Question

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

• A. $96s$
• B. $58s$
• C. $12s$
• D. $72s$

Answer 1

The correct option is C $12s$

We know $T=2\pi \sqrt{\frac{l}{g}}$

$T\propto \sqrt{l_1}$

Given data $T_1=1sec$

$T_2=2$

$l_1=l$

$l_2=44\%=100+44=144l$

$\therefore \frac{T_1}{T_2}=\sqrt{\frac{l_1}{l_2}}\Rightarrow \frac{1}{T_2}=\sqrt{\frac{1}{144}}$

$T_2=12sec$