First slide

Universal law of gravitation

Question

A simple pendulum has a time period T₁ when on the earth's surface and T₂ when taken to a height R above the earth's surface, where R is the radius of the earth. The value of T₂/T₁is:

Easy

A

1
B

$\sqrt{2}$
C

4
D

2

Solution

$T = 2 π \sqrt{\frac{l}{g}} \Rightarrow \frac{T_{2}}{T_{1}} = \sqrt{\frac{g_{1}}{g_{2}}} = \sqrt{\frac{g}{g / 4}} = 2$

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