Current Electricity — Page 12 | JEE Physics

Q111

Wheatstone bridge principle is used to measure the specific resistance

\left(S_1\right)

of given wire, having length

L

, radius

r

. If

X

is the resistance of wire, then specific resistance is ;

S_1=X\left(\dfrac{\pi r^2}{L}\right)

. If the length of the wire gets doubled then the value of specific resistance will be :

A

\dfrac{S_1}{4}

B

2 \mathrm{~S}_1

C

\dfrac{\mathrm{S}_1}{2}

D

S_1

Correct Answer

Option D

Solution

The specific resistance (or resistivity) of a material is a fundamental property that describes how much the material resists the flow of electric current.

The resistivity is typically denoted by the symbol $\rho$ (rho), and it can be calculated by using the resistance

X

of a uniform specimen of the material, along with its physical dimensions.

In the case of a wire, the resistivity formula in terms of its resistance

X

, length

L

, and cross-sectional area

A=\pi r^2

is given by:

\rho = \frac{X \cdot A}{L}

This formula is a reinterpretation of Ohm's law, and it states that the specific resistance is proportional to the area of the cross-section of the wire and inversely proportional to its length.

Given that the specific resistance of the wire

S_1

is determined using the formula:

S_1 = X \left(\frac{\pi r^2}{L}\right)

Now, let's see what happens to the specific resistance if the length of the wire is doubled.

If we denote the new length as

2L

, the resistance of the wire with the new length will change because resistance is directly proportional to the length of the wire.

However, resistivity (specific resistance) is an intrinsic property of the material and does not depend on its length or shape, only on its temperature.

Therefore, even if we change the length of the wire, the specific resistance should remain the same.

If we calculate the new resistance

X'

with the doubled length, it would be:

X' = X \left(\frac{2L}{L}\right) = 2X

Thus, we would use the new resistance

X'

and the new length

2L

to calculate the specific resistance again:

S_1' = X' \left(\frac{\pi r^2}{2L}\right) = 2X \left(\frac{\pi r^2}{2L}\right) = X \left(\frac{\pi r^2}{L}\right)

Since this formula is essentially the same as our original formula for

S_1

, we can conclude that:

S_1' = S_1

Therefore, the value of the specific resistance

S_1

will remain the same even if the length of the wire gets doubled. The correct answer to the question is: Option D:

S_1

Q112

The resistance per centimeter of a meter bridge wire is

r

, with

X \Omega

resistance in left gap. Balancing length from left end is at

40 \mathrm{~cm}

with

25 \Omega

resistance in right gap. Now the wire is replaced by another wire of

2 r

resistance per centimeter. The new balancing length for same settings will be at

A 10 cm

B 80 cm

C 40 cm

D 20 cm

Correct Answer

Option C

Solution

\begin{aligned} & \frac{25}{\mathrm{r} \ell_1}=\frac{\mathrm{X}}{\mathrm{r} \ell_2} \quad \text{.... (i)}\\ & \frac{25}{2 \mathrm{r} \ell_1^{\prime}}=\frac{\mathrm{X}}{2 \mathrm{r} \ell^{\prime}{ }_2} \quad \text{.... (ii)} \end{aligned}

From (i) and (ii)

\ell_2^{\prime}=\ell_2=40 \mathrm{~cm}

Q113

Two conductors have the same resistances at

0^{\circ} \mathrm{C}

but their temperature coefficients of resistance are

\alpha_1

and

\alpha_2

. The respective temperature coefficients for their series and parallel combinations are :

A

\alpha_1+\alpha_2, \dfrac{\alpha_1 \alpha_2}{\alpha_1+\alpha_2}

B

\dfrac{\alpha_1+\alpha_2}{2}, \dfrac{\alpha_1+\alpha_2}{2}

C

\alpha_1+\alpha_2, \dfrac{\alpha_1+\alpha_2}{2}

D

\dfrac{\alpha_1+\alpha_2}{2}, \alpha_1+\alpha_2

Correct Answer

Option B

Solution

Series :

\begin{aligned} & \mathrm{R}_{\mathrm{eq}}=\mathrm{R}_1+\mathrm{R}_2 \\ & 2 \mathrm{R}\left(1+\alpha_{\mathrm{eq}} \Delta \theta\right)=\mathrm{R}\left(1+\alpha_1 \Delta \theta\right)+\mathrm{R}\left(1+\alpha_2 \Delta \theta\right) \\ & 2 \mathrm{R}\left(1+\alpha_{\mathrm{eq}} \Delta \theta\right)=2 \mathrm{R}+\left(\alpha_1+\alpha_2\right) \mathrm{R} \Delta \theta \\ & \alpha_{\mathrm{eq}}=\frac{\alpha_1+\alpha_2}{2} \end{aligned}

