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