$$ is the resistance at temperature $$T$$ (measured from 0°C),

• $$R_0$$ is the resistance at 0°C, and

• $$\alpha$$ is the temperature coefficient.

However, the wire’s resistance is given at room temperature (20°C) as 10 Ω. So first, we find the resistance at 0°C:

Write the relation at 20°C:

$$ R(20) = R_0 (1+ \alpha \times 20) $$

Substitute the known values:

$$ 10 = R_0 (1+0.002\times20)= R_0 (1+0.04)= 1.04 \, R_0 $$

Solve for $$R_0$$:

$$ R_0 = \frac{10}{1.04}\approx 9.615 \, \Omega $$

Now, when the resistance increases by 10%, the new resistance becomes:

$$ R' = 10\,\Omega \times 1.10 = 11 \, \Omega $$

This new resistance at an unknown temperature $$T$$ (measured from 0°C) is given by:

$$ 11 = R_0 (1+\alpha T) $$

Substitute $$R_0 \approx 9.615 \, \Omega$$ and $$\alpha = 0.002$$:

$$ 11 = 9.615 \, (1+0.002T) $$

Now, solve for $$T$$:

Divide both sides by 9.615:

$$ 1+0.002T = \frac{11}{9.615}\approx 1.144 $$

Subtract 1 from both sides:

$$ 0.002T = 1.144-1 = 0.144 $$

Solve for $$T$$:

$$ T = \frac{0.144}{0.002} = 72\, ^\circ \text{C} $$

Thus, the new temperature of the wire is $$72^\circ \text{C}$$, which corresponds to Option D.

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