In a simple pendulum experiment for determination of acceleration due to gravity $(g)$, time taken for $20$ oscillations is measured by using a watch of $1$ second least count. The mean value of time taken comes out to be $30s$. The length of pendulum is measured by using a meter scale of least count $1\ mm$ and the value obtained is $55.0\ cm$. The percentage error in the determination of $g$ is close to $ ........... \%$
JEE MAIN 2019, Diffcult
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$T=\frac{30 \sec}{20}$
$\Delta T=\frac{1}{20} \sec$
$L=55 \ cm$
$\Delta L=1\ mm=0.1 \ cm$
$g=\frac{4 \pi^2 L}{T^2} $
percentage erroringis
$\frac{\Delta g}{g} \times 100=\left(\frac{\Delta L}{L}+\frac{2 \Delta T}{T}\right) 100 \%$
$=\left(\frac{0.1}{5.5}+2 \frac{\left(\frac{1}{20}\right)}{\frac{30}{20}}\right) 100 \% \simeq 6.8 \%$
$\Delta T=\frac{1}{20} \sec$
$L=55 \ cm$
$\Delta L=1\ mm=0.1 \ cm$
$g=\frac{4 \pi^2 L}{T^2} $
percentage erroringis
$\frac{\Delta g}{g} \times 100=\left(\frac{\Delta L}{L}+\frac{2 \Delta T}{T}\right) 100 \%$
$=\left(\frac{0.1}{5.5}+2 \frac{\left(\frac{1}{20}\right)}{\frac{30}{20}}\right) 100 \% \simeq 6.8 \%$
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