Current Electricity — Page 10 | JEE Physics

Q91

The number of turns of the coil of a moving coil galvanometer is increased in order to increase current sensitivity by

50 \%

. The percentage change in voltage sensitivity of the galvanometer will be :

A

0 \%

B

75 \%

C

100 \%

D

50 \%

Correct Answer

Option A

Solution

Current sensitivity $=$ Voltage sensitivity $\times R$ Current sensitivity is made $1.5$ times. $R$ also increase $1.5$ times.

Hence voltage sensitivity $=\dfrac{1.5 \times \text{ current sensitivity }}{1.5 \times R}$

=\text{ no change }

Q92

The

\mathrm{H}

amount of thermal energy is developed by a resistor in

10 \mathrm{~s}

when a current of

4 \mathrm{~A}

is passed through it. If the current is increased to

16 \mathrm{~A}

, the thermal energy developed by the resistor in

10 \mathrm{~s}

will be :

A

\dfrac{\mathrm{H}}{4}

B

16 \mathrm{H}

C H

D

4 \mathrm{H}

Correct Answer

Option B

Solution

$H \propto i^{2}$ for $t=$ constant $\Rightarrow$

\frac{H}{H^{\prime}}=\left(\frac{4}{16}\right)^{2}

$\Rightarrow$ $H^{\prime}=16 H$

Q93

The drift velocity of electrons for a conductor connected in an electrical circuit is

\mathrm{V}_{\mathrm{d}}

. The conductor in now replaced by another conductor with same material and same length but double the area of cross section. The applied voltage remains same. The new drift velocity of electrons will be

A

\dfrac{V_{d}}{4}

B

\mathrm{V}_{\mathrm{d}}

C

2 \mathrm{~V}_{\mathrm{d}}

D

\dfrac{V_{d}}{2}

Correct Answer

Option B

Solution

Drift velocity of electron, $V_d=\dfrac{-e E \tau}{m}$ As $E=\dfrac{V}{l}$ We can write, $V_d=\dfrac{-e V\tau}{m l}$ where, $l=$ length of conductor, $V$ is applied voltage As drift velocity does not depends upon area, so it will remain same, even after changing area of conductor.

Q94

The charge flowing in a conductor changes with time as

\mathrm{Q}(\mathrm{t})=\alpha \mathrm{t}-\beta \mathrm{t}^{2}+\gamma \mathrm{t}^{3}

. Where

\alpha, \beta

and

\gamma

are constants. Minimum value of current is :

A

\beta-\dfrac{\alpha^{2}}{3 \gamma}

B

\alpha-\dfrac{3 \beta^{2}}{\gamma}

C

\alpha-\dfrac{\beta^{2}}{3 \gamma}

D

\alpha-\dfrac{\gamma^{2}}{3 \beta}

Correct Answer

Option C

Solution

Q(t) = \alpha t - \beta {t^2} + \gamma {t^3}

i(t) = \alpha - 2\beta t + 3\gamma {t^2}

{{di} \over {dt}} = - 2\beta + 6\gamma t = 0

(for max/min of i) at

t = {\beta \over {3r}}

(i is minimum as i is an upward parabola)

i\left( {{\beta \over {3\gamma }}} \right) = \alpha - 2\beta \left( {{\beta \over {3\gamma }}} \right) + {{3\gamma {\beta ^2}} \over {9{\gamma ^2}}}

= \alpha {{ - {\beta ^2}} \over {3\gamma }}

Q95

With the help of potentiometer, we can determine the value of emf of a given cell. The sensitivity of the potentiometer is (A) directly proportional to the length of the potentiometer wire (B) directly proportional to the potential gradient of the wire (C) inversely proportional to the potential gradient of the wire (D) inversely proportional to the length of the potentiometer wire Choose the correct option for the above statements :

A A only

B B and D only

C A and C only

D inversely C only

Correct Answer

Option C

Solution

A potentiometer is an electrical instrument used to measure the electromotive force (EMF) of a cell.

The sensitivity of a potentiometer is defined as the change in potential difference per unit length of the wire.

It is directly proportional to the length of the potentiometer wire (Option A), because a longer wire has a larger potential difference that can be measured.

It is inversely proportional to the potential gradient of the wire (Option C), because a smaller potential gradient results in a smaller change in potential difference per unit length.

Therefore, the correct answer is (C) A and C only.

