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(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$,which implies $T \propto \sqrt{l}$.Let the initial length be $l_1 =

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$100$

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(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$,which implies $T \propto \sqrt{l}$.Let the initial length be $l_1 = l$ and the initial time period be $T_1$.If the length is increased by $300\%$,the new length $l_2 = l + 300\% 	ext{ of } l = l + 3l = 4l$.Using the proportionality $T \propto \sqrt{l}$,we have $\frac{T_2}{T_1} = \sqrt{\frac{l_2}{l_1}} = \sqrt{\frac{4l}{l}} = \sqrt{4} = 2$.Thus,$T_2 = 2T_1$.The percentage increase in the time period is given by $\frac{T_2 - T_1}{T_1} 	imes 100\% = \frac{2T_1 - T_1}{T_1} 	imes 100\% = 100\%$.Therefore,the time period increases by $100\%$.

If the length of a simple pendulum is increased by $300\%$,then the time period will be increased by ..... $\%$

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