The time period of a simple pendulum is given
by $T=2 \pi \sqrt{\frac{\ell}{g}}$. 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 ' $\mathrm{g}$ ' using this pendulum is ' $x$ '. The value of ' $x$ ' to the nearest integer is:-
Correct Option: , 2
$\mathrm{g}=\frac{4 \pi^{2} \ell}{\mathrm{T}^{2}}$
$\frac{\Delta \mathrm{g}}{\mathrm{g}}=\frac{\Delta \ell}{\ell}+2 \frac{\Delta \mathrm{T}}{\mathrm{T}}=\frac{0.1}{10}+2\left(\frac{\frac{1}{200}}{0.5}\right)$
$\frac{\Delta g}{g}=\frac{1}{100}+\frac{1}{50}$
$\frac{\Delta \mathrm{g}}{\mathrm{g}} \times 100=3 \%$
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