Category: Class 12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 7 Chemical Kinetics Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 7 Chemical Kinetics

12th Chemistry Guide Chemical Kinetics Text Book Questions and Answers

Part – I Text Book Evaluation

I. Choose the correct answer

Question 1.
For a first order reaction A → B the rate constant is x min^-1. If the initial concentration of A is 0.01 M, the concentration of A after one hour is given by the expression.
(a) 0.01 e^-x
(b) 1 × 10^-2 (1 – e^-60x)
(c) (1 × 10^-2) e^-60x
(d) none of these
Answer:
(c) (1 × 10^-2) e^-60x
Solutions:

In this case
k = x min^-1 and
[A₀] = 0.01 M = 1 × 10^-2 M
t = 1 hour = 60 min
[A] = 1 × 10^-2(e^-60x)

Question 2.
A zero order reaction X → Product. with an initial concentration 0.02M has a half life of 10 min. If one starts with concentration 0.04M, then the half life is …………….
(a) 10 s
(b) 5 min
(c) 20 min
(d) cannot be predicted using the given information
Answer:
(c) 20 min
Solutions:
Given,
[A₀] = 0.02 M ; t_1/2 = 10 min
[A₀] = 0.04 M ; t_1/2 = ?
Substitute in (1)
10 min ∝ 0.02 M ……………………..(2)
t_1/2 ∝ 0.04 M ……………………..(3)
Dividing Eq.(3) by Eq. (2) we get,
\(\frac { { t }^{ 1/2 } }{ 10min }\) = \(\frac { 0.04M }{ 0.02M }\)
t_1/2 = 2 × 10 min = 20 min

Question 3.
Among the following graphs showing variation of rate constant with temperature (T) for a reaction, the one that exhibits Arrhenius behavior over the entire temperature range is ……………

Answer:
b
Solution:
\(\text { Ae }\left(\frac{\mathrm{Ea}}{\mathrm{RT}}\right)\)
In k = In A – \(\left( \frac { { E }_{ a } }{ R } \right)\) \((\frac { 1 }{ T })\)
this equation is in the form of a straight line equation
y = c + m x
a plot of ink vs \((\frac { 1 }{ T })\) is a straight line with negative slope.

Question 4.
For a first order react ion A → product with initial concentration x mol L^-1, has a half life period of 2.5 hours. For the same reaction with initial concentration mol L^-1 the half life is
(a) (2.5 × 2) hours
(b) \((\frac { 2.5 }{ 2 })\) hours
(c) 2.5 hours
(d) Without knowing the rate constant, t_1/2 cannot be determined from the given data
Answer:
(c) 2.5 hours
Solutions:
For a first order reaction
t_1/2 = \(\frac { 0.693 }{ k }\) t_1/2 does not depend on the initial concentration and it remains constant (whatever may be the initial concentration)
t_1/2 = 2.5 hrs .

Question 5.
For the reaction, 2NH₃ → N₂ + 3H₂, if

then the relation between
k₁, k₂ and k₃ is
(a) k₁ = k₂ = k₃
(b) k₁ = 3k₂ = 2k₃
(c) 1.5k₁ = 3k₂ = k₃
(d) 2k₁ = k₂ = 3k₃
Answer:
(c) 1.5k₁ = 3 k₂ = k₃
Solution:

Question 6.
The decomposition of phosphine (PH₃) on tungsten at low pressure is a first order reaction. It is because the …………….
(a) rate is proportional to the surface coverage
(b) rate is inversely proportional to the surface coverage
(c) rate is independent of the surface coverage
(d) rate of decomposition is slow
Answer:
(c) rate is independent of the surface coverage
Solution:
Given:
At low pressure, the reaction follows first-order, therefore
Rate ∝ [reactant]¹
Rate ∝ (surface area)
At high pressure due to the complete coverage of the surface area, the reaction follows zero-order.
Rate ∝ [reactant]°.
Therefore the rate is independent of surface area.

Question 7.
For a reaction Rate = k [acetone]^3/2 then a unit of the rate constant and rate of reaction respectively is …………..
(a) (mol L^-1 s^-1), (mol^-1/2 L^1/2 s^-1)
(b) (mol^-1/2 L^1/2 s^-1), (mol L^-1 s^-1)
(c) (mol^1/2 L^1/2 s^-1), (mol L^-1 s^-1)
(d) (mol L s^-1), (mol^1/2 L^1/2 s)
Answer:
(b) (mol^1/2 L^1/2 s^-1), (mol L^-1 s^-1)
Solution:
Rate = k [A]ⁿ
Rate = \(\frac { -d[A] }{ dt } \)
unit of rate = \(\frac { mol{ L }^{ -1 } }{ s }\) = mol L^-1 s^-1
unit of rate constant = \(\frac { (mol{ L }^{ -1 }{ S }^{ -1 }) }{ { (mol{ L }^{ -1 }) }^{ n } }\)
= mol^1-n L^n-1 S^-1
rate = k [Acetone]^3/2
n = 3/2
mol^1-(3/2) L^(3/2)-1 s^-1
mol^-(1/2) L^(1/2) s^-1

Question 8.
The addition of a catalyst during a chemical reaction alters which of the following quantities?
(a) Enthalpy
(b) Activation energy
(c) Entropy
(d) Internal energy
Answer:
(b) Activation energy
Solution:
Activation energy:
A catalyst provides a new path to the reaction with low activation energy. i.e, it lowers the activation energy.

Question 9.
Consider the following statements:
(i) increase in the concentration of the reactant increases the rate of a zero-order reaction.
(ii) rate constant k is equal to collision frequency A if E_a = ∞
(iii) rate constant k is equal to collision frequency A if E_a = o
(iv) a plot of ln (k) vs T is a straight line.
(v) a plot of ln (k) vs \((\frac { 1 }{ T })\) is a straight line with a positive slope.

Correct statements are
(a) (ii) only
(b) (ii) and (iv)
(c) (ii) and (v)
(d) (i), (ii) and (v)
Answer:
(a) (ii) only
Solutions:
In zero order reactions, an increase in the concentration of reactant does not alter the rate, So statement (i) is wrong.

this equation is in the form of a straight line equation yc + mx. a plot of Ink vs \(\frac { 1 }{ T }\) is a straight line with negative slope so statements (iv) and (v) are wrong.

Question 10.
In a reversible reaction, the enthalpy change and the activation energy in the forward direction are respectively – x kJ mol^-1 and y kJ mol^-1. Therefore, the energy of activation in the backward direction is ………..
(a) (v – x) kJ mol^-1
(b) (x + y) J mol^-1
(c) (x – y)  kJ mol^-1
(d) (x + y) × 10³ J mol^-1
Answer:
(d) (x + y) × 10³ J mol^-1
Solution:

(x + y)kJmol^-1
(x + y) × 10³Jmol^-1

Question 11.
What is the activation energy for a reaction if its rate doubles when the temperature is raised from 200K to 400K? (R 8.314 JK^-1 mol^-1)
(a) 234.65 kJ mol^-1
(b) 434.65 kJ mol^-1
(c) 2.305 KJ mol^-1
(d) 334.65 J mol^-1
Answer:
(c) 2.305 KJ mol^-1
Solutions:

Question 12.

This reaction follows first order kinetics. The rate constant at particular temperature is 2.303 × 10^-2 hour^-1. The initial concentration of cyclopropane is 0.25 M. What will be the concentration of cyclopropane after 1806 minutes? (Log 2 = 0.30 10)
(a) 0.125 M
(b) 0.215 M
(c) 0.25 × 2.303 M
(d) 0.05 M
Answer:
(b) 0.2 15 M
Solution:

Question 13.
For a first-order reaction, the rate constant is 6.909 min^-1. The time taken for 75% conversion in minutes is …………
(a) \((\frac { 3 }{ 2 })\) log 2
(b) \((\frac { 3 }{ 2 })\) log 2
(c) \((\frac { 3 }{ 2 })\) log \((\frac { 3 }{ 4 })\)
(d) \((\frac { 2 }{ 3 })\) log \((\frac { 4 }{ 3 })\)
Answer:
(b) \((\frac { 3 }{ 2 })\) log 2
Solution:
k = \((\frac { 2.303 }{ t })\) log \(\left( \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \right)\)
[A₀]= 100; [A]=25
6.909 = \((\frac { 2.303 }{ t })\) log \((\frac { 100 }{ 25 })\)
t = \((\frac { 2.303 }{ 6.909 })\) log (4)
⇒ t = \((\frac { 1 }{ 3 })\) log 2²
t = \((\frac { 2 }{ 3 })\) log 2

Question 14.
In a first-order reaction x → y; if k is the rate constant and the initial concentration of the reactant x is 0.1 M, then, the half-life is ……..
(a) \((\frac { log2 }{ k })\)
(b) \((\frac { 0.693 }{ (0.1)k })\)
(c) \((\frac { In2 }{ k })\)
(d) none of these
Answer:
(c) \((\frac { In2 }{ k })\)
Solution:
k = \((\frac { 1 }{ t })\) In \(\left( \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \right)\)
[A₀] = 0.1
[A] = 0.05
k = \(\left( \frac { 1 }{ { t }_{ 1/2 } } \right)\) In \((\frac { 0.1 }{ 0.05 })\)
k = \(\left( \frac { 1 }{ { t }_{ 1/2 } } \right)\) In (2) ⇒ t_1/2 = \((\frac { In(2) }{ k })\)

Question 15.
Predict the rate law of the following reaction based on the data given below:
2A + B → C + 3D

(a) rate = k [A]² [B]
(b) rate = k [A][B]²
(c) rate = k [A][B]
(d) rate = k [A]^1/2 [B]^3/2
Answer:
(b) rate = k [A][B]²
Solution:
rate₁ = k [0.1]ⁿ [0.1]^m ……………(1)
rate₂ = k [0.2]ⁿ [0.1]^m …………(2)
Dividing Eq.(2) by Eq.(1)

\(\frac { 2x }{ x }\) = 2ⁿ
∴ n = 1
rate₃ = k [0.1]ⁿ [0.2]^m …………..(3)
rate₄ = k [0.2]ⁿ [0.2]^m …………..(4)
Dividing Eq.(4) by Eq.(2)

\(\frac { 8 }{ 2 } \) = 2^m
∴ m = 2
∴ rate = k [A]¹ [B]²

Question 16.
Assertion: rate of reaction doubles when the concentration of the reactant is doubled if it is a first-order reaction.
Reason: rate constant also doubles
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer:
(c) Assertion is true but the reason is false.
Solution:
For a first reaction, when the concentration of reactant is doubled, then the rate of reaction also doubled. Rate constant is independent of concentration and is a constant at a constant temperature, i.e., it depends on the temperature, and hence, it will not be doubled and when the concentration of the reactant is doubled.

Question 17.
The rate constant of a reaction is 5.8 × 10^-2 s^-1. The order of the reaction is ………….
(a) First order
(b) zero-order
(c) Second-order
(a) Third order
Answer:
(a) First order
Solution:
The unit of the rate constant is s^-1 and it indicates that the reaction is first order.

Question 18.
For the reaction N₂ O_5(g) → 2NO_2(g) +\(\frac { 1 }{ 2 }\) O_2(g) the value of rate of disappearance of N₂O₅ is given as 6.5 × 10^-2 mol L^-1s^-1 The rate of formation of NO₂ and O₂ is given respectively as
(a) (3.25 × 10^-2 mol L^-1s^-1) and (1.3 × 10^-2 mol L^-1s^-1)
(b) (1.3 × 10^-2 mol L^-1s^-1) and (3.25 × 10² mol L^-1s^-1)
(c) (1.3 × 10^-1 mol L^-1s^-1) and (3.25 × 10^-2 mol L^-1s^-1)
(d) None of these
Answer:
(c) (1.3 × 10^-1 mol L^-1s^-1) and (3.25 × 10^-2 mol L^-1s^-1)
Solution:

Question 19.
During the decomposition of H₂O₂ to give dioxygen, 48g O₂ is formed per minute at a certain point in time. The rate of formation of water at this point is …………….
(a) 0.75 mol min^-1
(b) 1.5 mol min^-1
(c) 2.25 mol min^-1
(d) 3.0 mol min^-1
Answer:
(d) 3.0 mol min^-1
Solution:

No. of moles of oxygen = \((\frac { 48 }{ 32 })\) = 1.5 mol
Rate of formation of oxygen = 2 × 1.5 = 3 mol min^-1

Question 20.
If the initial concentration of the reactant is doubled, the time for half-reaction is also doubled. Then the order of the reaction is …………
(a) Zero
(b) one
(c) Fraction
(d) none
Answer:
(a) Zero
Solution:
For a first order reaction t^1/2 is independent of initial concentration .i.e., n \(\neq\) 1 for such cases

Question 21.
In a homogeneous reaction A → B + C + D, the initial pressure was P₀ and after time t it was P. Expression for rate constant in terms of P₀, P and t will be

Answer:
(a)

Solution:

Question 22.
If 75% of a first-order reaction was completed in 60 minutes, 50% of the same reaction under the same conditions would be completed in ………
(a) 20 minutes
(b) 30 minutes
(c) 35 minutes
(d) 75 minutes
Answer:
(b) 30 minutes
Solution:

Question 23.
The half-life period of a radioactive element is 140 days. After 560 days, 1 g of the element will be reduced to
(a) \(\frac { 1 }{ 2 }\) g
(b) \(\frac { 1 }{ 4 }\) g
(c) \(\frac { 1 }{ 8 }\) g
(d) \(\frac { 1 }{ 16 }\) g
Answer:
(d) \(\frac { 1 }{ 16 }\) g
Solution:
in 140 days ⇒ initial concentration reduced to \(\frac { 1 }{ 2 }\) g
in 280 days ⇒ initial concentration reduced to \(\frac { 1 }{ 4 }\) g
in 420 days ⇒ initial concentration reduced to \(\frac { 1 }{ 8 }\) g
in 560 days ⇒ initial concentration reduced to \(\frac { 1 }{ 8 }\) g

Question 24.
The correct difference between first and second-order reactions is that …………
(a) A first-order reaction can be catalysed a second-order reaction cannot be catalysed.
(b) The half-life of a first-order reaction does not depend on [A₀] the half-life of a second-order reaction does depend on [A₀].
(c) The rate of a first-order reaction does not depend on reactant concentrations; the rate of a second-order reaction does depend on reactant concentrations.
(d) The rate of a first-order reaction does depend on reactant concentrations; the rate of a second-order reaction does not depend on reactant concentrations,
Answer:
(b) The half-life of a first-order reaction does not depend on [A₀]; the half-life of a second-order reaction does depend on [A₀].
Solution:
For a first order reaction
t_1/2 = \(\frac { 0.6932 }{ k }\)
For a second order reaction

Question 25.
After 2 hours, a radioactive substance becomes \((\frac { 1 }{ 16 })\)^th of original amount. Then the half life (in mm) is ………………
(a) 60 minutes
(b) 120 minutes
(c) 30 minutes
(d) 15 minutes
Answer:
(c) 30 minutes
Solution:

II. Answer the following questions:

Question 1.
Define average rate and instantaneous rate.
Answer:
1. Average rate:
The average rate of a reaction is defined as the rate of change of concentration of a reactant (or of a product) over a specified measurable period of time.

2. instantaneous rate:
Instantaneous rate of reaction gives the tendency of the reaction at a particular point of time during its course (or) The time derivative of the concentration of a reactant (or product) converted to a positive number is called the instantaneous rate of reaction.

Question 2.
Define rate law and rate constant.
Answer:
1. Rate law: The expression in which reaction rate is given in terms of molar concentration of the reactants with each term raised to some power, which may or may not be same as the Stoichiometric coefficient of the reacting species in a balanced chemical equation.
x A + y B → products
Rate = k [A]^m [B]^m
k = Rate constant

Rate constant is the proportionality constant equal to the rate of the reaction when the concentration of each reactant is unity.
In above rate law if [A] [B] = 1, rate constant k = Rate .

Question 3.
Derive integrated rate law for a zero-order reaction A → product.
Answer:
A reaction in which the rate is independent of the concentration of the reactant over a wide range of concentrations is called zero-order reactions. Such reactions are rare. Let us consider the following hypothetical zero-order reaction.
A → Product
The rate law can be written
Rate = k [A]°
(∴[A]° = 1)
– d [A] = k (1)
\(\frac { -d[A] }{ dt }\) = k(1)
-d[A] = k dt
Integrate the above equation between the limits of [A₀] at zero time and [A] at some later time ‘t’,

Question 4.
Define the half-life of a reaction. Show that for a first-order reaction half-life is independent of initial concentration.
Answer:
The half-life of a reaction is defined as the time required for the reactant concentration to reach one half of its initial value.
For a first-order reaction, the half-life is a constant i.e., it does not depend on the initial concentration. The rate constant for a first-order reaction is given by,
For a first-order reaction

Question 5.
What is an elementary reaction? Give the differences between the order and molecularity of a reaction.
Answer:
Elementary reaction – Each and every single step in a reaction mechanism is called an elementary reaction. Differences between order and molecularity:
Order of a reaction:

1. It is the sum of the powers of concentration terms involved in the experimentally determined rate law.
2. It can be zero (or) fractional (or) integer.
3. It is assigned for an overall reaction.

Molecularity of a reaction:

1. It is the total number of reactant species that are involved in an elementary step.
2. It is always a whole number, cannot be zero or a fractional number.
3. It is assigned for each elementary step of the mechanism.

Question 6.
Explain the rate-determining step with an example.
Answer:
Consider the decomposition of hydrogen peroxide catalysed by I^–
2H₂O_2(aq) → 2H₂O(l) + O₂(g)
Experimentally it is found that the reaction is first order with respect to both H₂O₂ and I^–, which indicates that I^– is also involved in the reaction.
The mechanism involves two steps.

These two are elementary reactions. Adding step 1 and step 2 gives the overall reaction.
Step 1: is the rate-determining slow step, since it involves both H2O2 and I. The overall reaction is bimolecular.

Question 7.
Describe the graphical representation of first order reaction.
Answer:
Rate constant for first order reaction is,
kt = ln\(\left( \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \right)\)
kt = In [A₀] – In [A]
In[A] = In [A₀] – kty
y = c + mx
If we follow the reaction by measuring the concentration of the reactants at regular time interval ‘t’, a plot of ln[A] against ‘t’ yields a straight line with a negative slope. From this, the rate constant is calculated.

Question 8.
Write the rate law for the following reactions.
a) A reaction that is 3/2 order in x and zero-order in y.
b) A reaction that is second order in NO and first order in Br₂.
Answer:
Rate =k[x]^3/2[y]^o
b) Rate = k[NO]²[Br₂]¹
Rate = k[x]^3/2

Question 9.
Explain the effect of a catalyst on reaction rate with an example.
Answer:

• Significant changes in the reaction can be brought out by the addition of a substance called a catalyst.
• A catalyst is a substance which alters the rate of a reaction without itself undergoing any permanent chemical change.
• They may participate in the reaction, but again regenerate and the end of the reaction.
• In the presence of a catalyst, the energy of activation is lowered and hence, the greater number of molecules can cross the energy barrier and change over to products, thereby increasing the rate of the reaction.
• For example, the decomposition of potassium chlorate is enhanced by the addition of MnO₂.
  

Question 10.
The rate law for a reaction of A, B, and C have been found to be rate = k[A]² [B][L]^3/2. How would the rate of reaction change when

1. The concentration of [L] is quadrupled
2. The concentration of both [A] and [B] are doubled
3. The concentration of [A] is halved
4. The concentration of [A] is reduced to(1/3) and the concentration of [L] is quadrupled.

Solution:
Rate = k [A]² [B] [L]^3/2 ………….(1)
1. when [L] = [4L]
Rate = k [A]² [B] [4L]^3/2
Rate = 8 (k[A]² [B] [L]^3/2) …………………..(2)
Comparing (1) and (3) rate is increased by 8 times.

2. when [A] = [2A] and [B] = [2B]
Rate = k[2A]² [2B ] [L]^3/2
Rate = 8 (k[A]² [B] [L]^3/2 …………….(3)
Comparing (1) and (3); rate is increased by 8 times.

3. when [A] = \([\frac { A }{ 2 }]\)
Rate = k \([\frac { A }{ 2 }]\)² [L]\(\frac { 3 }{ 2 }\)
Rate = \(\frac { 1 }{ 4 }\) (k[A]² [B] [L]^3/2) ……………..(4)
Comparing (1) and ( 4); rate is reduced to \(\frac { 1 }{ 4 }\) times.

4. when [A] = \([\frac { A }{ 3 }]\) and [L] = [4L]
Rate k\(\frac { A }{ 3 }\)² [B] [4L]^3/2
Rate = \([\frac { 8 }{ 9 }]\) (k[A]² [B] [L]^3/2) ……………….(5)
Comparing (1) and (5); rate is reduced to 8/9 times.

Question 11.
The rate of formation of a dimer in a second order reaction is 7.5 x 10^-3 mol L^-1s^-1 at 0.05 mol L^-1 monomer concentration. Calculate the rate constant.
Solution:
Let us consider the dimensation of a monomer M
2M → (M)₂
Rate = k [M]ⁿ
Given that n =2 and [M] = 0.05 mol L^-1
Rate = 7.5 x 10^-3 mol L^-1s^-1
Rate 7.5 x 10³ mol L^-1 s^-1
k = \(\frac { Rate }{ { \left[ M \right] }^{ n } }\)
k= =\(\frac { 7.5\times { 10 }^{ -3 } }{ { \left( 0.05 \right) }^{ 2 } }\) = 3 mol^-1 Ls^-1

Question 12.
For a reaction x +y + z → products, the rate law is given by rate = k [x]^3/2 [y]^1/2 what is the overall order of the reaction and what is the order of the reaction with respect to z.
Solution:
Rate = k [x]^3/2 [y]^1/2
overall order = \(\left( \frac { 3 }{ 2 } +\frac { 1 }{ 2 } \right)\) = 2
i.e., second order reaction.
Since the rate expression does not contain the concentration of Z , the reaction is zero order with respect to Z.

Question 13.
Explain briefly the collision theory of bimolecular reactions.
Answer:

• Collision theory was proposed independently by Max Trautz in 1916 and William Lewis in 1918.
• According to this theory, chemical reactions occur as a result of collisions between the reacting molecules.
• Consider the reaction A2(g) + B2(g) 4 2AB(8)
• The rate of this reaction would be proportional to the number of collisions per second between A₂ and B₂.
  Rate α number of molecules colliding per second (collision rate)
• The number of collisions is directly proportional to the concentration of both A₂ and B₂.
  Collision rate α [A₂] [B₂]
  Collision rate = Z [A₂] [B₂] where Z is a constant.
• Collision rate in gases can be calculated from kinetic theory of gases.
• For a gas at room temperature (298K) and 1 atm. pressure each molecule undergoes approximately 10⁹ collisions per second. i.e. 1 collision in 10^-9 second.
• Thus if every collision resulted in reaction, the reaction would he complete in 10^-9 second.
• But in actual practice this does not happen.
• It implies that all collisions are not effective to lead to the reaction.
• In order to react the colliding molecules must possess minimum energy called activation energy.
• The molecules that collide with less energy than activation energy will remain intact and no reaction occurs.
• Fraction of effective collisions f = ^Ea/RT
• For a reaction having activation energy 100 KJ mol^– at 300 K collision factor f is \(\mathrm{f}=\mathrm{e}^{-40} \cong 4 \times 10^{-18}=\frac{4}{10^{18}}\)
• Thus out of 10 collision only 4 collisions are sufficiently energetic to convert reactants to products.
• This fraction of collision is further reduced due to orientation factor.
• That is even if the reactant collide with sufficient energy, they will not react unless the orientation of the reactant molecules is suitable for the formation of the transition state.

Fraction of effective collisions (O having proper orientation is given by the steric factor P.
Rate = p x f x Collision rate
Rate = pe ^Ea/RT [A₂] [B₂] …………… (1)
Rate law is Rate = k [A₂] [B₂] ………………..(2)
Comparing (1) & (2) k = pze ^Ea/RT

Question 14.
Write Arrhenius equation and explains the terms involved.
Answer:
Arrhenius equation:
k = A\({ e }^{ \frac { { -E }_{ a } }{ RT } }\)
A = Arrhenius factor (frequency factor)
R = Gas constant
k = Rate constant
E_a = Activation energy
T = Absolute temperature (in K)

Question.15.
The decomposition of Cl₂O₇ at 500K in the gas phase to Cl₂ and O₂ is a first order reaction. After 1 minute at 500K, the pressure of Cl₂O₇ falls from 0.08 to 0.04 atm. Calculate the rate constant in s^-1.
Answer:
Solution:
k = \(\frac { 2.303 }{ t }\) log \(\frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}\)
k = \(\frac { 2.303 }{ 1 min }\) log \(\frac { [0.08] }{ [0.04] }\)
k = 2.303 log 2
k = 2.303 x 0.3010
k = 0.693 2 min^-1
k = \((\frac { 0.6932 }{ 60 })\) s-1
k = 1.153 x 10^-2 s^-1

Question 16.
Give examples for a zero-order reaction.
Answer:
Examples for a zero-order reaction:
1. Photochemical reaction between H₂ and Cl
H_2(g) + Cl_2(g) \(\underrightarrow { h\nu }\) 2HCI_(g)

2. Decomposition of N₂O on hot platinum surface
N₂ O_(g) \(\rightleftharpoons\) N_2(g) + \(\frac { 1 }{ 2 }\) O_2(g)

3. iodination of acetone in acid medium is zero-order with respect to iodine.
CH₃COCH₃ + I₂ \(\underrightarrow { { H }^{ + } } \) ICH₂COCH₃ + HI
Rate k [CH₃COCI₃] [H⁺]

Question 17.
Explain pseudo-first-order reaction with an example.
Answer:
A second-order reaction can be altered to a first-order reaction by taking one of the reactants in large excess, such reaction is called pseudo-first-order reaction. Let us consider the acid hydrolysis of an ester,
CH₃COOCH_3(aq) +H₂ O₍₁₎ \(\underrightarrow { { H }^{ + } } \) CH₃COOH_(aq) + CH₃OH_(aq)
Rate = k [CH₃COOCH₃] [H₂O]

If the reaction is carried out with a large excess of water, there is no significant change in the concentration of water during hydrolysis. i.e., the concentration of water remains almost constant. Now we can define k [H₂O] = k
∴ The above rate equation becomes
Rate k [CHCOOCH] Thus it follows first-order kinetics.

Question 18.
Identify the order for the following reactions
i) Rusting of Iron
ii) Radioactive disintegration of ₉₂U²³
iii) 2A+ B → products; rate = k [A]^1/2 [B]²
Answer:
i) Order of Rusting of iron.
Since Rusting of iron is a very slow reaction, it is difficult to determine its rate. Hence it is difficult to write the rate law and difficult to predict the order of rusting of iron.
ii) radioactive disintegrations ₉₂U²³⁸ first order reactions
iii) 2A + 3B → products:
rate = k[A]^1/2 [B]²
Order = \(\frac { 1 }{ 2 }\) + 2 = \(\frac { 5 }{ 2 }\) = 2.5

Question 19.
A gas phase reaction has energy of activation 200 kJ mol^-1. If the frequency factor of the reaction is 1.6 x 10¹³ s^-1. Calculate the rate constant at 600 K. (e^-40.09 = 3.8 x I0^-18 )
Solution:
Ea = 200KJmol^-1 = 200 x 10³Jmol^-1
A = 1.6 x 10¹³s^-1
T= 600K
R = 8.314Jkmol^-1
e^-40.09= 3.8 x 10^-18
k=Ae ^-Ea/RT
\(=1.6 \times 10^{13} \times \mathrm{e}^{\frac{-200 \times 10^{3}}{8.314 \times 600}}\)
= 1.6 x 10¹³ x e^-40.09
=1.6 x 10¹³ X 3.8 x 10^-18
k = 6.08 x 10^-5s^-1

Question 20.
For the reaction 2x +y → L find the rate law from the following data.

