The time period of a simple pendulum is given by $\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}} .$ The measured value of the length of pendulum is $10 \mathrm{~cm}$ known to a $1 \mathrm{~mm}$ accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1 s resolution. The percentage accuracy in the determination of ' $g$ ' using this pendulum is ' $x$ '. The value of ' $x$ ' to the nearest integer is:-
Correct Option: , 2
(2)
$g=\frac{4 \pi^{2} \ell}{T^{2}}$
$\frac{\Delta g}{g}=\frac{\Delta \ell}{\ell}+2 \frac{\Delta T}{T}=\frac{0.1}{10}+2\left(\frac{\frac{1}{200}}{0.5}\right)$
$\frac{\Delta g}{g}=\frac{1}{100}+\frac{1}{50}$
$\frac{\Delta g}{g} \times 100=3 \%$
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