Gravitation Question 1
Question 1 - 2024 (01 Feb Shift 1)
If $R$ is the radius of the earth and the acceleration due to gravity on the surface of earth is $g=\pi^{2} m / s^{2}$, then the length of the second’s pendulum at a height $h=2 R$ from the surface of earth will be,:
(1) $\frac{2}{9} m$
(2) $\frac{1}{9} m$
(3) $\frac{4}{9} m$
(4) $\frac{8}{9} m$
Show Answer
Answer: (2)
Solution:
$g^{\prime}=\frac{GMe}{(3 R)^{2}}=\frac{1}{9} g$
$T=2 \pi \sqrt{\frac{\ell}{g^{\prime}}}$
Since the time period of second pendulum is $2 sec$.
$T=2 sec$
$2=2 \pi \sqrt{\frac{\ell}{g} 9}$
$\ell=\frac{1}{9} m$
