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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 km
The acceleration due to gravity at a depth d below the surface of the Earth is given by

g=g(1dR)

The time period of a simple pendulum is given by

T=2π(lg)1/2

Hence, T is inversely proportional to g1/2 for the same pendulum.

TT=(gg)1/2 .....(1)

From question,

T=T+0.5100T=1.005T

putting all data in equation (1), we get

1.005TT=⎜ ⎜ ⎜ ⎜gg(1dR)⎟ ⎟ ⎟ ⎟1/2

1.005=1(1dR)1/2

Substituting, R=6400 km, we get

d=63.564 km

Hence, option (b) is correct.

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