedback allosteric inhibition negative feedback (end product) inhibition.
Answer:

Feedback Allosteric Inhibition

Negative feedback Inhibition

Allosteric inhibitors modify enzyme active sites (reversible)
E.g. Glucose  Hexokinase G-6 Phosphate
G.6. Phosphate – Inhibit Hexokinase
When end products accumulate they cause negative feedback or end product inhibition
After products get used up the enzyme reaction is switched on once again.

Question 11.
Tabulate other sugar compounds.
Answer:

Other polysaccharides

Structure

Functions

Inulin Polymer of fructose It is not metabolized in the human body and is readily filtered through the kidney
Hyaluronic acid Heteropolymer of d glucuronic acid and D-N acetyl glucosamine It accounts for the toughness and flexibility of cartilage and tendon
Agar Mucopolysaccharide from red algae Used as a solidifying agent in culture medium in a laboratory
Heparin Glucosamine glycan contains variably sulphated disaccharide unit present in liver Used as anticoagulant
Used as anticoagulant Sulphated glycosaminoglycan composed of altering sugars (N-acetylglucosamine and glucuronic acid) Dietary supplement for treatment of osteoarthritis
Keratan sulphate Sulphated glycosaminoglycan and is a structural carbohydrate Acts as cushion to absorb mechanical shock

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Samacheer Kalvi 11th Commerce Guide Chapter 30 Performance of Contract

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Commerce Guide Pdf Chapter 30 Performance of Contract Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Commerce Solutions Chapter 30 Performance of Contract

11th Commerce Guide Performance of Contract Text Book Back Questions and Answers

I. Choose the Correct Answer

Question 1.
On the valid performance of the contractual obligations by the parties, the contract
a. Is discharged
b. Become enforceable
c. Becomes void
d. Becomes legal
Answer:
a. Is discharged

Question 2.
Which of the following persons can perform the contract?
a. Promisor alone
b. Legal representatives of promisor
c. Agent of the promisor
d. All the above
Answer:
d. All the above

Samacheer Kalvi 11th Commerce Guide Chapter 30 Performance of Contract

Question 3.
A, B, C jointly promised to pay Rs.50,000 to D. Before performnce of the contract, C dies. Here, the contract
a. Becomes void on C’s death
b. Should be performed by A and B along with C’s legal representatives.
c. Should be performed by A and B alone.
d. Should be renewed between A, B and D.
Answer:
b. Should be performed by A and B along with C’s legal representatives.

Question 4.
Which of these parties cannot demand performance of promise?
a. Promisee
b. Any of the Joint Promisees
c. On the death of a Promisee, his Legal Representative.
d. Stranger to the Contract
Answer:
d. Stranger to the Contract

Samacheer Kalvi 11th Commerce Guide Chapter 30 Performance of Contract

Question 5.
A person is said to be a third person if he is not a
a. promisor
b. promise
c. agent
d. Legal Representative
Answer:
c. agent

II Very Short Answer Questions

Question 1.
State the ways of Performing a Contract.
Answer:
There are mainly two ways of performing a contract such as:

  1. Actual Performance
  2. Attempted Performance

Question 2.
Who is a Legal Representative?
Answer:
A person who oversees the legal affairs of another. Examples: The executor or administrator of an estate and a court-appointed guardian of a minor or incompetent person.

Samacheer Kalvi 11th Commerce Guide Chapter 30 Performance of Contract

Question 3.
Who is an agent?
Answer:
According to Para 2 of Section 40, the promisor may employ a competent person such as an agent to perform the promise if the contract is not formed on personal condition.

Question 4.
Define Reciprocal Promise.
Answer:
Promises which form consideration or part of the consideration for each other are called ‘reciprocal promise’.

Samacheer Kalvi 11th Commerce Guide Chapter 30 Performance of Contract

Question 5.
By whom must contracts be performed?
Answer:

  1. Promisor himself
  2. Agent
  3. Representations
  4. Third-person
  5. Joint Promisors

III. Short Answer Questions

Question 1.
What is a Valid tender?
Answer:
The tender which is in proper form, which contains the whole obligation and must not be in installments, which is unconditional is said to be a valid tender.

Question 2.
Who can execute and Perform a Contract?
Answer:
Promisor himself – Under Para 1 to Section 40, it is laid down that where it appears from the nature of the contract, the intention of the parties that any promise contained in it must be performed by the promisor himself or by his legal representatives or by any other competent person employed by him.

Question 3.
Who can demand performance?
Answer:
The following are those who can demand the performance of a valid contract:

  • Promise
  • Legal Representative
  • Third-Party

Question 4.
Write a note on the benefits of Reciprocal Promise.
Answer:
Meaning:
Promises which form the consideration or part of the consideration for each other are called reciprocal promises.
Benefits of Reciprocal Promise:

  • Since both the parties are working simultaneously, there is no chance of losses.
  • Each party are performing their promises independently to make the promise a successful one.
  • There is no loss raised due to lack of time.

Samacheer Kalvi 11th Commerce Guide Chapter 30 Performance of Contract

Question 5.
Who is a Joint Promisors?
Answer:
Joint Promisors:
(a) Devolution of Joint Liabilities (Section 42)
Section 42 of Indian Contract Act lays down that “When two or more persons have made a joint promise, then unless a contrary intention appears in the contract, all such persons, during their joint lives and after the death of the last survivor, representatives of all, jointly must fulfil the promise”

(b) Devolution of Joint Rights (Section 45)
“When a person has made a promise to two or more persons jointly, then unless there is a contract to the contrary, the right to claim performance rests as between him and them, with them during their joint lives and after the death of them with representatives of such deceased person jointly with survivors, and after the death of last survivor, with the representatives of all jointly”.

IV Long Answer Questions

Question 1.
Explain rules relating to place of performance of promise:
Under Section 47, specified time and place for performance: If the promise is to be performed on a certain day, the promisor may undertake to perform it without application of the promisee.

According to Section 47, In such a case the promisor may perform the promise at any time during the usual hours of business on such day and at the place at which the promise ought to be performed.

Question 2.
Elucidate the provision regarding time as a factor in performance.
Answer:
1. Under Section 46, performance within a reasonable time:
According to Section 46, a promisor is to perform his promise within a reasonable time. On the other hand, a reasonable time will depend upon the circumstance of the case, the usage of trade or on the intention of the parties entering into the contract.
Example: A has given an order of supply of books in July which should be performed within 4 to 5 days of the month of July.

2. Under Section 47, specified time and place for performance:
If the promise is to be performed on a certain day, the promisor may undertake to perform it without application of the promisee. According to Section 47, In such a case the promisor may perform the promise at any time during the usual hours of business on such day and at the place at which the promise ought to be performed.

Samacheer Kalvi 11th Commerce Guide Chapter 30 Performance of Contract

Question 3.
How do you think appropriation of payments takes place?
Answer:
Appropriation means application of payments The question of appropriation of payments arises when a debtor owes several debts to the same creditor and makes a payment that is not sufficient to discharge the whole indebtedness.
Appropriation of Payments:
Sometimes, a debtor owes several distinct debts to the same creditor and he makes a payment which is insufficient to satisfy all the debts. In such a case, a question arises as to which particular debt the payment is to be appropriated. Section 59 to 61 of the Act lay down the following rules as to appropriation of payments which provide an answer to this question. Therefore, where the debtor expressly states that the payment is to be applied to the discharge of a particular debt, the payment must be applied accordingly.

Application of payment:
Where debt to be discharged is not indicated [60] If section 60 is attracted, the creditor shall have the discretion to apply such payment for any lawful debt which is due to him from the person making the payment.

Application of payment:
Where neither party appropriates [61] The payment shall be applied in discharge of the debts in order of time whether they are or are not based by the limitation Act 1963 if the debt is of equal standing (i.e. payable on the same date) the payment shall be applied in discharge of each of this debt proportionately.

11th Commerce Guide Performance of Contract Additional Important Questions and Answers

I. Choose the Correct Answer:

Question 1.
Every promise and every set of promises, forming the consideration for each other is an
a. agreement
b. contract
c. offer
d. acceptance
Answer:
a. agreement

Question 2.
When, at the desire of the promisor, the promisee or any other person has done or abstained from doing or, does or abstain from doing or promises to do or to abstain from doing something, such act or abstinence or promise under section 2(d) is called
(a) Reciprocal promise
(b) consideration for the promise
(c) counteroffer
(d) acceptance
Answer:
(b) consideration for the promise

Samacheer Kalvi 11th Commerce Guide Chapter 30 Performance of Contract

Question 3.
Promises which form the consideration or part thereof, for each other under section 2(F) are called
(a) acceptances for different proposals
(b) agreements
(c) reciprocal promises
(d) consideration
Answer:
(c) reciprocal promises

Question 4.
In a valid contract, what comes first
a. enforceability
b. acceptance
c. promise
d. proposal
Answer:
a. enforceability

Samacheer Kalvi 11th Commerce Guide Chapter 30 Performance of Contract

Question 5.
The void agreement signifies ………..
a. agreement illegal in nature
b. agreement not enforceable by law
c. agreement violating legal procedure
d. agreement against public policy
Answer:
b. agreement not enforceable by law

II. Very Short Answer Questions:

Question 1.
What do you mean by the appropriation of payments?
Answer:
Appropriation means the application of payments. It arises when a debtor owes several debts to the same creditor and makes a payment that is not sufficient to discharge the whole indebtedness.

