**Notes**
**Concept map**
**Assignment**
**Assertion and Reason Type Question**
**Following question consider two statements one labelled as statement-1 and another as statement-2. Study both statements and mark your answer using the codes given blow.**
- Both Statement-1 and Statement-2 are true and Statement-2 is correct explanation of Statement-1.
- Both Statement-1 and Statement-2 are true and Statement-2 is not a correct explanation of Statement-1.
- Statement-1 is true but statement-2 is false.
- Statement-1 is false and statement-2 is true.

**Question-1**
**Statement-1 **Free electrons in conductor are always in state of continuous random motion.

**Statement-2 **The random motion of free electrons is due to thermal energy of the conductor.

**Question-2**
**Statement-1 **Drift velocity of electrons in a conductor increase on increasing the temperature of the conductor.

**Statement-2 **On increasing the temperature of a conductor, resistivity of the conductor increases.

**Question-3**
**Statement-1 **For metals relaxation time for electrons decreases with increase in temperature.

**Statement-2 **With the increase in temperature, number of collision per unit time that electrons made with lattice ion also increases.

Solution 1-3
1) a

2) d

3) b

## Multiple Choice Type Questions

**Question-4**
Resistivity of the material of a conductor having uniform area of cross-section varies along its length according to the relation

ρ= ρ

_{0 }(a+bx)

if L is the length of the conductor and be the area of cross-section then resistivity of the conductor given by relation

Solution
Consider an infinitesimally small length element dx of the conductor at a distance x from its one end. Then, the resistance dR of the length element dx of the conductor is

$dR=\rho \frac {dx}{A}= \rho _0 (a+bx) \frac {dx}{A}$

The resistance of the whole length of the conductor is

$R=\int_{0}^{L} dR= \int_{0}^{L} \rho _0 (a+bx) \frac {dx}{A} =\frac {\rho _0 }{A} (aL + \frac {bL^2}{2})$

obtained on solving the integral

**Question-5**
Given a current carrying wire of non-uniform cross-section. Which of the following is constant through out the wire.

**(A) **Current only

** **
**(B) **Current and drift speed

**(C) **Drift speed only

** **
**(D) **Current, drift speed and electric field

Solution
We know that I = neAv_{d }, Drift Velocity $v_d=\frac {1}{neA}_{ }\alpha \frac {1}{A}$ ,

For non-uniform cross-section drift speed will be different for different situations. Only current (rate of flow of charge) will same.

**Question-6**
When potential difference across a given copper wire is increase, drift velocity of charge carriers

**(A)** Decreases

** **
**(B) **Increases

**(C) **Remain same

** **
**(D) **Get reduced to zero

Solution
We know that

$v_d=\frac {eE}{m} \tau $

If L is the length of the copper wire and V is the potential difference across it then

$v_d=\frac {e}{m} \frac {V}{L} \tau $

$v_d _{ }\alpha V$ i.e. it potential difference is increase drift velocity of electrons also increases

**Question-7**
A material B has twice the specific resistance of the material A. A circular wire made of B has twice the diameter of the wire made of A. Then , for the two wires to have to have the same resistance, the ratio

L

_{a}/L

_{b} of their respective lengths must be

**(A)** ½

**(B) **2

**(C) **¼

**(D) **1

Solution
Let L_{a} and L_{b} be the lengths of two wires.

Given that ρ_{A}=ρ i.e. ρ_{B}=2ρ also diameter of wire A is D_{A}=D then for wire B D_{B}=2D,

The resistance of the wire made of material A,

$R_A= \frac {\rho _A L_A}{\pi (\frac {D_A}{2})^2} = \frac {\rho _A L_A}{\pi (\frac {D}{2})^2} = \frac {4\rho L_A}{\pi D^2}$

Resistance of the wire made of material B

$R_B= \frac {\rho _B L_B}{\pi (\frac {D_B}{2})^2} = \frac {\rho _B L_B}{\pi (\frac {2D}{2})^2} = \frac {2\rho L_B}{\pi D^2}$

Since R_{A}=R_{B}, we have

$\frac {4\rho L_A}{\pi D^2}=\frac {2\rho L_B}{\pi D^2}$

$\frac {L_A}{L_B} =\frac {1}{2} $

**Question-8**
What will be the equivalent resistance between A and D

**(A) **40

** Ω **
**(B) **20

**Ω**
**(C)** 30

** Ω **
**(D) **10

**Ω**
Solution
Resistance at the end B and C do not be the part of network of resistance between point A and D.

Therefore equivalent resistance between points A and D is

$R= 10 + \frac {1}{10+10} + \frac {1}{10+10} +10$

$R=30 \Omega $

**Question-9**
In the given circuit below the equivalent resistance between points A and C

**(A) **32

** **
**(B) **30.7

**(C)** 33.07

** **
**(D) **3.07

Solution
Resistance of the segment AB and BC are in series. Therefore effective resistance is

$R_1=9+5=14\Omega$

Similarly resistance of the segment AD and DC are in series. If R_{2 }is the effective resistance

$R_2 =3+6 =9 \Omega$

The resistance R_{1} and R_{2 }are in parallel with $ 7 \Omega $ resistor. Then equivalent resistance is

$\frac {1}{R_eq}= \frac {1}{14} + \frac {1}{9} + \frac {1}{7]$

R_{eq }=3.07 $\Omega$

**Question-10**
A circular ring having negligible resistance is used to connect four resistors as shown below in the figure.

The equivalent resistance between point A and B is

**(A) **10R

** **
**(B) **2R

**(C) **3R

** **
**(D)** R

Solution
Above network of resistors is equivalent to the arrangement shown below in the figure

If R

_{1} is equivalent resistance of parallel combination then

$\frac {1}{R_1 = \frac {1} {3R} + \frac {1} {3R} + \frac {1} {3R}$

R

_{1} = R

Therefore equivalent resistance of the network between A and B is

R

_{eq}=R+R=2R

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Class 12 Maths
Class 12 Physics