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(C) The circuit is a Wheatstone bridge. For no deflection in the galvanometer,the bridge must be balanced.Initially,the resistance of arm $BC$ is $R_{

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$-1 	imes 10^{-2} \ { }^{\circ} C^{-1}$

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(C) The circuit is a Wheatstone bridge. For no deflection in the galvanometer,the bridge must be balanced.Initially,the resistance of arm $BC$ is $R_{BC} = 3 \ m\Omega$. The other arms are $AB = 0.8 \ m\Omega$,$AD = 1 \ m\Omega$,and $DC$ is unknown. However,the balance condition for a Wheatstone bridge is $\frac{R_{AB}}{R_{AD}} = \frac{R_{BC}}{R_{DC}}$.From the figure,$R_{AB} = 0.8 \ m\Omega$,$R_{AD} = 1 \ m\Omega$,and $R_{BC} = 3 \ m\Omega$. Let $R_{DC} = x$. The bridge is balanced when $\frac{0.8}{1} = \frac{3}{x}$,so $x = 3.75 \ m\Omega$.After cooling for $10 \ s$ at $2^{\circ} C/s$,the temperature change is $\Delta T = -20^{\circ} C$.The new resistance of the semiconductor arm $BC$ is $R'_{BC} = 2.4 \ m\Omega$ (since $\frac{0.8}{1} = \frac{R'_{BC}}{3.75} \Rightarrow R'_{BC} = 3 	imes 0.8 = 2.4 \ m\Omega$).Using the formula $R' = R(1 + \alpha \Delta T)$:$2.4 = 3(1 + \alpha(-20))$$0.8 = 1 - 20\alpha$$20\alpha = 0.2$$\alpha = \frac{0.2}{20} = 0.01 = 10^{-2} \ { }^{\circ} C^{-1}$.Since it is a semiconductor,$\alpha$ is negative,so $\alpha = -1 	imes 10^{-2} \ { }^{\circ} C^{-1}$.

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