A metal rod of cross-sectional area $10^{-4} \,m ^{2}$ is hanging in a chamber kept at $20^{\circ} C$ with a weight attached to its free end. The coefficient of thermal expansion of the rod is $2.5 \times 10^{-6} \,K ^{-1}$ and its Young's modulus is $4 \times 10^{12} \,N / m ^{2}$. When the temperature of the chamber is lowered to $T$, then a weight of $5000 \,N$ needs to be attached to the rod, so that its length is unchanged. Then, $T$ is ............ $^{\circ} C$
KVPY 2019, Advanced
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$(a)$ 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 ${ }^{\circ} C$.
$\therefore 20-T=5^{\circ} C \text { or } T=15^{\circ} C$
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