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

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Professor PC Thomas Classes · Answer

(a) : $l_{2}=1.02 l_{1}$. Time period $(T)=2 \pi 	imes \sqrt{\frac{l}{g}} \propto \sqrt{l}$ 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 $1 \%$.\