Question 2.
What is the nature of the third person with regard to the performance of the contract?
Answer:
According to Section 41, if a promisee accepts the performance of the promise by a third person he cannot afterward enforce it against the promisor.

Samacheer Kalvi 11th Commerce Guide Chapter 30 Performance of Contract

III. Short Answer Questions:

Question 1.
Explain the ways of performing a contract:
Answer:
There are mainly two ways of performing a contract such as:

  1. Actual Performance: When the party has done what he had undertaken to do, it is called actual performance. In actual performance, the party is to fulfill all his obligations under the contract.
  2. Attempted Performance: When the party offers to perform his obligation, it is not accepted by the promisee. A valid tender of performance is considered to be the performance of a promise.

Question 2.
Explain the term Devolution of Joint Rights:
Answer:
According to section 45 of Indian Contract, Act Devolution means, “When a person has made a promise to two or more persons jointly, then unless there is a contract to the contrary. the right to claim performánce rests as between him and them, with them during their joint lives: and after the death of them with representatives of such deceased person jointly with survivors, and after the death of’ last survivor, with the representatives of all jointly”.

Samacheer Kalvi 11th Commerce Guide Chapter 30 Performance of Contract

IV. Long Answer Questions

Question 1.
State the essentials of a valid tender of performance:
Answer:
A tender, to be valid, must satisfy the following essential requirements

  • It must be unconditional
  • It must be for the whole obligation and must not be in installments if the contract requires in full. It must be by a person who is in a position and willing to perform the promise.
  • It must be at the proper time and place.
  • It must be in proper form.
  • It must be made to a proper person i.e. to the promisee or his authorized agent.
  • In case of the tender of goods, the promisee must be given a reasonable opportunity to inspect the goods.
  • It may be made to one of the several joint promisees.

Question 2.
Explain the types of Reciprocal Promises:
Answer:

  1. Mutual and Indepéndent: Where each party must perform his promise independently without the performance of the other, the promise is mutual and independent, For example, Ramu agrees to pay Somu the amount for the rice supplied on 10th June. Some promises to deliver rice on 18th June.
  2. Mutual and Dependent: Where the performance of the promise by one party depends upon the prior performance of the promise by the other party, the promises are conditional and dependent.
  3. For example, A agrees to construct a building for B. B agrees to supply cement for the construction. Hence A’s promise to perform depends on B’s promise.
  4. Mutual and Concurrent: Where the two promises are said to be performed simultaneously, they are said to be mutual and concurrent.

Samacheer Kalvi 11th Commerce Guide Chapter 30 Performance of Contract

Samacheer Kalvi 12th Maths Guide Chapter 9 Applications of Integration Ex 9.5

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 9 Applications of Integration Ex 9.5 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.5

Question 1.
Evaluate the following:
(i) \(\int_{0}^{π/2}\) \(\frac{dx}{1+5cos^2x}\)
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 9 Applications of Integration Ex 9.5 1

Samacheer Kalvi 12th Maths Guide Chapter 9 Applications of Integration Ex 9.5

(ii) \(\int_{0}^{π/2}\) \(\frac{dx}{5+4sin^2x}\)
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 9 Applications of Integration Ex 9.5 2

Samacheer Kalvi 12th Maths Guide Chapter 9 Applications of Integration Ex 9.5

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 3 Trigonometry Ex 3.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.4

Question 1.
If sin x = \(\frac{15}{17}\) and cos y = \(\frac{12}{13}\), 0 < x < \(\frac{\pi}{2}\), 0 < y <\(\frac{\pi}{2}\), find the value of
(i) sin(x + y)
(ii) cos(x – y)
(iii) tan(x + y)
Answer:
Given sin x = \(\frac{15}{17}\), 0 < x < \(\frac{\pi}{2}\)
we have cos2x + sin2x = 1
∴ cos2x = 1 – sin2x
= 1 – \(\left(\frac{15}{17}\right)^{2}\)
= 1 – \(\frac{225}{289}\)
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 1

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Given that 0 < x < \(\frac{\pi}{2}\), that is x lies in the first quadrant ∴ cos x is positive.
cos x = \(\frac{8}{17}\)
Also given cos y = \(\frac{12}{13}\), 0 < x < \(\frac{\pi}{2}\)
we have cos2y + sin2y = 1
sin2y = 1 – cos2y
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 2
Since 0 < y < \(\frac{\pi}{2}\), y lies in the first quadrant
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 3

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

(i) sin (x + y)
sin (x + y) = sin x cos y + cos x sin y
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 4

(ii) cos (x – y)
cos (x – y) = cos x cos y + sin x sin y
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 5

(iii) tan (x + y)
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 6

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 2.
If sin A = \(\frac{3}{5}\) and cos B = \(\frac{9}{41}\) 0 < A < \(\frac{\pi}{2}\), 0 < B < \(\frac{\pi}{2}\), find the value of
(i) sin (A + B)
(ii) cos (A – B)
Answer:
Given sin A = \(\frac{3}{5}\) 0 < A < \(\frac{\pi}{2}\)
we have sin2A + cos2A = 1
cos2A = 1 – sin2A
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 7
Since 0 < A < \(\frac{\pi}{2}\), A lies in the first quadrant cos A is positive. ∴ cos A = \(\frac{4}{5}\)
Also given cos B = \(\frac{9}{41}\), 0 < B < \(\frac{\pi}{2}\)
We have cos2B + sin2B = 1
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 8

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

(i) sin (A + B)
sin (A + B) = sin A cos B + cos A sin B
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 9

(ii) cos (A – B)
cos (A – B) = cos A cos B + sin A sin B
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 10

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 3.
Find cos(x – y),given that cos x = –\(\frac{4}{5}\) with π < x < \(\frac{3 \pi}{2}\) and sin y = –\(\frac{24}{25}\) with π < y < \(\frac{3 \pi}{2}\)
Answer:
Given cos x = –\(\frac{4}{5}\), π < x < \(\frac{3 \pi}{2}\)
we have cos2 x + sin2 x = 1
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 11
Since π < x < \(\frac{3 \pi}{2}\), x lies in the third quadrant.
Since x is negative in the third quadrant. ∴ sin x = –\(\frac{3}{5}\)
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 12

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 4.
Find sin(x – y) , given that sin x = \(\frac{8}{17}\) with 0< x < \(\frac{\pi}{2}\), and cos y = –\(\frac{24}{25}\), x < y < \(\frac{3 \pi}{2}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 13
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 14
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 15

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 5.
Find the value of
(i) cos 105°
Answer:
cos 105° = cos(90° + 150)
= -sin 15°
= -sin(45°- 30°)
= -[sin 45° cos 30° – cos 45° sin 30°]
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 16

(ii) sin 105°
Answer:
sin 105° = sin (90° + 15°)
= cos 15°
= cos (45° – 30°)
= cos 45° cos 30° + sin 45° sin 30°
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 17

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

(iii) tan \(\frac{7 \pi}{12}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 18
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 19

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 6.
prove that
(i) cos (30° + x) = \(\frac{\sqrt{3} \cos x-\sin x}{2}\)
(ii) cos (π + θ) = – cos θ
(iii) sin (π + θ) = – sin θ
Answer:
(i) cos (30° + x) = \(\frac{\sqrt{3} \cos x-\sin x}{2}\)
cos ( 30° + x) = cos 30°. cos x – sin 30° sin x
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 20

(ii) L.HS = cos(π + θ) = cos(180° + θ)
= cos 180° cos θ – sin 180° sin θ
= (-1) cos θ – (0) sin θ
= – cos θ = RHS

(iii) LHS = sin (π + θ) = sin π cos θ + cos π sin θ
= (0) cos θ + (-1) sin θ
= -sin θ = RHS

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 7.
Find a quadratic equation whose roots are sin 15° and cos 15°.
Answer:
sin 15° = sin (45° – 30°)
= sin 45°. cos 30° – cos 45°. sin 30°
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 21
cos 15° = cos(45° – 30°)
= cos 45° . cos 30° + sin 45° . sin 30°
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 22
The quadratic whose roots cos 15° and sin 15° is
x2 – (cos 15° + sin 15°)x + (cos 15°) (sin 15°) = 0 ——— (3)
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 23
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 24
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 25
Substituting in equation (3) we have
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 26

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 8.
Expand cos(A + B + C). Hence prove that cos A cos B cos C = sin A sin B cos C + sin B sin C cos A + sin C sin A cos B if A + B + C = \(\frac{\pi}{2}\)
Answer:
Taking A + B = X and C = Y
We get cos (X + Y) = cos X cos Y – sin X sin Y
(i.e) cos (A + B + C) = cos (A + B) cos C – sin (A + B) sin C
= (cos A cos B – sin A sin B) cos C – [sin A cos B + cos A sin B] sin C
cos (A + B + C) = cos A cos B cos C – sin A sin B cos C – sin A cos B sin C – cos A sin B sin C If (A + B + C) = π/2 then cos (A + B + C) = 0
⇒ cos A cos B cos C – sin A sin B cos C – sin A cos B sin C – cos A sin B sin C = 0
⇒ cos A cos B cos C = sin A sin B cos C + sin B sin C cos A + sin C sin A cos B

