Current Electricity — Page 11 | JEE Physics

Q101

A wire of resistance

160 ~\Omega

is melted and drawn in a wire of one-fourth of its length. The new resistance of the wire will be

A

640 ~\Omega

B

40 ~\Omega

C

16 ~\Omega

D

10 ~\Omega

Correct Answer

Option D

Solution

Let the original length of the wire be L and its cross-sectional area be A. Then, its resistance R is given by:

R = \frac{\rho L}{A}

where $\rho$ is the resistivity of the material of the wire.

When the wire is melted and drawn into a wire of one-fourth of its length, its new length is L/4 and its new cross-sectional area is 4A (since the same amount of material is now spread over a longer length).

Therefore, its new resistance R' is given by:

R' = \frac{\rho (L/4)}{4A} = \frac{R}{16}

Substituting the given value of R, we get:

R' = \frac{160}{16} = 10 ~\Omega

Therefore, the new resistance of the wire is

10 ~\Omega

.

Q102

Two identical heater filaments are connected first in parallel and then in series. At the same applied voltage, the ratio of heat produced in same time for parallel to series will be:

A 4 : 1

B 1 : 4

C 2 : 1

D 1 : 2

Correct Answer

Option A

Solution

Let's consider the power

P

dissipated by a single heater filament with resistance

R

when connected to a voltage

V

. The power dissipated by the heater filament is given by:

P = \frac{V^2}{R}

When the two identical heater filaments are connected in parallel, the equivalent resistance

R_P

is given by:

\frac{1}{R_P} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \Rightarrow R_P = \frac{R}{2}

The total power dissipated

P_P

by the two filaments connected in parallel is:

P_P = \frac{V^2}{R_P} = \frac{V^2}{\frac{R}{2}} = 2V^2 \cdot \frac{1}{R} = 2P

When the two identical heater filaments are connected in series, the equivalent resistance

R_S

is given by:

R_S = R + R = 2R

The total power dissipated

P_S

by the two filaments connected in series is:

P_S = \frac{V^2}{R_S} = \frac{V^2}{2R} = \frac{1}{2}V^2 \cdot \frac{1}{R} = \frac{1}{2}P

Now, let's find the ratio of the heat produced in the same time for parallel to series connection:

\frac{P_P}{P_S} = \frac{2P}{\frac{1}{2}P} = \frac{2P \times 2}{P} = \frac{4P}{P} = 4 : 1

Q103

The current sensitivity of moving coil galvanometer is increased by

25 \%

. This increase is achieved only by changing in the number of turns of coils and area of cross section of the wire while keeping the resistance of galvanometer coil constant. The percentage change in the voltage sensitivity will be:

A +25%

B

-

50%

C

-

25%

D Zero

Correct Answer

Option A

Solution

The current sensitivity ( $I_s$ ) of a moving coil galvanometer is given by:

I_s = \frac{nBA}{C}

where

n

is the number of turns in the coil,

B

is the magnetic field,

A

is the area of the coil, and

C

is the torsional constant of the suspension wire.

We are told that the current sensitivity is increased by 25% by changing only the number of turns and the area of the cross section while keeping the resistance of the galvanometer coil constant.

Let the new number of turns be

n'

, and the new area be

A'

. Since the current sensitivity is increased by 25%, we have:

I_s' = 1.25I_s = \frac{n' B A'}{C}

The resistance ( $R$ ) of the galvanometer coil is given by:

R = \rho \frac{l}{A}

where $\rho$ is the resistivity of the wire and

l

is the length of the wire. Since the resistance is kept constant, we can write:

R' = \rho \frac{l'}{A'} = R

Now, the voltage sensitivity ( $V_s$ ) of the galvanometer is given by:

V_s = I_s \times R

We want to find the percentage change in the voltage sensitivity. Let the new voltage sensitivity be

V_s'

:

V_s' = I_s' \times R'

Since

R = R'

, we can write:

V_s' = 1.25I_s \times R = 1.25V_s

The percentage change in the voltage sensitivity is:

\frac{V_s' - V_s}{V_s} \times 100\% = \frac{1.25V_s - V_s}{V_s} \times 100\% = 25\%

Q104

In a metre-bridge when a resistance in the left gap is

2 \Omega

and unknown resistance in the right gap, the balance length is found to be

40 \mathrm{~cm}

. On shunting the unknown resistance with

2 \Omega

, the balance length changes by :

A

62.5

B

22.5 \mathrm{~cm}

C

20 \mathrm{~cm}

D

65 \mathrm{~cm}

Correct Answer

Option B

Solution

To solve this problem, let's first understand that a meter bridge setup is based on the principle of a Wheatstone bridge, in which two unknown resistances are in such a configuration that if the bridge is balanced, the ratio of the resistances on one side is equal to the ratio of resistances on the other side.

In a balanced condition, no current flows through the galvanometer that is connected diagonally across the bridge.

