1.Units, Dimensions and Measurement
medium

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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