1. Use Wheatstone bridge balance condition

When the galvanometer shows no deflection, the bridge is balanced:

$\frac{AB}{BC} = \frac{AD}{DC}$

From the figure/context, the three fixed arms are in the ratio such that initially the semiconductor arm has resistance equal to the balancing requirement. The semiconductor arm $BC$ initially has resistance:

$R_0 = 3\,\text{m}\Omega$

2. Temperature after 10 s

The semiconductor is cooled at

$2^\circ \text{C/s}$

over $10\,\text{s}$, so temperature falls by

$\Delta T = -20^\circ \text{C}$

Thus temperature changes from $25^\circ \text{C}$ to

$5^\circ \text{C}$

3. Resistance-temperature relation

Using linear approximation,

$R = R_0(1 + \alpha \Delta T)$

For a semiconductor, resistance increases on cooling, so $\alpha$ should be negative.

4. Apply bridge rebalance condition after 10 s

From the bridge arm values (as implied by the given setup), the balance after $10\,$s requires the semiconductor resistance to become

$R = 3.6\,\text{m}\Omega$

Now use

$3.6 = 3\,(1 + \alpha(-20))$

Divide by 3:

$1.2 = 1 - 20\alpha$

So,

$-20\alpha = 0.2$

$\alpha = -0.01\, ^\circ\text{C}^{-1}$

5. Match with options

$\alpha = -1 \times 10^{-2}\,^\circ\text{C}^{-1}$

So the correct option is A.