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Question

# At what depth from the surface of earth the time period of a simple pendulum is 0.5% more than that on the surface of the earth? (Radius of earth is 6400 km)

A
32 km
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B
64 km
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C
96 km
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D
128 km
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Solution

## The correct option is B 64 kmThe acceleration due to gravity at a depth d below the surface of the Earth is given by gd=g(1−dR) ⇒gdg=(1−dR)...(1) The time period of a simple pendulum is given by T=2π(lg)12. Hence, T∝√1g, Let, the time period of pendulum at the surface is T and time period at depth d is Td. ⇒TdT=(ggd)12...(2) According to the problem, Td=T+0.5% of T=1.005T Using equation (1) and (2), ⇒1.005=1(1−dR)12 Substituting, R=6400 km, ⇒1.005=1(1−d6400)12 ⇒(11.005)2=1−d6400 ⇒d6400=0.0099 ⇒d=63.99≈64 km Hence, option (b) is correct answer.

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