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Q.
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
Oscillations
Solution:
Let initial length be $l_{1}$
Final length $l_{2}=l_{1} \times \frac{144}{100}$
$T_{1}=2 \pi \sqrt{\frac{l_{1}}{g}} $
$T_{2}=2 \pi \sqrt{\frac{l_{1}}{g} \times \frac{144}{100}}$
or $T_2 = 1.2 \,T_1$
$T_{1}=60\, s$
So $T_{2}=72 \,s$