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(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$.Taking the natural logarithm on both sides,we get $\ln T = \ln(2

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Increases by $2\%$

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(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$.Taking the natural logarithm on both sides,we get $\ln T = \ln(2\pi) + \frac{1}{2}\ln \ell - \frac{1}{2}\ln g$.Differentiating both sides,we get $\frac{dT}{T} = -\frac{1}{2} \frac{dg}{g}$.Given that the percentage decrease in $g$ is $4\%$,we have $\frac{dg}{g} 	imes 100 = -4\%$.Substituting this into the expression for the percentage change in $T$:$\frac{dT}{T} 	imes 100 = -\frac{1}{2} 	imes (-4\%) = +2\%$.Therefore,the time period increases by $2\%$.

If a simple pendulum is taken to a place where $g$ decreases by $4\%$,then the time period

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