The time period of oscillations of a simple pendulum is $1$ minute. If its length is increased by $44 \%$. then its new time period of oscillation will be ......... $s$
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(d)
Let initial length be $I_1$
Final length $I_2=I_1 \times \frac{144}{100}$
$T_1=2 \pi \sqrt{\frac{l_1}{g}}$
$T_2=2 \pi \sqrt{\frac{I_1}{g} \times \frac{144}{100}}$
or $T_2=1.2 T_1$
$T_1=60\,s$
So $T_2=72 \,s$
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