Question 9.
Prove that
(i) sin (45° + θ) – sin (45° – θ) = √2 sin θ
(ii) sin (30° + θ) + cos (60° + θ) = cos θ
Answer:
(i) sin (45° + θ) – sin (45° – θ) = √2 sin θ
sin(45° + θ) – sin(45° – θ) = (sin 45° cos θ + cos 45° sin θ) – (sin 45° cos θ + cos 45° sin θ)
= sin 45° cos θ + cos 45° sin θ – sin 45° cos θ + cos 45° sin θ
= 2 cos 45° sin θ
= 2 × \(\frac{1}{\sqrt{2}}\) sin θ
= \(\frac{2}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\) × sin θ
sin (45° + θ) – sin (45° – θ) = \(\frac{2 \sqrt{2}}{2}\) sin θ
= √2 sin θ

(ii) sin (30° + θ) + cos (60° + θ) = cos θ
sin (30° + θ) + cos (60° + θ)
= sin 30° cos θ + cos 30° sin θ + cos 60° cos θ – sin 60° sin θ
= \(\frac{1}{2}\) cos θ + \(\frac{\sqrt{3}}{2}\) sin θ + \(\frac{1}{2}\) cos θ – \(\frac{\sqrt{3}}{2}\) sin θ
= cos θ

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 10.
If a cos (x + y) = b cos(x – y), show that (a + b) tan x = (a – b) cot y.
Answer:
a cos (x + y) = b cos (x – y)
a [cos x cos y – sin x sin y] = b [cos x cos y + sin x sin y]
a cos x cos y – a sin x sin y = b cos x cos y + b sin x sin y
a cos x cos y – b cos x cos y = a sin x sin y + b sin x sin y
(a – b) cos x cos y = (a + b) sin x sin y
(a – b) \(\frac{\cos y}{\sin y}\) = (a + b) \(\frac{\sin x}{\cos x}\)
(a – b) cot y = (a + b) tan x
(a + b) tan x = (a – b) cot y .

Question 11.
Prove that sin 105° + cos 105° = cos 45°
Answer:
sin 105° + cos 105° = sin (90° + 15°) + cos ( 90° + 15°)
= cos 15° – sin 15°
= cos (45° – 30°) sin (45° – 30°)
= (cos 45° . cos30° + sin 45° sin 30°) – (sin 45° cos 30° – cos 45° sin 30°)
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 27

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 12.
Prove that sin 75° – sin 15° = cos 105° + cos 15°
Answer:
RHS = cos 105° + cos 15° = cos (90° + 15°) + cos (90° – 75°)
= -sin 15° + sin 75°
= sin 75° – sin 15°
= LHS

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 13.
Show that tan 75° + cot 75° = 4
Answer:
tan 75° = tan (45° + 30°)
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 29
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 30

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 14.
Prove that cos(A + B). cos C – cos(B + C) cos A = sin B sin (C – A)
Answer:
LHS = (cos A cos B – sin A sin B) cos C – (cos B cos C – sin B sin C) cos A
= cos A cos B cos C – sin A sin B cos C – cos A cos B cos C + cos A sin B sin C
= cos A sin B sin C – sin A sin B cos C
= sin B [sin C cos A – cos C sin A]
= sin B [sin (C – A)] = RHS

Question 15.
Prove that sin (n + 1)θ . sin(n – 1)θ + cos(n + 1)θ . cos(n – 1)θ = cos 2θ, n ∈ Z
Answer:
Taking (n + 1) θ = A and (n – 1) θ = B
LHS = sin A sin B + cos A cos B
= cos (A – B)
= cos[(n + 1) – (n – 1)]θ
= cos (n + 1 – n + 1)θ = cos 2θ = RHS

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 16.
If x cos θ = y cos\(\left(\theta+\frac{2 \pi}{3}\right)\) = z cos \(\left(\theta+\frac{4 \pi}{3}\right)\) find the value of xy + yz + zx.
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 31
= cos θ + cos (θ + 120°) + cos (θ + 240°)
= cos θ + cos θ cos 120° – sin θ sin 120° + cos θ . cos 240° – sin θ sin 240°
= cos θ + cos θ cos (180° – 60°) – sin θ sin( 180°- 60°) + cos θ cos (180°+ 60°) – sin θ sin (180° + 60°)
= cos θ + cos θ × – cos 60° – sin θ × sin 60° + cos θ × – cos 60°- sin θ (- sin 60°)
= cos θ – cos θ cos 60° – sin θ sin 60° – cos θ cos60° + sin θ sin 60°
= cos θ – 2 cos θ cos 60°
= cos θ – 2 cos θ × \(\frac { 1 }{ 2 }\) = cos θ – cos θ = 0
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 32

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 17.
Prove that
(i) sin(A + B) . sin(A – B) = sin2A – sin2B
Answer:
LHS = sin (A + B) sin (A – B)
= (sin A cos B + cos A sin B) (sin A cos B – cos A sin B)
= sin2A cos2 B – cos2 A sin2 B
= sin2 A (1 – sin2 B) – (1 – sin2 A) sin2 B
= sin2 A – sin2 A sin2 B – sin2 B + sin2 A sin2 B
= sin2 A – sin2 B = RHS

(ii) cos (A + B) . cos (A – B) = cos2A – sin2B = cos2B – sinA
Answer:
cos(A + B) . cos(A – B) = (cos A cos B – sin A sin B) (cos A cos B + sin A sin B)
= (cos A cos B )2 – (sin A sin B )2 = cos2 A cos2 B – sin2 A sin2 B
= cos2A(1 – sin2B) – (1 – cos2A) sin2B
= cos2A – cos2A sin2B – sin2B + cos2A sin2B
cos(A + B) . cos(A – B) = cos2A – sin2B
Also cos(A + B) . cos(A – B) = cos2A cos2B – sin2A sin2B
= (1 – sin2A)cos2B – sin2A(1 – cos2B)
= cos2B – sin2A cos2B – sin2A + sin2A cos2B
cos(A + B) . cos(A – B) = cos2B – sin2A

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

(iii) sin2 (A + B) – sin2(A – B) = sin 2 A sin 2 B
Answer:
sin2 A – sin2 B = sin (A + B) sin (A – B)
LHS = sin2 (A + B) – sin2 (A – B) = sin [(A + B) + (A – B)] [sin (A + B) – (A – B)]
= sin 2A sin 2B = RHS

(iv) cos 8θ . cos 2θ = cos2 5θ – sin2
Answer:
cos8θ . cos2θ = cos(5θ + 3θ) cos(5θ – 3θ)
= (cos 5θ . cos 3θ – sin 5θ sin 3θ) (cos 5θ . cos3θ + sin 5θ sin 3θ)
= (cos 5θ . cos 3θ)2 – (sin 5θ sin 3θ)2
= cos2 5θ cos23θ – sin2 5θ sin2
= cos2 5θ (1 – sin2 3θ) – (1 – cos2 5θ) sin2
= cos2 5θ – cos2 5θ sin2 3θ – sin2 3θ + cos2 5θ sin2
cos 8θ . cos 2θ = cos2 5θ – sin2

Question 18.
Show that cos2A + cos2B – 2 cos A cos B cos (A + B) = sin2(A + B)
Answer:
LHS = cos2 A + cos2 B – 2 cos A cos B [cos A cos B – sin A sin B]
= cos2 A + cos2 B – 2 cos2 A cos2 B + 2 sin A cos A sin B cos B
= (cos2 A – cos2 A cos2 B) + (cos2 B – cos2 A cos2 B) + 2 sin A cos A sin B cos B
= cos2 A (1 – cos2 B) + cos2 B (1 – cos2 A) + 2 sin A cos A sin B cos B
= cos2 A sin2 B + cos2 B sin2 A + 2 sin A cos B sin B cos A
= (sin A cos B + cos A sin B)2
= sin2 (A + B) = RHS

Question 19.
If cos (α – β) + cos(β – γ) + cos(γ – α) = [lαtex]-\frac{3}{2}[/lαtex],then prove thαt
cos α + cos β + cos γ = sin α + sin β + sin γ = 0
Answer:
Given cos( α – β) + cos (β – γ) + cos (γ – α) = [lαtex]-\frac{3}{2}[/lαtex]
cos α cos β + sin α sin β + cos β cos γ + sin β sin γ + cos γ cos α + sin γ sin α = [lαtex]-\frac{3}{2}[/lαtex]
2 [cos α cos β + sin α sin β + cos β cos γ + sin β sin γ + cos γ cos α + sin γ sin α] = – 3
(2 cos α cos β + 2 cos β cos γ + 2 cos γ cos α) + (2 sin α sin β + 2sin β sin γ + 2 sin γ sin α) + 3 = 0
(2 cos α cos β + 2 cos β cos γ + 2cos γ cos α) + (2 sin α sin β + 2 sin β sin γ + 2 sin γ sin α)+
(cos2 α + sin2 α) + (cos2β + sin2 β) + (cos2 γ + sin2 γ) = 0
(cos2 α + cos2β + cos2γ + 2 cos α cos β + 2 cos β cos γ + 2 cos γ cos α) + (sin2α + sin2β + sin2γ + 2 sin α sin β + 2 sin β sin γ + 2 sin γ sin α) = 0
(cos α + cos + cos γ)2 + (sin α + sin β + sin )2 = 0
cos α + cos β + cos γ = 0 and sin α + sin β + sin γ = 0
cos α + cos β + cos γ = sin α + sin + sin γ = 0

