Answer
a) $1.43 \hspace{1mm} cm^2$
b) $1.44 \hspace{1mm} cm^2$
Work Step by Step
a) At $25.0 \hspace{1mm} {}^0C$, diameter of the hole ($D$) is $1.35cm$.
Cross-sectional area $= \pi \frac{D^2}{4} \simeq 1.43 \hspace{1mm}cm^2$.
b) Coefficient of linear expansion of steel is $1.2\times10^{-5} \hspace{1mm}{}^0C^{-1}=\alpha$
Increase in temperature ($\Delta T$) $=(175-25) \hspace{1mm} {}^0C=150 \hspace{1mm}{}^0C$
Initial diameter ($L_o$) $=1.35cm$
Change in diameter ( $\Delta L$) $=L_o \alpha \Delta T$
Putting in all the values, $\Delta L$ $=2.43\times 10^{-3}cm$
New diameter $=L_o+\Delta L \simeq 1.3524 cm$
New cross-section using the new diameter $\simeq1.44 \hspace{1mm}cm^2$
