Question

On a planet, a body dropped from a height of 8 m, reaches the ground in 2 seconds. If the length of a simple pendulum on that planet is 1 m, the period of oscillation is:

• A. $\frac{1}{3}s$
• B. $\frac{\pi }{2}s$
• C. $\pi s$
• D. $\frac{1}{2}s$

Answer 1

The correct option is C $\pi s$

$S=Ut{+}^1{/}_{2}a{t}^2$
$U=0$
$\therefore 8{=}^1{/}_{2}a\times (2{)}^2$
$a=4$
$T=2\pi \sqrt{\frac{e}{a}}$
$=2\pi \sqrt{\frac{1}{4}}$
$=\pi sec$