Parallel :

\begin{aligned} & \frac{1}{R_{\text{eq }}}=\frac{1}{R_1}+\frac{1}{R_2} \\ & \frac{1}{\frac{R}{2}\left(1+\alpha_{\text{eq }} \Delta \theta\right)}=\frac{1}{R\left(1+\alpha_1 \Delta \theta\right)}+\frac{1}{\mathrm{R}\left(1+\alpha_2 \Delta \theta\right)} \end{aligned}

\begin{aligned} & \frac{2}{1+\alpha_{\mathrm{eq}} \Delta \theta}=\frac{1}{1+\alpha_1 \Delta \theta}+\frac{1}{1+\alpha_2 \Delta \theta} \\ & \frac{2}{1+\alpha_{\mathrm{eq}} \Delta \theta}=\frac{1+\alpha_2 \Delta \theta+1+\alpha_1 \Delta \theta}{\left(1+\alpha_1 \Delta \theta\right)\left(1+\alpha_2 \Delta \theta\right)} \\ & 2\left[\left(1+\alpha_1 \Delta \theta\right)\left(1+\alpha_2 \Delta \theta\right)\right] \\ & =\left[2+\left(\alpha_1+\alpha_2\right) \Delta \theta\right]\left[1+\alpha_{\mathrm{eq}} \Delta \theta\right] \\ & 2\left[1+\alpha_1 \Delta \theta+\alpha_2 \Delta \theta+\alpha_1 \alpha_2 \Delta \theta\right] \end{aligned}

\begin{aligned} & = \\ & 2+2 \alpha_{\mathrm{eq}} \Delta \theta+\left(\alpha_1+\alpha_2\right) \Delta \theta+\alpha_{\mathrm{eq}}\left(\alpha_1+\alpha_2\right) \Delta \theta^2 \end{aligned}

Neglecting small terms

\begin{aligned} & 2+2\left(\alpha_1+\alpha_2\right) \Delta \theta=2+2 \alpha_{\mathrm{eq}} \Delta \theta+\left(\alpha_1+\alpha_2\right) \Delta \theta \\ & \left(\alpha_1+\alpha_2\right) \Delta \theta=2 \alpha_{\mathrm{eq}} \Delta \theta \\ & \alpha_{\mathrm{eq}}=\frac{\alpha_1+\alpha_2}{2} \end{aligned}

Q114

A galvanometer having coil resistance

10 \Omega

shows a full scale deflection for a current of

3 \mathrm{~mA}

. For it to measure a current of

8 \mathrm{~A}

, the value of the shunt should be:

A

3.75 \times 10^{-3} \Omega

B

3 \times 10^{-3} \Omega

C

4.85 \times 10^{-3} \Omega

D

2.75 \times 10^{-3} \Omega

Correct Answer

Option A

Solution

To determine the value of the shunt resistor required to convert the galvanometer into an ammeter capable of measuring a current of

8 \mathrm{~A}

, we need to use the concept of shunting where the additional resistor (shunt) is placed in parallel with the galvanometer.

The formula for calculating the value of the shunt resistor $R_s$ is given by:

R_s = \frac{R_g \cdot I_g}{I - I_g}

Where: $R_g$ = resistance of the galvanometer =

10 \Omega

$I_g$ = full-scale deflection current of the galvanometer =

3 \mathrm{~mA}

=

0.003 \mathrm{~A}

$I$ = total current to be measured =

8 \mathrm{~A}

Substituting these values into the formula:

R_s = \frac{10 \Omega \cdot 0.003 \mathrm{~A}}{8 \mathrm{~A} - 0.003 \mathrm{~A}}

Perform the calculations:

R_s = \frac{0.03 \Omega \cdot \mathrm{~A}}{7.997 \mathrm{~A}}

R_s \approx 3.75 \times 10^{-3} \Omega

Thus, the value of the shunt resistor should be: Option A:

3.75 \times 10^{-3} \Omega

Q115

The electric current through a wire varies with time as

I=I_0+\beta t

, where

I_0=20 \mathrm{~A}

and

\beta=3 \mathrm{~A} / \mathrm{s}

. The amount of electric charge crossed through a section of the wire in

20 \mathrm{~s}

is :

A 80 C

B 800 C

C 1000 C

D 1600 C

Correct Answer

Option C

Solution

To calculate the amount of electric charge $Q$ that crosses through a section of the wire over a period of $20$ seconds, given the current varies with time as $I = I_0 + \beta t$ , where $I_0 = 20$ A (initial current) and $\beta = 3$ A/s (rate of change of current with time), we use the concept of integration from calculus because the current is not constant but changes linearly with time.

