If length of simple pendulum is increased by 44% , then change in time period of pendulum .....

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13.Oscillations

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If the length of the simple pendulum is increased by $44\%$, then what is the change in time period of pendulum ..... $\%$

A

$22$

B

$20$

C

$33$

D

$44$

Solution

(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\% $

Standard 11

Physics

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