First slide

Acceleration due to gravity and its variation

Question

A simple pendulum has a time period $T_{1}$ when on the earth’s surface and $T_{2}$ when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of $T_{2} / T_{1}$ is

Moderate

A

1
B

$\sqrt{2}$
C

4
D

2

Solution

If acceleration due to gravity is g at the surface of earth then at height R it value becomes $g' = g {(\frac{R}{R + h})}^{2} = \frac{g}{4}$
$T_{1} = 2 π \sqrt{\frac{l}{g}} and T_{2} = 2 π \sqrt{\frac{l}{g / 4}} ∴ \frac{T_{2}}{T_{1}} = 2$

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