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 20.
Show that
(i) Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 33
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 35

(ii) Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 34
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 36

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 21.
Prove that cot (A + B) = \(\frac{\cot A \cot B-1}{\cot A+\cot B}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 37
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 38

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 22.
If tan x = \(\frac{n}{n+1}\) and tan y = \(\frac{1}{2 n+1}\) find tan (x + y).
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 39

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 23.
Prove that tan \(\left(\frac{\pi}{4}+\theta\right)\) tan \(\left(\frac{3 \pi}{4}+\theta\right)\) = -1
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 40

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 24.
Find the value of tan (α + β), given that cot α = \(\frac{1}{2}\), α ∈ \(\left(\pi, \frac{3 \pi}{2}\right)\) and sec β = –\(\frac{5}{3}\) β ∈ \(\left(\frac{\pi}{2}, \pi\right)\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 42

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 25.
If θ + Φ = α and tan θ = k tan Φ, then prove that sin (θ – Φ ) = \(\frac{k-1}{k+1}\) sin α
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 43
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 44

Samacheer Kalvi 8th Social Science Guide Civics Chapter 5 Road Safety Rules and Regulations

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Social Science Guide Pdf Civics Chapter 5 Road Safety Rules and Regulations Text Book Back Questions and Answers, Important Questions, Notes.

Tamilnadu Samacheer Kalvi 8th Social Science Solutions Civics Chapter 5 Road Safety Rules and Regulations

Samacheer Kalvi 8th Social Science Road Safety Rules and Regulations Text Book Back Questions and Answers

I. Choose the correct answer:

1. At a red light
a) You can keep going if the path looks clear.
b) You must stop and wait for it turn green.
c) Quickly you can reply your friend’s text message.
d) You can attend call.
Answer:
b) You must stop and wait for it turn green

2. Pedestrians can cross the road only …………….
a) at anywhere
b) near the signals
c) at Zebra crossing
d) none
Answer:
c) at Zebra crossing

3. Road Safety Week is celebrated in the month of ……………. every year
a) December
b) January
c) March
d) May
Answer:
b) January

4. For emergency, call ……………. for ambulance service.
a) 108
b) 100
c) 106
d) 101
Arts: a) 108

5. What are the causes for the road accidents?
a) Over Speeding
b) Drunken Driving
c) Distraction to Drivers d) All of these Answer:
d) AH of these

6. The first category of traffic signs is …………….
a) Mandatory Signs
b) Cautionary Signs
c) Informatory Signs
d) None of these
Answer:
a) Mandatory Signs

7. ‘Setu Bharatam’, a program was launched in …………….
a) 2014
b) 2015
c) 2016
d) 2017
Answer:
c) 2016

8. Expand ABS:
a) Anti Brake start
b) Annual Base System
c) Anti – locking Brake System
d) None of these
Answer:
c) Anti-locking Brake System

9. Overtaking when approaching a bend is
a) permissible
b) not permissible
c) permissible with care
d) our wish
Answer:
b) not permissible

10. When the ambulance is approaching
a) allow passage if there are no vehicles from the front side
b) no preference need be given
c) allow free passage by drawing to the side of the road
d) drive behind the ambulance with great speed
Answer:
c) allow free passage by drawing to the side of the road

II. Fill in the Blanks.

1. Always keep ……………. while driving.
Answer:
to the left

2. Mandatory signs are exhibited in ……………. shape.
Answer:
circular

3. ……………. controls the speed of the vehicle.
Answer:
Brake

4. Higher the speed; ……………. the risk.
Answer:
greater

5. Use of ……………. in four-wheelers and ……………. for two-wheelers has been brought under the law.
Answer:
seat belt, helmet

III. Match the Following.

1. Pollution under control certificate a) Zebra crossing
2. One-time tax for a new car b) Comic book on road safety
3. Pedestrian c) 6 months
4. Brasilia Declaration d) 15 years
5. Swachha safer e) Global conference

Answer:

1. Pollution under control certificate c) 6 months
2. One-time tax for a new car d) 15 years
3. Pedestrian a) Zebra crossing
4. Brasilia Declaration e) Global conference
5. Swachha safer b) Comic book on road safety

IV. State true or false.

1. The problem of accidents lies with roads only.
Answer:
False

2. Check mirrors before changing lanes.
Ans.’True

3. Flashing yellow signal indicates to slow down and proceed with caution.
Answer:
True

4. On a two-wheeler only one pillion rider is allowed.
Answer:
True

5. The roads are one of the worst inventions of man.
Answer:
False

V. Consider the following statements and tick the appropriate answer.

1. Which of the following statement is/are not correct?
i) Maintain the right distance to the vehicle in the front.
ii) Maintain speed limit, never exceed the speed limit.
iii) Wearing seat belt is not necessary while driving.
iv) Don’t slow down on bend and turn in the road.
a) i, iii only
b) ii, iv only
c) i, ii only
d) iii, iv only
Answer:
d) iii, iv only

2. Assertion: Drunken driving causes accidents.
Reason: It hampers vision due to dizziness.
a) A is correct and R is the correct explanation of A.
b) A is correct and R is not the correct explanation of A.
c) A is wrong and R is correct.
d) Both are wrong.
Answer:
a) A is correct and R is the correct explanation of A.

What is the aim of observing Road Safety Week?
Answer:
The event aimed at raising public awareness about traffic rules and ultimately reducing casualties due to road accidents. It is organised by the National Safety Council of India (NSC).

7. Write any four traffic rules.
Answer:
i) On a two-way road, the driver must drive on the left side of the road.
ii) It is mandatory for the driver to slow down at all inter junctions and pedestrian crossing.
iii) Drivers should not use the horns in prohibited areas like hospital zones, school zones, etc.,
iv) It is our responsibility to give way to emergency vehicles such as Army convoy fire engine and ambulance.

8. How does alcohol affect driving?
Answer:
Because it reduces concentration, it hampers vision due to dizziness. This may cause an accident.

VII. Answer the following in detail.

1. Explain the factors that contributed to road accidents.
Answer:

  1. Drivers – Over speeding, rash driving, violation of rules, failure, to understand signs, fatigue and consumption of alcohol.
  2. Pedestrian – Carelessness, illiteracy, crossing at wrong places, moving on roads and jaywalkers.
  3. Passengers – Projecting their body outside tie vehicle by talking to drivers, travelling on footboards, catching a running bus etc.
  4. Vehicles – Failure of brakes or steering, tyre burst, insufficient headlights, overloading and projecting loads.
  5. Road Condition – Damaged road, potholes, eroded road merging of rural road with highways, diversion and illegal speed breakers.
  6. Weather Conditions – Fog, snow, heavy rainfall, wind, storms and hail storms.

2. Describe the steps taken by the Government of India to prevent road accidents.
Answer:

  • Always keep to the left – while driving, keep to the left, and allow vehicles to pass from the opposite direction.
  • Slow down on bends and turn. A very important thing to keep in mind is to become cautious and slow down on the bends.
  • Use helmets – Make it a habit of strapping the helmet before mounting the bikes.
  • Follow the road signs – Road signs are mostly pictorial so it is not hard to comprehend.
  • For pedestrians – Cross only at Zebra crossing.

3. What are the preventive measures for road accidents?
Answer:
Safety Measures:

  1. Always keep to the left-While driving, keep to the left, and allow vehicles to pass from the opposite direction.
  2. Use helmets – Make it a habit of strapping the helmet before mounting the bikes.
  3. Never exceed the speed limit – The speed limit is related to the traffic condition of the area. To maintain the speed limit.
  4. Maintain the right distance – Collisions occur because we do not maintain adequate distance from the vehicle in the front. To maintain a safe distance from the bigger vehicles.
  5. Park the vehicles only along the designated parking bays/zones, not on the sides of the highways.
  6. Follow the road signs – Road signs are mostly pictorial so it is not hard to comprehend.
  7. Some other safety tips are – Never ever drive when drunk, never use cell phone while driving, avoid listening to too loud music, check your mirrors before changing lanes.
  8. Never cross on a red and yellow light. Take a sidewalk and avoid walking on the roads.

VIII HOTS

1. What is the requirement for a two-wheeler rider during the night?
Answer:
Few tips for Riding (Motorcycle) two-wheeler during the night:

  1. Upgrade your bike’s lightings.
  2. Increase your visibility to others on the road.
  3. Be ensured that your vision is clear.
  4. Use the lights of other vehicles equally.
  5. Ride slowly.
  6. Keep controlling your speed.
  7. Take adequate rest.
  8. Proper planning.

2. Tabulate tips for avoiding driver fatigue.

  • Many people think fatigue is only a problem for long-distance drivers, however, it is just as relevant for short-distance drives.
  • People generally don’t become fatigued from driving. Usually, they are already tired when they get behind the wheel from long hours, lack of sleep apnoea, or physically demanding roles.
  • Take a 15 minutes power nap if you feel yourself becoming drowsy.

