If R is the radius of the earth and the acceleration due to gravity on the surface of earth is g = π 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:

Question

Aicharya · Accepted Answer

Acceleration due to gravity at the given height, $g' = \frac{G M}{{(3 R)}^{2}} = \frac{1}{9} \frac{G M}{{(R)}^{2}} = \frac{1}{9} g$ , where $g$ is acceleration due to gravity at the surface of the Earth.

Now, time period is given by, $T = 2 π \sqrt{\frac{l}{g'}}$

Also the time period of second pendulum is $2$ sec.

Therefore, $T = 2 \sec$

Then, $2 = 2 π \sqrt{\frac{l}{g} \times 9}$

$\Rightarrow l = \frac{1}{9} m$

If R is the radius of the earth and the acceleration due to gravity on the surface of earth is g = π 2 m s -…

Solution