First slide

Gravitational field intensity

Question

The length of seconds pendulum is 1 m on earth. If mass and diameter of the planet are doubled than that of earth, then length becomes

Moderate

A

1 m
B

2 m
C

0.5 m
D

4 m

Solution

For a second pendulum, T =2
$∴ 2 = 2 π \sqrt{(\frac{l}{g})} = 2 π \sqrt{(\frac{R^{2} l}{GM})}$
or $2 = 2 π \sqrt{(\frac{R^{2}}{GM})} (∵ l = 1 m)$ ……….(1)
At another planet
$2 = 2 π \sqrt{[\frac{4 R^{2} l_{p}}{G (2 M)}]} = 2 π \sqrt{[\frac{2 R^{2} l_{p}}{GM}]}$ …….(2)
From eqs. (l ) and (2), we get
$\sqrt{(\frac{R^{2}}{GM})} = \sqrt{(\frac{2 R^{2} l_{p}}{GM})} ∴ l_{p} = 0 .5 m$

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