The time period of a simple pendulum on the surface of the earth is “T”. What will be the time period of the same pendulum at a height of 2 times the radius of the earth?

(a) T

(b) 2T

(c) 3T

(d) 4T

This question was addressed to me during an online interview.

The query is from Acceleration due to Gravity below and above the Surface of Earth topic in section Gravitation of Physics – Class 11

Select the correct answer from above options

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The correct choice is (c) 3T

For explanation: Time period of the simple pendulum on the surface of the earth;

T = (2 x pi) x (l / g)^1/2

l = Length of the simple pendulum

The time period of the simple pendulum at a certain height above the earth’s surface;

T’ = (2 x pi) x (l / g’)^1/2

g’ = Acceleration due to gravity at a certain height above the surface of the earth

g’ for a height of 2R (R = Radius of the earth);

g’ = (G*M1)/(R + 2R)^2

= (G*M1)/(3R)^2

= (1/9) x g

T’ = (2 x pi) x (l / (1/9)g) ^½

= 3 x [(2 x pi) x (l / g)^½]

= 3T.