Solution:
Rate = k [x]ⁿ [y]^m
0.15 = k [0.2]ⁿ [0.02]^m ……………..(1)
0.30 = k [0.4]ⁿ [0.02]^m ……………… (2)
1.20 = k [0.4]ⁿ [0.08]^m ……………… (3)

Question 21.
How do concentrations of the reactant influence the rate of reaction?
Answer:
The rate of a reaction increases with the increase in the concentration of the reactants. The effect of concentration is explained on the basis of the collision theory of reaction rates.

According to this theory, the rate of a reaction depends upon the number of collisions between the reacting molecules. The higher the concentration, the greater is the possibility for collision and hence the rate.

Question 22.
How does the nature of the reactant influence rate of reaction?
Answer:
Rate ∝ Concentration of reactants.
Rate ∝ number of collisions between the reacting molecules.
∴ Concentration reactants a number of collisions.

As the concentration of reactants increases, the number of collisions between the reacting molecules increases, and hence the rate of reaction increases.
A chemical reaction involves breaking of certain existing bonds of the reactant and forming new bonds which lead to the product.

Net energy involved in this process is dependent on the nature of the reactant.
Hence the rates of reaction are different for different reactants.
(eg.) Redox reaction between ferrous ammonium sulphate and KMnO₄ is fast and the pink colour of KMnO₄ disappears immediately.

But the redox reaction between Oxalic acid and KMnO₄ is relatively slow compared to the above reaction. In fact, heat is required for this reaction and the reaction is carried out at 60°C. On heating, the pink colour of KMnO₄ disappears.
Hence the rate of the reaction depends on the nature of the reactant.

Question 23.
The rate constant for a first order reaction is 1.54 x 10 s^-1. Calculate its half life time.
Solution:
We know that, t_1/2 = 0.693 k
t_1/2 = 0.693/1.54 x 10^-3 = 450 s

Question 24.
The half life of the homogeneous gaseous reaction SO₂Cl₂ → SO₂ + Cl₂ which obeys first order kinetics Is 8.0 minutes. How long will it take for the concentration of SO₂Cl₂ to be reduced to 1% of the initial value?
Answer:
We know that, k = 0.693/ t_1/2
k = 0.693/8.0 minutes = 0.087 minutes ^-1
For a first order reaction,
k = \(\frac { 2.303 }{ k }\) log \(\left( \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \right)\)
t = \(\frac { 2.303 }{ 0.087{ min }^{ -1 } }\) log\(\frac { 100 }{ 1 }\)
t = 52.93 mm

Question 25.
The time for half change in the first-order decomposition of a substance A is 60 seconds. Calculate the rate constant. How much of A will be left after 180 seconds?
Answer:
1. Order of a reaction = 1
t_1/2 = 60
seconds, k = ?
k = \(\frac { 2.303 }{ 60 }\)
We know that, k = \(\frac { 2.303 }{ { t }_{ 1/2 } }\)
k = \(\frac { 2.303 }{ 60 }\) = 0.01155 s^-1

2. [A₀] = 100%
t = 180 s
k = 0.01155 seconds^-1
[A] = ?
For the first order reaction k = \(\frac { 2.303 }{ 60 }\) log \(\left( \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \right)\)
0.9207 = log 100 – log [A]
log [A] = log 100 – 0.9207
log [A] = 2 – 0.9207
log[A] = 1.0973
[A] = antilog of (1.0973)
[A] = 12.5%
After 180 seconds 12.5% of A will be left over.

Question 26.
A zero-order reaction is 20% complete in 20 minutes. Calculate the value of the rate constant. In what time will the reaction be 80% complete?
Answer:
1. A = 100%, x = 20%, Therefore, a – x =100 – 20 = 80
For the zero order reaction k= \((\frac { x }{ t })\) ⇒
k = \((\frac { 20 }{ 20 })\) = 1
Rate constant for a reaction = 1

2. To calculate the time for 80% of completion
k = 1, a = l00, x = 80%, t = ?
Therefore, t = \((\frac { x }{ k })\) = \((\frac { 80 }{ 1 })\) = 80 min

Question 27.
The activation energy of a reaction is 225 k cal mol^-1 and the value of rate constant at 40°C is 1.8 x 10^-5 s^-1. Calculate the frequency factor, A. Here, we arc given that
Answer:
E_a = 22.5 kcal mol^-1 = 22500 cal mol^-1
T = 40°C = 40 + 273 = 313 K
k = 1.8 x 10^-5 sec^-1
Substituting the values in the equation
log A = log k + \(\left( \frac { { E }_{ a } }{ 2.303RT } \right)\)
log A = log (l .8 x 10^-5) + \(\left( \frac { 22500 }{ 2.303\times 1.987\times 313 } \right)\)
log A = log (l.8) – 5 + (15.7089)
log A = (10.9642)
A = antilog ( 10.9642)
A = 9.208 x 10¹⁰ collisions s^-1

Question 28.
Benzene diazonium chloride in aqueous solution decomposes according to the equation C₆H₅N₂CI → C₆H₅CI + N₂. Starting with an initial concentration of 10 g L^-1 volume of N₂. gas obtained at 50°C at different intervals of time was found to be as under:

Show that the above reaction follows the first order kinetics. What is the value of the rate constant ?
Solution:
For a first order reaction
k = \(\frac { 2.303 }{ t }\) log \(\frac { a }{ (a-x) }\)
k = \(\frac { 2.303 }{ t }\) log \(\frac { { V }_{ \infty } }{ { V }_{ \infty }-{ V }_{ t } }\)
In this case, V_∞ = 58.3 ml
The value of k at different time can be calculated as follows:

Since the value of k comes out to be nearly constant, the given reaction is of the first order. The mean value of k = 0.0674 min^-1

Question 29.
From the following data, show that the decomposition of hydrogen peroxide is a reaction of the first order:

Where t is the volume of standard KMnO₄ solution required for titrating the same volume of the reaction mixture.
Solution:
Volume of KMnO₄ solution used Amount of H₂O₂ present. Hence if the given reaction is of the first order, it must obey the equation
k = \(\frac { 2.303 }{ t }\) log \(\frac { a }{ (a-x) }\)
k = \(\frac { 2.303 }{ t }\) log \(\frac { { V }_{ 0 } }{ { V }_{ t } }\)
In this case,V₀ = 46.1 ml
The value of k at each instant can be calculated as follows:

Thus, the value of k comes out to be nearly constant. Hence it is a reaction of the first order. The mean value of k = 0.04355min^-1

Question 30.
A first order reaction is 40% complete in 50 minutes. Calculate the value of the rate constant. in what time will the reaction be 80% complete?
Answer:
1. For the first order reaction k = \(\frac { 2.303 }{ t }\) log \(\frac { a }{ (a-x) }\)
Assume, a = 100 %, x = 40%, t = 50 minutes
Therefore, a – x = 100 – 40 = 60
k = (2.303/50) log (100/60)
k = 0.010216 min^-1
Hence the value of the rate constant is 0.010216 min^-1

2. t = ?, when x = 8O%
Therefore, a – x = 100 – 80 = 20
From above, k = 0.0102 16 min^-1
t = (2.303 / 0.010216) log (100 / 20)
t = 157.58 min
The time at which the reaction will be 80% complete is 157.58 min.

III. Evaluate Yourself

Question 1.
Write the rate expression for the following reactions, assuming them as elementary reactions.
i) 3A + 5B₂ → 4CD
ii) X₂ + Y₂ → 2XY
Answer:
1. 3A + 5B₂ → 4CD
Rate = – \(\frac { 1 }{ 3 }\) \(\frac { \triangle [A] }{ dt }\)
= – \(\frac { 1 }{ 5 }\) \(\frac { \triangle [{ B }_{ 2 }] }{ dt }\)
= + \(\frac { 1 }{ 4 }\) \(\frac { \triangle [CD] }{ dt }\)

2. X₂ + Y₂ → 2XY
Rate = – \(\frac { \triangle [{ X }_{ 2 }] }{ dt }\)
= + \(\frac { 1 }{ 2 }\) \( [latex]\frac { \triangle [{ XY }_{ 2 }] }{ dt }\)

Question 2.
Consider the decomposition of N₂O_5(g) to form NO_2(g) and O_2(g). At a particular instant N₂O₅ disappears at a rate of 2.5 x 10^-2 mol dm^-3 s^-1. At what rates are NO₂ and O₂ formed? What is the rate of the reaction?
Solution:
2N₂O_5(g) → 4NO_2(g) + O_2(g)
from the stoichiometry of the reaction.
– \(\frac { 1 }{ 2 }\) \(\frac { d[{ N }_{ 2 }{ O }_{ 5 }] }{ dt }\)
= \(\frac { 1 }{ 4 }\) \(\frac { d[{ N }{ O }_{ 2 }] }{ dt }\)
= –\(\frac { d[{ N }{ O }_{ 2 }] }{ dt }\)
= 2 –\(\frac { d[{ N }_{ 2 }{ O }_{ 5 }] }{ dt }\)
Rate of disappearance of N₂O₅ is 2.5 x 10^-2 mol dm^-3 s^-1
∴ The rate of formation of NO2 at this temperature is 2 x 2.5 x 10^-2 = 5 x 10^-2 mol dm^-3 s^-1.
– \(\frac { 1 }{ 2 }\) \(\frac { d[{ N }_{ 2 }{ O }_{ 5 }] }{ dt }\)
= – \(\frac { d[{ O }_{ 2 }] }{ dt }\)
∴ \(\frac { d[{ O }_{ 2 }] }{ dt }\) = \(\frac { 1 }{ 2 }\) x 2.5 x 10^-2 mol dm^-3 s^-1
= 1.25 x 10^-2 mol dm^-3 s^-1

Question 3.
For a reaction, X + Y → Product quadrupling [x], increases the rate by a factor of 8. Quailrupling both [x] and [y] increases the rate by a factor of 16. Find the order of the reaction with respect to x and y. what is the overall order of the reaction?
Solution:
X + Y → Product
Rate of Law is Rate = k[x]^m[y]ⁿ
Let us take rate = 1 ∴ k[x]^m [y]ⁿ …..(1)
Quailrupling [x] ie [4x], Rate = 8
∴ k[4x]^m[y]ⁿ = 8
∴ k^m4^m [x]^m[y]ⁿ = 8 ……………..(2)
Quadrupling [x] and [y] ie [4x] and [4y], Rate = 16
k[4x]^m[y]ⁿ = 16
k4^mx^m4ⁿ[y]ⁿ = 16 …………..(3)
Dividing (2) by (1)

(2²)^m = 2³
(2^2m ) = 2³
∴ 2m = 3
m = 3/2 = 1.5
∴ Order of the reaction w.r.t. x is 1.5
Dividing (3) by (2)

4ⁿ = 2
(2²)ⁿ = 2¹
2²ⁿ = 2¹
∴ 2n=1
n = 1/2
n =0.5
∴ Order of the reaction w.r.t. y is 0.5
∴ Overall order of the reaction = m + n
=1.5 + 0.5 = 2

Question 4.
Find the individual and overall order of the following reaction using the given data.
2NO(g) + Cl₂ (g) → 2NOCl(g)

Solution:
Rate = k [NO]^m [CI₂]
For experiment 1, the rate law is,
Rate₁ = k [NO]^m [CI₂]ⁿ
7.8 x 10^-5 k[0.1]^m [0.1]ⁿ ………………(1)

For experiment 2, the rate law is.
Rate₂ = k [NO]^m [CI₂]ⁿ
3.12 x 10^-4 = k[O.2]^m [0.1]ⁿ ……………….(2)

For experiment 3, the rate law is,
Rate₃ = k [NO]^m [CI₂]ⁿ
9.36 x 10^-4 = k [O.2]^m [0.3]^m ……………(3)
Dividing Eq (2) by Eq (l) we get,

4 = \(\frac { 0.2 }{ 0.1 }\)^m
⇒ 2² = 2^m
∴ m = 2
Therefore the reaction is secondary order with respect to NO.
Dividing Eq (3) by Eq (2) we get,

Therefore the reaction is first order with respect to Cl₂
The rate law is, Rate = k [NO]₂ [Cl₂]¹
The overall order of the reaction (2 +1) = 3.

Question 5.
In a first order reaction A → products, 60% of the given sample of A decomposes in 40 min. what is the half life of the reaction?
Solution:
k = \(\frac { 2.303 }{ t }\) log \(\frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}\)
k = \(\frac { 2.303 }{ 40min }\) log \(\frac { 100 }{ (100-60) }\)
k = 0.0575 (0.3979) ⇒ k = 0.02287 min^-1
t_1/2 = \(\frac { 0.6932 }{ k }\) log \(\frac { 0.6932 }{ 0.02287 }\)
t_1/2 = 30.31 min.

Question 6.
The rate constant for a first order reaction is 2.3 x 10^-4 s^-1. If the initial concentration of the reactant is 0.01 M. what concentration will remain after 1 hour?
Solution:

Question 7.
Hydrolysis of an ester in an aqueous solution was studied by titrating the liberated carboxylic acid against sodium hydroxide solution. The concentrations of the ester at different time intervals are given below.

Show that, the reaction follows first order kinetics.
Solution:

Since all the k values are constant
This value shows that reaction follws first order kinetics.

Question 8.
For a first order reaction the rate constant at 500K is 8 x 10^-4 s^-1. Calculate the frequency factor, if the energy of activation for the reaction is 190 kJ mol^-1.
k = 8 x 10^-4s
T = 500K
E_a = 190 kJ mol^-1 A = ?
Solution:

12th Chemistry Guide Chemical Kinetics Elements Additional Questions and Answers

I. Match the following 

Question 1.

Answer:
i. b. molL^-1s^-1
ii. c. s^-1
iii. d. mol^-1Ls^-1
iv. a. mol^-2L^-2s^-1

Question 2.

Answer:
i) c. lowers the activation energy
ii) a. can be negative
iii) d. can not be fraction
iv) b. independent of initial concentration

II. Assertion and Reason

i) Both A and R are correct and R explains A.
ii) A is correct but R is wrong.
iii) A is wrong but R is correct.
iv) Both A and R are correct but R does not explain A.

Question 1.
Assertion (A): Rate of a reaction can be
\(\text { written as }-\frac{\mathrm{d}[\text { Reactant }]}{\mathrm{dt}}\)
Reason (R): As time increases concentration of reactant decreases.
Answer:
(i) Both A and R are correct and R explains A.

Question 2.
Assertion (A) : For a gas phase reaction the unit of reaction rate is atm s^-1
Reason (R) : For a gas phase reaction, the concentration of the gaseous species is expressed in terms of mole per litre.
Answer:
(ii) A is correct but R is wrong.
Correct R: For a gas phase reaction, the concentration of the gaseous species is expressed in atm.

Question 3.
Assertion (A) : For a general reaction xA + yB → products, rate law is Rate = k[A]^m [B]ⁿ
Reason (R) : The values of m and n can be deduced from the stoichiometry of the reaction.
Answer:
(ii) A is correct but R is wrong.
Correct R: The values of m and n can be deduced from experiments.

Question 4.
Assertion (A) : Acid hydrolysis of an ester is a pseudo first order reaction.
Reason (R) : The rate of the reaction does not depend on the concentration of one of the reactants water.
Answer:
i) Both A and R are correct and R explains A.

III. Choose the correct statements.

Question 1.
(i) During the reaction the concentration of the reactant increases.
(ii) Unit of rate = Unit of concentration (iii) For a gas phase reaction the unit of rate is atm s^-1
(a) (i) & (ii) (b) (i) & (iii) c) (ii) &(iii) d) (iii) only
Answer:
(d) (iii) only
Correct statement: (i) During the reaction the concentration of the reactant decrease.
Unit of concentration Unit of time

Question 2.
(i) The rate of a reaction decreases with time as the reaction proceeds.
(ii) The average rate can be used to predict the rate of the reaction at any instant.
(iii) The rate of the reaction at a particular instant during the reaction is called the instantaneous rate.
(a) (i) & (ii)
(b) (i) & (iii)
c)) (ii) & (iii)
(d) (i), (ii) & (iii)
Answer:
(b) (i) & (iii)
Correct statement:
(ii) The average rate cannot be used to predict the rate of the reaction at any instant.

Question 3.
(i) Rate constant is a proportionality constant
(ii) Rate of a reaction depends on the initial concentration of the reactants.
(iii) Rate constant does not depend on the initial concentration of the reactants.
(a) (i) & (ii)
(b) (i) & (iii)
(c) (ii) & (iii)
(d) (i), (ii) & (iii)
Answer:
(d) (i), (ii) & (iii)

Question 4.
(i) Order is assigned for each elementary step of mechanism.
(ii) Order can be fractional.
(iii) Molecularity can not be fractional.
(a) (i) & (ii)
(b) (i) & (iii)
(c) (ii) & (iii)
(d) (i), (ii) & (iii)
Answer:
(c) (ii) & (iii)
Correct statement: (i) Molecularity is assigned for each elementary step of mechanism.

IV. Choose the incorrect statements.

Question 1.
(i) All radio active reactions are first order
reaction.
(ii) Half life of first order reaction depends on the initial concentration of the reactants.
(iii) For a first order reaction a plot of ln[A] against t gives a straight line with positive slope.
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (i), (ii) & (iii)
Answer:
(c) (ii) & (iii)
Correct statement:
(ii) Half life of first order reaction does not depend on the initial concentration of the reactants.
(iii) For a first order reaction a plot of ln[A] against t gives a straight line with negative slope.

Question 2.
(i) Acid hydrolysis of an ester is a second order reaction.
(ii) Iso merisation of cyclo propane to propene is a zero order reaction.
(iii) In acid hydrolysis of ester the concentration of ester remains constant.
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (i), (ii) & (iii)
Answer:
d) (i), (ii) & (iii)
Correct statement:
(i) Acid hydrolysis of an ester is a Pseudo first order reaction
(ii) Isomerisation of cyclo propane to propene is a first order reaction.
(iii) In acid hydrolysis of ester the concentration of water remains constant.

Question 3.
(i) A zero order reaction is independent of the
concentration of the reactants.
(ii) Iodination of acetone in acid medium is zero order with respect to acetone.
(iii) Zero order reactions are quite common.
(a) (i) & (ii)
(b) (i) & (iii)
(c) (ii) & (iii)
(d) (i), (ii) & (iii)
Answer:
(c) (ii) & (iii)
Correct statement:
(ii) Iodination of acetone in acid medium in zero order with respective to iodine.
(iii) Zero order reaction are rare.

Question 4.
(i) Rate of a reaction is inversely propor-tional
to the number of collisions per second.
(ii) Collision rate in gases can be calculated from kinetic theory of gases.
(iii) Number of collisions is inversely proportional to the concentration of the reactants.
(a) (i) & (ii)
(b) (i) & (iii)
(c) (ii) & (iii)
(d) (i), (ii) & (iii)
Answer:
(b) (i) & (iii)
Correct statement:
(i) Rate of a reaction is directly proportional to the number of collision per second.
(iii) Number of collisions is directly proport¬ional to the concentration of the reactants.

V. Choose the best Answer

Question 1.
As the reaction proceeds the concentration of the reactant
a) increases
b) decreases
c) remains the same
d) tends to be maximum
Answer:
b) decreases

Question 2.
For a general reaction xA + yB → Ic + mD regarding its rate which is incorrect

Answer:
b

Question 3.
The rate constant of a reaction is 2.3 x 10^-2 lit mol^-1 S^-1
The order of the reaction is
a) Zero order
b) First order
c) Second order
d) Third order
Answer:
c) Second order

Question 4.
The sum of exponential terms in the rate law is called as
a) Molecularity
b) Rate constant
c) Order
d) Rate
Answer:
c) order

Question 5.
The value of order of a reaction can be
a) Zero
b) Fractional
c) Integer
d) all the above
Answer:
d) all the above

Question 6.
The value of molecularity of a reaction can be
a) Zero
b) Fractional
c) whole number
d) all the above
Answer:
c) whole number

Question 7.
During the decomposition of H₂O₂ (H₂O₂ → 2H₂O + O₂). 48g of O₂ is formed per Minute at particular instant. The rate of formation of water at this instant in mol min^-1 is
a) 1
b) 2
c) 3
d) 4
Answer:
c) 3

Question 8.
If the rate law of a reaction is Rate = k[A]^3/2[B]¹ the order of the reaction with respect to A is
a) 1.5
b) 1
c) 2.5
d) 3
Answer:
a) 1.5

Question 9.
If the rate law of a reaction is Rate = k[A]^3/2[B]¹ the order of the reaction with respect to B is
a) 1.5
b) 1
c) 2.5
d) 3
Ans:
b) 1

Question 10.
If the rate law of a reaction is Rate = k[A]^3/2[B]¹ the overall order of the reaction is
a) 1.5
b) 1
c) 2.5
d) 3
Answer:
c) 2.5

Question 11.
If the rate law for a reaction is Rate = k[A]² [B]°, the unit of rate constant k for the
overall reaction is
(a) molL^-1V
(b) mol^-1LS^-1
(c) mol^-2L^-2S^-1
(d) S^-1
Answer:
(b) mol^-1LS^-1

Question 12.
Which among the following is the unit of k of a zero order reaction ?
(a) mol^-1L^-1S^-1
(b) mol^-1LS^-1
(c) mol^-2L^-2S^-1
(d) S^-1
Answer:
(a) mol^-1L^-1S^-1

Question 13.
The rate law of a reaction is Rate = k[A] [B], if the concentration of B is taken in large excess, the order of the reaction is
a) zero order
b) first order
c) pseudo first order
d) second order
Answer:
c) pseudo first order

Question 14.
The rate determining step of a reaction is ……………. step.
a) fast
b) slow
c) equilibrium
d) intermediate
Answer:
b) slow

Question 15.
From the following which is a second order rate constant?
a) k = 5.47 x 10^-4s^-1
b) k = 3.9 x 10^-3mol lit s^-1
c) k = 3.94 x 10^-4 lit mol ^-1 s^-1
d) k = 3.98 x 10^-5 lit mol^-2s^-1
Answer:
c) k = 3.94 x 10^-4 lit mol ^-1 s^-1

Question 16.
Total number of reactant species that are involved in an elementary step is called as
a) order
b) molecularity
c) rate
d) rate constant
Answer:
b) molecularity

Question 17.
For an elementary reaction 2A + B → C + D the molecularity is
a) zero
b) one
c) two
d) three
Answer:
d) three

Question 18.
Which is correct for first order reactions,
a) t 1/2 ∝ (concn)^-1
b) t 1/2 ∝ (concn)
c) t 1/2 ∝ (concn)⁰
d) t 14/2 ∝ (concn)^1/2
Answer:
c)

Question 19.
Time required for the reactant concentration to reach one half of its initial value is called
a) Half life period
b) First order
c) zero order
d) Second order
Answer:
a) Half life period

Question 20.
Molecularity can be determined by
a) Stoichiometry
b) experiment
c) mechanism
d) none of the above
Answer:
c) mechanism

Question 21.
Rate constant of a reaction is equal to the rate of the reaction when the concentration of the reactants is
a) zero
b) unity
c) infinity
d) fractional
Answer:
b) unity

Question 22.
For a first order reaction a plot of /n[A] against Y yields a straight line with a slope which is
a) positive
b) negative
c) zero
d) at infinity
Answer:
b) negative

Question 23
Which of the following is not a first order reaction ?
i) Decomposition of dinitrogen pentoxide.
ii) Decomposition of N20 on hot platinum surface.
iii) Decomposition of thionyl chloride.
a) (i) only
b) (i) & (ii)
c) (ii) only d) (ii) & (iii)
Answer :
c) (ii) only

Question 24.
Which of the following are zero order reaction ?
i) Isomerisation of cyclo propane to propene.
ii) Iodination of acetone in acid medium with respect to Iodine.
iii) Photo chemical reaction between H₂ and Cl₂
a) (i) only
b) (i) & (ii)
c) (ii) only
d) (ii) & (iii)
Answer:
d) (ii) & (iii)

Question 25.
Acid hydrolysis of an ester is an example for ……………. order reaction.
a) zero
b) first
c) pseudo first
d) second
Answer:
c) pseudo first

Question 26.
Base hydrolysis of an ester is an example for ………….. order reaction
a) zero
b) first
c) pseudo first
d) second
Answer:
d) Second

Question 27.
The half life period of a first order reaction is

Answer:
d) All the above

Question 28.
If the initial concentration of the reactant is doubled, the half life period of a first order reaction is
a) doubled
b) tripled
c) quadrupled
d) remains the same
Answer:
d) remains the same

Question 29.
The half life period of a zero order reaction is

Answer:
c)

Question 30.
If tiie initial concentration of the reactant is doubled, the half life period of a zero order reaction is
a) doubled
b) tripled
c) quadrupled
d) remains the same
Answer:
a) doubled

Question 31.
The half life period of a second order reaction is

Answer:
c)

Question 32.
The half life period of a first order reaction is 5 minutes, the time required for 99.9% completion is nearly equal to
a) 99.9 minu tes
b) 49.95 minutes
c) 50 minutes
d) 10 minutes
Answer:
c) 50 minutes

Question 33.
The time required for 99.9% completion of a first-order reaction is equal to
a) 2t_1/2
b) 5t_1/2
c) 10t_1/2
d) 100t_1/2
Answer:
c) 10t_1/2

Question 34.
Collision theory was proposed independently by
a) MaxTrautz
b) William Lewis
c) both (a) & (b)
d) none of the above
Answer:
c) both (a) & (b)

Question 35.
Collision theory is based on
a) Arrhenius theory
b) Kinetic theory of gases
c) Ostwald theory
d) Gas laws
Answer:
b) Kinetic theory of gases

Question 36.
In order to react the colliding molecules must possess a minimum energy called
a) Kinetic energy
b) activation energy
c) potential energy
d) bond energy
Answer:
b) activation energy

Question 37.
The rate of a reaction ………… with increasing temperature
a) increases
b) decreases
c) becomes zero
d) remains the same
Answer:
a) increases

Question 38.
For many reactions near room temperature, reaction rate tends to …………….. when the temperature is increased by 10 C.
a) decreases
b) double
c) triple
d) quaruple
Answer:
b) double

Question 39.
The slope for the straight line obtained from the plot of Ink Vs 1/T is T

Answer:
b)

Question 40.
The y intercept of the straight line obtained from the plot of Ink vs 1/T is
a) In k
b) \(-\frac{\mathrm{Ea}}{\mathrm{R}}\)
c) In A
d) \(-\frac{\mathrm{Ea}}{\mathrm{RT}}\)
Answer:
c) In A

Question 41.
Rate of a reaction …………….. with increase in concentration of the reactants
a) decreases
b) increases
c) remains the same
d) is slower
Answer:
b) increases

Question 42.
As concentration of the reactant increases the number of collisions between the molecules
a) decreases
b) increases
c) rem a ins the same
d) is slower
Answer:
b) increases

Question 43.
Increase in surface area of the reactant…………….the rate of the reaction
a) increases
b) decreases
c) no change
d) is slower
Answer:
a) increases

Question 44.
Increase in surface area of the reactant ………….. the number of collisions between the molecules.
a) increases
b) decreases
c) no change
d) is slower Ans :
a) increases

Question 45.
Generally addition of a catalyst ……………. the rate of a reaction
a)increases
b)decreases
c) inhibits
d) has no change
Answer:
a) increases

Question 46.
A catalyst ……………. the activation energy of a reaction.
a) increases
b) decreases
c) has no change
d) none of the above
Answer:
b) decreases

Question 47.
What is the rate law of the reaction 2A + 2B → C + 2D. If the concentration of A is doubled at constant [B] the rate of the reaction increases by factor 4. If the concentration of B is doubled at constant [A], the rate is doubled.
a) Rate = k [A] [B]²
b) Rate = k [A] [B]
c) Rate = k [A] ^1/2 [B]²
d) Rate = k [A]²[B]
Answer:
d) Rate = k [A]²[B]

Question 48.
The concentration of a reactant decreases from 0.5 M to 0.3 M in 10 minutes. The rate of the reaction is
a) 0.01mol L^-1 S^-1
b) 0.02 mol L^-1 S^-1
c) 0.08mol L^-1 S^-1
d) 0.15mol L^-1 S^-1
Answer:
b) 0.02 mol L^-1 S^-1

Question 49.
The rate of a reaction is doubled for every 10°C rise in temperature. The increase in reaction rate as a result of temperature rise from 10 C to 50°C is
a) 4
b) 8
c) 16
d) 32
Answer:
c) 16 Increase in reaction rate = ₂No 10°C rise
Froml0°C to 10°C = 4 times 10°C rise
∴ 2⁴ = 16

Question 50.
For the reaction 2N₂o₅ → 4N0₂ + O₂

a) k₁ = k₂ = k₃
b) k₁ = 2k₂ = 4k₃
c) 2k₁ = 4k₂ = k₃
d) 2k₁ = k₂ = 4k₃
Answer:
d) 2k₁ = k₂ = 4k₃

Question 51.
Rate law of a reaction A + B → C is Rate = k[A] [B] . If the concentration A and B are doubled at constant volume then the rate of the reaction will be increased by
a) two times
b) four times
c) eight times
d) sixteen times
Answer:
c) eight times
Rate = k[A][B]²
[A] & [B] are doubled
Rate = k[2A][2B]²
Rate = k₂[A]2²[B]²
Rate = 8 x k[A][B]²

Question 52.
If 60% of a first order reaction is complete in 60 minutes, its half life period is approximately (log4 = 0.6)
a) 50 minutes
b) 45 minutes
c) 60 minutes
d) 40 minutes
Answer:
b) 45 minutes

Question 53.
Half life period of a first order reaction is 1386 seconds. The rate constant of the reaction is
a) 5.0 x 10^-3s^-1
b) 0.5 x 10^-2s^-1
c) 0.5 x 10^-3s^-1
d) 5.0 x 10^-2s^-1
Answer:
c) 0.5 x 10^-3s^-1

Question 54.
For the reaction 2A + B → 3C + D
Which of the following does not express the reaction rate?

