Q.Find the change in length of a second's pendulum, if the acceleration due to gravity at the place changes from 9.75 m/s$^2$ to 9.80 m/s$^2$.

[!TLDR] For a second's pendulum, $L=g/\pi^2$. $\Delta L=\Delta g/\pi^2=(0.05)/\pi^2\approx5.07$ mm. Since g INCREASES (9.75 to 9.80 m/s$^2$ as stated), L must correspondingly INCREASE by 5.07 mm -- the printed answer's 'decreases' appears to be a sign slip re …

For a second's pendulum, $T=2$ s always, so from $T=2\pi\sqrt{L/g}$, $L=\frac{gT^2}{4\pi^2}=\frac{g}{\pi^2}$ (Eq. 5.30, using $T^2=4$). Differentiating, $\Delta L=\frac{\Delta g}{\pi^2}$. With $\Delta g=9.80-9.75=0.05$ m/s$^2$, $\Delta L=\frac{0.05}{\pi^2}=\frac{0.05}{9.8696}\approx0.005066\text{ m}=5.07\text{ mm}$ Since g INCREASES from 9.75 to 9.80 m/s$^2$ as literally stated in the problem, and $L\propto g$ for a fixed period, the length must correspondingly INCREASE by 5.07 mm to keep the period at exactly 2 s. The source answer key states 'decreases by 5.07 mm', which is inconsistent with the g values exactly as printed (an increasing g requires an increasing L); the magnitude 5.07 mm is correct, but the sign disagrees …

Differentiate $L=g/\pi^2$ (fixed T=2s) to get $\Delta L=\Delta g/\pi^2$, and track the SIGN of $\Delta g$ ca …

Assuming, without checking, that the change is a 'decrease' just because the answer key says so, without verifyi …