a
$(a)$ Length of rod given remains unchanged. This means contraction due to cooling is equals to elongation due to hanging of weight.
As, thermal strain $=$ strain caused by
$\alpha \Delta \theta=\frac{\Delta l}{l}$
$\Rightarrow \Delta \theta=\frac{\Delta l}{l\alpha}=\frac{F}{Y A} \quad\left(\because Y=\frac{F l }{A\Delta l}\right)$
$\Rightarrow \Delta \theta=\frac{5000}{4 \times 10^{12} \times 10^{-4} \times 2.5 \times 10^{-6}}=5^{\circ} C$
Note $\Delta \theta$ in $K$ is same as $\Delta \theta$ in ${ }^{\circ} C$.
$\therefore 20-T=5^{\circ} C \text { or } T=15^{\circ} C$