Answer:
c

Question 55.
The half life period of a first order reaction is 10 minutes. If the initial concentration is 0.08 mol L^-1 at what time the concentration will become 0.01 mol L ^-1
a) 10 minutes
b) 20 minutes
c) 30 minutes
d) 40 minutes
Answer:
c) 30 minutes
t_1/2 = 10minutes

Question 56.
For a reaction the rate law is Rate = k [A]^3/2[B]^-1/2 , the over all order of reaction is
a) 2
b) 1
c) -1/2
d) 3/2
Answer:
b) 1

Question 57.
A first order reaction is half completed in 45 minutes . How long does it need 99.9% of the reaction to be completed?
a) 5 hours
b) 7.5 hours
c) 10 hours
d) 20 hours
Answer:
b) 7.5 hours
t_99.9% = 10t_1/2 = 10 x 45
= 450 minutes = 450/60 = 7.5hours

Question 58.
If the activation energy for a reaction at TK is 2.303 RT J mol^-1, the ratio of rate constant to frequency factor is
a) 2 x 10^-3
b) 2 x 10^-2
c) 10^-1
d) 10^-2
Answer:
c) 10^-1
logk = log A – \(\frac{\mathrm{Ea}}{2.303 \mathrm{RT}}\)
l°gk = logA – \(\frac{2.303 \mathrm{RT}}{2.303 \mathrm{RT}}\)
logk = log A – 1
logk – log A = -1
\(\log _{10}\left(\frac{\mathrm{k}}{\mathrm{A}}\right)=-1\)
\(\therefore\left(\frac{\mathrm{k}}{\mathrm{A}}\right)=10^{-1}\)

Question 59.
Chemical reactions with very high Ea values are generally
a) very fast
b) very slow
c) moderately fast
d) spontaneous
Answer:
b) very slow

Question 60.
If the activation energy of a reaction is zero, the rate constant of this reaction
a) increases with increase in temperature
b) decreases with increase in temperature
c) decreases with decrease in temperature
d) is independent of temperature
Answer:
d)is independent of temperature
Ea = 0; k = \(\mathrm{Ae}^{\frac{-\mathrm{Ea}}{\mathrm{RT}}}\)
k = \(\mathrm{Ae}^{\frac{-\mathrm{O}}{\mathrm{RT}}}\)
k = Ae°
k = A.
Hence k is independent of temperature

VI. Two mark Questions

Question 1.
Define rate of a reaction.
Answer:
The change in concentration of a species involved in a chemical reaction per unit time is called the rate of the reaction.

Question 2.
Write the rate expression of the following reaction.
N₂ + 3H₂ → 2NH₃
Answer:

Question 3.
Why a negative sign is introduced in the rate expression ?
Answer:

During the reaction, the concentration of the reactant decreases ie. [A₂] < [A₁] and hence the change in concentration [A₂] – [A₁] gives a negative value. By convention the reaction rate is positive one and hence a negative sign is introduced in the rate expression.

Question 4.
Define Order of a reaction.
Answer:
Order of a reaction is the sum of powers of concentration terms involved in the experimentally determined rate law.

Question 5.
Define molecularity of a reaction.
Answer:
Molecularity of a reaction is the total number of reactant species that are involved in an elementary step of a reaction.

Question 6.
Mention the factors that affect the rate of the reaction.
Answer:

1. Nature and state of the reactant
2. Concentration of the reactant
3. Surface area of the reactant
4. Temperature of the reaction
5. Presence of a catalyst.

VII. Three mark Questions

Question 1.
Write the differences between rate and rate constant of a reaction.
Answer:
Rate of a reaction :

1. It represents the speed | at which the reactants I are converted into products at any instant.
2. It is measured as decre-ase in the concentration of the reactants or increase in the concentration of products.
3. It depends on the initial concentration of reactants.

Rate constant of a reaction:

1. It is a proportionality constant.
2. It is equal to the rate of reaction, when the concentration of each of the reactants is unity.
3. It does not depend on the initial concentration of reactants.

Question 2.
Give three examples for first order reaction.
i) Decomposition of dinitrogen pentoxide
N₂O_5(g) → 2NO_2(g) + \(\frac{1}{2}\)O_2(g)

ii) Decomposition of sulphuryl chloride
SO₂Cl_2(I) → SO_2(g) + Cl_(g)

iii) Decomposition of hydrogen peroxide in aqueous solution
H₂O_{2( aq)} → H₂O_2(I) + \(\frac{1}{2}\)O₂

Question 3.
Calculate the half -life period of zero order reaction.
Answer:
Let us calculate the half life period for a zero order reation.

Question 4.
For the general reaction A → B, Plot of concentration of A Vs time is given in the graph below.

Answer the following questions on the basis of this graph.
(i) What is the order of reaction?
(ii) What is the slope of the curve?
(iii) What is the unit of rate constant?
Answer:
(i) It is a zero order reaction as the graph is satisfying the equation [A] = [A₀] – kt
(ii) The slope of the curve is the negative of the rate constant that is denoted bv -k.
(iii) Unit of rate constant is Ms^-1 or mol L^-1 s^-1

VIII.Five mark Questions

Question 1.
Derive the integrated rate law for a first order reaction.
Answer:
Let us consider a first order reaction.
A → products
The rate of first order reaction depends on the reactant concentration raised to the first power.
∴ Rate law can be written as. Rate = k [A]¹
k is the first order rate constant

Integrating between the limits t = 0 to t t [A]
and concentration [A₀] to [A]


Converting natural logarithm to usual logarithm

Question 2.
Derive Arrhenius equation to calculate activation energy from the rate constant k₁ and k₂ temperature T₁ and T₂ respectively.
Answer:

• The rate of reaction generally increases with increase in temperature.
• For many reactions near room temperature, reaction rate tends to double when the temperature is increased by 10°C.
• Arrhenius suggested that the rates of most reactions vary with temperature.

He proposed that the rate constant is directly proportional to e RT
\\(\mathrm{e} \frac{-\mathrm{Ea}}{\mathrm{RT}}\)
k ∝ \(\mathrm{e} \frac{-\mathrm{Ea}}{\mathrm{RT}}\)
k = A\(\mathrm{e} \frac{-\mathrm{Ea}}{\mathrm{RT}}\)
k = Rate constant
A = Frequency factor
Ea = activation energy
R = Gas constant
T = Temperature in KelvinFrequency factor A is related to the frequency of collisions between the reactant molecules.

• Frequency factor A does not vary significantly with temperature and hence it may be taken as a constant.
• Ea the activation energy is the minimum energy that a molecule must have to posses to react.

k = A\(\mathrm{e} \frac{-\mathrm{Ea}}{\mathrm{RT}}\) ……… (1)
Taking logarithm Ln k = lnA+ lne \(\mathrm{e} \frac{-\mathrm{Ea}}{\mathrm{RT}}\)
lnk= ln A – \(\frac{-\mathrm{Ea}}{\mathrm{RT}}\) (∵ lne = 1)
T = T₁ , k = k₁
T = T₂ , k = k₂

This equation can be used to calculate F.a from rate constant k₁ and k₂ at temperature T₁ and T₂

Question 3.
Explain the factors affecting reaction rate.
Answer:
i) Nature and state of the reactant.

• A chemical reaction involves breaking of certain existing bonds of the reactant and forming new bonds which lead to the product. The net energy involved in this process is dependent on the nature of the reactant and hence the rates are different for different reactants.
• (eg)- Redox reaction between ferrous ammonium sulphate and KMnO₄ is fast and the pink colour of KMnO₄ disappears on cold.
• Redox reaction between oxalic acid and KMn0₄ is slow and on heating the reaction proceeds faster and the pink colour of KMn04 disappears on heating.
• Physical state of the reactant plays an important role on the rate of the reaction. Gas phase reactions are faster as compared to the reactions involving solid or liquid reactants.
• (eg). Sodium metal with iodine vapours react faster than solid sodium and solid iodine. Solid lead nitrate and solid potassium iodide react slowly than aqueous solution of both which gives yellow precipitate immediately.

ii) Concentration of the reactants.

• Rate of the reaction increases with increase in the concentration of the reactants.
• This can be explained on the basis of collision theory of reaction rates.
• Rate of reaction depends upon the number of collisions between the reacting molecules.
• As the concentration increases, the number of collisions increases and hence rate increases.

iii) Effect of surface area of the reactant.

• When the particle size decreases surface area increases.
• Increase in surface area of reactant leads to more collisions per litre per second.
• Hence the rate of reaction is increased.
• (eg) Powdered calcium carbonate reacts much faster with dilute HC1 than with the same mass of CaCO3 marble.

iv) Effect of temperature

• The rate of reaction generally increases with increase in temperature.
• For many reactions near room temperature reaction rate tends to double when the temperature is increased by 10°C.
• (cg) Magnesium reacts readily with hot water resulting in a basic solution indicated by the pink colour of phcnolphthalein than cold water.

v) Effect of Catalyst.

• A catalyst is a substance which alters the rate of a reaction without itself undergoing any permanent chemical change.
• A catalyst may participate in the reaction, but again regenerated at the end of the reaction.
• A catalyst lowers the activation energy hence greater number of molecules can cross the energy barrier and change over to products. Thus the rate of the reaction is increased.

IX. Problems based on rate constant

Question 1.
From the following data on N₂O₅ decomposition in CCl₄ at 298K show that the reaction is first order. Also evaluate the rate constant of the reaction.

Solution:


Since k values are nearly constant for different timings, this is a first order reaction.

Question 2.
From the following data, show that decomposition of H₂O₂ in aqueous solution follows first order reaction, what is the value of rate constant ?

Answer:

Since all the k values are nearly constant the reaction is first order.

Question 3.
In a first order reaction, it takes the reactant 40.5 minutes to be 25% decomposed. Calculate the rate constant
of the reaction.
Solution:
[A₀] = 100%
[A] = 100 – 25 = 75%
t = 40.5 mins
k = ?

Question 4.
The rate of formation of a dimer in a second order reaction is 7.5 x 10^-3 mol L^-1S^-1 at 0.05 mol L^-1 monomer concentration. Calculate the rate constant.
Solution:
Let us consider the dimerisation of a monomer M
2M → (M)₂
Rate = k[M]ⁿ
Given that n = 2 and [M]=0.05 mol L^-1
Rate = 7.5 x 10^-3mol L^-1 S^-1

X. Problems based on half life period

At 25°C the rate constant of a first order reaction is 0.45 s^-1. What is its half life? Calculate the time required for 12.5% of the reactant to remain.
Solution:
k = 0.45 s^-1
t_1/2 = ?
\(\mathrm{t}_{1 / 2}=\frac{0.6932}{\mathrm{k}}=\frac{0.6932}{0.45}=1.54 \mathrm{sec}\)

time needed to 12.5% of reactant to remain= 3t_1/2 = 3 x 1.54 = 4.62 sec

Question 2.
The rate constant of a first order reaction is 1.54 x 10^-3 s^-1. What is its half life?
k = 1.54 x 10^-3 sec^-1, t_1/2 = ?
Solution:
t_1/2 = \(\frac{0.6932}{1.54 \times 10^{-3}}\) = 450 sec

Question 3.
The half life period of a first order reaction is 10 mins, what percentage of the reactant will remain after one hour?
Solution:
t_1/2 = 10mins
t = 1hour = 60 mins = 6×10 min = 6 x t_1/2

After one hour 1.5625% of the reactant will remain.

Question 4.
A first order reaction is 25% complete in 100 minutes. Calculate its rate constant and half life.
Solution:
t =100mins
[A₀] = 100%
[A] = 100 – 25 = 75%
k = ?
t_1/2 = ?

Question 5.
75% of a first order reaction is completed in 48 minutes. What is its half life ?
Solution:
t = 48 mins
[A₀] =100%
[A] = 100-75 = 25% k = 25%
k = ?
t_1/2 = ?

Question 6.
Show that for a first order reaction the time required for 99% completion of the reaction is twice the time required for 90% completion of the reaction.
Solution:

XI. Problems based on activation energy

Question 1.
For a first order reaction if the rate constant at 25°C is 3.46 x 10^-5 s^-1 and the rate constant at 35°C is 13.50 x 10^-5 s^-1, calculate the activation energy Ea and frequency factor A.
Solution:
T₁ = 25°C + 273 = 298K
k₁ = 3.46 x 10^-5s^-1
T₁ = 35°C + 273 = 308K
k₁ = 3.46 x 10^-5s^-1

= log3.46 x log 10^-5 + 18.21
= log 3.46 + log 10^-5 + 18.21
= Iog3.46 – 51og10 + 18.21
= 0.5391-5 + 18.21
logA = 13.7491
A = Anti log 13.7491
A = 5.611 x 10¹³

Question 2.
The activation energy of a certain reaction is 100 kj mol^-1. What is the change in the rate constant of the reaction if the temperature is changed from 25°C to 35°C ? Let the rate constants at 25°C be k₁ and at 35°C be k₂ respectively.
Answer:
T₁ = 25°C + 273 = 298K =>K = k₁
T₂ = 35°C + 273 = 308K =>K = k₂
E_a = 100kjmol^-1 = 100 x 10³ Jmol^-1
R = 8.314Jk^-1mol^-1

∴ k₂, the rate constant at 35°C will be 3.70 times the rate constant k₁ at 25°c.

Question 3.
Decomposition of ethyl bromide and propyl bromide follow first order kinetics and have the same frequency factor ‘A’. The rate constant for decomposition of ethyl bromide at 390°C is same as that for propyl bromide at 320’C. If Ea of ethyl bromide reaction is 230 kj mol^-1. What is the Ea of propyl bromide reaction?
Solution:
log k = log A – \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}\)
For ethyl bromide decomposition at 390°C ie 663k
log k = log A – \(\frac{230 \times 10^{3}}{2.303 \times 8.314 \times 663}\) ………….(1)
For propyl bromide decomposition at 320°C ie 593k
logk = logA – \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \times 8.314 \times 593} …………………. (2)
Since k values for both reactions are equal equation (1) & (2) are equal.

Question 4.
The rate constant k values of a reaction were determined at several temperatures. A plot of In k vs 1/T gave a straight line with the slope – 2.6 x 10⁴ K. What is the activation energy Ea of the reaction?
Solution:
slope = [latex]-\frac{E_{a}}{R}\)
E_a = -Rslope
= -8.3 14 x (-2.6 x 10⁴)
= 21.61 x 10⁴ Jmol^-1
E_a = 216.1 kJmol^-1

Question 5.
The rate constant for a first order reaction at 45°C is twice that at 35°C. Find the activation energy of the reaction.
At T₁ = 35°C + 273 = 308K
rate constant k₁
T₂ =45°C + 273 = 318K rateconstanf k₂ =2k₁
R= 8.3 14Jk^-1 mol^-1  E_a = ?
Solution:

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Physics Guide Pdf Chapter 2 Current Electricity Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity

12th Physics Guide Current Electricity Text Book Back Questions and Answers

Part – 1

Text Book Evaluation

I. Multiple choice questions:

Question 1.
The following graph shows the current versus voltage values of some unknown conductor. What is the resistance of this conductor?

a) 2 ohm
b) 4 ohm
c) 8 ohm
d) 1 ohm
Answer:
a) 2 ohm
Solution:
V = IR
R = \(\frac{\mathrm{V}}{\mathrm{I}}\)
From graph,
\(\text { Slope }=R=\frac{\Delta V}{\Delta I}=\frac{(4-0)}{(2-0)}=\frac{4}{2}=2 \Omega\)

Question 2.
A wire of resistance 2 ohms per meter is bent to form a circle of radius 1m. The equivalent resistance between its two diametrically opposite points, A and B as shown in the figure is

a) πΩ
b) \(\frac{\pi}{2} \Omega\)
c) 2πΩ
d) \(\frac{\pi}{4} \Omega\)
Answer:
a) πΩ
Solution:
Circumference of circle = 2πr = 2 × π × 1 = 2π
Resistance of wire = 2 × 2π
Resistance of each section = \(\frac{4 \pi}{2}\) = 2π
Equivalent Resistance = \(\begin{array}{l}
\frac{2 \pi \times 2 \pi}{2 \pi+2 \pi} \\
\frac{4 \pi^{2}}{4 \pi}=\pi \text { ohm }
\end{array}\)

Question 3.
A toaster operating at 240 V has a resistance of 120 Ω. The power is
a) 400 W
b) 2 W
c) 480 W
d) 240 W
Ans :
c) 480 W
Solution:
P = VI
= \(\frac{V^{2}}{R}=\frac{240 \times 24 \not b}{12 \phi}\)
= 240 × 2 = 480W
P = 480W

Question 4.
A carbon resistor of (47 ± 4.7 ) kΩ to be marked with rings of different colours for its identification. The colour code sequence will be
a) Yellow — Green — Violet — Gold
b) Yellow  — Violet — Orange — Silver
c) Violet — Yellow — Orange — Silver
d) Green — Orange — Violet — Gold
Answer:
b) Yellow — Violet — Orange — Silver
Solution:
Colour code:
B B R O Y G B V G W
0 1 2 3 4 5 6 7 8 9
4 — yellow
7 — violet
1kΩ = 103 = Orange
b) is correct answer.

Question 5.
What is the value of resistance of the following resistor?

a) 100 kΩ
b) 10 kΩ
c) 1 kΩ
d) 1000 kΩ
Answer:
a) 100 kΩ
Solution:
Brown – 1
Black – 0
Yellow – 10⁴
10 × 10⁴ = 100 × 10³ = 100 kΩ

Question 6.
Two wires of A and B with circular cross-section are made up of the same material with equal lengths. Suppose R_A = 3R_B, then what is the ratio of radius of wire A to that of B?
a) 3
b) √3
c) \(\frac{1}{\sqrt{3}}\)
d) \(\frac{1}{2}\)
Answer:
c) \(\frac{1}{\sqrt{3}}\)
Solution:
Given R_A = 3_B
\(\begin{array}{l}
\qquad \rho=\frac{\pi r^{2} R}{l} \\
R \alpha \frac{1}{r^{2}} \\
R_{A}=\frac{1}{r_{1}^{2}} \quad R_{B}=\frac{1}{r_{2}^{2}} \\
\frac{1}{r_{1}^{2}}=3 \frac{1}{r_{2}^{2}} \Rightarrow \frac{r_{1}^{2}}{r_{2}^{2}}=\frac{1}{3} \\
\frac{r_{1}}{r_{2}}=\frac{1}{\sqrt{3}}
\end{array}\)

Question 7.
A wire connected to a power supply of 230 V has power dissipation P₁. Suppose the wire is cut into two equal pieces and connected parallel to the same power supply. In this case, power dissipation is P₂. The ratio of \(\frac{P_{2}}{P_{1}}\) is
a) 1
b) 2
c) 3
d) 4
Answer:
d) 4
Solution:

Question 8.
In India electricity is supplied for domestic use at 220 V. It is supplied at 110 V in the USA. If the resistance of a 60W bulb for use in India is R, the resistance of a 60W bulb for use in the USA will be
a) R
b) 2R
c) \(\frac{\mathrm{R}}{4}\)
d) \(\frac{R}{2}\)
Answer:
c) \(\frac{\mathrm{R}}{4}\)
Solution:
\(P=\frac{V^{2}}{R}\)
India, V = 220V, P = 60W, R = R
USA, V = 110V, P’ = 60W, R’ = ?
∴ R = \(\frac{\mathrm{V}^{2}}{\mathrm{P}}\)
P = P’
∴ \(\mathrm{R}^{\prime}=\frac{\mathrm{V}^{2}}{\mathrm{~V}^{2}}=\frac{110 \times 110}{220 \times 220}=\frac{1}{4} \quad {\mathrm{R}^{\prime}=\frac{\mathrm{R}}{4}}\)

Question 9.
In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W, and 1 heater of 1 kW are connected. The voltage of electric mains is 220 V. The minimum capacity of the main fuse of the building will be
a) 14 A
b) 8 A
c) 10 A
d) 12 A
Answer:
d) 12A
Solution:

Question 10.
There is a current of 1.0 A in the circuit shown below. What is the resistance of P?

a) 1.5 Ω
b) 2.5 Ω
c) 3.5 Ω
d) 4.5 Ω
Answer:
c) 3.5 Ω
Solution:
By Kirchhoff’s voltage law,
9 = (3 × 1) + (2.5 × 1) + (P × 1)
9 =3 + 2.5 + P
9 = 5.5 + P
P = 9 – 5.5
P = 3.5 Ω

Question 11.
What is the current drawn out from the battery?

a) 1 A
b) 2 A
c) 3 A
d) 4 A
Solution:

Question 12.
The temperature deficient of resistance of a wire is 0.00125 per°C At 20°C, its resistance is 1Ω. The resistance of the wire will be 2Ω at
a) 800°C
b) 700°C
c) 850°C
d) 820°C
Answer:
d) 820°C

Question 13.
The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω is
a) 0.2 Ω
b) 0.5 Ω
c) 0.8 Ω
d) 1.0 Ω
Answer:
b) 0.5 Ω
Solution:
\(r=\left(\frac{E-V}{V}\right) R\)
E = 2.1 V
V = IR = 0.2 × 10
V = 2 V
\(\begin{aligned}
r &=\left(\frac{2.1-2}{2}\right) 10 \\
&=\frac{0.1}{2} \times 10
\end{aligned}\)
r = 0.5 Ω

Question 14.
A piece of copper and another of germanium are cooled from room temperature to 80 K. The resistance of
a) each of them increases
b) each of them decreases
c) copper increases and germanium decreases
d) copper decreases and germanium increases
Answer:
d) copper decreases and germanium increases
Solution:
Copper is a positive temperature coefficient of resistance, i.e their resistance decreases with a decrease in temperature. Germanium is a negative temperature coefficient of resistance, i.e their resistance increases with a decrease in temperature.

Question 15.
In Joule’s heating law, when R and t are constant, if the H is taken along the y axis and I² along the x-axis, the graph is _______
a) straight line
b) parabola
c) circle
d) ellipse
Answer:
a) straight line
Solution:
H = I²RT
H α I²
∴ The graph is a straight line.

II. Short Answer Questions:

Question 1.
Why current is a scalar?
Answer:
The current I is defined as the scalar product of current density and area vector in which the charges cross.
I = \(\vec { j } \) . \(\vec { A } \)
The dot product of two vector quantity is a scalar form. Hence, the current is called a scalar quantity.

Question 2.
Define Current density.
Answer:
The current density is defined as the current per unit area of the cross-section of the conductor.
J = \(\mathrm{I} / \mathrm{A}\)
S.I unit is A/m²

Question 3.
Distinguish between drift velocity and mobility.
Answer:

Question 4.
State microscopic form of Ohm’s law.
Answer:
The macroscopic form of Ohm’s Law relates voltage, current, and resistance. Ohm’s Law states that the current through an object is proportional to the voltage across it and inversely proportional to the object’s resistance.
V = IR.

Question 5.
State macroscopic form of Ohm’s law.
Answer:
The macroscopic form of Ohm’s law is V = IR
Where V → Potential difference
I → Current
R → Resistance.

Question 6.
What are the ohmic and non-ohmic devices?
Answer:
Materials for which the current against voltage graph is a straight line through the origin, are said to obey Ohm’s law and their behaviour is said to be ohmic. Materials or devices that do not follow Ohm’s law are said to be non-ohmic.

Question 7.
Define electrical resistivity.
Answer:
The electrical resistivity of a material is defined as the resistance offered to current flow by a conductor of unit length having a unit area of cross-section.
\(\rho=\frac{R A}{L}\)
Its unit is ohm-meter (Ωm)

Question 8.
Define temperature coefficient of resistance.
Answer:
The ability of certain metals, their compounds, and alloys to conduct electricity with zero resistance at very low temperatures is called superconductivity.

Question 9.
What are superconductors?
Answer:

1. The resistance of certain materials becomes zero below a certain temperature, known as critical or transition temperature.
2. The materials that exhibit this property are known as superconductors and the phenomenon is known as superconductivity.

Question 10.
What are electric power and electric energy?
Answer:

Question 11.
Derive the expression for power P = VI in an electrical circuit.
Answer:
Power is defined as the rate at which the electrical potential energy is delivered.
\(\begin{array}{l}

Question 12.
Write down the various forms of expression for power in electrical circuit.
Answer:
The electric power P is the rate at which the electrical potential energy is delivered,
P = [latex]\frac { dU }{ dt }\) = \(\frac { 1 }{ dt }\) (V.dQ) = V.\(\frac { dQ }{ dt }\)
[dU = V.dQ]
The electric power delivered by the battery to any electrical system.
P = VI
The electric power delivered to the resistance R is expressed in other forms.
P = VI = I(IR) = I²R
P = IV = \((\frac { V }{ R })\) V = \(\frac {{ V }^{2}}{ R }\).

