JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Units & Measurement

For a simple pendulum, $T=2\pi \sqrt{\ell /g}$, so $g=\frac{4{\pi }^2\ell }{T^2}$, giving the error relation

$\frac{\Delta g}{g}=\frac{\Delta \ell }{\ell }+\frac{2\Delta T}{T}$

Length error: $\frac{\Delta \ell }{\ell }=\frac{0.2}{20}=0.01$

For the period, 50 oscillations take 40 s with a resolution of 1 s. Since $T=t_{50}/50$, dividing both the reading and its uncertainty by the same constant leaves the fractional error unchanged:

$\frac{\Delta T}{T}=\frac{\Delta t_{50}}{t_{50}}=\frac{1}{40}$

So

$\frac{\Delta g}{g}=0.01+2\times \frac{1}{40}=0.01+0.05=0.06$

The percentage error in g is 6%, so N = 6.