The time \(T\) of oscillation of a simple pendulum of length \(L\) is governed by \(T=2 \pi…
The time \(T\) of oscillation of a simple pendulum of length \(L\) is governed by \(T=2 \pi \sqrt{\frac{L}{g}}\), where \(g\) is constant. The percentage by which the length be changed in order to correct an error of loss equal to 2 minutes of time per day is
\(-\frac{5}{18}\)
\(-\frac{2}{9}\)
\(\frac{1}{6}\)
\(\frac{1}{9}\)
Solution
For simple pendulum of length \(L\), the time of oscillation is
\(\begin{aligned}
& \quad T=2 \pi \sqrt{\frac{L}{g}} \Rightarrow \frac{\Delta T}{T} \times 100=\frac{1}{2} \frac{\Delta L}{L} \\
& \Rightarrow \quad \frac{\Delta L}{L} \%=-\frac{2 \times 2}{24 \times 60} \times 100(\because \Delta T=2 \mathrm{~min}) \\
& =-\frac{10}{36}=\frac{-5}{18}
\end{aligned}\)
\(\therefore \frac{\Delta L}{L}=-\frac{5}{18}\) percentage.
Hence, option (1) is correct.