Q96

Ratio of thermal energy released in two resistors R and 3R connected in parallel in an electric circuit is :

A 1 : 1

B 1 : 27

C 1 : 3

D 3 : 1

Correct Answer

Option D

Solution

For parallel connection, potential difference is same $(v)$

\begin{aligned} & P_{1}=\left(\frac{v^{2}}{R_{1}}\right) \\\\ & P_{2}=\left(\frac{v^{2}}{R_{2}}\right) \\\\ & \frac{P_{1}}{P_{2}}=\frac{H_{1}}{H_{2}}=\left(\frac{R_{2}}{R_{1}}\right)=\frac{3 R}{R}=(3: 1) \end{aligned}

Q97

The resistance of a wire is 5

\Omega

. It's new resistance in ohm if stretched to 5 times of it's original length will be :

A 25

B 625

C 5

D 125

Correct Answer

Option D

Solution

\mathrm{R}_{\text{initial }}=\frac{\rho \ell}{A}=5 \Omega

$\because$ Volume of wire is constant in stretching

\begin{aligned} & \mathrm{V}_{\mathrm{i}}=\mathrm{V}_{\mathrm{f}} \\\\ & \mathrm{A}_{\mathrm{i}} \ell_{\mathrm{i}}=\mathrm{A}_{\mathrm{f}} \ell_{\mathrm{f}} \\\\ & \mathrm{A} \ell=\mathrm{A}^{\prime}(5 \ell) \\\\ & \mathrm{A}^{\prime}=\frac{\mathrm{A}}{5} \\\\ & \mathrm{R}_{\mathrm{f}}=\frac{\rho \ell_{\mathrm{f}}}{\mathrm{A}_{\mathrm{f}}}=\frac{\rho(5 \ell)}{\left(\frac{\mathrm{A}}{5}\right)} \\\\ & =25\left(\frac{\rho \ell}{\mathrm{A}}\right) \\\\ & =25 \times 5=125 \Omega \end{aligned}

Q98

A uniform metallic wire carries a current 2 A, when 3.4 V battery is connected across it. The mass of uniform metallic wire is 8.92

\times

10

^{-3}

kg, density is 8.92

\times

10

^{3}

kg/m

^3

and resistivity is 1.7

\times

10

^{-8}~\Omega

-

\mathrm{m}

. The length of wire is :

A

l=100

m

B

l=6.8

m

C

l=5

m

D

l=10

m

Correct Answer

Option D

Solution

$m=8.92 \times 10^{-3} \mathrm{~kg}$ Density $=8.92 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ Volume $=\dfrac{8.92 \times 10^{-3}}{8.92 \times 10^{3}}=\left(10^{-6}\right) \mathrm{m}^{3}$ Resistance $=\dfrac{3.4}{2}=1.7 \Omega=\left(\dfrac{\rho l}{A}\right)$ $1.7=\dfrac{\rho l^{2}}{(A l)}$ $\Rightarrow 1.7=\dfrac{1.7 \times 10^{-8} \times l^{2}}{10^{-6}}$ $l^2=100$ $l=10 \mathrm{~m}$

Q99

Given below are two statements: Statement I : The equivalent resistance of resistors in a series combination is smaller than least resistance used in the combination. Statement II : The resistivity of the material is independent of temperature. In the light of the above statements, choose the correct answer from the options given below :

A Statement I is true but Statement II is false

B Both Statement I and Statement II are true

C Both Statement I and Statement II are false

D Statement I is false but Statement II is true

Correct Answer

Option C

Solution

Statement I is incorrect, as the equivalent resistance of resistors in a series combination is always greater than the largest individual resistance used in the combination.

This is because the total voltage drop across the resistors in series is equal to the sum of the voltage drops across each resistor, and the current through each resistor is the same.

Therefore, the resistance of the entire series combination is the sum of the individual resistances.

Statement II is also incorrect, as the resistivity of most materials changes with temperature.

This is because temperature affects the mobility of charge carriers in a material, which in turn affects its resistivity.

For example, the resistivity of metals generally increases with temperature, while the resistivity of semiconductors generally decreases with temperature.

Therefore, the correct answer is Both Statement I and Statement II are false

Q100

The deflection in moving coil galvanometer falls from 25 divisions to 5 division when a shunt of

24 \Omega

is applied. The resistance of galvanometer coil will be :

A

48 \Omega

B

100 \Omega

C

96 \Omega

D

12 \Omega

Correct Answer

Option C

Solution

Let x = current/division After applying shunt Now

5 \mathrm{x} \times \mathrm{G}=20 \mathrm{x} \times 24

\begin{aligned} & \mathrm{G}=4 \times 24 \\ & \mathrm{G}=96 \Omega \end{aligned}