We can express the condition of the balance as follows:

\frac{R_1}{R_2} = \frac{L_1}{L_2}

where $R_1$ is the known resistance $2 Ω$ , $R_2$ is the unknown resistance, $L_1$ is the balance length $40 cm$ and $L_2$ is the length of the remaining wire on the meter bridge (100 cm - 40 cm = 60 cm).

Let's calculate the initial unknown resistance ( $R_2$ ) using the balance condition:

\frac{2}{R_2} = \frac{40}{60}

\Rightarrow R_2 = \frac{2 \times 60}{40} = 3 \Omega

Now, when the unknown resistance $R_2$ is shunted with a 2 Ω resistor, the new combined resistance ( $R'_2$ ) can be calculated using the parallel resistance formula:

\frac{1}{R'_2} = \frac{1}{R_2} + \frac{1}{2}

\Rightarrow \frac{1}{R'_2} = \frac{1}{3} + \frac{1}{2} = \frac{2 + 3}{6} = \frac{5}{6}

Therefore, the new combined resistance ( $R'_2$ ) is:

R'_2 = \frac{6}{5} \Omega

If $L'_1$ is the new balance length and $L'_2$ is the remaining length, we now have:

\frac{2}{\frac{6}{5}} = \frac{L'_1}{100 - L'_1}

\Rightarrow \frac{L'_1}{100 - L'_1} = \frac{5}{3}

Now, solve for (L'_1):

3L'_1 = 5(100 - L'_1)

\Rightarrow 3L'_1 = 500 - 5L'_1

\Rightarrow 8L'_1 = 500

\Rightarrow L'_1 = 62.5 \mathrm{~cm}

The balance length has changed from 40 cm to 62.5 cm, so the change by $L'_1 - L_1$ is:

62.5 - 40 = 22.5 \mathrm{~cm}

Therefore, the balance length changes by 22.5 cm, which corresponds to Option B.

Q105

In an ammeter,

5 \%

of the main current passes through the galvanometer. If resistance of the galvanometer is

\mathrm{G}

, the resistance of ammeter will be :

A

199 \mathrm{~G}

B

200 \mathrm{~G}

C

\dfrac{G}{20}

D

\dfrac{\mathrm{G}}{199}

Correct Answer

Option C

Solution

When an ammeter is designed, a shunt resistor ( $R_s$ ) is placed in parallel with the galvanometer to ensure that only a small fraction of the total current passes through the galvanometer itself.

This is done because the galvanometer is usually a sensitive instrument designed to measure small currents, and passing a large current through it could damage it.

In this scenario, we are told that $5\%$ of the main current passes through the galvanometer.

This means that the remaining $95\%$ of the current must pass through the shunt.

Let's denote the total current as $I$ , the current through the galvanometer as $I_g$ , and the current through the shunt as $I_s$ .

Therefore, we have:

I_g = \frac{5}{100} I

I_s = I - I_g = I - \frac{5}{100} I = \frac{95}{100} I

Since the galvanometer and shunt are in parallel, the voltage across each must be the same:

V_g = V_s

According to Ohm's law, $V = IR$ , where $V$ is the voltage, $I$ is the current, and $R$ is the resistance.

Hence, for the galvanometer and the shunt:

I_g G = I_s R_s

By substituting $I_g$ and $I_s$ from the above proportionality, we get:

\left(\frac{5}{100} I\right) G = \left(\frac{95}{100} I\right) R_s

Let's solve for $R_s$ :

R_s = \frac{5}{95} G

Further simplifying this:

R_s = \frac{G}{19}

The total resistance of the ammeter $R_a$ can be found using the parallel resistance formula:

\frac{1}{R_a} = \frac{1}{G} + \frac{1}{R_s}

Substitute $R_s$ with $\dfrac{G}{19}$ :

\frac{1}{R_a} = \frac{1}{G} + \frac{19}{G}

\frac{1}{R_a} = \frac{20}{G}

Thus the resistance of the ammeter $R_a$ is:

R_a = \frac{G}{20}

Hence, the correct answer is Option C: $\dfrac{G}{20}$ .

Q106

A motor operating on 100 V draws a current of 1 A . If the efficiency of the motor is

91.6 \%

, then the loss of power in units of

\mathrm{cal} / \mathrm{s}

is

A 6.2

B 2

C 8.4

D 4

Correct Answer

Option B

Solution

The motor operates at 100 V and draws 1 A, so the total electrical power input is:

P_{\text{in}} = IV = 100 \, \text{V} \times 1 \, \text{A} = 100 \, \text{W}.

With an efficiency of 91.6%, only 91.6% of the input power is used for the motor's work, while the rest is lost as heat.

The output power is:

P_{\text{out}} = 0.916 \times 100 \, \text{W} = 91.6 \, \text{W}.

