A simple pendulum is being used to determine the value of gravitational acceleration $\mathrm{g}$ at a certain place. The length of the pendulum is $25.0\; \mathrm{cm}$ and a stop watch with $1\; \mathrm{s}$ resolution measures the time taken for $40$ oscillations to be $50\; s$. The accuracy in $g$ is ....... $\%$
A$3.40$
B$5.40 $
C$4.40 $
D$2.40 $
JEE MAIN 2020, Medium
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C$4.40 $
c $\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}$
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