If a simple pendulum is taken to place where g decreases by $2\%$, then the time period
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(d) $T = 2\pi \sqrt {\frac{l}{g}} $ ==> $T \propto \frac{l}{{\sqrt g }}$
$ \Rightarrow \frac{{\Delta T}}{T} \times 100 = - \frac{1}{2}\left( {\frac{{\Delta g}}{g}} \right) \times 100$$ = - \frac{1}{2}( - \,2\% ) = 1\% .$
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