Question 13.
State Kirchhoff’s current rule.
Answer:

1. Kirchhoff’s current rule states that the algebraic sum of the currents at any junction of a circuit is zero.
   i.e Σ_junction i = 0
2. It is a statement of conservation of electric charge.

Question 14.
State Kirchhoff’s voltage rule.
Answer:
It states that in a closed circuit the algebraic sum of the products of the current and resistance of each part of the circuit is equal to the total emf included in the circuit. This rule follows from the law of conservation of energy for an isolated system.

Question 15.
State the principle of the potentiometer.
Answer:
The emf (E) of the cell is directly proportional to the balancing length (l)
ξ ∝ l
ξ = Irl
Where I →current
r → resistance per unit length of the wire.

Question 16.
What do you mean by the internal resistance of a cell?
Answer:
The resistance offered by the electrolyte of a cell to the flow of current between its electrodes is called internal resistance of the cell. An ideal battery has zero internal resistance and the potential difference across the battery equal to its emf. But a real battery is made of electrodes and electrolytes, there is resistance to the flow of charges within the battery. A freshly prepared cell has low internal resistance and it increases with aging.

Question 17.
State Joule’s law of heating.
Answer:
The heat developed in an electrical circuit due to the flow of current varies directly as

• the square of the current
• the resistance of the circuit and
• the time of flow H = I²Rt

Question 18.
What is the Seebeck effect?
Answer:
Seebeck discovered that in a closed circuit consisting of two dissimilar metals, when the junctions are maintained at different temperatures, an emf (potential difference) is developed.

Question 19.
What is the Thomson effect?
Answer:

1. When two points in a conductor are at different temperatures, the density of electrons at these points will differ and as a result, me potential difference is created between these points.
2. This is known as the Thomson effect. Thomson effect is also reversible.

Question 20.
What is the Peltier effect?
Answer:
When an electric current is passed through a circuit of a thermocouple, heat is evolved at one junction and absorbed at the other junction. This is known as the Peltier effect.

Question 21.
State the applications of the Seebeck effect.
Answer:

1. The Seebeck effect is used in thermoelectric generators (Seebeck generators). These thermoelectric generators are used in power plants to convert waste heat into electricity.
2. This effect is utilised in automobiles as automotive thermoelectric generators for increasing fuel efficiency.
3. The Seebeck effect is used in thermocouples and thermopiles to measure the temperature difference between the two objects.

III. Long Answer Questions:

Question 1.
Describe the microscopic model of current and obtain the general form of Ohm’s law.
Answer:

1. XY is a conductor of an area of cross-section A
2. E is the applied electric field.
3. n is the number of electrons per unit volume with the same drift velocity (v_d)

Let electrons move through a distance dx in time interval dt.

Microscopic model of the current
∴ \(v d=\frac{d x}{d t}\)
dx = v_d . dt …………(1)

Electrons available in the given volume
= volume × number per unit volume
= A. dx × n
= A . v_d dt × n [From (1)]

Total charge in volume dQ = charge × number of electrons
dQ = (e) (Av_d dt) n

Question 2.
Obtain the macroscopic form of Ohm’s law from its microscopic form and discuss its limitation.
Answer:

1. Consider a segment of wire of length l and area of cross section A
2. Electric field is created when a potential difference V is applied
3. If ‘E’ is uniform, then, V = El
4. Current density, J = σ E = σ\(\frac{\mathrm{V}}{l}\)

Since J = \(\frac{\mathrm{I}}{\mathrm{A}}\)
\(\frac{\mathrm{I}}{\mathrm{A}}=\sigma \frac{\mathrm{V}}{l}\)
Rearranging the above equation, Current through the conductor

Limitations:
There are certain materials and devices where the proportionality of V and I does not hold good.

Question 3.
Explain the equivalent resistance of a series and parallel resistor network.
Answer:
Resistors in series:

1. Let V₁, V₂, V₃ be the potential difference across R₁, R₂, R₃
2. In the series network, the current is the same.
3. The net potential difference V = V₁ + V₂ + V₃
4. By Ohm’s law: V₁ = IR₁, V₂ = IR₂, V₃= IR₃
5. The equivalent resistance is R_S and V = IR_S in series.
6. IR_S = IR₁ + IR₂ + IR₃
7. R_S = R₁ + R₂ + R₃
8. The equivalent resistance is the sum of the individual resistances.

Resistors in Parallel:

1. In a parallel network, the potential difference across each resistor is the same.
2. current I is divided into I₁, I₂, I₃, same that
   I = I₁ + I₂ + I₃ ………..(1)
   By ohm’s law
   
3. In parallel network, the reciprocal of the equivalent resistance is equal to the sum of the reciprocal of the resistance of the individual resistors.

Question 4.
Explain the determination of the internal resistance of a cell using voltmeter.
Answer:

1. The emf of the cell ξ is found by connecting a high resistance voltmeter across it. Without connecting the external resistance R.
2. Since the voltmeter draws very little current for deflection, the circuit is called an open circuit.
3. Voltmeter reading gives the emf of the cell.
4. Resistance ‘R is included in the circuit and current I is established in the circuit.
5. The potential drop across the resistor.
   R is V = IR …………(1)
6.  Due to internal resistance (r) the voltmeter reads a value V, less than the emf of cell (E).
7.  V = ξ – Ir
8. Ir = ξ – V …………(2)
9. Dividing (2) by (1)
   
10. Internal resistance of the cell
    \(\frac{\mathrm{Ir}}{\mathrm{IR}}=\frac{\xi-\mathrm{V}}{\mathrm{V}}\)
    \(\mathrm{r}=\left(\frac{\xi-\mathrm{V}}{\mathrm{V}}\right) \mathrm{R}\)
11. Since ξ, V and R are known, internal resistance r can be calculated

Question 5.
State and explain Kirchhoff’s rules.
Answer:
Kirchhoff’s rules are used to find current and voltage for more complex circuits. There are 2 rules.

• Kirchhoff’s current rule.
• Kirchhoff’s voltage rule.

Kirchhoff’s current rule:
The algebraic sum of the currents at any function of a circuit is zero.

Sign convention:

1. Current flowing towards a junction is positive
2. Current flowing away from a junction is negative.
3. From the figure
   I₁ + (-I₂) + (-I₃) + I₄ + I₅ = 0 (or)
   I₁ + I₄ + I₅ = I₂ + I₃

Kirchhoff’s second [voltage rule or Loop rule]:
In a closed circuit, the algebraic sum of the products of the current and resistance of each part of the circuit is equal to the total emf included in the circuit.

Sign convention:

1. The product of current and resistance is positive if the current direction is followed (a)
2. If current is opposite to the direction of the loop, then the product of current and resistance is negative (h)
3. The emf is positive when moving from negative to the positive terminal (c) & (d).

Question 6.
Obtain the condition for bridge balance in Wheatstone’s bridge.
Answer:

1. Let P, Q, R, and S be the four resistances in the form of a bridge.
2. A cell of emf E is connected between points A and C
3. The current I is divided into I₁, I₂, I_3, and I₄ across the four branches.
   By Kirchhoff’s current rule
4. At junction B
   I₁ – I_g – I₃ = 0 ………..(1)
   
5. At junction D
   I₂ + I_g – I₄ = 0 …………..(2)
   By Kirchoff’s voltage rule
6. At closed path ABDA
   I₁P + I_gG – I₂R = 0 ……….(3)
7. At closed path ABCDA
   I₁P + I₃Q – I₄S – I₂R = 0 ………(4)
8. When the galvanometer shows zero deflection, the points B and D are at the same potential, and I_g = 0
9. Substitute I_g = 0 in (1), (2) and (3)
   I₁ = I₃ …………….(5)
   I₂ = I₄ …………….(6)
   I₁P = I₂R …………….(7)
10. Substitute (5) and (6) in (4)
    I₁P + I₁Q – I₂S – I₂R = 0
    I₁(P + Q) = I₂(R + S) …………….(8)
    Dividing (8) by (7),
    
    This is the condition of bridge balance.

Question 7.
Explain the determination of unknown resistance using a meter bridge.
Answer:

1. It consists of a uniform manganin wire AB of 1-meter length.
2. The wire is stretched along a meter scale on a wooden board between 2 copper strips C and D
3. Between the 2 copper strips, another copper strip E is mounted to enclose gaps G₁ and G₂
4. An unknown resistance P is connected in G₁
5. A standard resistance Q is connected in G₂.
6. A jockey is connected to E through a galvanometer (G) and a high resistance (HR)
7. The exact position of the jockey can be read on the scale.
8. A Lechianche cell and a key (k) are connected across the ends of the wire.
9. By measuring the radius r and length
10. l of the wire P₁ then specific resistance of the wire is found.
    P = Resistance × \(\frac{A}{l}\)
11. If l is unknown resistance specific Resistance
    \(P=\frac{P \cdot \pi r^{2}}{l}\)
    
12. The jockey is adjusted for zero deflection. Let the point be J.
13. Lengths AJ and JB replace resistances R and S in Wheatstone’s bridge
14. Then, \(\frac{P}{Q}=\frac{R}{S}=\frac{\left.R^{\prime} A\right]}{R^{\prime} J B}\)
    Where R’ → resistance per unit length of wire
    \(\frac{P}{Q}=\frac{A J}{J B}=\frac{l_{1}}{l_{2}}\)
    \(P=Q \cdot \frac{l_{1}}{l_{2}}\)
15. Due to the soldering of copper strips, there occurs an error called end resistance.
16. This can be eliminated by taking another set of readings by interchanging P and Q.
17. By measuring the radius a and length l of the wire P, then specific resistance of the wire is found
    \(\rho=\text { Resistance } \times \frac{A}{l}\)
    If \(\rho\) is unknown resistance equation becomes
    \(\rho=\frac{P \cdot \pi a^{2}}{l}\)

Question 8.
How the emf of two cells are compared using a potentiometer?
Answer:
The primary circuit consists of battery (Bt), Key (k), and rheostat (Rh)

1. The end C is connected to M of a DPPT switch.
2. The terminal N is connected to the jockey (J) through a galvanometer (G) and high resistance (HR)
3. cell 1 is connected between M₁ and N₁
4. cell 2 is connected between M₂ and N₂
5. The jockey is adjusted for zero deflection.
6. The DPDT has pressed towards ξ₁ The potential difference across balancing length l₁ is Irl₁
7. ξ₁ = Irl₁ ………..(1)
8. DPDT switch is pressed towards ξ₂
9. ξ₂ = Irl₂ ………..(2)
   
   Comparison of EMF of two cells
10. Dividing (1) by (2)
    \(\frac{\xi 1}{\xi_{2}}=\frac{l_{1}}{l_{2}}\)
11. If ξ₁ is known, then unknown emf
    \(\xi_{2}=\xi_{1} \cdot \frac{l_{2}}{l_{1}}\)
12. The experiment can be repeated several times by changing the current emf of flowing through it.

IV. Numeric Problems:

Question 1.
The following graphs represent the current versus voltage and voltage versus current for the six conductors A, B, C, D, E, and F. Which conductor has the least resistance and which has maximum resistance?

Answer:

Question 2.
Lightning is a very good example of a natural current. In typical lightning, there is 10⁹ J energy transfer across the potential difference of 5 × 10⁷ V during a time interval of 0.2 s. Using this information, estimate the following quantities:
(a) the total amount of charge transferred between cloud and ground
(b) the current in the lightning bolt
(c) the power delivered in 0.2 s.

Answer:
Given data: E = 10⁹J, V = 5 × 10⁷ V, t = 0.2s

Question 3.
A copper wire of 10^-6 m² area of cross-section, carries a current of 2 A. If the number of electrons per cubic meter is 8 × 10²⁸, calculate the current density and average drift velocity.
Answer:
Given data:
A = 10^-6 m², I = 2A, n = 8 × 10²⁸
Formula:
Current density, J = \(J=\frac{I}{A}\)

Question 4.
The resistance of a nichrome wire at 0°C is 10Ω. If its temperature coefficient of resistivity of nichrome is 0.004/ °C, find its resistance of the wire at boiling point of water. Comment on the result.
Answer:
Given data:
T₁ = 10°C, R₀ = 10Ω
The boiling point of water T = 100°C
α = 0.0041°C
R_T=?
Formula:
R_T = R0 [l + α(T – T₀)]
R_T = 10 [1+0.004(100 – 20)]
= 10(1 + 0.32)
= 10 (1.32)
RT = 13.2Ω
As the temperature increase the resistance of the wire also increases.

Question 5.
The rod given in the figure is made up of two different materials.

Both have square cross-sections of 3 mm side. The resistivity of the first material is 4 × 10^-3Ω and that of the second material has a resistivity of 5 × 10^-3Ωm. What is the resistivity of the rod between its ends?
Answer:

Question 6.
Three identical lamps each having a resistance R are connected to the battery of emf E as shown in the figure. Suddenly the switch S is closed.
(a) Calculate the current in the circuit when S is open and closed
(b) What happens to the intensities of the bulbs A, B, and C.
(c) Calculate the voltage across the three bulbs when S is open and closed
(d) Calculate the power delivered to the circuit when S is opened and closed
(e) Does the power delivered to the circuit decreases, increases or remain the same?

Answer:

Question 7.
An electronics hobbyist is building a radio which requires 150Ω in her circuit, but she has only 220Ω, 79Ω, and 92Ω resistors available. How can she connect the available resistors to get the desired value of resistance?
Answer:
Available resistances = 220Ω, 79Ω 92Ω
Case I:
If 3 resistors are connected in series, then
R_S = R₁ + R₂ + R₃ = 220 + 79 + 92 = 391
This value is greater than the required resistance so it is not possible.
Case II:
If 3 resistors are connected in parallel, then

Question 8.
A cell supplies a current of 0.9 A through a 2Ω resistor and a current of 0.3 A through a 7Ω resistor. Calculate the internal resistance of the cell.
Answer:

ξ = 0.3(7 + r)
Since ‘ξ’ is constant
0.9(2+r) = 0.3(7+r)
1.8 + 0.9r = 2.1 + 0.3r
O.6r = 0.3
r = \(\frac{0.3}{0.6}\)
r = \(\frac{1}{2}\) = O.5Ω

Question 9.
Calculate the currents in the following circuit.

Answer:
At junction B, applying current law,
I₁ – I₂ – I₃ = 0
I₁ = I₂ + I₃ ………..(1)
Kirchhoff’s voltage law in loop ABEFA
100I₃ + 100I₁ = 15
Using (1), we get,
100I₃ + 100 (I₂ + I₃) = 15
100I₃ + 100I₂ + 100₃ = 15
100I₂ + 200₃ = 15 ……………(2)
Voltage law in loop BCDEB
100I₂ – 100I₃ = -9
100I₃ – 100I₂ = 9 ……………(3)
Adding (1) & (2), we get
\(\begin{array}{l}
200 \mathrm{I}_{3}+100 \mathrm{I}_{2}=15 \\
100 \mathrm{I}_{3}-100 \mathrm{I}_{2}=9 \\
300 \mathrm{I}_{3} \quad=24 \quad \mathrm{I}_{3}=\frac{24}{300}=0.08
\end{array}\)
I₃ = 0.08A
(3) → 100I₃ – 100I₂ = 9
Substituting the value of I₃, we get
100 × 0.08 – 100I₂ = 9
8 – 100I₂ = 9
-100I₂ = 1
\(\Rightarrow \mathrm{I}_{2}=\frac{-1}{100}=-0.01 \mathrm{~A}\)
I₂ = -0.01A
From (1),
I₁ = I₂ + I₃
= -0.01 + 0.08
I₁ = 0.07A

Question 10.
A potentiometer wire has a length of 4 m and a resistance of 20Ω. It is connected in series with a resistance of 2980Ω and a cell of emf 4 V. Calculate the potential along the wire.
Answer:
Given data:
l = 4m of R = 20Ω
In series with R’ = 2980Ω
E = 4 V

Question 11.
Determine the current flowing through the galvanometer (G) as shown in the figure.

Answer:
I₂ = I – I₁
Current flowing through the circuit I = 2A
Applying Kirchhoffs II law to PQSP
5I₁ + 10I_g – 15I₂ = 0
5I₁ + 10I_g – 15(I – I₁) = 0
5I₁ + 10I_g – 15I + 15I₁ = 0
20I₁ + 10I_g = 15 × I
20I₁ + 10I_g = 30
2I₁ + I_g = 3 ………..(1)
Applying Kirchhoffs II law to QRSQ
10(I₁ – I_g) – 20 (I₂ + I_g) – 10 I_g = 0
10I₁ – 10 I_g – 20 (I – I₁ + I_g) – 10 I_g = 0
10I₁ – 10 I_g – 20I + 20I₁ – 20I_g – 10I_g = 0
30I₁ – 40I_g = 20I
30I₁ – 40I_g = 20 × 2
3I₁ – 4I_g = 4 ………..(2)
(1) × 3 ⇒ 6I₁ + 3I_g = 9 …………..(3)
(2) × 2 ⇒ 6I₁ – 8I_g = 8 ……………(4)
Solving (3) and (4)
+ 11 I_g = 1
\(\mathrm{I}_{\mathrm{g}}=\frac{1}{11} \mathrm{~A}\)

Question 12.
Two cells each of 5V are connected in series across an 8Ω resistor and three parallel resistors of 4Ω, 6Ω, and 12Ω. Draw a circuit diagram for the above arrangement. Calculate
(i) the current drawn from the cell
(ii) current through each resistor
Answer:
Circuit Diagram:

i) Current drawn from the cell
E_eq = 5+5 = 10V
R_eff = R_s+ R_p
R_s = 8Ω
\(\frac{1}{R_{P}}=\frac{1}{4}+\frac{1}{6}+\frac{1}{12}=\frac{18+12+6}{72}=\frac{36}{72}\)
∴R_p = \(\frac{72}{36}\) = 2Ω
R_eff = 8 + 2 = 10Ω
\(I=\frac{E_{e g}}{R_{e f f}}=\frac{10}{10}=1 A\)
∴ I = 1A
Voltage drop V= IR =1 × 2 = 2V
V = 2V

ii) Current through each resistor.
∴ current in 4Ω resistor
I = \(\frac{2}{4}\) = O.5A
I = O.5A
current in 6Ω resistor, I = \(\frac{2}{6}\) = 0.33A
I = 0.33A
current in 12Ω resistor, I = \(\frac{2}{12}\) = \(\frac{1}{6}\) = 0.17A
I = 0.17A

Question 13.
Four light bulbs P, Q, R, S are connected in a circuit of unknown arrangement. When each bulb is removed one at a time and replaced, the following behavior is observed.

Draw the circuit diagram for these bulbs.
Answer:

Question 14.
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63 cm, what is the emf of the second cell?
Answer:
E₁ = 1.25 V
l₁ = 35 cm
E₂ =?
l₂ = 63cm
img 81
Emf of the second cell is 2.25 V

Part II:

12th Physics Guide Current Electricity Additional Questions and Answers

I(a). Match the following:

Answer:
a. iv
b. iii
c. i
d. ii

I(b). Match the following:

Answer:
a. iii
b. i
c. ii
d. iv

I(c). Match the following:

Answer:
a. iv
b. iii
c. i
d. ii

I(d). Match the following:

Answer:
a. ii
b. iv
c. i
d. iii

II. Fill in the Blanks.
Question 1.
The graph plotted with current against the voltage that obeys Ohm’s law is _______.
Answer:
Straight line

Question 2.
Current is a _______ quantity.
Answer:
Scalar

Question 3.
________ is used in complicated circuits.
Answer:
Kirchhoff’s laws

Question 4.
Mercury exhibits superconductor behaviour at _______.
Answer:
4.2K

III. Choose the Odd One Out:

Question 1.
a) Current
b) current density
c) drift velocity
d) Electric field
Answer:
a) Current – it is scalar and others are vector.

Question 2.
a) Current
b) resistance
c) time
d) electromotive force
Answer:
d) electromotive force – because heat dissipated depends only on the first 3 factors.

Question 3.
a) Current rule
b) Voltage rule
c) Wheatstone’s bridge
d) Joule’s law
Answer:
d) Joule’s law – Because the first 3 comes under Kirchhoff’s laws.

Question 4.
a) 1kWh
b) 1Js^-1
c) 1000Wh
d) 3.6 × 10⁶J
Answer:
b) 1Js^-1 – Others are equivalent of 1 unit of electrical energy

IV. Choose the Incorrect Pair:

Question 1.
a) Ohmic material – Straight line
b) non – Ohmic material – non-linear
c) macroscope form – V = IR
d) macroscope form – V = El
Answer:
d) macroscope form – V = El

Question 2.
a) Electrical resistivity – Ωm
b) Electrical conductivity – Ω^-1 m^-1
c) Internal resistance – Ωm^-1
d) Electrical resistance – Ω
Answer:
c) Internal resistance – Ωm^-1

Question 3.
a) Resistors in series – R_s = R₁ + R₂
b) Resistors in parallel – R_p = \(\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}\)
c) Internal resistance – r = \(\left(\frac{\mathrm{E}-\mathrm{V}}{\mathrm{V}}\right) \mathrm{R}\)
d) Energy – E = VI
Answer:
b) Resistors in parallel – R = R_p = \(\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}\)

Question 4.
a) Electromotive force – force acting
b) Electric energy – work done
c) Electric power – rate of change in electric energy
d) Internal resistance – resistance offered
Answer:
a) Electromotive force – force acting because emf is the potential difference

V. Choose the Correct Pair:

Question 1.
a) Household appliances – Parallel network
b) HousehoLd appliances – Series network
c) \(\frac{1}{\mathrm{R}_{\mathrm{P}}}\) – R₁ + R₂
d) RR_p – \(\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}\)
Answer:
a) Household appliances — Parallel network

Question 2.
Colour code in carbon resistor
a) Brown ring – 0
b) Yellow ring – 1
c) Silver ring – 10%
d) Colourless – 5%
Answer:
a) Silver – 10%

Question 3.
a) Voltmeter – used to measure current
b) Ammeter – used to measure voltage
c) Multimeter – used to measure voltage current and resistance
Answer:
c) Multimeter – used to measure voltage current and resistance

Question 4.
a) Electric heater – tungsten
b) Electric fuse – trippers
c) Electric furnace – Nichrorne
d) Electric lamp – Carbon arc
Answer:
b) Electric fuse – trippers

VI. Assertion and Reason:

Question 1.
A: Fuses are connected in series to protect electric devices.
R: It melts and breaks the circuit if the current exceeds a certain value.
a) R does not explain A
b) R explains A
c) A is correct, R is wrong
d) Both A and R are wrong
Answer:
b) R explains A

Question 2.
A: Positive charge flows from higher electric potential to lower electric potential.
R: These positive charges constitute the flow of current.
a) R does not explain A
b) R explains A
c) A is correct, R is wrong
d) Both A and R are wrong
Answer:
c) A is correct, R is wrong

Question 3.
A: Current is a scalar quantity.
R: Current does not obey the law of vector addition.
a) R does not explain A
b) R explains A
c) A is correct, R is wrong
d) Both A and R are wrong
Answer:
b) R explains A

Question 4.
A: A semiconductor with a negative temperature coefficient of resistance is called a thermopile.
R: The value of a remains the same for all materials
a) R does not explain A
b) R explains A
c) A is correct, R is wrong
d) Both A and R are wrong
Answer:
d) Both A and R are wrong

VII. Choose the Correct Statement:

Question 1.
a) All carbon resistors have a tolerance value.
b) The positive charges constitute the electric current
c) Allessandro Volta invented the telephone
d) Ohm’s law is used for complicated circuits.
Answer:
a) All carbon resistors have a tolerance value

Question 2.
a) The innermost electrons in the atom is called the free electrons.
b) All atoms are neutral with an equal number of protons and electrons.
c) Conventional current flows from negative to positive charges.
d) The outermost electrons are core electrons.
Answer:
b) All atoms are neutral with an equal number of protons and electrons

Question 3.
a) Charging the battery on my mobile.
b) My mobile phone battery has no charge
c) Battery has no charge.
d) My mobile is charging.
Answer:
d) My mobile is charging.

Question 4.
a) Ohmic materials do not have constant resistance.
b) The V – I graph for ohmic materials is non-linear.
c) Non-ohmic materials do not have constant resistance.
d) The V – I graph for non-ohmic materials is linear. constant resistance.
Answer:
c) Non-ohmic materials do not have constant resistance.

VIII. Choose the Incorrect Statement:

Question 1.
a) The value of the equivalent resistance in series connection is lesser than each individual resistance.
b) The value of equivalent resistance in the parallel network is lesser than individual resistance.
c) The value of equivalent resistance in the series network is R_s = R₁ + R₂
d) The value of equivalent resistance in parallel network is \(\frac{1}{\mathrm{R}_{\mathrm{P}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}\)
Answer:
a) The value of the equivalent resistance in series connection is lesser than each individual resistance.

Question 2.
a) Superconductivity was discovered by H.K. Onnes.
b) The resistance of certain materials becomes zero.
c) That temperature is called critical temperature.
d) Experiments showed Ag exhibits superconductivity at 4.2K
Answer:
d) Experiments showed Ag exhibits superconductivity at 4.2K

Question 3.
a) Electric power produced by a resistor is I“R.
b) It depends on the square of the current.
c) If the current is doubled, the power will increase by 8 times.
d) The SI unit of power is a watt.
Answer:
c) If the current is doubled, the power will increase by 8 times.

Question 4.
a) Kirchhoff’s current law obeys the laws of conservation of charges
b) Kirchhoff’s voltage law obeys the laws of conservation of energy
c) Important application of Kirchhoff s rules is Wheatstone’s bridge
d) This rule is used for simple electrical circuits.
Answer:
d) This rule is used for simple electrical circuits.

IX. Choose the best answer:

Question 1.
When current I flows through a wire, the drift velocity of the electrons is v. When current 21 flows through another wire of the same material having double the length and area of cross-section, the drift velocity of the electrons will be-
(a) \(\frac { v }{ 4 }\)
(b) \(\frac { v }{ 2 }\)
(c) v
(d) 2 v
Answer:
(c) v
Hint:
V_d = \(\frac { 1 }{ nAe }\); v’_d = \(\frac { 2I }{ (2A)ne }\) = v_d

Question 2.
The resistance of the wire varies inversely as _______.
a) Area of the cross-section
b) Resistivity
c) Length
d) Temperature
Answer:
a) Area of the cross-section

Question 3.
The curve representing Ohms law is a _______.
a) linear
b) cosme function
c) parabola
d) Hyperbola
Answer:
a) linear

Question 4.
A 10 m long wire of resistance 20Ω is connected in series with a battery of emf 3V and a resistance of 10 Ω. The potential gradient along the wire in volt per meter is
(a) 6.02
(b) 0.1
(c) 0.2
(d) 1.2
Answer:
(c) 0.2
Hint:
Potential difference across the wire = \(\frac { 20 }{ 3 }\) x 3 = 2 V
Potential gradient = \(\frac { v }{ l }\) = \(\frac { 2 }{ 10 }\) = 0.2 V/m

Question 5.
To produce an electric current what is the requirement?
a) A voltage source
b) a source of energy that moves charges
c) an electric field
d) Any of the above
Answer:
d) Any of the above

Question 6.
In Joule’s heating law, when R and t are constant, if the H is taken along the y-axis and I along the x-axis, the graph is:
a) straight line
b) parabola
c) circle
d) ellipse
Answer:
a) straight line

Question 7.
A series circuit consists of 3 resistors with 140Ω, 250Ω, and 220Ω. The total resistance is _______.
a) 330Ω
b) 610Ω
c) 720Ω
d) None of the above
Answer:
b) 610Ω

Question 8.
A cell has an emf of 1.5 V. When short-circuited, it gives a current of 3A. The internal resistance of the cell is
(a) 0.5 Ω
(b) 2.0 Ω
(c) 4.5 Ω
(d) \(\frac { 1 }{ 4.5 }\) Ω
Answer:
(a) 0.5 Ω
Hint:
r = \(\frac { ξ }{ I }\) = \(\frac { 1.5 }{ 3 }\) = 0.5 Ω.

