If the length of the simple pendulum is increased by $44\%$, then what is the change in time period of pendulum ..... $\%$
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(b) $T = 2\,\pi \sqrt {\frac{l}{g}} $
$ \Rightarrow \frac{{{T_2}}}{{{T_1}}} = \sqrt {\frac{{{l_2}}}{{{l_1}}}} $
$ = \sqrt {\frac{{144}}{{100}}} = \frac{{12}}{{10}}$
==> $T_2 = 1.2\, T_1$
Hence $\%$ increase $ = \frac{{{T_2} - {T_1}}}{{{T_1}}} \times 100 = 20\% $
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