The electric charge $Q$ is the integral of current $I$ over the time interval from $0$ to $20$ seconds.

The formula for $Q$ is given by: $Q = \int_{0}^{T} I(t) \, dt$ Where $I(t) = I_0 + \beta t$ and $T = 20$ s.

Substituting the given values: $Q = \int_{0}^{20} (20 + 3t) \, dt$ This integral can be solved in two parts: 1.

The integral of the constant term $20$ : $\int 20 \, dt = 20t$ 2.

The integral of the linear term $3t$ : $\int 3t \, dt = \dfrac{3}{2}t^2$ Thus, combining these and evaluating from $0$ to $20$ seconds: $Q = \left[20t + \dfrac{3}{2}t^2\right]_{0}^{20}$ $Q = \left[20(20) + \dfrac{3}{2}(20)^2\right] - \left[20(0) + \dfrac{3}{2}(0)^2\right]$ $Q = 400 + 600 = 1000$ C Therefore, the amount of electric charge that crosses through a section of the wire in $20$ seconds is $1000$ C.

Q116

When a potential difference

V

is applied across a wire of resistance

R

, it dissipates energy at a rate

W

. If the wire is cut into two halves and these halves are connected mutually parallel across the same supply, the energy dissipation rate will become:

A 1/2W

B 4W

C 1/4W

D 2W

Correct Answer

Option B

Solution

\begin{aligned} & \frac{\mathrm{v}^2}{\mathrm{R}}=\mathrm{W} \qquad \text{.... (i)}\\ & \frac{\mathrm{v}^2}{\frac{1}{2}\left(\frac{\mathrm{R}}{2}\right)}=\mathrm{W}^{\prime} \quad \text{.... (ii)} \end{aligned}

From (i) & (ii), we get

W^{\prime}=4 W

Q117

An electric toaster has resistance of

60 \Omega

at room temperature

\left(27^{\circ} \mathrm{C}\right)

. The toaster is connected to a

220 \mathrm{~V}

supply. If the current flowing through it reaches

2.75 \mathrm{~A}

, the temperature attained by toaster is around : ( if

\alpha=2 \times 10^{-4}

/

^\circ \mathrm{C}

)

A 1235

^\circ

C

B 1667

^\circ

C

C 694

^\circ

C

D 1694

^\circ

C

Correct Answer

Option D

Solution

\begin{aligned} & \mathrm{R}_{\mathrm{T}-27}=60 \Omega, R_T=\frac{220}{2.75}=80 \Omega \\ & \mathrm{R}=\mathrm{R}_0(1+\alpha \Delta \mathrm{T}) \\ & 80=60\left[1+2 \times 10^{-4}(\mathrm{~T}-27)\right] \\ & \mathrm{T} \approx 1694^{\circ} \mathrm{C} \end{aligned}

Q118

Given below are two statements : Statement-I : The equivalent emf of two nonideal batteries connected in parallel is smaller than either of the two emfs. Statement-II : The equivalent internal resistance of two nonideal batteries connected in parallel is smaller than the internal resistance of either of the two batteries. In the light of the above statements, choose the correct answer from the options given below.

A Statement-I is false but Statement-II is true

B Statement-I is true but Statement-II is false

C Both Statement-I and Statement-II are false

D Both Statement-I and Statement-II are true

Correct Answer

Option A

Solution

When two non-ideal batteries are connected in parallel, their equivalent emf is

{E_{eq}} = {{{{{E_1}} \over {{r_1}}} + {{{E_2}} \over {{r_2}}}} \over {{1 \over {{r_1}}} + {1 \over {{r_2}}}}} \Rightarrow {E_{eq}} = {{{E_1}{r_2} + {E_2}{r_1}} \over {{r_1} + {r_2}}}

If emf of both cells is same say E,

{E_{eq}} = {{E{r_2} + E{r_1}} \over {{r_1} + {r_2}}} = {{E({r_1} + {r_2})} \over {({r_1} + {r_2})}} = E

Hence, equivalent emf is equal to emf of the single cell.

Also, when two non-ideal batteries are connected in parallel, their equivalent resistance is

{r_{eq}} = {{{r_1}{r_2}} \over {{r_1} + {r_2}}}

Hence, the equivalent internal resistance is smaller than either of the two internal resistances.

Therefore 1 is correct.