IX. Project and Activity.

1.  Tabulate a few basic road safety rules for children.
Answer:

  1. Know your signals
  2. Stop, look and cross
  3. Pay attention – Listen
  4. Don’t Run on Roads
  5. Always use sidewalks
  6. Crossroads and Pedestrian crossing
  7. Never crossroad at bends
  8. Staying safe on a bicycle

2. Demonstrate road safety techniques.
Answer:
Road Safety Techniques

  • A day keeps the accident away
  • A constant glance at the mirrors.
  • Properly hold the steering wheel.
  • Regularly check the speed control.
  • Use Signals
  • Keep your feed on the breakers
  • Wrapping up

3. Prepare a Road Safety Awareness pocket guide and circulate in school.
Answer:
Slogan:

  • Start Early
  • Drive Slowly
  • Reach Safely.

Samacheer Kalvi 8th Social Science Road Safety Rules and Regulations Additional Important Questions and Answers

I. Choose the Correct answer.

1. For emergency calls ………….. for traffic accidents.
a) 103
b) 104
c) 102
d) 100
Answer:
a) 103

2. Totally ………….. cautionary traffic signs in our Roadways and Transport Department
a) 30
b) 40
c) 50
d) 20
Answer:
b) 40

3. In which year, the motor vehicle act came into force?
a) 1988
b) 1989
c) 1990
d) 1987
Answer:
b) 1989

4. The police will file a criminal case under this section
a) 300A
b) 302 A
c) 303A
d) 303 A
Answer:
d) 304 A

5. A multi – Pronged strategy has been adopted based on …………..
a) E
b) Three E’S
c) Four E’S
d) Two E’S
Answer:
c) Four E’S

6. The government had made of mandatory for a person who wants to obtain ………….. to be well versed in a traffic signal.
a) Bank Passbook
b) Aadhar number
c) Voters ID
d) Driving License
Answer:
c) Driving License

7. ………….. colour number plate used in the vehicle for the president of India and
Governor of states
a) Red
b) blue c) White
d) yellow
Answer:
a) Red

8. Quick response ambulances are parked in every ………….. on the National Highways.
a) 40 km
b) 50 km
c) 10 Km
d) 20 km
Answer:
b) 50 km

9. Decade of action for road safety from the year 2011 to ………….. proclaimed by the Un General Assembly.
a) 2015
b) 2020
c) 2000
d) 2010
Answer:
b) 2020

10. In which year Indian signed in
a) 2014
b) 2016
c) 2017
d) 2015
Answer:
d) 2015

II. Fill in the Blanks.

1. ……………………… is Random break test illegal and dangerous habit
Answer:
Tailgating

2. ……………………… is done to detect the consumption of alcohol.
Answer:
Random breath Test

3. The major distraction now a day is talking on ……………………… while driving.
Answer:
mobile phone

4. We are not allowed to overtake and to stay with in the lane mentioned by ………………………
Answer:
Solid white line

5. ……………………… ministry is responsible to prevent Road accidents.
Answer:
Road Transport and Highways

6. ……………………… made mandatory for two wheelers to make them more conspicuous..
Answer:
Automatic headline.

7. A set of comic books on road safety released for children are ………………………, ………………………
Answer:
Swacha, suvarshit

8. A public charitable trust that in workers to improve road safety is ………………………
Answer:
Save life foundation

9. ……………………… suggested 10 points a few Road safety.
Answer:
IPRC

10. ……………………… is for Commercial Vehicle.
Answer:
Yellow

11. The main distribution of road accident is ………………………
Answer:
adjusting mirror

12. ……………………… are mostly pictorial. so it’s easy to comprehend.
Answer:
Essential documents

13. Never ever drive when ………………………
Answer:
drunk

14. We never cross the roads when the ……………………… and lights are flashed.
Answer:
Red, yellow

15. ……………………… lines are used in areas where visibility is low.
Answer:
Single solid yellow

III. Match the Following.

1. Mandatory signs a) Triangle Shape
2. Cautionary signs b) Signalling device
3. Informatory signs c) Circular in shape
4. Traffic digits d) Complete Stop
5. Flashing red signal e) Rectangular shape

Answer:

1. Mandatory signs c) Circular in shape
2. Cautionary signs a) Triangle Shape
3. Informatory signs e) Rectangular shape
4. Traffic digits b) Signalling device
5. Flashing red signal d) Complete Stop

IV. State true or false.

1. Violation of any mandatory traffic sign is an offence punishable by law by the transport Department.
Answer:
True

2. Informatory signs never provide information to the drivers in board
Answer:
False

3. On a two-wheeler only one pillion ride is allowed.
Answer:
True

4. Never exceed the speed limit
Answer:
True

5. Pedastrinas cross only at a zebra crossing
Answer:
True

V. Consider the following statements and tick the appropriate answer.

1. i) Road accidents are desired events.
ii) Wearing a helmet while drivers is not mandatory.
iii) Road safety primarily means the protection and security of all road users.
iv) The ministry of road transport and highways has taken a number of steps to prevent road accidents.
a) i and ii correct
b) i and iii correct
c) ii and iv correct
d) iii and iv correct
Answer:
d) iii and iv correct

VI. Answer the following in one or two sentences.

1. Write a Short note on -108 Emergency Response Service.
Answer:

  • It is a free emergency service providing integrated medical (ambulance), police, and fire services.
  • If you find any victims on road don’t panic.
  • Call 108 for help and 103 for traffic accidents.

2. What is the importance of Indian Penal code 304A?
Answer:
The police will file a criminal case under this section, which deals with offences relating to death due to rashness and negligence of the driver.

3. Different colour number plates used in India – explain.
Answer:
In India, four different colour number plates are used. They are

  • Red Number Plate – is used in the vehicle for the President of India and Governor of States.
  • Blue Number Plate – is given to a vehicle that is used by foreign delegates/ ambassadors.
  • White Number Plate – it means that the car belongs to a common citizen.
  • Yellow Number Plate – is for commercial vehicles.

4. What are the essential Documents Possessed by a driver?
Answer:
Driving License, Registration Certificate of the Vehicle, Taxation Certificate, Insurance Certificate, Fitness Certificate, and Permit.

5. Write Short note on Setu Bharatam.
Answer:

  • A program was launched in 2016 for building bridges for safe and seamless travel on the National Highways.
  • It aims to make all National Highways free of railway level crossing by 2019.

6. Explain – Save LIFE Foundation.
Answer:
Save LIFE Foundation is an independent, non-profit, non-governmental, and public charitable trust that is working to improve road safety and emergency medical care across India.

VII. Answer the following in Detail.

1. What are the Significances of the lines marked on Indian roads?
Answer:

  • Broken white line – basic marking on roads, you may change lanes, and are allowed to overtake a vehicle or take U-turn of it is safe to do so.
  • Solid white line – seen on areas of strategic importance. This implies that you are not allowed to overtake and to stay within the lane.
  • Single solid Yellow lines – used in areas where visibility is low. It implies that you can overtake And should drive on your side.
  • Double solid yellow lines – used on dangerous roads or for two-way traffic. It strictly prohibits anybody from crossing over into the lane. You can overtake inside your own lane.
  • Stop line – This is marked before the pedestrian crossing and sets the deadline where the car should stop before the traffic signal.
  • Solid and Broken lines – if you are driving on the side with the broken line you are allowed to overtake and if you are driving on the side of the solid line you are not.

2. What are Traffic Signals?
Answer:
Traffic signals are there to regulate traffic, warn about hazards, and to guide the road users. Importance of Traffic Signals:

  • Red: This signal indicates to stop behind the stop line.
  • Yellow: This signal indicates to stop. Do not pass through or start until green shows.
  • Green: This signal indicates you may go if the way is clear.
  • Steady Green Arrow Signal: This signal may be provided in addition to the full green signal. This indicates to proceed with caution in the direction the arrow points.
  • Flashing Red Signal: It means to come to a complete stop. Proceed only when the way is clear.
  • Flashing Yellow Signal: It indicates to slow down and proceed with caution.

Samacheer Kalvi 8th Social Science Guide History Chapter 6 Development of Industries in India

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Social Science Guide Pdf History Chapter 6 Development of Industries in India Text Book Back Questions and Answers, Important Questions, Notes.

