l with a shunt $R$, the potential difference across it (for a balance length $l$ on potentiometer) is:

$$ V = k l $$

where $k$ is the potential gradient.

Current through the external resistance $R$:

$$ I = \frac{E}{R + r} $$

Potential difference across $R$:

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

At balance, this $V$ is equal to $k l$:

$$ k l = \frac{E R}{R + r} $$

Step 2: Write equations for both conditions

Case 1: $R_1 = 5\,\Omega$, $l_1 = 250$ cm

$$ k \cdot 250 = \frac{E \cdot 5}{5 + r} \tag{1} $$

Case 2: $R_2 = 20\,\Omega$, $l_2 = 400$ cm

$$ k \cdot 400 = \frac{E \cdot 20}{20 + r} \tag{2} $$

Step 3: Divide (1) by (2) to eliminate $k$ and $E$

$$ \frac{k \cdot 250}{k \cdot 400} = \frac{\frac{E \cdot 5}{5 + r}}{\frac{E \cdot 20}{20 + r}} $$

$$ \frac{250}{400} = \frac{5}{5 + r} \times \frac{20 + r}{20} $$

$$ \frac{5}{8} = \frac{5 (20 + r)}{20 (5 + r)} $$

Step 4: Simplify and solve for $r$

Multiply both sides by $20(5 + r)$:

$$ \frac{5}{8} \times 20(5 + r) = 5(20 + r) $$

$$ \frac{100}{8}(5 + r) = 100 + 5r $$

$$ 12.5(5 + r) = 100 + 5r $$

$$ 62.5 + 12.5r = 100 + 5r $$

Subtract $5r$ from both sides:

$$ 62.5 + 7.5r = 100 $$

Subtract $62.5$:

$$ 7.5r = 37.5 $$

$$ r = \frac{37.5}{7.5} = 5\,\Omega $$

Final Answer:

$\boxed{5\,\Omega}$

So, the correct answer is Option C: $5\,\Omega$.

Current Electricity

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