Question 9.
The instrument used for measuring electric current is ________.
a) galvanometer
b) ammeter
c) voltmeter
d) potentiometer
Answer:
b) ammeter

Question 10.
What is the most commonly used conductor in electronics?
a) copper
b) aluminum
c) gold
d) silver
Answer:
a) copper

Question 11.
Kirchhoff’s two laws for electrical circuits are manifestations of the conservation of
(a) charge only
(b) both energy and momentum
(c) energy only
(d) both charge and energy
Answer:
(d) both charge and energy

Question 12.
Why does a circuit require a battery?
a) measure current
b) maintain a potential difference
c) oppose the current
d) measure potential
Answer:
b) maintain a potential difference

Question 13.
In the above circuit, the equivalent resistance between A and B is

a) \(\frac{20}{3}\) Ω
b)10Ω
c)16Ω
d)20Ω
Answer:
c) 16Ω

Question 14.
A flow of 10⁷ electrons per second in a conduction wire constitutes a current of
(a) 1.6 x 10^-26 A
(b) 1.6 x 10¹² A
(c) 1.6 x 10^-12 A
(d) 1.6 x 10²⁶ A
Answer:
(c) 1.6 x 10^-12 A
Hint:
I = \(\frac { Q }{ t }\) = \(\frac{10^{7} \times 1.6 \times 10^{-19}}{1}\) = 1.6 x 10¹² A.

Question 15.
A short circuit has _______.
a) no resistance
b) no conductance
c) low current
d) None of the above
Answer:
b) no conductance

Question 16.
A current of 2A flows through a 12V bulb then calculates the resistance?
a) 6Ω
b) 0.16Ω
c) 32Ω
d) 0.32Ω
Answer:
c) 32Ω

Question 17.
A galvanometer is converted into an ammeter when we connect a
(a) high resistance in series
(b) high resistance in parallel
(c) low resistance in series
(d) low resistance in parallel
Answer:
(d) low resistance in parallel

Question 18.
The resistors each on resistance 1Ω is connected as shown in the figure. The resultant resistance between A and B is

a) \(\frac{11}{5} \Omega\)
b) \(\frac{5}{11} \Omega\)
c) \(\frac{6}{5} \Omega\)
d) \(\frac{5}{6} \Omega\)
Answer:
b) \(\frac{5}{11} \Omega\)

Question 19.
The rings in the carbon resistor are red red red silver. What is its value?
a) 22 × 10² ± 10%
b) 22 × 10³ ± 10%
c) 22 × 10² ± 20%
d) 22 × 10³ ± 20%
Answer:
a) 22 × 10² ± 10%

Question 20.
The electrical resistivity of a thin copper wire and a thick copper rod are respectively ρ₁m and ρ₂m then _______
a) \(\rho_{1}>\rho_{2}\)
b) \(\rho_{2}^{>} \rho_{1}\)
c) \(\rho_{1}=\rho_{2}\)
d) \(\frac{\rho_{2}}{\rho_{1}}=\infty\)
Answer:
c) \(\rho_{1}=\rho_{2}\)

Question 21.
A material with a negative temperature coefficient of resistance is called _______.
a) metal
b) alloy
c) thermistor
d) thermometer
Answer:
c) thermistor

Question 22.
Kirchoff’s first law is a consequence of conservation of ________
a) current
b) charges
c) energy
d) power
Answer:
b) charges

Question 23.
Condition for bridge balance of Wheatstone’s bridge is _______.
a) \(\frac{P}{Q}=\frac{S}{R}\)
b) \(\frac{P}{Q}=R S\)
c) \(\frac{P}{Q}=\frac{R}{S}\)
d) \(\frac{Q}{P}=\frac{R}{S}\)
Answer:
c) \(\frac{P}{Q}=\frac{R}{S}\)

Question 24.
Five cells, each of emf E, are joined in parallel. The total emf of the combination is
(a) 5E
(b) \(\frac { E }{ 5 }\)
(c) E
(d) \(\frac { 5E }{ 2 }\)
Answer:
(c) E

Question 25.
Nichrome wire is used as the heating element because it has ________
a) low specific resistance
b) low melting point
c) high specific resistance
d) high conductivity
Answer:
c) high specific resistance

Question 26.
Peltier effect is the converse of ________
a) Joule effect
b) Raman effect
c) Thomson effect
d) Seebeck effect
Answer:
d) Seebeck effect

Question 27.
The potential difference between the points A and B in the given figure is ________.

a) -3V
b) +3V
c) +6V
d) +9V
Answer:
d) +9V

Question 28.
Potentiometer measures potential more accurately because
(a) It measures potential in the open circuit.
(b) It uses a sensitive galvanometer for null detection.
(c) It uses high resistance potentiometer wire.
(d) It measures potential in the closed circuit.
Answer:
(a) It measures potential in the open circuit.

X. Two Mark Questions:

Question 1.
Define current?
Answer:
Current is defined as a net charge Q passes through any cross-section of a conductor in time t
then, I = \(\frac { Q }{ t }\).

Question 2.
Define 1 ampere.
Answer:
1A of current is equivalent to 1 Coulomb of charge passing through a perpendicular cross-section in 1 second.

Question 3.
Does lightning produce an electric current? How?
Answer:

1. The lightning bolt produces an enormous electric current in a short time.
2. During lightning, a very high potential difference is created between the clouds and ground so charges flow between the clouds and ground.

Question 4.
What is meant by transition temperature?
Answer:
The resistance of certain materials becomes zero below a certain temperature T_c. This temperature is known as critical temperature or transition temperature.

Question 5.
Define mobility.
Answer:

1. The mobility of the electron is defined as the magnitude of the drift velocity per unit electric field.
   \(\mu=\frac{\left|\vec{V}_{d}\right|}{|E|}\)
2. The SI unit of mobility is \(\frac{\mathrm{m}^{2}}{\mathrm{~V}_{\mathrm{S}}}\)

Question 6.
Define resistance.
Answer:
The resistance is the ratio of potential difference across the given conductor to the current passing through the conductor.

Question 7.
Three bulbs 40w, 60w, and 100w are connected to 220 V mains which bulb will glow brightly. If they are joined in series?
Answer:
From the relation = \(\frac{\mathrm{V}^{2}}{\mathrm{p}}\)
It follows that resistance of 40w bulb will be maximum.

Question 8.
What is a thermistor?
Answer:
A material with a negative temperature coefficient is called a thermistor.
Eg:

1. Insulator
2. Semiconductor.

Question 9.
Define equivalent resistance when resistors are connected in series.
Answer:
When several resistances are connected in series, the total or equivalent resistance is the sum of the individual) resistances.
R_S = R₁ + R₂ + R₃

Question 10.
Define equivalent resistance in a parallel resistance network.
Answer:
When a number of resistors are connected in parallel, the sum of the reciprocal of the values of resistance of the individual resistor is equal to the reciprocal of the effective resistance of the combination
\(\frac{1}{\mathrm{R}_{\mathrm{P}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}+\frac{1}{\mathrm{R}_{3}}\)

Question 11.
Household appliances are always connected in parallel, why?
Answer:
Household appliances are always connected in parallel so that even if one is switched off, the other devices could function properly.

Question 12.
What is a carbon resistor?
Answer:

1. Carbon resistors consist of a ceramic core, on which a thin layer of crystalline Carbon is deposited.
2. These resistors are inexpensive, stable, and compact in size.

Question 13.
What is a multimeter?
Answer:
A multimeter is a very useful electronic instrument used to measure voltage, current, resistance, and capacitance. In fact, it can also measure AC voltage and AC current.

Question 14.
What is a thermistor?
Answer:
A semiconductor with a negative temperature coefficient of resistance is called a thermistor.

Question 15.
Define the internal resistance of a battery.
Answer:
A real battery is made of electrodes and electrolytes, there is resistance to the flow of charges within the battery.

Question 16.
The headlights of a car turn dim on starting the car, why?
Answer:
When the car engine is started with headlights turned on, they sometimes become dim. This is due to the internal resistance of the car battery.

Question 17.
Why is a potentiometer preferred over a voltmeter for comparison of emf of cells?
Answer:
The potentiometer reads actual emf as it does not draw any current from the cell whereas the voltmeter measures terminal potential difference as it draws current from the cell.

Question 18.
What are the factors that affect the heat produced in an electrical circuit?
Answer:

1. Current I flows through a conductor kept across a potential difference V for a time t, the work done or the electric potential energy spent is W = Vlt
2. In the absence of any other external effect, this energy is spent heating the conductor.

The amount of heat(H) produced is H = VIt
For a resistance R,
H = I²Rt
This is known as Joule’s law of heating.
The heat developed in an electrical circuit due to the flow of current varies directly as

• • the square of the current
  • the resistance of the circuit and
  • the time of flow.

Question 19.
Why nichrome is used as a heating element?
Answer:
Nichrome has a high specific resistance arid can be heated to very high temperatures without oxidation.

Question 20.
What is the use of trippers in houses?
Answer:
Whenever there is an excessive current produced due to a faulty wire connection, the circuit breaker switch opens. After repairing the faulty connection, we can close the circuit breaker switch.

Question 21.
What are the applications of the heating effect of current?
Answer:
Electric discharge lamps, electric welding, and electric arc also utilize the heating effect of current

Question 22.
What is the thermoelectric effect?
Answer:
Conversion of temperature differences into electrical voltage and vice versa is known as the thermoelectric effect.

Question 23.
A wire of resistance 10Ω is stretched uniformly, to thrice its original length calculate the resistance of the stretched wire.
Answer:
R = 10Ω ⇒ R = \(\frac{\rho l}{A}\)
Original length = l
New length = l’ =3l
Original area = A
If length increases, area decreases

Question 24.
Calculate the internal resistance of a 2.1V cell which gives a current of 0.2A through a resistance of 10Ω
Answer:
\(\mathrm{r}=\left(\frac{\xi-\mathrm{V}}{\mathrm{V}}\right) \mathrm{R}\)
V = IR = 0.2 × 10 = 2V
\(r=\left(\frac{2.1-2}{2}\right) 10\) = 0.5Ω
r = 0.5Ω

Question 25.
Differentiate between the Joule heating effect and the Peltier effect.
Answer:

Question 26.
What is the transition or critical temperature in superconductivity?
Answer:
The temperature at which the resistance of the material becomes zero and the material changes from normal to superconductor is known as critical or transition temperature.

Question 27.
Write the formula for electrical conductivity and write its unit?
Answer:

1. Electrical conductivity,
   \(\sigma=\frac{\mathrm{ne}^{2} \tau}{\mathrm{m}}\)
2. It’s unit is Ω^-1 m^-1

Question 28.
Calculate the current in the wire, if a charge of 180C flows through a wire for 1 minute.
Answer:
Current I = \(\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{180}{60}\) = 3A
I = 3A

Question 29.
A parallel combination of two cells of emf’s E₁ and E₂, internal resistances r₁ and r₂ are used to supply current to load of resistance R – Write the expression for the current through the load in terms E₁, E₂, r_1, and r₂.
Answer:

Question 30.
A wire with uniform cross-section A, length l, and resistance R is bent into a complete circle. Calculate the resistance between two diametrically opposite points.
Answer:

The resistance between two diametrically opposite points is R/4Ω.

XI. Three Mark Questions.

Question 1.
Derive a relation between drift velocity and mobility.
Answer:

1. The drift velocity is the average velocity acquired by the electrons inside the conductor
2. The average time between successive collisions is called the mean free time denoted by τ.
3. The acceleration \(\overrightarrow{\mathrm{a}\) experienced by the electron in an electric field \(\overrightarrow{\mathrm{E}\) is given by

Question 2.
Explain the concept of colour code for carbon resistors.
Answer:
Color code for Carbon resistors:
Carbon resistors consist of a ceramic core, on which a thin layer of crystalline carbon is deposited. These resistors are inexpensive, stable, and compact in size. Color rings are used to indicate the value of the resistance according to the rules.

Three coloured rings are used to indicate the values of a resistor: the first two rings are significant figures of resistances, the third ring indicates the decimal multiplier after them. The fourth color, silver or gold, shows the tolerance of the resistor at 10% or 5%. If there is no fourth ring, the tolerance is 20%. For the resistor, the first digit = 5 (green), the second digit = 6 (blue), decimal multiplier = 10³ (orange) and tolerance = 5% (gold). The value of resistance = 56 x 10³ Q or 56 kΩ with a tolerance value of 5%.

Question 3.
Draw the current versus voltage graph for the ohmic and non-ohmic devices. Give one example for each.
Answer:

• Ohmic devices – Conductors.
• Non-ohmic devices – Diode.

Question 4.
Repairing the electrical connection with the wet skin is always dangerous why?
Answer:

1. The human body contains a large amount of water which has a low resistance of around 200Ω and the dry skin has high resistance of around 500 kΩ.
2. But when the skin is wet, the resistance is reduced to around 1000Ω. This is the reason, repairing the electrical connection with the wet skin is always dangerous.

Question 5.
Under what condition, is the heat produced in an electric circuit
(i) directly proportional and
(ii) inversely proportional to the resistance of the circuit?
Answer:
H = I²Rt
Its constant current is flowing through a circuit then heat produced is directly proportional to resistance R
But H = I²Rt
\(\begin{array}{l}
I=\frac{V}{R} \\
\text { then } H=\frac{V^{2}}{R^{2}} \times k t=\frac{V^{2}}{R} t
\end{array}\)
If a constant voltage is flowing through a circuit then heat produced is inversely proportional to resistance R.

Question 6.
How are batteries formed?
Answer:

1. An electric cell converts chemical energy into electrical energy to produce electricity. It contains two electrodes immersed in an electrolyte.
2. Several electric cells connected together form a battery.
3. When a cell or battery is connected to a circuit, electrons flow from the negative terminal to the positive terminal through the circuit.
4. By using chemical reactions, a battery produces a potential difference across its terminals.
5. This potential difference provides the energy to move the electrons through the circuit.

Question 7.
How will you determine the current in a circuit when 2 or more cells are connected in series.
Answer:

1. In a series connection, the negative terminal of one cell is connected to the positive terminal of the second cell, the negative terminal of the second cell is connected to the positive terminal of the third cell, and so on.
2. The free positive terminal of the first cell and the free negative terminal of the last cell become the terminals of the battery.
3. Suppose n cells, each of emf ξ volts and internal resistance r ohms is connected in series with an external resistance R
4. The total emf of the battery = nξ
5. The total resistance in the circuit = nr + R. By Ohm’s law, the current in the circuit is
   
6. It is the current due to a single cell. That is, current due to the whole battery is the same as that due to a single cell and hence there is no advantage in connecting several cells.

Question 8.
What is end resistance and how will you eliminate it.
Answer:

1. The bridge wire is soldered at the ends of the copper strips.
2. Due to imperfect contact, some resistance might be introduced at the contact. These are called end resistances.
3. This error can be eliminated, if another set of readings are taken with P and Q interchanged and the average value of P is found.

Question 9.
Write a note on the electric furnace.
Answer:

1. Furnaces are used to manufacture a large number of technologically important materials such as steel, silicon carbide, quartz, gallium arsenide, etc.
2. To produce temperatures up to 1500°C, a molybdenum-nichrome wire wound on a silica tube is used.
3. Carbon arc furnaces produce temperatures up to 3000 °C.

Question 10.
Compare the emf and potential difference
Answer:

Question 11.
The colours of the carbon resistors are orange, orange, orange. What is the value of the resistor?
Answer:

1. The first orange ring corresponds to 3. The next orange ring corresponds to 3.
2. The next orange ring corresponds to 3, that is the 3rd orange ring corresponds to 10³.
3. There is no coloured ring at the end. So tolerance is 20%.
4. Total resistance is 33 × 10³Ω ± 20%

Question 12.
In the given circuit, the heat produced per second in 6Ω resistor is 32 J. Then calculate the heat produced per second in 3Ω resistor.

Answer:

Question 13.
Find the value of unknown resistance x in the given circuit.

Answer:
P.d across 1Ω = P.d across x-axis
6 × 1 = (11 – 9)x-axis
6 = 2x
[∴11 = 3 + 6 + x, x = (11 – 9)]
x = 3Ω

Question 14.
Aluminum and copper of equal length are found to have some resistance. If the ratio of their radii is 1:3, calculate the ratio of their resistivity.
Answer:

Question 15.
The resistance of a metal wire of length AB to 2Ω. Another wire of length PQ of the same material with twice the diameter of AB is found to have the same resistance of 2Ω. what is the length of PQ?
Answer:

Question 16.
A filament bulb (500W, 100V) is to be used in a 230V main supply, when a resistance R is connected in series, it works perfectly and the bulb consumes 500W. What is the value of R?

Answer:

Question 17.
Derive an expression for Joule’s law.
Answer:
Current I flows through a conductor kept across a potential difference V for a time t, the work done or the electric potential energy spent is
W = Vit
In the absence of any other external effect, this energy is spent heating the conductor.
The amount of heat(H) produced is
H = Vit
For a resistance R,
H = I²Rt
This is known as Joule’s law of heating. The heat developed in an electrical circuit due to the flow of current varies directly as

• the square of the current
• the resistance of the circuit and
• the time of flow.

Question 18.
Write a note on electric fuses.
Answer:

1. Fuses are connected in series in a circuit to protect the electric devices from the heat developed by the passage of excessive current.
2. It is a short length of a wire made of a low melting point material.
3. It melts and breaks the circuit if the current exceeds a certain value.
4. The only disadvantage with the above fuses is that once fuse wire is burnt due to excessive current, they need to be replaced. Nowadays in houses, circuit breakers (trippers) are also used instead of fuses.

Question 19.
Distinguish between Positive and Negative Thomson effect.
Answer:

Question 20.
V = I graph for a metallic wire at two different temperatures T₁ and T₂ is shown in the Fig which of the two temperatures T₁ and T₂ is higher and why?

Answer:
If the same potential difference V is applied current equal to I₁ and I₂ will flow through the wire, when its temperature is kept at T₁ and T₂ respectively since R = V/I it follows that resistance of the wire is more at temperature T₂ further, as in case of the metals the resistance increases with rising of temperature T₂ > T₁.

Question 21.
Which material is used for the meter bridge wire and why?
Answer:
The wire used in the meter bridge is made up of all manganin due to its high specific resistance copper strips are used at the ends because it is a good conductor of electricity.

XII. Five Mark Questions:

Question 1.
Establish a relation between drift velocity and current density.
Answer:

1. XY is a conductor of the area of cross-section A
2. E is the applied electric field.
3. n is the number of electrons per unit volume with the same drift velocity (vd)
4. Let electrons move through a distance dx in time interval dt.
   \(\mathrm{v}_{\mathrm{d}}=\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}} ; \mathrm{d} \mathrm{x}=\mathrm{v}_{\mathrm{d}} \mathrm{dt} \ldots \ldots \ldots \ldots \ldots(1)\)
   
5. Electrons available in the given volume = volume × number per unit volume
   = Adx × n
   = A v_d × n[from (1)]
   Total charge in volume dQ = charge × number of electrons
   dQ = (e)(A v_d)n
   Hence the current,
   \(\begin{array}{l}
   I=\frac{d Q}{d t} \\
   I=\frac{n e A v_{d} d t}{d t}
   \end{array}\)
   I = neAv_ddt.

Question 2.
Deduce ohm’s law using the concept of current density?

Answer:

1. Consider a segment of wire of length l and area of cross section A.
2. Electric field is created when a potential difference V is applied.
3. If ‘E’ is uniform, then, V = El
4. Current density, J = σE = σ\(\frac{V}{l}\)
5. Since J = \(\frac{I}{A}\)
   \(\frac{I}{A}=\sigma \frac{V}{l}\)
   Rearranging the above equation,
   \(\mathrm{V}=\mathrm{I}\left(\frac{I}{\sigma A}\right)\)
   Where \(\left(\frac{I}{\sigma A}\right)\) – resistance of conductor (R)
   :. R α \(\left(\frac{I}{\sigma A}\right)\)
   :. The macroscopic form of ohm’s law is
   V = IR
   V α I.

Question 3.
Explain how the internal resistance of a cell can be determined using a potentiometer.
Answer:

Measurement of internal resistance

1. To measure the internal resistance of a cell, the circuit connections are made as shown in Figure.
2. The end C of the potentiometer wire is connected to the positive terminal of battery B and the negative terminal of the battery is connected to the end D through a key K₁ which forms the primary circuit.

Measurement of internal resistance:

1. The positive terminal of the cell ξ whose internal resistance is to be determined is also connected to the end C of the wire.
2. The negative terminal of the cell £ is connected to a jockey through a galvanometer and a high resistance.
3. A resistance box R and key K₂ are connected across the cell ξ
4. With K₂ open, the balancing point J is obtained and the balancing length CJ= l is measured since the cell is in an open circuit, its emf
   ξ α l₁ ………..(1)
5. With the help of resistance box R, introduce resistance, say R and then key K₂ is closed.
6. The current passing through the cell and the resistance R is given by
   \(I=\frac{\xi}{R+r}\)
7. The potential difference across R is
   \(V=\frac{\xi R}{R+r}\)
8. When this potential difference is balanced on the potentiometer wire, let l₂ be the balancing length, then
   
   \(\mathrm{r}=\mathrm{R}\left(\frac{l_{1}-l_{2}}{l_{2}}\right)\)
9. The internal resistance of the cell is not constant but increases with the increase of external resistance connected across its terminals.

Question 4.
Define electric power and electric potential energy. A potential difference V is applied across a resistance R. Find a relation for the power consumed.
Answer:
1. The electrical power P, is the rate at which the electrical potential energy is delivered.
2. If the charge q, flows between two points having a potential difference V, then the work done in moving the charge is dW=VdQ.
i.e dW=V dQ

3. Consider a circuit in which a battery of voltage V is connected to the resistor.
4. Assume that a positive charge of dQ moves from point a to b through the battery and moves from point c to d through the resistor and back to point a.
5. When the charge moves from point a to b, it gains potential energy dU =V dQ, and the chemical potential energy of the battery decreases by the same amount.
6. When this charge dQ passes through the resistor it loses the potential energy dU=V dQ due to collision with atoms in the resistor and again reaches point a.
7. This process occurs continuously till the battery is connected in the circuit.
8. The electrical power P, is the rate at which the electrical potential energy is delivered,
\(P=\frac{d W}{d t}\)
W.K.T dW = V dQ
\(P=\frac{d}{d t}(V d Q)=V\left(\frac{d Q}{d t}\right)\)
P = VI (∴\(I=\frac{d Q}{d t}\))
SI unit = Watt
9. The bigger units of electric power are kilowatt = 1KW = 10³ W
Megawatt = IMW = 10⁶ W
10.Expression for electric power in terms of I and R. P = I²R (∴ V=IR)
11. Expression for electric power interms of V and R.
\(\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}\left(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}\right)\)

Question 5.
Explain the principle of potentiometer.
Answer:

Potentiometer:

1. A steady current is maintained across the wire CD by a battery.
2. The battery, key, and potentiometer wire are connected in series from the primary circuit.
3. The positive terminal of a primary cell of emf ξ is connected to point c and negative terminal is connected to the Jockey through a galvanometer G and a high resistance HR. This forms the secondary circuit.
4. Let contact be made at any point J on the wire by a jockey.
5. If the potential difference across CJ is equal to the emf of the cell then no current will flow through the galvanometer and it will show zero deflection.
6. CJ is the balancing length L.
7. The potential difference across CJ is equal to IrL
8. ξ = IrL
9. Where I – current, r – resistance per unit length of the wire.
10. Since I and r are constants, ξ α l
11. The emf of the cell is directly proportional to the balancing length.

Question 6.
Explain the Thomson effect.
Answer:
Thomson effect:
If two points in a conductor are at different temperatures, the density of electrons at these points will differ and as a result, the potential difference is created between these points. Thomson effect is also reversible.

Positive Thomson effect:

1. If the current is passed through a copper bar AB which is heated at the middle point C, the point C will be at a higher potential.
2. This indicates that the heat is absorbed along with AC and evolved along with the CB of the conductor. Thus heat is transferred due to the current flow in the direction of the current. It is called the positive Thomson effect.
3. A similar effect is observed in metals like silver, zinc, and cadmium.

Negative Thomson effect:

1. When the copper bar is replaced by an iron bar, heat is evolved along CA and absorbed along BC.
2. Thus heat is transferred due to the current flow in the direction opposite to the direction of the current. It is called the negative Thomson effect.
3. A similar effect is observed in metals like platinum, nickel, cobalt, and mercury.

Question 7.
Give a short account on carbon resistor
Answer:

1. Carbon resistors consist of a ceramic core, on which a thin layer of crystalline Carbon is deposited.
2. These resistors are inexpensive, stable, and compact in size.
3. Three coloured rings are used to indicate the values of a resistor: the first two rings are significant figures of resistances, the third ring indicates the decimal multiplier after them. The fourth colour, silver or gold shows the tolerance of the resistor at 10% or 5%. If there is no fourth ring, the tolerance is 20%.
4. While reading the colour code, hold the resistor with colour bands to your left. Resistors never start with a metallic band on the left.

Question 8a.
Define electric power
Answer:
Electric power is the rate at which electrical potential energy is delivered.
(i.e) \(\frac{\mathrm{dW}}{\mathrm{dt}}=\mathrm{P}=\mathrm{Vi}\)
unit → watt

Question 8b.
An electric iron of resistance of 80Ω is operated at 200 V for 2 hours, find the electrical energy consumed.
Answer:
Given data: R = 80Ω t = 2hours v = 200V
P = VI
P = \(\frac{\mathrm{V}^{2}}{\mathrm{R}}\)
= \(\frac{200 \times 200}{80}=500 \mathrm{~W}\)
Energy consumed in 2 hour = Power in watt × time in hour
E = 500 × 2 = 1000 Wh
E = 1 kWh

Question 9.
To balance the Wheatstone’s bridge shown in the figure, determine an additional resistance that has to be connected with 15Ω
Answer:

Given data
P = 4.2Ω, Q = 3Ω, R = 2Ω, S = 15Ω

Question 10.
Determine the effective resistance of the given circuit between points A and B.

Answer:


The combined resistance between the points A and B is
R = R’ + R’ + R’
= 7.5 + 7.5 + 7.5
R = 22.5Ω

Question 11.
Find the heat energy produced in a resistance of 10Ω when 5A current flows through it for 5 minutes.
Answer:
R = 10Ω, I = 5A, t = 5 minutes = 5 x 60s
H = I²Rt
= 5² × 10 × 5 × 60
= 25 × 10 × 300
= 25 x 3000
H = 75000J or 75 KJ.