Q119

A galvanometer has a coil of resistance

200 \Omega

with a full scale deflection at

20 \mu \mathrm{A}

. The value of resistance to be added to use it as an ammeter of range

(0-20) \mathrm{mA}

is :

A

0.40\Omega

B

0.10\Omega

C

0.20\Omega

D

0.50\Omega

Correct Answer

Option C

Solution

To convert a galvanometer into an ammeter, we need to add a shunt resistance in parallel with the galvanometer's coil.

The purpose of the shunt resistance is to bypass the majority of the current while allowing only a small fraction of it to pass through the galvanometer, thereby preventing it from being damaged by high currents.

Given parameters: Resistance of the galvanometer's coil,

R_g = 200 \Omega

Full-scale deflection current of the galvanometer,

I_g = 20 \mu \mathrm{A} = 20 \times 10^{-6} \mathrm{A}

Desired ammeter range,

I = 20 \mathrm{mA} = 20 \times 10^{-3} \mathrm{A}

The shunt resistance,

R_s

, can be calculated using the formula:

\frac{R_s}{R_g + R_s} = \frac{I_g}{I}

Solving for

R_s

:

R_s = R_g \left(\frac{I_g}{I - I_g}\right)

Substituting the given values:

R_s = 200 \left(\frac{20 \times 10^{-6}}{20 \times 10^{-3} - 20 \times 10^{-6}}\right)

Simplifying the expression:

R_s = 200 \left(\frac{20 \times 10^{-6}}{19.98 \times 10^{-3}}\right)

R_s \approx 200 \left(\frac{20 \times 10^{-6}}{20 \times 10^{-3}}\right) = 200 \left(\frac{1}{1000}\right) = 0.20 \Omega

Therefore, the value of the resistance to be added to use the galvanometer as an ammeter of range

(0-20) \mathrm{mA}

is

0.20 \Omega

. Thus, the correct answer is: Option C:

0.20 \Omega

Q120

To measure the internal resistance of a battery, potentiometer is used. For

R=10 \Omega

, the balance point is observed at

l=500 \mathrm{~cm}

and for

\mathrm{R}=1 \Omega

the balance point is observed at

l=400 \mathrm{~cm}

. The internal resistance of the battery is approximately :

A

0.1 \Omega

B

0.3 \Omega

C

0.2 \Omega

D

0.4 \Omega

Correct Answer

Option B

Solution

To measure the internal resistance of a battery using a potentiometer, we need to understand the principle behind it.

The potentiometer is used to measure the voltage across the battery (emf) under different conditions.

The balance length corresponds to the emf of the battery when no current is drawn, whereas under load, it corresponds to the terminal voltage.

Let's denote the emf of the battery by

E

and the internal resistance by

r

. According to Ohm's law, when a resistance

R

is connected across the battery, the terminal voltage

V

is given by:

V = E - Ir

where

I

is the current through the circuit. From the problem, the balance length

l

is proportional to the voltage across the potentiometer wire. Thus, we can write:

\frac{V_1}{V_2} = \frac{l_1}{l_2}

In the first condition, when

R = 10 \Omega

and the balance length

l_1 = 500 \, \text{cm}

:

V_1 = E - I_1 r

For the second condition, when

R = 1 \Omega

and the balance length

l_2 = 400 \, \text{cm}

:

V_2 = E - I_2 r

Given that:

\frac{V_1}{V_2} = \frac{500}{400} = \frac{5}{4}

Now let's denote the internal emf of the battery as

E

. We can use Ohm’s Law in the calculation of

I_1

and

I_2

:

I_1 = \frac{E}{R_1 + r} = \frac{E}{10 + r}

I_2 = \frac{E}{R_2 + r} = \frac{E}{1 + r}

Now, rewriting the values of

V_1

and

V_2

we have:

V_1 = E - I_1 r = E \left(1 - \frac{r}{10 + r}\right) = E \frac{10}{10 + r}

V_2 = E - I_2 r = E \left(1 - \frac{r}{1 + r}\right) = E \frac{1}{1 + r}

Using the ratio:

\frac{V_1}{V_2} = \frac{\frac{10E}{10 + r}}{\frac{E}{1 + r}} = \frac{10 \cdot (1 + r)}{1 \cdot (10 + r)} = \frac{10 + 10r}{10 + r}

Given:

\frac{V_1}{V_2} = \frac{5}{4}

So, we can set up the equation:

\frac{10 + 10r}{10 + r} = \frac{5}{4}

Cross multiplying gives:

4(10 + 10r) = 5(10 + r)

Expanding both sides:

40 + 40r = 50 + 5r

Rearranging terms to solve for

r

:

40r - 5r = 50 - 40

35r = 10

r = \frac{10}{35} = \frac{2}{7} \approx 0.285 \, \Omega

Since this value is closest to

0.3 \Omega

, the correct option is: Option B:

0.3 \Omega