JEE Main · 2015 · 04 Apr 2015 · Offline · Units & Measurement

Since $g=\frac{4{\pi }^{2}L}{T^2}$, the fractional errors add as $\frac{\Delta g}{g}=\frac{\Delta L}{L}+2\frac{\Delta T}{T}$, the period entering twice because it is squared. The length is good to $\frac{0.1}{20}=0.005$. The watch reads $90$s for $100$ swings with a $1$s resolution, so the fractional error in the period is $\frac{1}{90}\approx 0.011$. Hence $\frac{\Delta g}{g}\approx 0.005+2(0.011)=0.027$, and the accuracy in $g$ is close to $3\%$ -- notice the timing error, doubled, is what dominates.