Question

On a planet a freely falling body takes $2$ sec when it is dropped from a height of $8\text{m}$, the time period of simple pendulum of length $1\text{m}$ on that planet is

• A. 3.14
• B. 3.1
• C. 3

Answer 1

Answer: (A), (B), (C)

Length of the pendulum $=1\text{m}$
On a planet, if a body dropped initial velocity $(u=0)$ from a height $h$ and takes time $t$ to reach the ground then,

$h=\frac{1}{2}{g}_p{t}^2\Rightarrow g_p=\frac{2h}{t^2}=\frac{2\times 8}{4}=4\text{m}/{\text{s}}^2$
$\text{Using}T=2\pi \sqrt{\frac{l}{g_p}}$
$\Rightarrow T=2\pi \sqrt{\frac{1}{4}}=\pi =3.14\text{sec}$