Question 12.
The resistance of a nichrome wire ar 0°C is 10Ω. If its temperature coefficient of resistance is 0.004/°C find its resistance at the boiling point of water. Comment on the result.
Answer:
Given data:
R₀ = 10Ω, α = 0.004/°C, t = 100°C
R_t = R₀ (1 + α t)
=10 (1 + 0.004 × 100)
= 10 (1 + 0.4)
= 10 (1.4)
R_t = 14Ω
As the temperature increases, the resistance of the wire increases.

Question 13.
When two resistors connected in series and parallel their equivalent resistances are 15Ω, 56/15 Ω respectively. Find the two resistances.
Answer:
R_S = R₁ + R₂ = 15Ω ………….(1)
\(R_{p}=\frac{R_{1} R_{2}}{R_{1}+R_{2}}=\frac{56}{15} \Omega\) ………..(2)
From equation (1) substituting R₁ + R₂ in equation (2)
\(\frac{R_{1} R_{2}}{15}=\frac{56}{15} \Omega\)
∴R₁R₂ = 56
R₂ = \(\frac{56}{15}\) = 15 …………..(3)
Substituting for R₂ in equation (1) from equation (3)
R₁ = \(\frac{56}{\mathrm{R}_{1}}\) = 15
Then, \(\frac{R_{1}^{2}+56}{R_{1}}=15\)
R₁² + 56 = 15R₁
R₁² – 15 R₁ + 56 = 0
The above equation can be solved using the simple mathematics
R₁² – 8R₁ – 7R₁ + 56=0
R₁ (R₁ – 8) – 7 (R₁ – 8) = 0
(R₁ – 8) (R₁ – 7) = 0
(R₁ = 8Ω) or R₁ = 7Ω
If R₁ = 8Ω
using in equation (1)
8 + R₂ = 15
R₂ = 15 – 8 = 7Ω
R₂ = 7Ω i.e, (when R₁ = 8Ω; R₂ = 7Ω)
If R₁ = 7Ω,
Substituting the equation (1)
7 + R₂ = 15
R₂ = 8Ω i.e, (when R₁ = 7Ω, R₂ = 8Ω)

Question 14.
In the circuit, find the current through each branch of the circuit and the potential drop across the 10Ω resistor.
Answer:

ApIying kirchhof’s second rule for loop AFDCA
10 (I₁ + I₂)+ 2 I₁ = 4
12 I₁ + 10 I₂ = 4 ……….(1)
For loop AFEBA
10 (I₁+ I₂) + 4 I₂ = 5
FE I₂4V 5V A B
10 I₁ + 14I₂ = 5 …………(2)
(1) × 1o ⇒ 120 I₁ + 100 I₂ = 40 ………….(3)
(2) × 12 ⇒ 120 I₁ + 168 I₂ = 60 …………..(4)
(4) – (3) 68 I₁₂ = 20
\(I_{2}=\frac{20}{68}=\frac{10}{34}=0.2941\)
I₂ = 0.2941A
Substituting I₂ in eqn (1),
12 I₁ + 10(0.2941) = 4
12I₁ = 4 – 2.941 = 1.059
I₁ = \(\frac{1.059}{12}\) = 0.08825A
I₁ = 0.08825A
The voltage drop across 10Ω resistor is
V = IR
V=(I₁ + I₂)R
= (0.008825 + 0.2941) × 10
V = 3.824 V

Question 15.
Obtain the condition for maximum current through a resistor, when a number of cells are connected cells in series
(i) in series and
(ii) in parallel suppose n cells, each of emf ξ volts and internal resistance r ohms is connected in series with on external resistance R as shown in the figure.
Answer:
Cells in series:
The total emf of the cells = nξ
The total internal resistance of the cells = r + r + r +……..n times = nr

Therefore, total resistance of the circuit = R+nr
If I is current flowing through the circuit then,
\(I=\frac{\text { total emf }}{\text { total resistance }}\)
\(I=\frac{n \xi}{n r+R}\)
total resistance nr + R
1) When r << R, I = \(\frac{n \xi}{R}\) = nI,
I₁, 5 the current due to a single cell

2) When r<< R, I = \(\frac{n \xi}{nr}\) ≈ \(\frac{\xi}{r}\) current due to single cell.
From the above discussion it maybe concluded that in order to have maximum current, the cells should be connected in series, when the total internal resistance of the cells is negligible as compared to the external resistance in the circuit.

Cells in parallel:
Let n cells be connected in parallel between the points A and B and a resistance R is connected between the points A and B as shown in figure.

Let ξ be the emf and r the internal resistance of each cell.
The total emf of the cells = ξ
The total internal resistance of the cells is given by,
\(\frac{1}{\mathrm{r}_{\mathrm{eq}}}=\frac{1}{\mathrm{r}}+\frac{1}{\mathrm{r}}+\frac{1}
{\mathrm{r}}+\ldots . . \mathrm{n} \text { times }\)
\(\begin{aligned} \frac{1}{\mathrm{r}_{\mathrm{eq}}} &=\frac{\mathrm{n}}{\mathrm{r}} \\ \mathrm{r}_{\mathrm{eq}} &=\frac{\mathrm{r}}{\mathrm{n}} \end{aligned}\)
The total resistance of the circuit = R + \(\frac{\mathrm{r}}{\mathrm{n}}\)
If I is the current flowing through the circuit then,

Form the above discussion, it may be concluded that in order to have maximum current, the cells should be connected in parallel, when the external resistance in the circuit is negligible as compared to the total internal resistance of the cells.

Question 16.
When an inductor is connected to a 230 V d.c. source, a current of 2A passes through it. When the same conductor is connected to a 230 V, 50 HZ. a.c.source, the amount of current decreases (i.elA)). Why?
Answer:

1. X_L = 2πfL
2. In the case of a DC source, f will be equal to zero.
   ∴ X_L also becomes zero.
3. When the resistance is less, more current 2A passes through the circuit.
4. In the case of an AC source with f= 50 HZ, there will be some resistance added to the circuit, which reduces the overall current to 1A.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 10 Ordinary Differential Equations Ex 10.7 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.7

Question 1.
Solve the following Linear differential equations.
cos x \(\frac { dy }{ dx }\) + y sin x = 1
Solution:
The given differential equation can be written as

This is the form \(\frac { dy }{ dx }\) + py = Q
where P = tan x
Q = sec x
Thus, the given differential equation is linear.
I.F = e^{∫ pdx} = e^{∫ tan x dx} = e^{log (sec x)} = sec x
So, the required solution is given by
[y × I.F] = ∫ [Q × IF] dx + c
y × sec x = ∫ sec x × sec x dx + c
y sec x = ∫ sec² x dx + c
y sec x = tan x + c

= sin x + c cos x
y = sin x + c cos x is the required solution.

Question 2.
(1 – x²)\(\frac { dy }{ dx }\) – xy = 1
Solution:
The given differential equation can be written as

Thus, the given differential equation is linear.

Which is a required solution.

Question 3.
\(\frac { dy }{ dx }\) + \(\frac { y }{ x }\) = sin x
Solution:
The given differential equation can be written as
\(\frac { dy }{ dx }\) + (\(\frac { 1 }{ x }\))y = sin x
This is of the form \(\frac { dy }{ dx }\) + Py = Q
where P = \(\frac { 1 }{ x }\)
Q = sin x
Thus, the given differential equation is linear.
I.F= e^{∫ pdx} = e^{∫ \(\frac { 1 }{ x }\) dx} = e^{log x} = x
So, the required solution is given by
yx I.F = ∫ (Q × I.F) dx + c
yx = ∫ sin x × x dx + c
= x (-cos x) – (1) (-sin x) + c
yx = -x cos x + sin x + c
yx + x cos x = sin x + c
(y + cos x) x = sin x + c is a required solution.

Question 4.
(x² + 1)\(\frac { dy }{ dx }\) + 2xy = \(\sqrt { x^2+4 }\)
Solution:
The given differential equation may be written as

Thus, the given differential equation is linear.
I.F= e^{∫ pdx} = e^{∫ \(\frac { 2x }{ x^+1 }\) dx} = e^{log (x² + 2)} = x² + 1
So, the required solution is given by

Is the required solution.

Question 5.
(2x – 10y³) dy + y dx = 0
Solution:
The given differential equation may be written as
y dx = -(2x – 10y³) dy

So, the required solution is
x × I.F = ∫ (Q × I.F)dy + c
xy² = ∫ 10 y² × y² dy + c
= ∫ 10 y⁴ dy + c
= \(\frac { 10y^5 }{ 5 }\) + c = 2y⁵ + c
xy² = 2y⁵ + c is a required solution.

Question 6.
x sin x \(\frac { dy }{ dx }\) + (x cos x + sin x) y = sin x
Solution:
The given differential equation can be written as

Thus, the given differential equation is linear.
I.F = e^{∫ pdx} = e^{∫ (cot x + 1/x)dx} = e^{log sin x + log x}
= e^{log (x sin x)} = x sin x
So, the solution of the given differential equation is given by
y × I.F = ∫(Q × I.F) dx + c
y (x sin x) = ∫ \(\frac { 1 }{ x }\) x sin x dx + c
= ∫ sin x dx + c
y (x sin x) = -cos x + c
xy sin x + cos x = c is the required solution.

Question 7.
Solve (y – e^sin-1x) \(\frac { dx }{ dy }\) + \(\sqrt { 1-x^2 }\) = 0
Solution:

Thus, the given differential equation is Linear

Question 8.
\(\frac { dy }{ dx }\) + \(\frac { y }{ (1-x)√x }\) = 1 – √x
Solution:
The given linear differential equation is of the form
\(\frac { dy }{ dx }\) + py = Q

Question 9.
(1 + x + xy²) \(\frac { dy }{ dx }\) + (y + y³) = 0
Solution:

So, the solution of the equation is given by

xy + tan^-1 y = c
Which is the required solution.

Question 10.
\(\frac { dy }{ dx }\) + \(\frac { y }{ x log x }\) = \(\frac { sin 2x }{ log x }\)
Solution:
The given differential equation may be written as
\(\frac { dy }{ dx }\) + \(\frac { 1 }{ x log x }\)y = \(\frac { sin 2x }{ log x }\)
This is of the form \(\frac { dy }{ dx }\) + Py = Q
Where P = \(\frac { 1 }{ x log x }\); Q = \(\frac { sin 2x }{ log x }\)
Thus, the differential equation is linear.
I.F = e^∫pdx
= e^{∫ \(\frac { 1 }{ x log x }\) dx}
= e^{∫ \(\frac { 1 }{ t }\) dt}
= e^{log t}
= log x
So, the solution of the given differential equation is
y × I.F = ∫ (Q × I.F) dx + c
y × log x = ∫ \(\frac { sin 2x }{ log x }\) × log x dx + c
= ∫ sin 2x dx + c
y log x = \(\frac { -cos 2x }{ 2 }\) + c
y log x + \(\frac { cos 2x }{ 2 }\) = c is a required solution.

Question 11.
(x + a) \(\frac { dy }{ dx }\) – 2y = (x + a)⁴
Solution:

= -2 log (x + a)
= log (x + a)^-2

Question 12.
\(\frac { dy }{ dx }\) = \(\frac { sin^x }{ 1+x^3 }\) – \(\frac { 3x^2 }{ 1+x^3 }\)y
Solution:
The equation can be written as

Question 13.
x \(\frac { dy }{ dx }\) + y = x log x
Solution:
The given differential equation may be written as

Thus, the given differential equation is linear.
I.F = e^∫pdx = e^{∫\(\frac { 1 }{ x }\) dx} = e^{log x} = x
So, the solution of the given differential equation is given by
y × I.F = ∫(Q × I.F)dx + c
yx = ∫ log x x dx + c
yx = ∫ x log x dx + c
u = log x ∫dv = ∫x dx

Multiply by 4
4yx = 2x² log x – x² + 4c
4xy = 2x² log x – x² + 4c is a required solution.

Question 14.
x \(\frac { dy }{ dx }\) + 2y – x² log x = 0
Solution:
The given differential equation may be written as
\(\frac { x }{ x }\) \(\frac { dy }{ dx }\) + \(\frac { 2y }{ x }\) = \(\frac { x log x }{ x }\)
This is of the form \(\frac { dy }{ dx }\) + Py = Q
where P = \(\frac { 2 }{ x }\); Q = x log x
Thus, the given equation is linear.
I.F = e^∫pdx = e^{∫\(\frac { 2 }{ x }\) dx}
e^{log x} = e^{log x²} = x²
So the required solution is
y × I.F = ∫(Q × I.F) dx + c
yx² = ∫x log x x² dx + c
yx² = ∫x³ logx dx + c

Multiply by 16
16x²y = 4x⁴ log x – x⁴ + 16c is a required solution

Question 15.
\(\frac { dy }{ dx }\) + \(\frac { 3y }{ x }\) = \(\frac { 1 }{ x^2 }\), given that y = 2 when x = 1
Solution:
The given differential equation can be written as

So, its solution is given by
y × I.F = ∫(Q × I.F) dx + c
yx³ = ∫ \(\frac { 1 }{ x^2 }\) + x³ dx + c
= ∫x dx + c
y x³ = \(\frac { x^2 }{ 2 }\) + c
2yx³ = x² + c
Given that y = 2 when x = 1
2 (2) (1)³ = 1 + c
4 – 1 = c
c = 3
∴ 2yx³ = x² + 3 is a required solution.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 15 Chemistry in Everyday Life Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 15 Chemistry in Everyday Life

12th Chemistry Guide Chemistry in Everyday Life Text Book Back Questions and Answers

Part – I Text Book Evaluation

I. Choose the Correct Answer

Question 1.
Which of the following is an analgesic?
a) Streptomycin
b) Chloromycetin
c) Asprin
d) Penicillin
Answer:
c) Asprin

Question 2.
Dettol is the mixture of
a) Chloroxylenol and bithionol
b) Chloroxylenol and a-terpineol
c) phenol and iodine
d) terpineol and bithionol
Answer:
b) Chloroxylenol and a-terpineol

Question 3.
Antiseptics and disinfectants either kill or prevent growth of microorganisms. Identify which of the following statement is not true.
a) dilute solutions of boric acid and hydrogen peroxide are strong antiseptics.
b) Disinfectants harm the living tissues.
c) A 0.2% solution of phenol is an antiseptic while 1% solution acts as a disinfectant.
d) Chlorine and iodine are used as strong disinfectants.
Answer:
a) dilute solutions of boric acid and hydrogen peroxide are strong antiseptics.

Question 4.
Saccharin, an artificial sweetener is manufactured from
a) cellulose
b) toluene
c) cyclohexene
d) starch
Answer:
b) toluene

Question 5.
Drugs that bind to the receptor site and inhibit its natural function are called
a) antagonists
b) agonists
c) enzymes
d) molecular targets
Answer:
a) antagonists

Question 6.
Aspirin is a/an
a) acetylsalicylic acid
b) benzoyl salicylic acid
c) chlorobenzoic acid
d) anthranilic acid
Answer:
a) acetylsalicylic acid

Question 7.
Which one of the following structures represents nylon 6,6 polymer?

Answer:
d)

Question 8.
Natural rubber has
a) alternate cis- and trans-configuration
b) random cis- and trans-configuration
c) all cis-configuration
d) all trans-configuration
Answer:
c) all cis-configuration

Question 9.
Nylon is an example of
a) polyamide
b) polythene
c) polyester
d) poly saccharide
Answer:
a) polyamide

Question 10.
Terylene is an example of
a) polyamide
b) polythene
c) polyester
d) polysaccharide
Answer:
c) polyester

Question 11.
Which is the monomer of neoprene in the following?


Answer:

Question 12.
Which one of the following is a biodegradable polymer?
a) HDPE
b) PVC
c) Nylon 6
d) PHBV
Answer:
d) PHBV

Question 13.
Non stick cook wares generally have a coating of a polymer, whose monomer is
a) ethane
b) prop-2-enenitrile
c) chloroethene
d) 1,1,2,2-tetrafluoroethane
Answer:
d) 1,1,2,2-tetrafluoroethane

Question 14.
Assertion : 2-methyl-I ,3-butadiene is the monomer of natural rubber
Reason : Natural rubber is formed through anionic addition polymerisation.
a) If both assertion and reason are true and reason is the correct explanation of assertion.
b) if both assertion and reason are true but reason is not the correct explanation of assertion.
c) assertion is true but reason is false.
d) both assertion and reason are false.
Answer:
c) assertion is true but reason is false.

Question 15.
An example of antifertility drug is
a) novestrol
b) seldane
c) salvarsan
d) Chioramphenicol
Answer:
a) novestrol

Question 16.
The drug used to induce sleep is
a) paracetamol
b) bithional
c) chloroquine
d) equanil
Answer:
d) equanil

Question 17.
Which of the following is a co-polymer?
a) Orlon
b) PVC
c) Teflon
d) PHBV
Answer:
d) PHBV

Question 18.
The polymer used in making blankets (artificial wool) is
a) polystyrene
b) PAN
c) polyester
d) polythene
Answer:
b) PAN

Question 19.
Regarding cross-linked or network polymers, which of the following statement is incorrect? (NEET)
a) Examples are Bakelite and melamine
b) They are formed from bi and tri-functional monomers
c) They contain covalent bonds between various linear polymer chains
d) They contain strong covalent bonds in their polymer chain
Answer:
d) They contain strong covalent bonds in their polymer chain

Question 20.
A mixture of chloroxylenol and terpineol acts as (NEET)
a) antiseptic
b) antipyretic
c) antibiotic
d) analgesic
Answer:
a) antiseptic

II. Short Answer

Question 1.
Which chemical is responsible for the antiseptic properties of Dettol.
Answer:
1. Chloroxylenol and
2. Terpineol are the chemicals responsible for the antiseptic properties of Dettol. But among these two, chloroxylenol plays a more important role. Chloroxylenol is an antiseptic and disinfectant which is used for skin disinfection and cleaning surgical instruments.

Question 2.
What are antibiotics?
Answer:
Medicines that have the ability to kill pathogenic bacteria are called antibiotics
Ex.: amoxicillin, cefixime.

Question 3.
Name one substance which can act as both analgesic and antipyretic.
Answer:
Aspirin (acetylsalicylic acid) is a chemical substance which lowers body temperature (to normal) and also reduces body pain. Therefore it acts as both antipyretic and analgesic.

Question 4.
Write a note on synthetic detergents.
Answer:

• Synthetic detergents are formulated products containing either sodium salts of alkyl hydrogen sulphates or sodium salts of long chain alkyl benzene sulphonic acids.
• There are three types.
  
  

Question 5.
How do antiseptics differ from disinfectants?
Answer:

Question 6.
What are food preservatives?
Answer:
Food preservatives are chemical substances are capable of inhibiting, regarding, or arresting the process of fermentation, acidification or other decomposition of food by growth of microorganisms.
Examples:

1. Acetic acid is used mainly as a preservative for the preparation of pickles.
2. Sodium meta suiphite is used as a preservative for fresh vegetables and fruits.
3. Sodium benzoate is used as a preservative for juices.

Question 7.
Why do soaps not work in hard water?
Answer:

• Hard water contains calcium and magnesium ions.
• When soaps are dissolved in hard water, calcium and magnesium ions displace sodium or potassium ion from soaps forming insoluble calcium or magnesium salts of fatty acids.
• These insoluble salts separate as scum.
• So soaps do not work in hard water.

Question 8.
What are drugs? How are they classified?
Answer:
A drug is a substance that is used to modify or explore physiological systems or pathological states for the benefits of the recipient. it is used for the purpose of diagnosis, prevention cure or relief of a disease.

Classification of drugs:
Drugs care classified based on their proportions such as chemical structure, pharmacological effect, target system, site of action etc.

1. Classification based on the chemical structure:
In this classification, drugs with a common chemical skelton are classified into a single group. For example, ampicillin, amoxicillin, methiceillin etc. all have similar structure and are classified into a single group called penicillin.

Similarly we have other group of drugs such as opiates, steroids, catecholamines etc. Compounds having similar chemical structure are expected to have similar chemical properties. However, their biological actions are not always similar.

2. Classification based on the pharmacological effect:

• In this classification, the drugs are grouped based on their biological effect that they produce on the recipient.
• For example, the medicines that have the ability to kill the pathogenic bacteria are grouped as antibiotics.
• This kind of grouping will provide the full range of drugs that can be used for a particular disease.

3. Classification based on the target system:

• In this classification, the drugs are grouped based on the biological system (or) process that they target in the recipient.
• This classification is more specific than the pharmacological classification.
• For example, the antibiotics streptomycin and erythromycin inhibit the protein synthesis (target) in bacteria and are classified in a same group. However their mode of action is different.
• Streptomycin inhibits the initiation of protein synthesis, while erythromycin prevents the incorporation of new aminoacids to the protein.

4. Classification based on the site of action:

1. The drug molecule interacts with biomolecules such as enzymes, receptors etc, which are referred as drug targets.
2. We can classify the drug based on the drug target with which it binds.
3. This classification is highly specific compared to the others. These compounds often have a common mechanism of action, as the target is the same.

Question 9.
How the tranquilizers work in body.
Answer:
They are neurologically active drugs tranquilizers act on the central nervous system by blocking the neurotransmitter dopamine in the brain.
Ex: Major tranquilizers : Haloperidol, Clozapine.
Mine tranquilizers : Diozepam (Valium), alprazolam.

Question 10.
Write the structural formula of aspirin. Structural formula of aspirin :
(Acetyl Salicylic Acid) C₉ H₈ O₄
Answer:

Question 11.
Explain the mechanism of cleansing action of soaps and detergents
Answer:

• An example for soap is sodium palmitate.
• The cleansing action of soap is directly related to the structure of carboxylate ions (palmitate ions) present in soap.
• The structure of palmitate exhibit dual polarity
  
• Hydrocarbon portion is non polar, hydrophobic and soluble in oils and greases but not in water.
• Carboxyl portion is polar, hydrophilic and soluble in water.
• Dirt in the cloth is due to the presence of dust particles intact or grease which stick.
• When soap is added to an oily or greasy part of the cloth, the hydrocarbon part of the soap dissolve in the grease, leaving the negatively charged carboxylate end exposed on the grease surface.
• At the same time the negatively charged carboxylate groups are strongly attracted by water, forming small droplets called micelles and grease is floated away from the solid object.
• When clothes are rinsed with water, the grease goes with it.
• As a result, the doth gets free from dirt and the droplets are washed away with water.
• The micelles do not combine into large drops because their surfaces are all negatively charged and repel each other.
• The cleansing action of soap depends upon its tendency to act as an emulsifying agent between water and water insoluble greases.
• The cleansing action of detergents are similar to the cleansing action of soaps.

Question 12.
Which sweetening agent are used to prepare sweets for a diabetic patient?
Answer:
Artificial sweetening agents such as saccharin, Alitame, Aspartame can be used in preparing sweets for diabetic patients.

Question 13.
What are narcotic and non – narcotic drugs. Give examples.
Answer:

Question 14.
What are anti-fertility drugs? Give examples.
Answer:
Artificially drugs are chemical substances which suppress the action of hormones that promote pregnancy. These drugs actually reduce the chances of pregnancy and act as a protection. Antifertility drugs are made up of derivatives of synthetic progesterone or a combination of derivatives of oestrogen and progesterone. Example : Ethynylestradiol, menstranol and norcthynodrel etc.

Question 15.
Write a note on copolymer
Answer:

• A polymer containing two or more kinds of monomer units is called a copolymer.
• Ex: SBR rubber (Buna-S). This contains styrene and butadiene monomer units.
• Co-polymers have properties quite different from homopolymers.
• Mixture of styrene and 1, 3 butadiene to form a copolymer (Buna-S).

Question 16.
What are biodegradable polymers? Give examples.
Answer:

1. The materials that are readily decomposed by microorganisms in the environment are called biodegradable.
2. Natural polymers degrade on their own after certain period of time but the synthetic polymers do not.
3. The biopolymers which disintergrates by themselves in biological systems during a
4. certain period of time by enzymatic hydrolysis and to some extent by oxidation are called biodegradable polymers.

