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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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