If R is the radius of the earth and the acceleration due to gravity on the surface of earth is , then the length of the second's pendulum at a height h=2R from the surface of earth will be

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Solution

Q.
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 [2024]

1 $\frac{2}{9} m$

2 $\frac{1}{9} m$

3 $\frac{4}{9} m$

4 $\frac{8}{9} m$

Ans.
(2)

$g = \frac{G M}{R^{2}} = π^{2}$

$g' = \frac{G M}{9 R^{2}} = \frac{g}{9} = \frac{π^{2}}{9}$

We know, $T = 2 π \sqrt{\frac{l}{g}}$

Since the time period of second pendulum is 2 sec.

$2 = 2 π \sqrt{\frac{l \times 9}{π^{2}}}$

On solving, $l = \frac{1}{9} m$

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