To measure the internal resistance of a battery, potentiometer is used. For R=10 \Omega , the balance point is observed at l=500 ~cm and for R=1 \Omega the balance point is observed at l=400 ~cm . The internal resistance of the battery is approximately :

Current Electricity PYQ — JEE Main 2024 · 4 Apr · Shift 1 · Q79 with Solution

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Correct answer: B

Principle used

In a potentiometer experiment for finding internal resistance of a cell:

When the cell is on open circuit, balance length $l_1 \propto E$ (emf).
When the cell is connected across an external resistance $R$ , balance length $l_2 \propto V$ (terminal voltage).

Hence,

\frac{E}{V} = \frac{l_1}{l_2}

Also,

V = \frac{ER}{R+r}

where $r$ is the internal resistance.

So,

\frac{E}{V} = \frac{R+r}{R}

Therefore,

\frac{l_1}{l_2} = \frac{R+r}{R}

Use the two given observations

For $R=10\,\Omega$ , balance length is $500\,\text{cm}$ .

For $R=1\,\Omega$ , balance length is $400\,\text{cm}$ .

Since the same potentiometer current is maintained, balance length is proportional to terminal voltage. Thus,

\frac{V_{10}}{V_1} = \frac{500}{400} = \frac{5}{4}

Now,

V_{10} = \frac{E\cdot 10}{10+r}, \qquad V_1 = \frac{E\cdot 1}{1+r}

So,

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

Given this ratio is $\frac{5}{4}$ , hence

\frac{10(1+r)}{10+r} = \frac{5}{4}

Solve for $r$

Cross-multiplying,

40(1+r) = 5(10+r)

40 + 40r = 50 + 5r

35r = 10

r = \frac{10}{35} = \frac{2}{7} \approx 0.286\,\Omega

Thus, the internal resistance is approximately

r \approx 0.3\,\Omega

Check options

A: $0.1\,\Omega$ ❌
B: $0.3\,\Omega$ ✅
C: $0.2\,\Omega$ ❌
D: $0.4\,\Omega$ ❌

Therefore, the correct option is B.

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View written solutionFree

Correct answer: B

Principle used

In a potentiometer experiment for finding internal resistance of a cell:

When the cell is on open circuit, balance length $l_1 \propto E$ (emf).
When the cell is connected across an external resistance $R$ , balance length $l_2 \propto V$ (terminal voltage).

Hence,

\frac{E}{V} = \frac{l_1}{l_2}

Also,

V = \frac{ER}{R+r}

where $r$ is the internal resistance.

So,

\frac{E}{V} = \frac{R+r}{R}

Therefore,

\frac{l_1}{l_2} = \frac{R+r}{R}

Use the two given observations

For $R=10\,\Omega$ , balance length is $500\,\text{cm}$ .

For $R=1\,\Omega$ , balance length is $400\,\text{cm}$ .

Since the same potentiometer current is maintained, balance length is proportional to terminal voltage. Thus,

\frac{V_{10}}{V_1} = \frac{500}{400} = \frac{5}{4}

Now,

V_{10} = \frac{E\cdot 10}{10+r}, \qquad V_1 = \frac{E\cdot 1}{1+r}

So,

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

Given this ratio is $\frac{5}{4}$ , hence

\frac{10(1+r)}{10+r} = \frac{5}{4}

Solve for $r$

Cross-multiplying,

40(1+r) = 5(10+r)

40 + 40r = 50 + 5r

35r = 10

r = \frac{10}{35} = \frac{2}{7} \approx 0.286\,\Omega

Thus, the internal resistance is approximately

r \approx 0.3\,\Omega

Check options

A: $0.1\,\Omega$ ❌
B: $0.3\,\Omega$ ✅
C: $0.2\,\Omega$ ❌
D: $0.4\,\Omega$ ❌

Therefore, the correct option is B.

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