Question

If length of a simple pendulum is increased by $69\%$, then the percentage increase in its time period is

• A. $69\%$
• B. $30\%$
• C. $50\%$
• D. $10\%$

Answer 1

Answer: (B)

$30\%$

$T=2\pi \sqrt{\frac{l}{g}}$

$\begin{array}{c}\text{Taking log both side}\\ \log T=\log \left(2\pi {\left(\frac{\lambda }{g}\right)}^{1/2}\right)\\ \log T=\log 2+\log \pi +\frac{1}{2}\log l-\frac{1}{2}\log g\end{array}$

$\begin{array}{c}\text{Differentiating both side.}\\ \frac{\delta T}{T}=0+0+\frac{1}{2}\frac{\delta l}{l}-0\\ \text{Error always added}\\ \frac{\delta T}{T}=\frac{1}{2}\frac{\delta l}{l}+0\end{array}$

$\begin{array}{c}As\frac{\delta l}{l}=\frac{60}{100}\\ \frac{\delta T}{T}=\frac{1}{2}\times \frac{60}{100}\end{array}$

$\frac{\delta T}{T}\times 100=30\%$