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  4. A simple pendulum is being used to determine the value of gravitational acceleration g at a certain place. The length of the pendulum is 25.0 cm and a stop watch with 1 s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in g is :

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Q. A simple pendulum is being used to determine the value of gravitational acceleration $g$ at a certain place. The length of the pendulum is $25.0\, cm$ and a stop watch with $1\, s$ resolution measures the time taken for $40$ oscillations to be $50\, s$. The accuracy in $g$ is :

JEE MainJEE Main 2020Physical World, Units and Measurements

Solution:

$T=2\pi\sqrt{\frac{\ell}{g}}$
$g=\frac{4\pi^{2}\ell}{T^{2}}$
$\frac{\Delta g}{g}=\frac{\Delta\ell}{\ell}+\frac{2\Delta T}{T}$
$=\frac{0.1}{05}+\frac{2\times1}{50}$
$\frac{\Delta g}{g}=4.4\%$