The power lost is:

P_{\text{loss}} = P_{\text{in}} - P_{\text{out}} = 100 \, \text{W} - 91.6 \, \text{W} = 8.4 \, \text{W}.

Since $1 \, \text{W} = 1 \, \text{J/s}$ and $1 \, \text{cal} \approx 4.2 \, \text{J}$ , we convert the lost power into calories per second:

P_{\text{loss}} (\text{in cal/s}) = \frac{8.4 \, \text{J/s}}{4.2 \, \text{J/cal}} = 2 \, \text{cal/s}.

Thus, the correct answer is: Option B: 2

Q107

A galvanometer has a resistance of

50 ~\Omega

and it allows maximum current of

5 \mathrm{~mA}

. It can be converted into voltmeter to measure upto

100 \mathrm{~V}

by connecting in series a resistor of resistance :

A

19500 \Omega

B

5975 \Omega

C

20050 \Omega

D

19950 \Omega

Correct Answer

Option D

Solution

To convert a galvanometer into a voltmeter to measure higher voltages, you need to add a series resistance to it.

Let's figure out the required resistance value.

First, let's find the maximum voltage that can be directly measured by the galvanometer without any additional resistance.

We know the maximum current, $I$ , that the galvanometer can safely measure is $5 \text{ mA}$ , and the resistance of the galvanometer, $R_g$ , is $50 \Omega$ .

Using Ohm's law $V = I \times R$ , the maximum voltage $V_g$ the galvanometer can measure is: $V_g = I \times R_g$ $V_g = 5 \times 10^{-3} \text{ A} \times 50 \Omega$ $V_g = 0.25 \text{ V}$ Next, to measure up to $100 \text{ V}$ , we need to add a series resistor $R_s$ so that the total voltage drop when the maximum current is flowing is $100 \text{ V}$ .

The voltage drop across the additional resistor $R_s$ when the maximum current flows would be the total voltage minus the voltage across the galvanometer: $V_s = V_{\text{total}} - V_g$ $V_s = 100 \text{ V} - 0.25 \text{ V}$ $V_s = 99.75 \text{ V}$ Applying Ohm's law to the series resistor to find its resistance value: $R_s = \dfrac{V_s}{I}$ $R_s = \dfrac{99.75 \text{ V}}{5 \times 10^{-3} \text{ A}}$ $R_s = 19950 \Omega$ Therefore, the resistance of the series resistor required to convert the galvanometer into a voltmeter that can measure up to $100 \text{ V}$ is $19950 \Omega$ .

The correct option is D: $19950 \Omega$ .

Q108

A wire of length

10 \mathrm{~cm}

and radius

\sqrt{7} \times 10^{-4} \mathrm{~m}

connected across the right gap of a meter bridge. When a resistance of

4.5 \Omega

is connected on the left gap by using a resistance box, the balance length is found to be at

60 \mathrm{~cm}

from the left end. If the resistivity of the wire is

\mathrm{R} \times 10^{-7} \Omega \mathrm{m}

, then value of

\mathrm{R}

is :

A 63

B 70

C 66

D 35

Correct Answer

Option C

Solution

For null point,

\begin{aligned} & \frac{4.5}{60}=\frac{R}{40} \\ & \text{ Also, } R=\frac{\rho \ell}{A}=\frac{\rho \ell}{\pi r^2} \\ & 4.5 \times 40=\rho \times \frac{0.1}{\pi \times 7 \times 10^{-8}} \times 60 \\ & \rho=66 \times 10^{-7} \Omega \times \mathrm{m} \end{aligned}

Q109

A wire of resistance

\mathrm{R}

and length

\mathrm{L}

is cut into 5 equal parts. If these parts are joined parallely, then resultant resistance will be :

A

\dfrac{1}{25} \mathrm{R}

B

\dfrac{1}{5} R

C 25 R

D 5 R

Correct Answer

Option A

Solution

Resistance of each part

=\frac{R}{5}

Total resistance

=\frac{1}{5} \times \frac{\mathrm{R}}{5}=\frac{\mathrm{R}}{25}

Q110

A current of

200 \mu \mathrm{A}

deflects the coil of a moving coil galvanometer through

60^{\circ}

. The current to cause deflection through

\dfrac{\pi}{10}

radian is :

A 120

\mu

A

B 180

\mu

A

C 30

\mu

A

D 60

\mu

A

Correct Answer

Option D

Solution

\mathrm{i} \propto \theta

(angle of deflection)

\begin{aligned} & \therefore \frac{\mathrm{i}_2}{\mathrm{i}_1}=\frac{\theta_2}{\theta_1} \Rightarrow \frac{\mathrm{i}_2}{200 \mu \mathrm{A}}=\frac{\pi / 10}{\pi / 3}=\frac{3}{10} \\ & \Rightarrow \mathrm{i}_2=60 \mu \mathrm{A} \end{aligned}