The acceleration due to gravity at height $R$ above the surface of the earth is $g / 4 .$ The periodic time of a simple pendulum in an artificial satellite at this height will be :

T = 4 π \sqrt{l / g}

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T = 2 π \sqrt{l / 2 g}

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Solution

The correct option is A $T = 4 π \sqrt{l / g}$
Lets consider $l =$ length of simple pendulum
The acceleration due to gravity at height $R$ above the surface of the earth is $g = \frac{g}{4}$ .
The periodic time of simple pendulum in an artificial satellite at the $R$ height is given by
$T = 2 π \sqrt{\frac{l}{g^{'}}} = 2 π \sqrt{\frac{l}{g / 4}}$
$T = 2 π \sqrt{\frac{4 l}{g}}$
$T = 4 π \sqrt{\frac{l}{g}}$
The correct option is A.

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The acceleration due to gravity at height R above the surface of the earth is g/4. The periodic time of a simple pendulum in an artificial satellite at this height will be :

The acceleration due to gravity at height $R$ above the surface of the earth is $g / 4 .$ The periodic time of a simple pendulum in an artificial satellite at this height will be :