The original volume of the aluminium pot \(=V_{ o }\), Let the new volume \(=V\)
Since there is a decrease of \(0.20 \%\) in volume,
\(\begin{aligned} V &=V_{0}-\left(\frac{0.20}{100}\right) V_{0}=V_{0}-0.002 V_{0} \\ &=0.998 V _{0} \\ \gamma &=3 \alpha=3\left(2.3 \times 10^{-5}\right)=6.9 \times 10^{-5} \end{aligned}\)hange in temperature \(=\Delta T\)
\(1=V_{0}(1+\gamma \Delta T )\)
\(0.998) K =K\left(1+6.9 \times 10^{-5} \Delta T\right)\)
\begin{aligned}
\Delta T &=\frac{0.998-1}{6.9 \times 10^{-5}}=\frac{0.002}{6.9 \times 10^{-5}} \\
&=-28.980^{\circ} C \\
&=-20^{\circ} C
\end{aligned}