If the length of the simple pendulum is incre | vedclass

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Question

(b) $T = 2\,\pi \sqrt {\frac{l}{g}} $

$ \Rightarrow \frac{{{T_2}}}{{{T_1}}} = \sqrt {\frac{{{l_2}}}{{{l_1}}}} $

$ = \sqrt {\frac{{144}}{{100}}}

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(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}}} 	imes 100 = 20\% $

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