1. Use the linear relation for resistance with temperature

For a metallic wire over a moderate temperature range, $R_t = R_0(1 + \alpha t)$ where:

• $R_t$ = resistance at temperature $t^ \circ C$
• $R_0$ = resistance at $0^\circ C$
• $\alpha$ = temperature coefficient of resistance

2. Write the two given conditions

At $50^\circ C$: $R_{50} = R_0(1 + 50\alpha) = 5$

At $100^\circ C$: $R_{100} = R_0(1 + 100\alpha) = 6$

So we have: $R_0(1 + 50\alpha) = 5 \quad ...(1)$ $R_0(1 + 100\alpha) = 6 \quad ...(2)$

3. Divide equation (2) by equation (1)

$\frac{1 + 100\alpha}{1 + 50\alpha} = \frac{6}{5}$

Cross-multiplying: $5(1 + 100\alpha) = 6(1 + 50\alpha)$

$5 + 500\alpha = 6 + 300\alpha$

$200\alpha = 1$

$\alpha = \frac{1}{200}$

4. Substitute back to find $R_0$

Using equation (1): $R_0\left(1 + 50 \cdot \frac{1}{200}\right) = 5$

$R_0\left(1 + \frac{1}{4}\right) = 5$

$R_0 \cdot \frac{5}{4} = 5$

$R_0 = 4\,\Omega$

5. Check with second condition

At $100^\circ C$: $R_{100} = 4\left(1 + 100\cdot \frac{1}{200}\right) = 4\left(1 + \frac{1}{2}\right) = 4 \cdot \frac{3}{2} = 6\,\Omega$

This matches the given value, so the result is correct.

6. Option evaluation

• A: $3\,\Omega$ ❌
• B: $2\,\Omega$ ❌
• C: $1\,\Omega$ ❌
• D: $4\,\Omega$ ✅

Hence, the resistance at $0^\circ C$ is $4\,\Omega$.