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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 $30\,s$. 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 ........... $\%$
$0.7$
$3.5$
$6.8$
$0.2$
Solution
$\begin{array}{l}
T = \frac{{30\,\sec }}{{20}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta T = \frac{1}{{20}}\sec ,\\
\,\,\,\,\,\,\,L = 55\,cm\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta L = 1\,mm = 0.1\,cm\\
\,\,\,\,\,\,\,g = \frac{{4{\pi ^2}L}}{{{T^2}}}\\
\,\,\,\,\,\,\,\,percentage\,error\,in\,g\,is\\
\,\,\,\,\,\,\,\,\,\frac{{\Delta g}}{g} \times 100 = \left( {\frac{{\Delta L}}{L} + \frac{{2\Delta T}}{T}} \right)100\% \\
\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{0.1}}{{5.5}} + 2\frac{{\left( {\frac{1}{{20}}} \right)}}{{\frac{{30}}{{20}}}}} \right)100\% \simeq 6.8\% .
\end{array}$
