JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Units & Measurement

The period of one oscillation is $T=\frac{30}{20}=1.5\text{s}$, and since the stopwatch reads to $1\text{s}$ over 20 oscillations, the uncertainty per oscillation is $\Delta T=\frac{1}{20}\text{s}$. The length is $L=55.0\text{cm}$ with $\Delta L=0.1\text{cm}$. Because $g=\frac{4{\pi }^{2}L}{T^2}$, the length enters to the first power and the period to the second, so the fractional errors combine as $\frac{\Delta g}{g}=\frac{\Delta L}{L}+2\frac{\Delta T}{T}$. Putting in the numbers, $\frac{0.1}{55}+2\times \frac{1/20}{1.5}=0.0018+0.0667\approx 0.068$, which is close to $6.8\%$. The timing term dominates, both because it is doubled and because the clock is the cruder instrument here.