Question

A simple pendulum is suspended in a lift which is going up with an acceleration of $5m/s^2$. An electric field of magnitude $5N/C$ and directed vertically upward is also present in the lift. The charge of the bob is $1\mu C$ and mass is $1mg$. Taking $g={\pi }^2$ and length of the simple pendulum $1m$, the time period of the simple pendulum (in sec) is____

Answer 1

Answer: 2

$T=2\pi \sqrt{\frac{l}{g_{eff}}}$

$g_{\text{eff}}=g-\frac{qE}{M}+5=15-\frac{1\times 5\times 10^{-6}}{1\times 10^{-6}}$
$g_{\text{eff}}=10={\pi }^2$
$T=2\sec$