Question

If the length of a simple pendulum is increased by $2 \%$, then the time period

• A. increases by 1%
• B. decreases by 1%
• C. increases by 2 %
• D. decreases by 2 %

Answer 1

Answer: (A)

$l_{2}=1.02 l_{1}$.
Time period $(T)=2 \pi \times \sqrt{\frac{l}{g}} \propto \sqrt{l}$
Therefore $\frac{T_{2}}{T_{1}}=\sqrt{\frac{l_{2}}{l_{1}}}=\sqrt{\frac{1.02 l_{1}}{l_{1}}}=1.01$.
Thus time period increased by $1 \%$.