In an experiment to find acceleration due to gravity ( $g$ ) using simple pendulum, time period of 0

Question

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 ________ .\

Professor PC Thomas Classes · Answer

(5) : Given : $\Delta T=\frac{1}{100}=0.01$ second\ $L=10 \mathrm{~cm} ; \Delta L=1 \mathrm{~mm}=0.1 \mathrm{~cm} ; T=0.5 \mathrm{sec}$ Also, $g=\frac{4 \pi^{2} L}{T^{2}}$, \% age error in ' $g$ ' is,\ $\frac{\Delta g}{g} 	imes 100 \%=\left(\frac{\Delta L}{L}+\frac{2 \Delta T}{T}ight) 100 \%$\ $=\left(\frac{0.1}{10}+2 	imes \frac{0.01}{0.5}ight) 100 \%$\ $=(0.01+0.04) 100 \%=(0.05) 100 \%=5 \%$\