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1.Units, Dimensions and Measurement
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In an experiment to find acceleration due to gravity $(g)$ using simple pendulum, time period of $0.5\,s$ is measured from time of $100$ oscillation with a watch of $1\;s$ resolution. If measured value of length is $10\; cm$ known to $1\; mm$ accuracy. The accuracy in the determination of $g$ is found to be $x \%$. The value of $x$ is
A
$4$
B
$5$
C
$3$
D
$2$
(JEE MAIN-2022)
Solution
$T=2 \pi \sqrt{\frac{\ell}{g}}$
$g=\frac{1}{4 \pi^{2}} \frac{T^{2}}{\ell}$
$\frac{\Delta g}{g}=\frac{2 \Delta T }{ T }+\frac{\Delta \ell}{\ell}$
$\frac{\Delta g}{g}=2 \cdot \frac{1}{100 \times 0.5}+\frac{1\,mm }{10\,cm }$
$\frac{\Delta g}{g}=\frac{5}{100}$
Standard 11
Physics