Examples:

1. Polyhydroxy butyrate (PHB)
2. Polyhydroxy butyrate – co – hydroxyl valerate (PHBV)
3. Polylactic acid (PLA)

Question 17.
How is terylene prepared?
Answer:

•  Monomers – Ethylene glycol and terepathalic acid (or) dimethyl terephthalate.
• Catalyst – Zinc acetate / antimony trioxide.
• Temperature – 500K
• Product – Terylene

Question 18.
Write a note on vulcanization of rubber
Answer:

1. Natural rubber is very soft and brisky. it has high water absorption capacity and low tensile strength. Its properties can be improved by a process called vulcanization.
2. Natural rubber is mixed with 3 – 5% sulphur and heated at 100 – 150°C causes cross-linking of the cis – 1, 4 polyisoprene chains through disuiphide ( – S – S – ) bonds.
3. The physical properties of rubber can be altered by controlling the amount of sulphur that is used for vulcanization. When 3 to 10% sulphur is used the resultant rubber is somewhat harder but flexible.
4. Following properties of rubber CH₃

Question 19.
Classify the following as linear, branched or cross-linked polymers
a) Bakelite
b) Nylon
c) polythene
Answer:

Question 20.
Differentiate thermoplastic and thermosetting plastics.
Answer:

12th Chemistry Guide Chemistry in Everyday Life Additional Questions and Answers

Part – II – Additional Questions

I. Choose the best answer

Question 1.
An ideal drug is the one which is
(i) non-toxic
(ii) bio-compatible
(iii) bio-degradable
(iv) should not have any side effects .
a) (i) & (ii)
b) (i) & (iii)
c) (i), (ii) & (iv)
d) (i), (ii), (iii) & (iv)
Answer:
d) (i), (ii), (iii) & (iv)

Question 2.
A drag is safer when its therapeutic index is ………………..
a) lower
b) higher
c) one
d) zero
Answer:
b) higher

Question 3.
Which among the following belongs to the group penicillin?
a) Ampicillin
b) Amoxicillin
c) Methicillin
d) All the above
Answer:
d) All the above

Question 4.
Which among the following is an antibiotic?
a) Atenolol
b) Cefixime
c) Amlodipine
d) All the above
Answer:
b) Cefixime

Question 5.
Which among the following is an anti-hypertensive drug?
a) Ampicillin
b) Cefixime
c) Atenolol
d) Cefpodoxime
Answer:
c) Atenolol

Question 6.
Proteins which act as biological catalysts are called.
a) Antibiotics
b) Enzymes
c) Tranquilizers
d) Analgesics
Answer:
b) Enzymes

Question 7.
Penicillin is an example of
a) Analgesics
b) Antibiotic
c) Anaesthetic
d) Antacid
Answer:
b) Antibiotic

Question 8.
Drugs which act on the central nervous system by blocking the neurotransmitter dopamine in the brain are called.
a) Analgesics
b) Antacids
c) Tranquilizers
d) Antihistamines
Answer:
c) Tranquilizers

Question 9.
Diazepam is an example of ……………..
a) analgesics
b) antacids
c) tranquilizer
d) antihistamine
Answer:
c) tranquilizer

Question 10.
Which among the following is not a non-narcotic analgesic?
a) Aspirin
b) Ibuprofen
c) Acetaminophen
d) Morphine
Answer:
d) Morphine – A narcotic analgesic

Question 11.
Which among the following is an inhalational general anesthetics?
a) lidocaine
b) Isoflurane
c) Propofol
d) Procaine
Answer:
b) Isoflurane

Question 12.
Which among the following is used as an antacid?
a) Sodium hydroxide
b) Potassium hydroxide
c) Aluminium hydroxide
d) Calcium carbonate
Answer:
c) Aluminium hydroxide

Question 13.
The drugs which provide relief from allergic effects are called.
a) Antimicrobials
b) Antacids
c) Opioids
d) Antihistamines
Answer:
d) Antihistamines

Question 14.
Neomycin belongs to …………………
a) beta-lactams
b) aminoglycosides
c) tetracyclines
d) fluoroquinolones
Answer:
b) aminoglycosides

Question 15.
Benzalkonium chloride is used as an ………………………
a) disinfectant
b) antibiotic
c) antiseptic
d) antifertility drug
Answer:
c) antiseptic

Question 16.
Which drug suppresses ovulation/fertilisation?
a) Oxytetracycline
b) Terfenadine
c) Menstranol
d) Minocycline
Answer:
c) Menstranol

Question 17.
The drug which inhibits bacterial cell wall synthesis is ……………….
a) macrolide
b) beta-lactams
c) fluoroquinolone
d) tetracycline
Answer:
b) beta-lactams

Question 18.
The drug which targets bacterial ribosomes and prevents protein production is …………………
a) macrolide
b) beta-lactam
c) fluoroquinolone
d) tetracycline
Answer:
a) macrolide

Question 19.
Which among the following are food additives?
(i) Food colours
(ii) Antioxidants
(iii) Stabilizers
(iv) Preservatives
a) (i), (ii) & (iv)
b) (i), (iii) & (iv)
c) (i), (ii) & (iii)
d) (i), (ii), (iii) & (iv)
Answer:
d) (i), (ii), (iii) & (iv)

Question 20.
The preservative used for the preparation of pickles is ………………..
a) benzoic acid
b) sorbic acid
c) acetic acid
d) hydroxybenzoic acid
Answer:
c) acetic acid

Question 21.
Which of the following not correctly matched


Answer:

Question 22.
Which among the following is used as an emulsifier?
a) Sodium metabisulphite
b) Alkyl ester of hydroxybenzoic acid
c) Sucrose ester with palmitic acid
d) All the above
Answer:
c) Sucrose ester with palmitic acid

Question 23.
Antioxidants are …………….
a) butyl hydroxy toluene
b) butylated hydroxy anisole
c) both (a) & (b)
d) none of the above
Answer:
c) both (a) & (b)

Question 24.
Sulphur dioxide acts as ……………….
a) antimicrobial agent
b) antioxidant
c) enzyme inhibitor
d) all the above
Answer:
d) all the above

Question 25.
Artificial sweetening agent is ……………….
a) Butyl hydroxytoluene
b) Butylated hydroxyanisole
c) alitame
d) all the above
Answer:
c) alitame

Question 26.
Softsoap is ……………..
a) Sodium palmitate
b) Potassium palmitate
c) Sodium methylbenzene sulphonate
d) Methyl hydrogen sulphate
Answer:
b) Potassium palmitate

Question 27.
For a good soap, its TFM value should be ………………..
a) higher
b) lower
c) minimum
d) zero
Answer:
a) higher

Question 28.
As per BIS standards, grade -1 soaps should have a minimum TFM of
a) 60%
b) 70%
c) 76%
d) 50
Answer:
c) 76%

Question 29.
An example of anionic detergent is ………………..
a) n-hexadecyltrimethylammonium chloride
b) Sodium lauryl sulphate
c) Pentaerythrityl stearate
d) Sodium stearate
Answer:
b) Sodium lauryl sulphate

Question 30.
Which among the following is a natural polymer?
a) Cellulose diacetate
b) Cellulose
c) Polythene
d) Viscose rayon
Answer:
b) Cellulose

Question 31.
Which among the following is a synthetic polymer?
a) Cellulose diacetate
b) Cellulose
c) Polythene
d) Viscose rayon
Answer:
c) Polythene

Question 32.
Which among the following is a semi-synthetic polymer?
a) Cellulose diacetate
b) Cellulose
c) Polythene
d) Polyvinyl chloride
Answer:
a) Cellulose diacetate

Question 33.
Which among the following is a linear polymer?
a) LDPE
b) HDPE
c) Polypropylene
d) Bakelite
Answer:
b) HDPE

Question 34.
Which among the following is a branched polymer?
a) LDPE
b) HOPE
c) PVC
d) Bakelite
Answer:
a) LDPE

Question 35.
Which among the following is a cross-linked polymer?
a) LDPE
b) HDPE
c) PVC
d) Bakelite
Answer:
d) Bakelite

Question 36.
Which among the following is an elastomer?
a) Nylon 6,6
b) Polythene
c) Neoprene
d) Polystyrene
Answer:
c) Neoprene

Question 37.
Which among the following polymer belongs to fibers?
a) Nylon 6,6
b) Polythene
c) Neoprene
d) Polystyrene
Answer:
a) Nylon 6,6

Question 38.
Which among the following is a thermoplastic?
a) Nylon 6,6
b) Polythene
c) Neoprene
d) Buna-N
Answer:
b) Polythene

Question 39.
Which among the following is a thermosetting plastic?
a) Nylon 6,6
b) Polythene
c) Neoprene
d) Bakelite
Answer:
d) Bakelite

Question 40.
Which among the following is a condensation polymer?
a) PVC
b) Nylon 6,6
c) Polythene
d) Teflon
Answer:
b) Nylon 6,6

Question 41.
Which is used as a free radical initiator?
a) Benzyl acetate
b) Benzyl alcohol
c) Benzoyl peroxide
d) Benzyl nitrate
Answer:
c) Benzoyl peroxide

Question 42.
The catalyst used in the preparation of LDPE is …………………
a) Hydrogen
b) Oxygen
c) Zeigler-Natta catalyst
d) Wilkinson catalyst
Answer:
b) Oxygen

Question 43.
The catalyst used in the preparation of HDPE is ………………….
a) Hydrogen
b) Oxygen
c) Zeigler-Natta catalyst
d) Wilkinson catalyst
Answer:
c) Zeiglar-Natta catalyst

Question 44.
Zeiglar-Natta catalyst is ………………….
a) ZnCl₂+(C₂H₅)₃Al
b) TiCl₄+(C₂H₅)₃Al
c) AlCl₃+(C₂₅)₃Al
d) CuCl₂+(C₂H₅)₃Al
Answer:
b) TiCl₄+(C₂H₂)₃Al

Question 45.
The monomer used in the manufacture of Teflon is ………………..
a) Tetra fluoro methane
b) Tetra chloromethane
c) Tetra fluoro ethylene d) Vinyl cyanide
Answer:
c) Tetra fluoro ethylene

Question 46.
The chemical name of orlon is ……………….
a) Polyethylene terephthalate
b) Polyacrylonitrile
c) Poly hexamethylene adipamide
d) Poly tetrafluoroethylene
Answer:
b) Polyacrylonitrile

Question 47.
The monomer of nylon 6 is ………………
a) Hexan-l,6-dioic acid
b) Hexan-1, 6-diamine
c) Caprolactam
d) both (a) & (b)
Answer:
c) Caprolactam

Question 48.
The monomer of nylon 6,6 is …………………..
a) Hexan-1,6-dioic acid
b) Hexan-1, 6-diamine
c) Caprolactam
d) both (a) & (b)
Answer:
d) both (a) & (b)

Question 49.
Terylene is an example of ………………….
a) Poly amide
b) Polyester
c) Polyene
d) Poly acid
Answer:
b) Polyester

Question 50.
The polymer used to make unbreakable crockery is ………………
a) Novolac
b) Bakelite
c) Melamine formaldehyde
d) Urea formaldehyde
Answer:
c) Melamine formaldehyde

Question 51.
Natural rubber is a polymer of ………………..
a) Chloroprene
b) 2-methyl buta-1,3-diene
c) 2-chloro buta-1, 3-diene
d) acrylonitrile
Answer:
b) 2-methyl buta-1, 3-diene

Question 52.
Which is used in vulcanization of rubber?
a) Carbon
b) Nitrogen
c) Sulphur
d) Chlorine
Answer:
c) Sulphur

Question 53.
The polymer used in the manufacture of conveyer belts is ………………….
a) Buna – N
b) Buna – S
c) Neoprene
d) Chloroprene
Answer:
c) Neoprene

Question 54.
Which among the following is not a copolymer?
a) Neoprene
b) Buna – N
c) Buna – S
d) all the above
Answer:
a) Neoprene

Question 55.
Which among the following is a biodegradable polymer?
a) PCA
b) Nylon-6
c) PVC
d) HDPE
Answer:
a) PGA

II. Pick out the correct statements

Question 1.
(i) All drugs belonging to penicillin group have same biological action.
(ii) The drug molecule interacts with biomolecules such as enzymes, receptors etc., which are referred as drug targets.
(iii) In all living systems, the biochemical reactions are catalysed by hormones.
(iv) The substrate molecule binds to the active site of the enzymes by means of strong covalent bonding.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
a) (i) & (ii)
Correct Statements:
(iii) In all living systems, the biochemical reactions are catalysed by enzymes.
(iv) The substrate molecule binds to the active site of the enzymes by means of weak hydrogen bonding.

Question 2.
(i) Many drugs exert their physiological effects by binding to a specific molecule called a messenger.
(ii) Receptors bind to the active site of messengers.
(iii) Compounds that carry messages to cells are called chemical messengers.
(iv) Most receptors are chiral and hence different enantiomers of a drug can have different effect.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
c) (iii) & (iv)
Correct Statements:
(i) Many drugs exert their physiological effects by binding to a specific molecule called a receptor.
(ii) Messengers bind to the active site of receptors.

Question 3.
(i) Aspirin prevents platelet coagulation and hence useful in the prevention of heart attacks.
(ii) Lidocaine is used for major surgical procedures.
(iii) Procaine block pain perception that is transmitted via peripheral nerve fibres to the brain.
(iv) Fluoro quinolones increase the activity of bacterial enzyme DNA gyrase.
a) (i) & (ii)
b) (ii) & (iii)
c) (i) & (iii)
d) (i) & (iv)
Answer:
c) (i) & (iii)
Correct Statements:
(ii) Lidocaine is used for minor surgical procedures.
(iv) Fluoro quinolones inhibits the activity of bacterial enzyme DNA gyrase.

Question 4.
(i) In soap manufacture common salt is added to the reaction mixture to increase the solubility of soap which helps the precipitation of soap from the aqueous solution.
(ii) In soap the hydrocarbon portion is non-polar and the carboxyl portion is polar
(iii) In soap the hydrocarbon portion is water soluble and the carboxyl portion is soluble in oil or grease.
(iv) The cleansing ability of a soap depends upon its tendency to act as an emulsifying agent between water and water insoluble grease.
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (i) & (iv)
Answer:
c) (ii) & (iv)
Correct Statements:
(i) In soap manufacture common salt is added to the reaction mixture to decrease the solubility of soap which helps the precipitation of soap from the aqueous solution.
(iii) In soap the hydrocarbon portion is soluble in oil or grease and the carboxyl portion is water soluble.

III. Pick out the incorrect statements

Question 1.
(i) Natural polymers are obtained from plants or animals.
(ii) Natural polymers modified by chemical treatment are called synthetic polymers.
(iii) Addition polymers are formed by the polymerisation of monomers without the elimination of any by product.
(iv) Thermosetting plastics can be remoulded,
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (i) & (iii)
Answer:
c) (ii) & (iv)
Correct Statements:
(ii) Natural polymers modified by chemical treatment are called semi-synthetic polymers
(iv) Thermo plastics can be remoulded (or) Thermosetting plastics cannot be remoulded.

Question 2.
(i) Polyethene is obtained by condensation polymerisation.
(ii) In the preparation of HDPE peroxides formed from oxygen acts as a free radical initiator.
(iii) The monomer of teflon is tetrafluoroethylene.
(iv) Orion is polyacrylonitrile.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
a) (i) & (ii)
Correct Statements:
(i) Polyethene is obtained by addition polymerisation.
(ii) In the preparation of LDPE peroxides formed from oxygen acts as a free radical initiator.

Question 3.
(i) Proteins which act as biological catalysts are called enzymes.
(ii) Proteins which are important for communication systems are called receptors.
(iii) Allosteric inhibitors do not change the active site geometry of enzymes.
(iv) When allosteric inhibitors bind to the allosteric site, the substrate can also bind to the enzyme.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
c) (iii) & (iv)
Correct Statements:
(iii) Allosteric inhibitors change the active site geometry of enzymes.
(iv) When allosteric inhibitors bind to the allosteric site, the substrate can not bind to the enzyme.

Question 4.
(i) For the treatment of stress, anxiety, depression tranquilizers are used.
(ii) Non-narcotic analgesics are mainly used for post operative pain, pain of terminal cancer.
(iii) In the treatment of peptic ulcer aminoglycosides are used.
(iv) Rabeprazole is used as an antacid,
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)
Correct Statements:
(ii) Narcotic analgesics are mainly used for post operative pain, pain of terminal cancer.
(iii) In the treatment of peptic ulcer tetracyclines are used.

IV. Assertion and Reason

Question 1.
Assertion : The quality of soap is described in terms of total fatty matter (TFM value)
Reason : As per BIS Standards, Grade-I soaps should have 60% minimum TFM value.
a) Both assertion and reason are true and reason is correct explanation of assertion.
b) Both assertion and reason are true but reason is not the correct explanation of assertion.
c) Assertion is true but reason is false
d) Both assertion and reason are false.
Answer:
c) Assertion is true but reason is false

Question 2.
Assertion : Paracetamol and aspirin are both antipyretics.
Reason : Both paracetamol and aspirin are controlled and reversible loss of concionsness by affecting central nervous system.
a) Both assertion and reason are true and reason is correct explanation of assertion.
b) Both assertion and reason are true and reason is not the correct explanation for assertion,
c) Assertion is true but reason is false
d) Both assertion and reason are false.
Answer:
c) Assertion is true but reason is false

Question 3.
Assertion (A): Soaps are superior to detergents.
Reason (R): Detergents can be used even in hard water,
a) Both (A) and (R) are correct, (R) explains (A)
b) Both (A) and (R) are correct, but (R) does not explain (A)
c) (A) is correct but (R) is wrong
d) (A) is wrong but (R) is correct
Answer:
d) (A) is wrong but (R) is correct Correct Assertion (A): Detergents are superior to soaps.

Question 4.
Assertion (A): Nylon 6,6 is a condensation polymer.
Reason (R): Addition polymers are formed by the polymerisation of monomers without the elimination of any product.
a) Both (A) and (R) are correct, (R) explains (A)
b) Both (A) and (R) are correct, but (R) does not explain (A)
c) (A) is correct but (R) is wrong
d) (A) is wrong but (R) is correct
Answer:
b) Both (A) and (R) are correct, but (R) does not explain (A)

V. Match the following

Question 1.

Answer:
1. –  d. Procaine
2. –  c. Lidocaine
3. –  a. Propofol
4. – b. Isoflurane

Question 2.

Answers:
1. – d. Alprazolam
2. – f. Ibuprofen
3. – g. Codeine
4. – e. Cetirizine
5. – a. Erythromycin
6. – c. Benzalkonium chloride
7. – b. Menstranol

Question 3.

Answers:
1. – c. Buna-S
2. – d. PHBV
3. – e. PVC
4. – b. Buna-N
5. – a. Bakelite

VI. Two mark questions

Question 1.
What is called as medicine?
Answer:
The drug which interacts with macromolecular targets such as proteins to produce a therapeutic and useful biological response is called medicine.

Question 2.
What are the characteristics of an ideal drug?
Answer:
Characteristics of an ideal drug:

• Non-toxic > Bio-compatible .
• Bio-degradable
• Should not have any side effects
• Bio-compatible .

Question 3.
What is therapeutic index? How is it related to the safety of the drug.
Answer:

• Therapeutic index is defined as the ratio between the maximum tolerated dose of a drug (above which it become toxic) and the minimum curative dose (below which the drug is ineffective).
• Therapeutic index = \(\frac {Maximum tolerated dose of the drug}{Minimum curative dose of the drug}\)
• Higher the value of therapeutic index, safer is the drug.

Question 4.
What are competitive inhibitors?
Answer:

• When a drug molecule that has a similar shape as the substrate is administered, it can also bind to the enzyme and inhibit its activity.
• In other words, the drug acts as an inhibitor to the enzyme catalyst.
• These type of inhibitors are called competitive inhibitors.

Question 5.
Write the mode of action and uses of antacids. Give example.
Answer:

• Substances which neutralize the acid in the stomach that causes acidity are called antacids.
• Uses: To relieve symptoms such as burning sensation in the chest/throat area (heartburns) caused by acid reflux.
• Ex.: Milk of Magnesia, Aluminium hydroxide

Question 6.
What are antihistamines? Give example.
Answer:
Drugs which block histamine release from histamine-1-receptors.

Question 7.
What are food additives?
Answer:
The substances which are not naturally a part of the food and added to improve the quality of food are called food additives.

Question 8.
Write the important categories of food additives.
Answer:

• Aroma compounds
• Preservatives
• Artificial Sweeteners
• Buffering substances
• Vitamins and minerals

Question 9.
What are the advantages of food additives?
Answer:

• Preservatives reduce the product  spoilage and extend the shelf-life of food.
• Vitamins and minerals reduce the mall nutrient.
• Flavouring agents enhance the aroma of the food.
• Antioxidants prevent the formation of potentially toxic oxidation products of lipids and other food constituents.

Question 10.
What are sugar substituents? Give example.
Answer:

• Compounds that are used like sugars for sweetening, but are metabolised without the influence of insulin are called sugar substituents.
• Ex.: Sorbitol, Xylitol, Mannitol

Question 11.
What are artificial sweetening agents? Give example.
Answer:

• Synthetic compounds which imprint a sweet sensation and possess no or negligible nutritional value are called artificial sweeteners.
• Ex.: Saccharin, Aspartame, sucralose, alitame

Question 12.
Define soap and detergent.
Answer:

• Soap is the sodium or potassium salt of higher fatty acids.
• Detergent is the sodium salt of alkyl hydrogen sulphates or alkyl benzene sulphonic acids.

Question 13.
How is teflon prepared? Mention its use.
Answer:

• Reaction : Addition polymerisation
• Monomer: Tetrafluroethylene
• Catalyst: Oxygen (or) ammonium persulphate
• Polymer : Teflon
• Use : Coating non-stick utensils.

Question 14.
How is orlon prepared? Mention its use.
Answer:

• Reaction : Addition polymerisation
• Monomer : Vinylcyanide (acrylonitrile)
• Catalyst: Peroxide initiator
• Polymer : Polyacrylonitrile – PAN (Orlon)
• Use : Substitute of wool for making blankets, sweaters etc…

Question 15.
Differentiate addition polymerisation and condensation polymerisation.
Answer:

Question 16.
What are Antiseptics ? Give an example.
Answer:

• Chemicals that stop or slowdown the growth of microorganisms are called antiseptics.
• They are applied to living tissues.
• Ex.: Hydrogen peroxide, povidone – iodine, benzalkonium chloride.

VII. Three mark questions

Question 1.
Write notes on antagonists and agonists.
Answer:

• Many drugs exert their physiological effects by binding to a specific molecule called a receptor.
• Drugs which bind to the receptor site inhibiting its natural functions are called antagonists.
• Drugs which mimic the natural messenger by switching on the receptor are called agonists. Agonists are used when there is lack of chemical messenger.
• Ex .: when adenosine binds to the adenosine receptors, it induces sleepiness. But the antagonist drug caffeine binds to the adenosine receptor making it inactive. This results in reduced sleepiness.
• The agonist drug, morphine used as a pain killer, binds to the opioid receptors activating them. This suppresses the neurotransmitters that cause pain.

Question 2.
Write about antioxidants.
Answer:

• Antioxidants are substances which retard the oxidative deteriorations of food.
• Food containing fats and oils is easily oxidized and turn rancid.
• To prevent the oxidation of the fats and oils, BHT (butylhydroxytoluene), BHA (Butylated hydroxyanisole) are added as food additives.
• They are generally called antioxidants.
• These materials readily undergo oxidation by reacting with free radicals generated by the oxidation of oils, thereby stop the chain reaction of oxidation of food.
• Sulphur dioxide and sulphites are also used as food additives.
• They act as anti-microbial agents, antioxidants and enzyme inhibitors

Question 3.
Write about the preparation of soaps.
Answer:

• Animal fats or vegetable oils contain glyceryl esters of long-chain fatty acids.
• When the glycerides are heated with sodium hydroxide or potassium hydroxide soap (sodium or potassium salt of fatty acids) and glycerol are obtained.
• This is known as saponification.
• Common salt is added to the reaction mixture to decrease the solubility of soap and it helps to precipitate out soap from the aqueous solution.
• Soap is then mixed with desired colours, perfumes and chemicals of medicinal importance.

Question 4.
Write a note on total fatty matter (TFM) of soap
Answer:

• The quality of a soap is described in terms of total fatty matter (TFM value).
• It is defined as the total amount of fatty matter that can be separated from a sample after splitting with mineral acids.
• Higher the TFM value of the soap better is its quality.
• As per BIS standards, minimum TFM value of Grade-1 soaps is 76% , Grade-2 soap is 70%, Grade-3 soap is 60%.

Question 5.
Write the preparation of LDPE and HDPE.
Answer:

Question 6.
How is nylon 6,6 prepared? Mention its use.
Answer:

• Reaction : Condensation polymerisation
• Monomers : Equimolar adipic acid and hexamethylene – diamine
• Intermediate : Nylon salt which on heating eliminate a water molecule.
• Polymer : Poly hexamethylene adipamide (Nylon 6,6)
•  Uses : Textiles, manufacture of cards

Question 7.
How is nylon 6 prepared? Mention its use. Nylon -6
Answer:

• Capro lactam (monomer) on heating at 533K in an inert atmosphere with traces of water gives ∈ amino carproic acid which polymerises to give nylon – 6

• it is used in the manufacture of tyrecards fabrics etc…

Question 8.
How is bakelite prepared? Mention its use.
Answer:

• Reaction: Condensation polymerization
• Monomers: Phenol and formaldehyde
• Catalyst : acid or base
• Intermediate: Ortho or para hydroxy methyl phenol which on reaction with phenol gives linear polymer novolac.
  Novalac on heating with formaldehyde undergo cross linkages to form bakelite.
• Polymer: Bakelite
• Uses : Navolac – in paints. Soft bakelites – making glue for binding laminated wooden planks, in varinishes Hard bakelites – combs, pens etc..
  

Question 9.
How is melamine formaldehyde prepared? Mention its use.
Answer:

• Reaction : Condensation polymerization
• Monomers : Melamine and formaldehyde
• Polymer : Melamine formaldehyde
• Uses : Making unbreakable crockery
  

Question 10.
How is urea formaldehyde polymer is prepared?
Answer:

• Reaction : Condensation polymerization
• Monomers : Urea and formaldehyde
• Polymer: Urea formaldehyde polymer
  

Question 11.
Differentiate natural rubber and synthetic rubber.
Answer:

Question 12.
Write about Neoprene preparation? Mention its use.
Answer:

• Reaction : Free radical polymeristion
• Monomer : 2-chloro buta-1, 3-diene (chloroprene)
• Polymer : Neoprene (Superior to rubber and resistant to chemical action)
• Uses : Manufacture of chemical containers, conveyer belts.
  

Question 13.
What is PHBV? How is it prepared? Mention its use.
Answer:

• PHBV is Poly- β -hydroxy butyrate-co- β -hydroxy valerate
• It is a co polymer.
• Monomers : 3-hydroxy butanoic acid and 3-hydroxy pentanoic acid
• Monomer units are joined by ester linkages.
• Uses : in ortho paedic devices, in controlled release of drugs.
  

Question 14.
How is Nylon-2-Nylon-6 prepared?
Answer:

• It is a co polymer.
• Obtained by condensation polymerisation.
• Monomers – glycine and E – amino caproic acid.

VIII. Five mark questions

Question 1.
How polymers are classified on the basis of structure and molecular formula and molecular
forces gives example one.
Answer:
(A) Classification based on structure of polymers
(i) Linear polymers:
The monomers in these are linked together to form along chain. Eg: HDPE, PVC
(ii) Branch chain polymers:
One main chain with small chains as branches. Eg: polypropylene, LDPE
(iii) Crosslinked or Network polymers:
Monomers are linked together to form a three-dimensional network. The monomers contain strong
covalent bonds as they are composed of bi-functional and tri-functional in nature. These polymers are brittle and hard.
Ex: Bakelite (used in electrical insulators), Melamine formaldehyde etc.,

(B) Classification Based on Molecular Forces
Intra molecular forces are the forces that hold atoms together within a molecule. In Polymers, strong covalent bonds join atoms to each other in individual polymer molecules. Intermolecular forces (between the molecules) attract polymer molecules towards each other. Polymers can be classified into 4 types :

i) Elastomers : (soft and stretched)
Elastomers are rubber like solid polymers, that are elastic in nature. The polymer chains are hold by the weakest intermolecular forces, hence allowing the polymer to be stretched. But as you notice removing that stress also results in the rubber band taking up its original form.
Eg: Buna-S, Buna-N neoprene.

ii) Thermoplastics : They become soft on heating and hard cooling they can be remolded . Eg: Polythene, PVC, Polystyrene

iii) Thermosetting : Do not become soft on heating but set to an infusible mass upon heating.
Eg: Bakelite, melamine formaldehyde.

iv) Fibre : They have strong inter molecules forces between the chains giving them less elasticity and high tensile strength. The intermolecular force may be hydrogen bonds or dipole-dipole interaction Eg: Nylon-6,6, Terylene.

Question 2.
Explain various types of anaesthetics.
Answer:
Drugs which produce loss of sensation are called anaesthetics.
Local anaesthetics:

• They cause loss of sensation, in the area in which they are applied without losing consciousness.
• They block pain perception that is transmitted via peripheral nerve fibres to the brain.
• Often used during minor surgical procedures.
• Ex.: Ester-linked local anaesthetic – Procaine, Amide-linked local anaesthetic – Lidocaine

General anaesthetics:

• Cause a controlled and reversible loss of consciousness by affecting central nervous system.
• They are often used for major surgical procedures.
• Ex.: Intravenous general anaesthetics: Propofol
• Inhalational general anaesthetics: Isoflurane

Question 3.
Explain about antimicrobials.
Answer:
Beta-Lactams:

• Inhibits bacterial cell wall biosynthesis.
• Used to treat skin infections, dental infections, ear infections, respiratory tract infections, pneumonia, urinary tract infections, and gonorrhoea.
• Ex.: Penicillins, ampicillin, cephalosporins.

Macrolides:

• Targets bacterial ribosomes and prevent protein production.
• Used to treat respiratory tract infections, genital, gastrointestinal tract and skin infections.
• Ex.: Erythromycin, azithromycin

Fluoroquinolones:

• Inhibits bacterial enzyme DNA gyrase.
• Used to treat urinary tract infections, skin infections, respiratory infections (sinusitis, pneumonia, bronchitis), pulmonary infections in cystic fibrosis.
• Ex.: Ciprofloxacin, levofloxacin

Tetracyclines:

• Inhibit the bacterial protein synthesis via interaction with the 30S subunit of the bacterial ribosome.
• Used to treat peptic ulcer, respiratory tract infections, cholera, acne vulgaris.
• Ex.: Doxycycline, minocycline, oxytetracycline

Aminoglycosides:

• Bind to the 30S subunit of the bacterial ribosome, thus stopping bacteria from making proteins.
• Used to treat infections caused by gram-negative bacteria
• Ex.: Kanamycin, gentamycin, neomycin

Question 4.
Explain the steps involved in free radical polymerisation.
Answer:
When alkenes are heated with free radical initiator such as benzoyl peroxide, they undergo addition polymerisation through free radical mechanism.
1. Initiation – formation of free radical

2. Propagation step

Chain growth will continue with the successive addition of several thousands of monomer units.