Tamilnadu Samacheer Kalvi 8th Social Science Solutions History Chapter 6 Development of Industries in India

Samacheer Kalvi 8th Social Science Development of Industries in India Text Book Back Questions and Answers

I. Choose the correct answer.
1. Which of the following activities of the people will not come under handicraft?
a) Carving statues out of stone
b) Making bangles with glass
c) Weaving silk sarees
d) Smelting of iron
Answer:
d) Smelting of iron

2. The oldest industry in India was industry.
a) Textile
b) Steel
c) Electrical
d) Fertilizers
Answer:
a) Textile

3. The woollen and leather factories became prominent in
a) Bombay
b) Ahmadabad
c) Kanpur
d) Dacca
Answer:
c) Kanpur

4. What was the aim of first Three Five year Plans of India?
a) To control population growth
b) To reduce illiteracy rate
c) To built a strong industrial base
d) To empower the women
Answer:
c) To built a strong industrial base

5. What was not the reason for the decline of Indian Industries?
a) Loss of royal patronage
b) Competition of machine made goods
c) Industrial policy of India
d) Trading policy of British
Answer:
c) Industrial policy of India

II. Fill in the blanks.
1. ………………. was the integral part in the life of the people.
Answer:
crafts

2. Industrial revolution took place in ………………. .
Answer:
England

3. The Assam Tea Company was founded in ………………. .
Answer:
1839

4. Jute industry was started in the Hoogly Valley at ………………. near Calcutta.
Answer:
Rishra

5. ………………. shortened the distance between Europe and India.
Answer:
Suez canal

III. Match the following.

1. Tavernier a) Drain Theory
2. Dacca b) Paper mill
3. Dadabai Naoroji c) Artisan
4. Ballygunj d) Muslin
5. Smiths e) French traveller

Answer:

1. Tavernier e) French traveller
2. Dacca d) Muslin
3. Dadabai Naoroji a) Drain Theory
4. Ballygunj b) Paper mill
5. Smiths c) Artisan

IV. State True or False.
1. India was famous for cotton and silk cloths.
Answer:
True

2. The railway was introduced in India by the British.
Answer:
True

3. Steel was first manufactured by modem methods at Jamshedpur.
Answer:
False

4. The Industrial policy of 1948, brought mixed economy in industrial sector.
Answer:
True

5. The tenth and eleventh five-year plans witnessed a high growth rate of Agricultural production.
Answer:
False

V. Consider the following statements and tick the appropriate answer.
1. Which of the following statements are correct?
i) According to Edward Baines, ‘The birthplace of cotton manufacture is in England’.
ii) Before mechanised industry handicrafts were the second-largest source of employment in rural India.
iii) Saurashtra was known for the tin industry.
iv) Construction of the Suez Canal made the British goods cheaper in India,
a) i and ii are correct
b) ii and iv are correct
c) iii and iv are correct
d) i, ii, and iii are correct
Answer:
b) ii and iv are correct

2. Assertion (A): Indian handicrafts collapsed under colonial rule.
Reason (R): British made India the producer of raw materials and markets for their finished
products.
a) A is correct R is the correct explanation of A
b) A is correct and R is not the correct explanation of A
c) Both A and R are correct
d) Both A and R is wrong
Answer:
A is correct R is the correct explanation of A

3. Which one of the following is wrongly matched?
a) Bernier – Shajahan
b) Cotton mill – Ahmadabad
c) TISCO – Jamshedpur
d) Economic Liberalisation – 1980
Answer:
d) Economic Liberalisation -1980

VI. Answer the following in one or two sentences.
1. What are the traditional handicraft industries of India?
Answer:

  1. The traditional handicrafts industries of India are textiles, woodwork, ivory, stone cutting, leather, fragrance wood, metalwork, and jewellery.
  2. The village artisans such as potters, weavers, smiths produced articles and utensils.

2. Write about the drain theory.
Answer:
Dadabai Naoroji was the first to acknowledge that the poverty of the Indian people was due to the British exploitation of India’s resources and the drain of India’s wealth to Britain.

3. Name the inventions which made the production of textiles on a large scale
Answer:
The invention of cotton gin, flying shuttle, spinning jenny and steam engine in England, which made the production of textiles on large scale.

4. Write a short note on the Confederation of Indian Industry.
Answer:

  • It is a business association in India. CII is a non-government, not for profit, industry-led, and industry- managed organisation.
  • It was founded in 1985.

5. What is de-industrialization?
Answer:
The process of disruption of traditional Indian crafts and a decline in national income has been referred to as de-industrialisation.

VII. Answer the following.
1. How was the trading policy of the British caused the decline of Indian Industries?
Answer:
The decline of Indian Industries:

  • Loss of Royal Patronage.
  • Transition from producer to exporter of raw materials.
  • Competition of Machine-Made goods.
  • The trading policy of the British.
  • De-industrialization.

2. Write in detail about the plantation industries.
Answer:
Plantation industries:

  1. The plantation industry was the first to attract Europeans. This provides jobs on a large scale.
  2. In reality, it could meet the increasing demands for tea, coffee and indigo by the British Society.
  3. The Assam Tea Company was founded in 1839.
  4. The coffee plantations also started simultaneously.
  5. As the tea plantation was the most important industry of Eastern India, the coffee plantation became the center of activities in South India.
  6. The Third important plantation, which gave birth to the factory was jute.
  7. All these Industries were controlled by many former employees of the British East India Company.

3. Explain Industrial development after the 1991 reforms.
Answer:
Industrial development after the 1991 reforms:

  • The year 1991 ushered a new era of economic liberalisation.
  • India took a major decision to improve the performance of the industrial sector.
  • The tenth and Eleventh Five-Year plans witnessed a high growth rate of industrial production.
  • The abolition of industrial licensing, dismantling of price controls, dilution of reservation of small- scale industries, and the virtual abolition of monopoly law enabled the Indian industry to flourish.
  • The new policy welcomes foreign investments.

VIII. HOTs.
1. How do handicraft products differ from machine-made products?
Answer:
Handicraft:
Something you make with your own hands, especially an ornament or decoration, is a handicraft. Instead, items made by artisans like pottery, handwoven blankets, handmade jewellery, and quilts stitched by hand are all examples of handicrafts.

Machine-made Products:
Machine-made products are produced faster and all are exactly the same. Machine manufacturing is faster and more economical. Also, machine-made goods are cheaper than hand made goods.

X. Project and Activity.
1. Name the industries in your state and divide them into agro-based metal-based, and forest-based. Agro-based industries:
Answer:
Cotton textile industries, Jute industry, Sugar industries, etc are agro-based industries.
Eg: Coimbatore

Metal-based industries:

  • Mineral-based industries use both metallic and nonmetallic as raw material. Eg: Chennai.
  • Forest-based industries: India has a rich diversity of forest resources. The most important industry is the paper industry. Eg: Chennai.

Samacheer Kalvi 8th Social Science Development of Industries in India Additional Important Questions and Answers

I. Choose the Correct answer.
1. The established of cotton textile industry ……………. in 1854
a) Calcutta
b) Hyderabad
c) Gujarat
d) Bombay
Answer:
d) Bombay

2. The first paper mill was started in Ballyguni in the year
a) 1870
b) 1871
c) 1872
d) 1950
Answer:
19478

3. The industrial policy Resolution act introduced in the year
a) 1947
b) 1948
c) 1949
d) 1950
Answer:
b) 1948

4. India has emerged as the ……………. largest produces of electricity in Asia
a) First
b) Second
c) Third
d) fourth
Answer:
c) Third

5. Introduction of ……………. plan the most important innovations in the industrial field
a) Annual
b) Four – year
c) Five – Year
d) Six-year
Answer:
c) Five – Year

6. On the basis of raw material used industries can be classified into two types of sectors are ……………., ……………..
a) Public and private
b) Mineral and private
c) basic and important
d) agro and mineral
Answer:
d) agro and mineral

7. The period can be considered as the period of the industrial recovery
a) 1980
b) 1981
c) 1982
d) 1983
Answer:
a) 1980

8. Steel was the first manufactured by modern methods in 1874
a) Jamshedpur
b) Kulti
c) Bally guni
d) None of these
Answer:
b) Kulti

9. The openings of Suez canal also shortened the distance between Europe and India by
about
a) 4800 km
b) 4890 Km
c) 4830 km
d) 4000 Km
Answer:
c) 4830 km

10. The consideration of Indian Industry was founded in
a) 1980
b) 1982
c) 1985
d) 1985
Answer:
c) 1985

II. Fill in the blanks.
1. “The birth place of cotton manufacture is India,” told by, ………………..
Answer:
Baines

2. ………………. cloth was used to Preserved Mummies in Egyptian.
Answer:
The Muslin of Dacca

3. ………………. is was identified with muslin cloths.
Answer:
Dacca

4. ………………. was known for tin Industry
Answer:
Bengal

5. The drain of India’s wealth to Britain acknowledge by ………………..
Answer:
Dadabai Naoroji

6. The period of the 1980s can be considered as the period of the ………………. recovery
Answer:
Industrial

7. The year ………………. where a new era of the economic liberalisation.
Answer:
1991

8. India ………………. has become one of the largest in the world.
Answer:
Road network

9. The industrial expansion over the plan period presents a ………………. picture
Answer:
mixed economic

10. In India a fourth sector industries are ……………….
Answer:
Information related industries

III. Match the following.

1. 1839 a) industrial recovery
2. 1854 b) Steel industry
3. 1874 c) Tata Iron and steel company
4. 1907 d) Tea company
5. 1980 e) Cotton textile industry

Answer:

1. 1839 d) Tea company
2. 1854 e) Cotton textile industry
3. 1-874 b) Steel industry
4. 1907 c) Tata Iron and steel company
5. 1980 a) industrial recovery

IV. State True or False.
1. Dacca was identified with Muslim clothes.
Answer
True

2. The plantation industry did not attract the European in the beginning.
Answer
False

3. Textile was the oldest industries in India.
Answer
True

4. The woolen and leather factories became prominent in Kanpur
Answer
True

5. The credit for the development of large – scale manufacture of steel in India goes to Jansheji Tata
Answer
True

V. Consider the following statements and tick the appropriate answer.
1. i) Textile was the oldest industry in India.
ii) India became the market for the finished products of Britain.
iii) The Indian goods made with primitive techniques could not compute with Industrial goods made in England
iv) The English good enterious India were charged.
a) i and ii are correct
b) ii and iv are correct
c) iii and iv are correct
d) i, ii and iii are correct
Answer:
b) ii and iv are correct

2. Assertion: The Indian road network has been one of the largest in the world
Reason: The Indian road network has directly contributed to industrial growth
a) A is correct R is the correct explanation of A
b) A is correct and R is not the correct explanation of A
c) Both A and R are correct
d) Both A and R is wrong
Answer:
c) Both A and R are correct

VI. Answer the following in one or two sentences.
1. Give an account of Muslin Cloth.
Answer:

  • Muslin was famous under the Hindu kings.
  • Early these cloths are light in nature. And also so costliest in the early period.
  • After the Muslim invasion, the muslin preparation was unknown by the other rulers.

