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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)
Solution
$\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}$
$\mathrm{g}=\frac{4 \pi^{2} \ell}{\mathrm{T}^{2}}$
$\frac{\Delta g}{g}=\frac{\Delta \ell}{\ell}+\frac{2 \Delta T}{T}$
$=\frac{0.1}{25}+\frac{2 \times 1}{50}$
$\frac{\Delta g}{g}=4.4 \%$
Standard 11
Physics
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