Termination:
The above chain reaction can be stopped by stopping the supply of monomer or by coupling of two chains or reaction with an impurity such as oxygen.

Question 5.
Write about buna rubbers.
Answer:
Buna-N:

• It is a co-polymer.
• Monomers : acrylonitrile, buta-1, 3-diene.
• Catalyst: Sodium
• Uses : Manufacture of hoses and tank linings.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Commerce Guide Pdf Chapter 7 Stock Exchange Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Commerce Solutions Chapter 7 Stock Exchange

12th Commerce Guide Stock Exchange Text Book Back Questions and Answers

I. Choose The Correct Answer.

Question 1.
…………. is the oldest stock exchange in the world.
a) London Stock Exchange
b) Bombay Stock Exchange
c) National Stock Exchange
d) Amsterdam Stock Exchange
Answer :
d) Amsterdam Stock Exchange

Question 2.
There are ………………. stock exchanges in the country.
a) 21
b) 24
c) 20
d) 25
Answer :
a) 21

Question 3.
Stock exchanges deal in
a) Goods
b) Serves
c) Financial Securities
d) Country’s Currency
Answer:
c) Financial Securities

Question 4.
Stock exchange allow trading in
a) All types of Shares of any Company
b) Bonds issued by the Govt.
c) Listed Securities
d) Unlisted Securities
Answer:
c) Listed Securities

Question 5.
Jobbers transact in a stock exchange
a) For their Clients
b) For their Own Transactions
c) For other Brokers
d) For other Members
Answer:
b) For their Own Transaction

Question 6.
A pessimistic speculator is
a) Stag
b) Bear
c) Bull
d) Lame Duck
Answer:
b) Bear

Question 7.
An optimistic speculator is
a) Bull
b) Bear
c) Stag
d) Lame Duck
Answer:
a) Bull

Question 8.
A bull operator believes in
a) Increase in Prices
b) Decrease in Prices
c) Stability in Prices
d) No change in Prices
Answer:
a) Increase in Prices

Question 9.
…………………… means the price at which securities are bought and sold are recorded and made public.
a) Market Quotations
b) Trade Quotations
c) Business Quotations
d) Buyers Quotations
Answer:
a) Market Quotations

Question 10.
The rules and regulations of Stock exchange is framed by ……………………. guide lines.
a) RBI
b) Central Government
c) SEBI
d) BSE
Answer:
c) SEBI

II. Very Short Answer Questions.

Question 1.
What is meant by Stock Exchange?
Answer:
Stock Exchange is an organized market for the purchase and sale of industrial and financial security. It is also called the stock market or share market.

Question 2.
Define Stock Exchange.
Answer:
Stock Exchange:

• “An association, organization or a body of individuals, whether incorporated or not, established for the purpose of assisting, regulating and controlling business in buying, selling and dealing in securities.
•  Indian securities contracts [Regulation] Act of 1956.

Question 3.
Write any 5 Stock Exchanges in India.
Answer:

1. The Bombay Stock Exchange
2. Bangalore Stock Exchange Ltd
3. The Madras Stock Exchange Ltd
4. The Hyderabad Stock Exchange Ltd
5. The Cochin Stock Exchange Ltd

Question 4.
What is meant by Remisier?
Answer:

• He acts as an Agent of a member of a Stock Exchange.
• He obtains business for his principal.
• For that service, he gets a commission.

Question 5.
Who is called a Broker?
Answer:
The broker is a commission agent. He acts as an intermediary between buyers and sellers of securities. He charges a commission for his services from both parties.

Question 6.
What are the types of Speculator?
Answer:

• Bull – Tejiwala
• Bear – Mandiwala
• Stag – Premium hunter
• Lame Duck

Question 7.
Mention the Recent Development in Stock Exchange?
Answer:
A commodity exchange is an exchange where commodities are purchased and sold. Commodities are listed here: Metals, agricultural products.

III. Short Answer Questions.

Question 1.
What are the limitations of Stock exchange? AL FAN
Answer:

•  Absence of restriction on the membership of Stock Exchange.
•  Lack of uniformity and control of Stock Exchanges.
•  Failure to control unhealthy speculation.
•  Allowing more than one exchange in a place.
•  No proper regulation on the listing of securities.

Question 2.
Explain Bull and Bear.
Answer:
Speculators in a stock market are of different types based on animal behaviour.
They are:

1. Bull: A Bull or Tejiwala is an operator who expects a rise in prices of securities in the future. He purchases the securities expecting the price of rise in future.
2. Bear: A bear or Mandiwala speculator expects prices to fall in future and sells securities at present with a view to purchase them at lower price.

Question 3.
Explain Stag and Lame Duck.
Answer:
Stag: [Premium Hunter]

• He is a Cautious Speculator.
• He applies for shares in new companies and expects to sell them at a premium.
• If he gets an allotment, he sells the shares before being called to pay the allotment money.

Lame Duck:

• When a Bear finds it difficult to fulfill his commitment, he is said to be struggling like a lame duck.
• Bear contracts to sell securities at a later date.
• At the appointed time he is not able to get the securities as the holders are not willing to part with them.
• In such situations, he feels concerned.

Question 4.
Explain National Stock Market System. (NSMS)
Answer:
The national stock market system was advocated by the High Powered Group. It is headed by Shri. Pherwani (popularly known as Pherwani Committee). At present the National Stock Market comprises the following:

1. National Stock Exchange of India Limited (NSE)
2. Stock Holding Corporation of India Limited (SHCIL)
3. Securities Trading Corporation of India (STCI)

Question 5.
Explain National Stock Exchange (NSE)
Answer:
National Stock Exchanges:

• NSE was incorporated in November 1992.
• It is a Countrywide, Screen-based, Online, and order-driven trading system.
• It uses a satellite link to spread trading throughout the country thereby connecting members scattered all over India.
• NSE has two segments Debt and Capital Segments.
• It has revolutionized stock trading in India.
• Through a computer network, member’s orders for buying and selling within prescribed prices are matched by the central computers with each other and instantly communicate to the trading member.

IV. Long Answer Questions.

Question 1.
Explain the functions of the stock Exchange.
Answer:

1. Ready and Continuous Market: Stock exchange helps investors to sell their securities easily. And also he can convert his cash into securities.
2. Correct Evaluation of Securities: The prices at which securities are bought and sold are recorded and informed to the public. These prices are called “market quotations”
3. Protection to Investors: All dealings in stock exchange are in accordance with well- defined rules and regulations. Any malpractice will be severely punished.
4. Proper Channelisation of Capital: People like to invest in the shares of companies which yield good profits. Also, people invest in the companies which are giving good dividends.
5. Facilities for Speculation: Speculation is an integral part of stock exchange operations. As a result of speculation, demand for and supply of securities are equalized.

Question 2.
Explain the features of the Stock Exchange.
Answer:
Features:
Market for Securities:
Stock Exchange is a market, where securities of corporate, bodies Government and Semi-Government bodies are bought and sold.

Association of Persons:
It is an association of persons or body of individuals whether incorporated or not.

Deals in second-hand securities:
In the secondary market (Stock Exchange) Shares, Debentures, Bonds already issued by the companies are (re-sale) traded.

Regulates Trade in Securities :
It regulates the trade activities so as to ensure free and fair trade.

Allow dealings only in Listed Securities:

•  It maintains the list of companies (securities) that could be bought and sold on its floor.
•  Unlisted Securities can’t be traded in the stock exchange.

Specific location:

• Stock exchange is a particular place where authorized brokers come together daily (on working days – working hours) on the floor of market.

Question 3.
Explain the Benefits of Stock Exchange.
Answer:
The benefits of the Stock Exchange are classified into benefits to the Community, Company, and Investors.
Benefits to the Community:

1. Economic Development: It increases economic development by ensuring a steady flow of savings for production.
2. Fund Raising Platform: It enables well-managed companies to raise funds by issue of shares.

Benefits to the Company:

1. Enhances Goodwill or Reputation: Companies whose shares are quoted on a stock exchange enjoy greater goodwill and credit standing.
2. Raises huge funds: Stock exchange helps the companies to raise huge funds by issue of shares and debentures.

Benefits to Investors:

1. Liquidity: Stock exchange helps to convert his shares into cash quickly.
2. Investor protection: The stock exchange safeguards, investor’s interests by following strict rules and regulations.

Question 3.
Benefits to the Investors [MLA].
Answer:

• The mechanism to trade security – It provides a mechanism by which listed securities can be bought and sold within few minutes.
• Liquidity – An investor can convert his. securities into cash and cash into securities quickly and easily.
  Adding collateral value -It is good collateral security for obtaining loans from banks.

Question 4.
Distinguish between Stock Exchange and Commodity Exchange. MP SOFA
Answer:

12th Commerce Guide Stock Exchange Additional Important Questions and Answers

I. Choose the correct answer.

Question 1.
A cautious speculator is……….
a) Bull
b) Bear
c) Stag
d) Lame Duck
Answer:
c) Stag

Question 2.
A Bear operator believes in ………….
a) Increase in Price
b) Decrease in Price
c) Stability in Price
d) No Change in Price
Answer:
b) Decrease in Price

Question 3.
Which is not a foreign Stock exchange?
(a) London Stock Exchange
(b) Bombay Stock Exchange
(c) Tokyo Stock Exchange
(d) New York Stock Exchange
Answer:
(b) Bombay Stock Exchange

Question 4.
Which one of the following is not correctly matched?
a) Jobbers – India pendent operators
b) Brokers – Intermediaries
c) Authorised clerks – Employee
d) Bull – Mandiwala
Answer:
d) Bull – Mandiwala

Question 5.
Pick the odd one out:
a) Bull
b) Bear
c) Stag
d) Swan
Answer:
d) Swan

Question 6.
Pick the odd one out:
a) Khazimar Street
b) Dalai Street
c) Wall Street
d) Lombard Street
Answer:
a) Khazimar Street

Question 7.
Choose the correct statement.
i) Future market is an Auction Market
ii) Right to sell a security is called “Put Option”
iii) Right to buy a security is called “Call Option”
a) (i) is correct
b) (ii) is correct
c) (iii) is correct
d) All (i) (ii) and (iii) are correct
Answer:
d) All (i) (ii) and (iii) are correct

II. Match the following

1. Match List I with List – II

Answer:
a) (i) 4, (ii) 3, (iii) 2, (iv) 1

Question 2.

Answer:
a) (i) 2, (ii) 3. (iii) 4, (iv) 1

III. Assertion and Reason

Question 1.
Assertion (A): Bear or Mandiwala speculator expects prices to fall in future and sells the securities at present with a view to purchase them at lower prices.
Reason (R): A Bear usually presses its viction down to down to the ground. Similarly, the Bear speculator tends to force down prices of securities,
a) (A) is True (R) is False
b) (A) is False (R) is True
c) Both (A) and (R) are False
d) Both (A) and (R) are True
Answer:
c) Both (A) and (R) are True

IV. Very Short Answer Questions.

Question 1.
What is Future Market?
Answer:

• A “Future Market” is an auction market in which participants Buy and Sell commodities and future contracts for delivery on a Specified Future Date.
• (e.g) York Mercantile Exchange.

Question 2.
What is Options Market?
Answer:
An “Option” is a type of Security that can be Bought and Sold as a specified price within a specified period of time in exchange for a non-refundable upfront deposit.

Question 3.
What is leverage?
Answer:

•  Options help you profit from changes in share prices without puffing down the full price of the share.
•  You get control over the shares without buying them outright.

Question 4.
What is hedging?
Answer:

•  They can also be used to protect yourself from fluctuations in the price of a share.
•  Letting you buy or sell the shares at a pre-determined price for a specified period of time.

Question 5.
What is Sensex?
Answer:

• Sensex in an Index of the Stocks in BSE.
• It has a list of 30 stocks.
• BSE decides the stocks that are to be listed on sensex.
• The criteria for picking a stock to be listed on Sensex is volume of the trade of that stock and the total volume of stock in BSE.

Question 6.
What is NIFTY?
Answer:

•  NIFTY derived from two words. “National” and “Fifty”.
•  It means the index of the 50 most actively traded stocks from across all sectors.

Question 7.
What is meant by Commodity Exchange?
Answer:
Commodity Exchange is an Exchange where Commodities are traded.

Question 8.
What is the stock trading time in India?
Answer:
Equity Market:

•  Normal Trading Time – 9.15 a.m to 3.30 p.m [Monday to Friday]
•  Closed on – All Saturday Sunday and National Holidays.

Commodity Market:

•  Normal Trading Time – 10.00 a.m to 11.30 a.m [Monday to Friday]
• Closed on – All Saturdays and Sundays and National Holidays.

Question 9.
Explain Dalai Street.
Answer:

•  DalaI Street is an area in down town Mumbai, India that houses the Bombay Stock Exchange (BSE).
•  The largest Stock Exchange of India.
•  It received the name Dalai Street after the BSE moved to the area in 1874.

V. Long Answer Questions.

Question 1.
Explain Lombard street and Wall street.
Answer:
Lombard Street:

• Lombard street in London is notable for its connections with the city of London’s Merchant, Banking, Insurance Industries stretching back to medieval times.
• From Bank Junction,where NINE streets coverage by the Bank of England.
• It runs south east for a short distance before bearing left into a most easterly direction and terminates at a junction with Grace church Street and Fenchruch Street.

Wall Street:

•  Wall Street in New York is a street in Lower Manhattan that is the original home of New York Stock Exchange.
• The Historic Head Quarters of the largest US brokerages and investment Banks.
• It is the collective name for the financial and Investment community which includes stock exchanges and – banks, Brokerages, Securities and underwriting firms and big Businesses.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 6 Solid State Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 6 Solid State

12th Chemistry Guide Solid State Text Book Questions and Answers

Part – I Text Book Evaluation

I. Choose the correct answer

Question 1.
Graphite and diamond are ………..
(a) Covalent and molecular crystals
(b) ionic and covalent
(c) both covalent crystals
(d) both molecular crystals
Answer:
(c) both covalent crystals

Question 2.
An ionic compound Ax By crystallizes in fee type crystal structure with B ions at the centre of each face and A ion occupying centre of the cube, the correct formula of A B is ………..
(a) AB
(b) AB₃
(c) A₃B
(d) A₈B₆
Answer:
(b) AB₃

Question 3.
The rano of close packed atoms to tetrahedral hole in cubic packing is ………..
(a) 1:1
(b) 1:2
(c) 2:1
(d) 1:4
Answer:
(b) 1:2

Question 4.
Solid CO₂ is an example of ………..
(a) Covalent solid
(b) metallic solid
(c) molecular solid
(d) ionic solid
Answer:
(c) molecular solid

Question 5.
Assertion: monoclinic sulphur is an example of monoclinic crystal system.
Reason: for a monoclinic system, a \(\neq\) b \(\neq\) c and α = γ = 90° , β \(\neq\) 90°.
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer:
(a) Both assertion and reason are true and reason is the correct explanation of assertion

Question 6.
In calcium fluoride, having the flurite structure the coordination number of Ca²⁺ ion and F Ion are ………..
(a) 4 and 2
(b) 6 and 6
(c) 8 and 4
(d) 4 and 8
Answer:
(c) 8 and 4

Question 7.
The number of unit cells in 8gm of an element X (atomic mass 40) which crystallizes in bcc pattern is (NA is the Avogadro number)
(a) 6.023 × 10²³
(b) 6.023 × 10²²
(c) 60.23 × 10²³
(d) \(\left( \frac { 6.023\times { 10 }^{ 23 } }{ 8\times 40 } \right)\)
Answer:
(b) 6.023 × 10²²

Question 8.
In a solid atom M occupies ccp lattice and \(\left( \frac { 1 }{ 3 } \right)\) of tetrahedral voids are occupied by atom N. Find the formula of solid formed by M and N.
(a) MN
(b) M₃N
(C) MN₃
(d) M₃N₂
Answer:
(d) M₃N₂

Question 9
The ionic radii of A+ and B- are 0.98 × 10^-10 m and 1.81 × 10^-10 m , the coordination number of each ion in AB is ………..
(a) 8
(b) 2
(c) 6
(d) 4
Answer:
(c) 6

Question 10.
CsCl has bcc arrangement, its unit cell edge length is 400pm, its inter atomic distance is ………..
(a) 400pm
(b) 800pm
(c) \(g\sqrt { 3 } \) × 100pm
(d) \(\left( \frac { \sqrt { 3 } }{ 2 } \right)\) × 400 pm
Answer:
(d) \(\left( \frac { \sqrt { 3 } }{ 2 } \right)\) × 400 pm

Quesiton 11.
A solid compound XY has NaCl structure, if the radius of the cation is 100pm , the radius of the anion will be ………..
(a) \(\left( \frac { 100 }{ 0.414 } \right)\)
(b) \(\left( \frac { 0.732 }{ 100 } \right)\)
(c) 100 × 0.414
(d) \(\left( \frac { 0.414 }{ 100 } \right)\)
Answer:
(a) \(\left( \frac { 100 }{ 0.414 } \right)\)

Question 12.
The vacant space in bcc lattice unit cell is ………..
(a) 48%
(b) 23%
(c) 32%
(d) 26%
Answer:
(c) 32%

Question 13.
The radius of an atom is 300pm, if it crystallizes in a face centered cubic lattice, the length of the edge of the unit cell is ………..
(a) 488.5pm
(b) 848.5pm
(c) 884.5pm
(d) 484.5pm
Answer:
(b) 848.5pm

Question 14.
The fraction of total volume occupied by the atoms in a simple cubic is ………..
(a) \(\left( \frac { \pi }{ 4\sqrt { 2 } } \right)\)
(b) \(\left( \frac { \pi }{ 6 } \right)\)
(c) \(\left( \frac { \pi }{ 4 } \right)\)
(d) \(\left( \frac { \pi }{ 3\sqrt { 2 } } \right)\)
Answer:
(b) \(\left( \frac { \pi }{ 6 } \right)\)

Question 15.
The yellow colour in NaCl crystal is due to ………..
(a) excitation of electrons in F centers
(b) reflection of light from Cl- ion on the surface
(c) refraction of light from Na+ ion
(d) all of the above
Answer:
(a) excitation of electrons in F centers

Question 16.
If ’a’ stands for the edge length of the cubic system; sc ,bcc, and fcc. Then the ratio of radii of spheres in these systems wilL be respectively.

Answer:
\(\left(\frac{1}{2} a: \frac{\sqrt{3}}{4} a: \frac{1}{2 \sqrt{2}} a\right)\)

Question 17.
If a is the length of the side of the cube, the distance between the body centered atom and one comer atom in the cube will be ………..
(a) \(\left( \frac { 2 }{ \sqrt { 3 } } \right) a\)
(b) \(\left( \frac { 4 }{ \sqrt { 3 } } \right) a\)
(c) \(\left( \frac { \sqrt { 3 } }{ 4 } \right) a\)
(d) \(\left( \frac { \sqrt { 3 } }{ 2 } \right) a\)
Answer:
(d) \(\left( \frac { \sqrt { 3 } }{ 2 } \right) a\)

Question 18.
Potassium has a bcc structure with nearest neighbor distance 4.52 A. its atomic weight is 39. Its density will be ………..
(a) 915 kg m^-3
(b) 2142 kg m^-3
(c) 452 kg m^-3
(d) 390 kg m^-3
Answer:
(a) 915 kg m^-3

Question 19.
Schottky defect in a crystal is observed when ………..
(a) unequal number of anions and anions are missing from the lattice
(b) equal number of anions and anions are missing from the lattice
(e) an ion leaves its normal site and occupies an interstitial site
(d) no ion is missing from its lattice.
Answer:
(b) equal number of anions and anions are missing from the lattice

Question 20.
The cation leaves its normal position in the crystal and moves to some interstitial position, the defect in the crystal is known as ………..
(a) Schottky defect
(b) F center
(c) Frenkel defect
(d) non-stoichiometric defect
Answer:
(c) Frenkel defect

Question 21.
Assertion – due to Frenkel defect, density of the crystalline solid decreases.
Reason – in Frenkel defect cation and anion leaves the crystal.
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false
Answer:
(d) Both assertion and reason are false

Question 22.
The crystal with a metal deficiency defect is ………..
(a) NaCI
(b) FeO
(c) ZnO
(a) KCI
Answer:
(b) FeO

Question 23.
A two dimensional solid pattern formed by two different atoms X and Y is shown below. The black and white squares represent atoms X and Y respectively. The simplest formula for the compound based on the unit cell from the pattern is ………..

(a) XY₈
(b) X₄Y₉
(c) XY₂
(d) XY₄
Answer:
(a) XY₈

II. Answer the following questions:

Question 1.
Define unit cell.
Answer:
A basic repeating structural unit of a crystalline solid is called a unit ccli.

Question 2.
Give any three characteristics of ionic crystals.
Answer:

1. Ionic solids have high melting points.
2. These solids do not conduct electricity, because the ions are fixed in their lattice positions.
3. They do conduct electricity in a molten state (or) when dissolved in water because the ions are free to move in the molten state or solution.

Question 3.
Differentiate crystalline solids and amorphous solids.
Answer:
Crystalline solids:

1. A long-range orderly arrangement of constituents
2. Definite shape
3. Generally, crystalline solids are anisotropic in nature
4. They are true solids
5. Definite Heat of fusion
6. They have sharp melting points.
7. Examples: NaCl, diamond etc.,

Amorphous solids:

1. The short-range, random arrangement of constituents
2. Irregular shape
3. They are isotropic like liquids
4. They are considered as pseudo solids (or) supercooled liquids
5. Heat of fusion is not definite
6. Gradually soften over a range of temperatures and so can be moulded.
7. Examples: Rubber, plastics, glass etc

Question 4.
Classify the following solids.
a) P₄
b) Brass
c) Diamond
d) NaCI
e) iodine
Answer:
P₄ – Molecular solid
Brass – Metallic solid
Diamond –
NaCl – Ionic solid
Iodine – Molecular solid

Question 5.
Explain briefly seven types of unit cell.
Answer:
Seven types of unit cell:

Cubic – NaCl
Tetragonal – TiO₂
Orthorhombic – BaSO₄
Hexagonal – ZnO
Monoclinic – PbCrO₄
Triclinic – H₃BO₃
Rhombohedral – Cinnabar Cubic

They differ in the arrangements of their crystallographic axes and angles. Corresponding to the above seven, Bravis defined 14 possible crystal system as shown in the figure.

Question 6.
Distinguish between hexagonal close packing and cubic close packing.
Answer:
Hexagonal close-packing

1. aba arrangement
2. In this case, the spheres of the third layer are exactly aligned with those of the first layer
3. In HCP, tetrahedral voids of the second layer may be covered by the spheres of the third layer

Cubic close packing:

1. abc arrangement
2. In this case, the spheres of the third layer are not aligned with those of the first layer or second layer. Only when the fourth layer is placed, its spheres are aligned with the first layer.
3. In cep third layer may be placed above the second layer in a manner such that its sphere covers the octahedral voids

Question 7.
Distinguish tetrahedral and octahedral voids.
Answer:
Tetrahedral void

1. A single triangular void in a crystal is surrounded by four (4) spheres and is called a tetrahedral void
2. A sphere of second layer is above the void of the first layer, a tetrahedral void is formed
3. This constitutes four spheres, three in the lower and one in upper layer. When the centres of these four spheres are joined a tetrahedron is formed
4. The radius of the sphere which can be accommodated in an octahedral hole without disturbing the structure should not exceed 0.414 times that of the structure forming sphere
5. Radius of an tetrahedral void \(\frac { r }{ R }\) = 0.225

Octahedral void

1. A double triangular void like c is surrounded by six(6) spheres and is called an octahedral void
2. The voids in the first layer are partially covered by the spheres of layer now such a void is called a octahedral void
3. This constitutes six spheres, three in the lower layer and three in the upper layer. When the centers of these six spheres are joined an octahedron is formed
4. The sphere which can be placed in a tetrahedral hole without disturbing the close-packed structure should not have a radius larger than 0.225 times the radius of the sphere-forming the structure
5. Radius of a octahedral void \(\frac { r }{ R } \) = 0.414

Question 8.
What are point defects?
Answer:
If the deviation occurs due to missing atoms, displaced atoms or extra atoms the imperfection is named as a point defect. Such defects arise due to imperfect packing during the original crystallisation or they may arise from thermal vibrations of atoms at elevated temperatures.

Question 9.
Explain Schottky defect.
Answer:
Schottky defect arises due to the missing of equal number of cations and anions from the crystal lattice. This effect does not change the stoichiometry of the crystal.Ionic solids in which the cation and anion are of almost of similar size show schottky defect.
Example: NaCl.

Presence of large number of schottky defects in a crystal, lowers its density. For example, the theoretical density of vanadium monoxide (VO) calculated using the edge length of the unit cell is 6.5 g cm-3 but the actual experimental density is 5.6 gcm3. It indicates that there is approximately 14% Schottky defect in VO crystal. Presence of Schottky defect in the crystal provides a simple way by which atoms or ions can move within the crystal lattice.

Question 10.
Write short note on metal excess and metal deficiency defect with an example. Metal excess defect.
Answer:
Metal excess defect arises due to the presence of more number of metal ions as compared to anions.Alkali metal halides NaCl, KCl show this type of defect.The electrical neutrality of the crystal can be maintained by the presence of anionic vacancies equal to the excess metal ions (or) by the presence of extra cation and electron present in interstitial position.

For example, when NaCl crystals are heated in the presence of sodium vapour, Na⁺ ions are formed and are deposited on the surface of the crystal. Chloride ions (Cl^–) diffuse to the surface from the lattice point and combines with Na⁺ ion.

The electron lost by the sodium vapour diffuse into the crystal lattice and occupies the vacancy created by the Cl^– ions. Such anionic vacancies which are occupied by unpaired electrons are called F centers. Hence, the formula of NaCl which contains excess Na+ ions can be written as Na_1+xCl.

Metal deficiency defect:
Metal deficiency defect arises due to the presence of less number of cations than the anions. This defect is observed in a crystal in which, the cations have variable oxidation states. For example, in FeO crystal, some of the Fe²⁺ ions are missing from the crystal lattice.

To maintain the electrical neutrality, twice the number of other Fe²⁺ ions in the crystal is oxidized to Fe³⁺ ions. In such cases, overall number of Fe2²⁺ and Fe³⁺ ions is less than the O^2- ions. It was experimentally found that the general formula of ferrous oxide is FexO, where x ranges from 0.93 to 0.98.

Question 11.
Calculate the number of atoms in a fee unit cell.
Answer:
Number of atoms in a fee unit cell,
= \(\frac { { N }_{ c } }{ 8 }\) + \(\frac { { N }_{ f } }{ 2 }\)
= \(\frac { 8 }{ 8 } \) + \(\frac { 6 }{ 2 } \) = 1 + 3 = 4

Question 12.
Explain AAAA and ABABA and ABCABC type of three dimensional packing with the help of neat diagram.
Answer:
1. AAAA type of three dimensional packing
This type of three dimensional packing arrangements can be obtained by repeating the AAAA type two dimensional arrangements in three dimensions, i.e., spheres in one layer sitting directly on the top of those in the previous layer so that all layers are identical.