>Let $R_1 = 5\,\Omega$ and $R_2 = 15\,\Omega$.

Concept:

At null point, potential difference across $R$ (shunt) is balanced by potential across $l$ cm of the potentiometer wire.

So,

$ \frac{E \times R_1}{R_1 + r} \propto l_1 $

$ \frac{E \times R_2}{R_2 + r} \propto l_2 $

Taking ratio:

$ \frac{\left(\frac{E R_1}{R_1 + r}\right)}{\left(\frac{E R_2}{R_2 + r}\right)} = \frac{l_1}{l_2} $

$ \frac{R_1}{R_1 + r} \cdot \frac{R_2 + r}{R_2} = \frac{l_1}{l_2} $

Now, cross-multiplied:

$ \frac{R_1 (R_2 + r)}{R_2 (R_1 + r)} = \frac{l_1}{l_2} $

Substitute the values:

$ \frac{5 (15 + r)}{15 (5 + r)} = \frac{200}{300} = \frac{2}{3} $

Now, solve for $r$:

Cross-multiplied:

$ 3 \times 5 (15 + r) = 2 \times 15 (5 + r) $

$ 15 (15 + r) = 10 (5 + r) $

$ 225 + 15 r = 50 + 10 r $

$ 15 r - 10 r = 50 - 225 $

$ 5 r = -175 $

$ r = -35 $

Check calculation:

Let's try again, carefully:

$ 3 \times 5 (15 + r) = 2 \times 15 (5 + r) $

$ 15 (15 + r) = 30 (5 + r) $

$ 225 + 15 r = 150 + 30 r $

$ 225 - 150 = 30r - 15r $

$ 75 = 15r $

$ r = \frac{75}{15} = 5 \Omega $

Final Answer:

The internal resistance of the cell is $\boxed{5\,\Omega}$.

Correct Option: Option C

Current Electricity

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