2. What are the classification Industries?
Answer:

  • On the basis of raw materials used, industries can be classified into agro-based and mineral-based.
  • According to their role, it can be classified into basic and key industries.
  • On the basis of ownership, it can be classified into the public sector, private sector, joint sector, and co-operative sector.

3. How the Indian road network used in industrial growth?
Answer:

  • The Indian road network has become one of the largest in the world.
  • Government efforts led to the expansion of the network of National Highways, State highways, and major district roads, which in turn has directly contributed to industrial growth.

VII. Answer the following.
1. Explain the growth of Heavy industries in the beginning period.
Answer:

  • The heavy industries included the iron and steel industry, Steel was first manufactured by modem methods at Kulti in 1874.
  • Iron and steel industries began rooted in the Indian soil at the beginning of the 20th century.
  • However, the credit for the development of large-scale manufacture of steel in India goes to Jamshedji Tata.
  • The Tata Iron and Steel Company (TISCO) was set up in 1907 at Jamshedpur.
  • It started producing pig iron in 1911 and steel ingots in 1912.

2. Explain the classification of industries per the industrial policy Resolution 1956.
Answer:
As per the Industrial Policy Resolution 1956, industries were classified into three categories:
Schedule A:
Only the Government can handle these industries. Some of these are atomic energy, electrical, iron and steel, and others.

Schedule B:
These comprise road and sea transportation, machine tools, aluminium, chemicals including plastics and fertilisers, ferroalloys, and certain types of mining.

Schedule C:
Under this category, the remaining industries and left to the private sector.

3. What are the phases of Industrial development in India?
Answer:
Industrial development during 1950 – 1965:

  • During this phase, a majority of consumer goods were produced in India.
  • As a result, this phase witnessed a strong acceleration in the growth rate of production.

Industrial development during 1965 – 1980:

  • It mostly focused on the development of the capital goods sector, the consumer goods sector was neglected.
  • So this period is marked as the period of structural retrogression.

Industrial development during 1980 – 1991:
This period witnessed quite a healthy industrial growth.

Industrial development post-1991 Reforms:

  • The year 1991 a new era of economic liberalisation.
  • India took a major decision to improve the performance of the industrial sector.
  • The new policy welcomes foreign investments.

Samacheer Kalvi 5th Tamil Guide Chapter 6.2 விதைத் திருவிழா

Tamilnadu State Board New Syllabus Samacheer Kalvi 5th Tamil Guide Pdf Chapter 6.2 விதைத் திருவிழா Text Book Back Questions and Answers, Summary, Notes.

Tamilnadu Samacheer Kalvi 5th Tamil Solutions Chapter 6.2 விதைத் திருவிழா

மதிப்பீடு 

படிப்போம்! சிந்திப்போம்! எழுதுவோம்!

அ. சரியான சொல்லைத் தெரிவு செய்து எழுதுக.

Question 1.
அனுமதி – இச்சொல் குறிக்கும் பொருள் ..adscanadian………
அ) கட்டளை
ஆ) இசைவு
இ) வழிவிடு
ஈ) உரிமை
Answer:
ஆ) இசைவு

Samacheer Kalvi 5th Tamil Guide Chapter 6.2 விதைத் திருவிழா

Question 2.
விளம்பரத்தாள்கள் – இச்சொல்லைப் பிரித்து எழுதக் கிடைப்பது ……………………..
அ) விளம்பர + தாள்கள்
ஆ) விளம்பரத்து + தாள்கள்
இ) விளம்பரம் + தாள்கள்
ஈ) விளம்பு + தாள்கள்
Answer:
இ) விளம்பரம் + தாள்கள்

Question 3.
ஆலோசித்தல் – இச்சொல்லுக்குரிய பொருள் ………………………
அ) பேசுதல்
ஆ) படித்தல்
இ) எழுதுதல்
ஈ) சிந்தித்தல்
Answer:
ஈ) சிந்தித்தல்

Samacheer Kalvi 5th Tamil Guide Chapter 6.2 விதைத் திருவிழா

Question 4.
தோட்டம் + கலை – இச்சொற்களைச் சேர்த்து எழுதக் கிடைப்பது …………………..
அ) தோட்டம்கலை
ஆ) தோட்டக்கலை
இ) தோட்டங்கலை
ஈ) தோட்டகலை
Answer:
ஆ) தோட்டக்கலை

Question 5.
பழங்காலம் இச்சொல்லுக்குரிய எதிர்ச்சொல் …………………….
அ) பழைய காலம்
ஆ) பிற்காலம்
இ) புதிய காலம்
ஈ) இடைக்காலம்
Answer:
இ) புதிய காலம்

ஆ. கீழ்க்காணும் சொற்களைச் சேர்த்து எழுதுக.
அ) வழிபாடு + கூட்டம் = ……………………..
ஆ) வீடு + தோட்டம் = ……………………..
Answer:
அ) வழிபாடு + கூட்டம் – வழிபாட்டுக்கூட்டம்
ஆ) வீடு + தோட்டம் – வீட்டுத்தோட்டம்

Samacheer Kalvi 5th Tamil Guide Chapter 6.2 விதைத் திருவிழா

இ. கீழ்க்காணும் சொற்களைப் பிரித்து எழுதுக.
அ) அழைப்பிதழ்- …………. + ……………..
ஆ) விதைத்திருவிழா- …………. + ……………..
Answer:
அ) அழைப்பிதழ் – அழைப்பு + இதழ்
ஆ) விதைத்திருவிழா – விதை + திருவிழா

Samacheer Kalvi 5th Tamil Guide Chapter 6.2 விதைத் திருவிழா

ஈ. கோடிட்ட இடத்தை நிரப்புக.

Question 1.
விதைத்திருவிழாவில் அமைக்கப்பட்டிருந்த அரங்குகளின் எண்ணிக்கை………..
Answer:
27

Question 2.
விதைகள் …………… ஆனவையாக இருத்தல் வேண்டும்.
Answer:
தரம்

Samacheer Kalvi 5th Tamil Guide Chapter 6.2 விதைத் திருவிழா

Question 3.
கொண்டைக்கடலை என்பது ……………….. ஒன்று.
Answer:
நவதானியங்களுள்

உ. வினாக்களுக்கு விடையளிக்க.

Question 1.
மாணவர்களை எங்கே அழைத்துச் செல்வதாகத் தலைமையாசிரியர் கூறினார்?
Answer:
மாணவர்களை அருகிலுள்ள மேல்நிலைப் பள்ளியில் நடைபெறும் விதைத் திருவிழாவிற்கு அடுத்த வாரம் அழைத்துச் செல்வதாக தலைமையாசிரியர் கூறினார்.

Samacheer Kalvi 5th Tamil Guide Chapter 6.2 விதைத் திருவிழா

Question 2.
ஆசிரியர் வழங்கிய துண்டு விளம்பரத்தாளில் என்ன செய்தி இருந்தது?
Answer:
ஆசிரியர் வழங்கிய துண்டு விளம்பரத்தாளில் இருந்த செய்தி விதைத்திருவிழா தொடர்பான செய்தி’ ஆகும்.

Question 3.
‘பாதிப்பு’ என்று எழுதப்பட்ட அரங்கத்தில் என்ன செய்தி சொல்லப்பட்டது?
Answer:
இரசாயன விதைகள், இரசாயனப் பூச்சி மருந்துகள் பயன்படுத்துவதால் ஏற்படும் விளைவுகளைத்தாம் ‘பாதிப்பு’ என்று சொல்கிறார்கள். இதனால், மண்ணின் தன்மை கெடுகிறது. இதனைக் தடுக்கும் வகையில் இயற்கை முறையில் வேளாண்மை செய்ய வேண்டும் என்பதே அதன் பொருள்.

Samacheer Kalvi 5th Tamil Guide Chapter 6.2 விதைத் திருவிழா

Question 4.
நவதானியங்களுள் ஐந்தின் பெயரை எழுதுக.
Answer:

  • கொண்டைக்கடலை
  • தட்டைப்பயறு
  • மொச்சை
  • பாசிப்பயறு
  • கோதுமை.

Samacheer Kalvi 5th Tamil Guide Chapter 6.2 விதைத் திருவிழா

ஊ. சிந்தனை வினாக்கள்.

