In an experiment of the measurement of 'g' using simple pendulum, the time period was measured with an…
In an experiment of the measurement of 'g' using simple pendulum, the time period was measured with an accuracy of $0 \cdot 2 \%$ while the length was measured with an accuracy of $0.5 \%$. The percentage accuracy in the value of ' $\mathrm{g}$ ' thus obtained is
$0 \cdot 7 \%$
$0 \cdot 3 \%$
$0 \cdot 9 \%$
$0 \cdot 1 \%$
Solution
Time period of the pendulum is given by \(T=2 \pi \sqrt{ } \frac{\overline{1}}{\mathrm{~g}}\)
\(\begin{aligned}
& \text {org }=\left(\frac{2 \pi}{T}\right)^2 \cdot l \\
& \Rightarrow \frac{\Delta g}{g}=\frac{\Delta l}{l}+\frac{2 \Delta T}{T}=0.5+2 \times 0.2
\end{aligned}\)
Thus percentage error in the value \(\frac{\Delta \mathrm{g}}{\mathrm{g}} \times 100=0.9 \%\).