செயற்கை உரங்கள், மண்ணின் வளத்தைக் கெடுக்கும் எனில், அதற்கு மாற்றாக நாம் என்ன செய்யலாம்?
Answer:
செயற்கை உரங்கள், மண்ணன் வளத்தைக் கெடுக்கும் எனில், அதற்கு மாற்றாக நாம் செய்ய வேண்டுவன:

  1. இயற்கை வேளாண்மையைக் கடைப்பிடிக்க வேண்டும்.
  2. மண்புழு வளர்த்தல்.
  3. கால்நடைகள் வளர்த்து அவற்றின் சாணங்களை எருவாக்குதல்.
  4. அவுரிச் செடிகளை வளர்த்து வயலுக்கு எருவாக்குதல்.

Samacheer Kalvi 5th Tamil Guide Chapter 6.2 விதைத் திருவிழா

கற்பவை கற்றபின்

Question 1.
இயற்கை வேளாண்மையின் சிறப்புகளைப் பற்றி, வழிபாட்டுக்கூட்டத்தில் பேசுக.
Answer:
வணக்கம். வேளாண்மையில் செயற்கையான வேதிப்பொருட்களைப் பயன்படுத்தாமல் இயற்கையோடு இணைந்து வேளாண்மை செய்வது இயற்கை வேளாண்மை ஆகும். இம்முறையைப் பயன்படுத்துவதால் மண், நீர், காற்று மற்றும் ஆகாயம் ஆகியவை பாதுகாக்கப்படுகின்றன.

பயனீட்டாளர்களுக்கும் உடல்நலத்திற்கேற்ற உணவு கிடைக்கிறது. விவசாயிகளும் அதிக விளைச்சலுடன் லாபத்தையும் பெறுகின்றனர். முக்கியமாக நாம் நமது அடுத்த தலைமுறைக்கு மாசற்ற வேளாண் முறையைத் தருவதோடு ஆரோக்கியமான உணவுக்கும் வழிவகை செய்கின்றோம்.

Question 2.
இயற்கை உணவுப் பொருள்களின் படங்களைத் திரட்டித் தொகுப்பேடு உருவாக்குக.
Answer:
மாணவர்கள் தாங்களாகவே செய்ய வேண்டும்.

Samacheer Kalvi 5th Tamil Guide Chapter 6.2 விதைத் திருவிழா

Question 3.
‘இயற்கை உரம் பயன்படுத்துவோம், இனிமையாய் வாழ்வோம்’ என்னும் தலைப்பில் கட்டுரை எழுதுக.
Answer:
முன்னுரை :
வளர்ந்துவரும் நவீனயுகத்தில் நாம் அனைவரும் இயற்கையை மறந்து செயற்கையைப் போற்றியதால் பல தீமைகளை எதிர்கொள்கிறோம். இந்நிலையை மாற்றுவதே இயற்கை வேளாண்மை. அதனைப் பற்றி இக்கட்டுரையில் காண்போம்.

மாசடைந்த நிலங்கள் :
நல்ல விளைச்சல், பார்ப்பதற்குப் பெரிய பெரிய காய்கறிகள், கனிகள், குறுகிய காலத்தில் நிறைய இலாபம் இதனை மட்டுமே கருத்தில் கொண்டு செயற்கை உரங்களையும் பூச்சிக்கொல்லிகளையும் பயன்படுத்தினோம். அதனால் வளம் இழந்தது மண் மட்டுமா? நாமும்தான். நீர்வளம், நிலவளம் குறைந்தது போல நமக்கும் புதிய புதிய நோய்கள் வந்து வலுவிழந்துவிட்டோம்.

Samacheer Kalvi 5th Tamil Guide Chapter 6.2 விதைத் திருவிழா

இயற்கை உரங்கள் :
மண்புழுக்களை உற்பத்தி செய்து உரமாகப் பயன்படுத்துதல், கால்நடைகளின் சாணங்களை எருவாக்குதல், பண்ணையில் வளர்க்கப்படும் பறவை, விலங்குகளின் கழிவுகளை உரமாக்குதல், ஆமணக்கு, நிலக்கடலை, எள், பருத்தி, தேங்காய் ஆகியவற்றிலிருந்து கிடைக்கும் பிண்ணாக்குகளை உரமாக்குதல். இவையனைத்தும் இயற்கை உரங்கள். இவற்றைப் பயன்படுத்தி விளைச்சலைப் பெருக்குவோம்.

நன்மைகள் :
இயற்கை உரங்களைப் பயன்படுத்துவதால் விளைநிலங்களில் நன்மை செய்யும் பூச்சிகள், நுண்ணுயிரிகளின் எண்ணிக்கைபெருகுகிறது. பயிர்கள் நோய் எதிர்ப்புத்திறனைப் பெறுகின்றன. பயிர்கள் சத்துகளை எளிதாக எடுத்துக் கொள்கின்றன. பயிர்கள் சீராக விளைகின்றன. தரமான விளைச்சல் கிடைக்கின்றது. மிகவும் இன்றியமையாத நன்மை எதுவெனில் சுற்றுப்புறச் சூழல் தூய்மை ஆகிறது. உழவர்கள் உரங்களைத் தாங்களே தயாரிப்பதால் செலவும் குறைகிறது.

Samacheer Kalvi 5th Tamil Guide Chapter 6.2 விதைத் திருவிழா

முடிவுரை :
இயற்கை உரங்களைப் பயன்படுத்துவதால் நல்ல சத்தான உணவுகள் கிடைக்கின்றன. அதனை உண்பதனால் நாம் நோயின்றி வாழலாம். ஆரோக்கியமான வாழ்வைப் பெறலாம்.

Question 4.
உங்கள் இருப்பிடத்திற்கு அருகிலுள்ள இயற்கை விதைப் பண்ணைகளுக்குச் சென்று, செய்தி திரட்டுக.
Answer:
மாணவர்கள் தாங்களாகவே செய்ய வேண்டும்.

Samacheer Kalvi 5th Tamil Guide Chapter 6.2 விதைத் திருவிழா

Question 5.
உங்கள் பள்ளியில் நடைபெறும் ஏதேனும் ஒரு விழாவுக்கு மாதிரி அழைப்பிதழ்/துண்டு விளம்பரம் உருவாக்கி மகிழ்க.
Answer:
Samacheer Kalvi 5th Tamil Guide Chapter 6.2 விதைத் திருவிழா - 1

Samacheer Kalvi 11th Tamil Guide Chapter 2.3 காவியம்

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Tamil Guide Pdf Chapter 2.3 காவியம் Questions and Answers, Summary, Notes.

Tamilnadu Samacheer Kalvi 11th Tamil Solutions Chapter 2.3 காவியம்

Samacheer Kalvi 11th Tamil Guide Chapter 2.3 காவியம்

குறுவினா

Question 1.
காற்றின் தீராத பக்கங்களில், எது எதனை எழுதிச் சென்றது?
Answer:
சிறகிலிருந்து பிரிந்த இறகு ஒன்று, ஒரு பறவையின் யாழ்வை எழுதிச் சென்றது.

சிறுவினாக்கள் – கூடுதல் வினாக்கள்

Question 1.
இறகு எழுதியது காவியமானதைப் பிரமிள் பார்வையில் விளக்குக.
Answer:

  • நிலத்துக்கும் வானுக்கும் இடையில், காற்றுட இடைவிடாது தழுவி, மண்ணில் விழாமல் காக்கிறது. அதனால், அந்த இறகு, பறவையின் வாழ்வை எழுதுவதுபோல் உள்ளது.
  • காவியங்களுள் பொதுவான பால் பொருள் வாழ்வுதானே! அதனால், பிரமிள் பார்வையில் சிறகின் இடையறாத இருப்பு நிரந்தர வாழ்வாகிறது.

Question 2.
பிரமிள் குறித்து நீ அறிவன வை?
Answer:
இலங்கையில் பிறந்த சிவராமலிங்கம், ‘பிரமிள் ‘ என்னும் பெயரில் எழுதினார். பானுசந்திரன், அரூப்சிவராம், தரமுசவராம் எனப் பல புனைபெயர்களில் எழுதிவந்தார்.

புதுக்கவிதை, விமாசனம், சிறுகதை, நாடகம், மொழியாக்கம் என விரிந்த தளங்களில் இயங்கினார். ஓவியம், சிறடம் ஆகியவற்றிலும் ஈடுபாடு காட்டினார்.

இவர் விதைகள் அனைத்தும், ‘பிரமிள் கவிதைகள்’ என்னும் பெயரில் தொகுத்து வெளியிடப் பட்டேளது. ‘லங்காபுரி ராஜா’ உள்ளிட்ட சிறுகதைத் தொகுப்புகளும், ‘நக்ஷத்திரவாசி’ என்னும் நாடகமும், ‘வெயிலும் நிழலும்’ உள்ளிட்ட கட்டுரைத் தொகுப்புகளும் வெளிவந்துள்ளன.

பலவுள் தெரிக (கூடுதல்)

Question 1.
ஒரு பறவையின் வாழ்வை எழுதிச் சென்றது………………..
அ) சிறகு
1. அ சரி
ஆ) இறகு
2. ஆ சரி
இ) காற்று
3. இ சரி
ஈ) ஒரு பறவை
4. ஈ தவறு
Answer:
2. ஆ சரி

Question 2.
‘பிரமிள்’ என்னும் பெயரில் எழுதியவர்………………
அ) இராசேந்திரன்
ஆ) அரவிந்தன்
இ) சிவராமலிங்கம்
ஈ) விருத்தாசலம்
Answer:
இ) சிவராமலிங்கம்