A pendulum that beats seconds on the surface of the earth were taken to a depth of (1/4)th the radius of the earth. Its time period of oscillation will be
A
2.3s
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B
2.8s
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C
1.9s
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D
1.7s
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Solution
The correct option is A2.3s Gravitational acceleration at a depth d from the surface of earth is given by
g′=g(1−dR)
here d=R4
⟹g′=3g4
Time period of oscillation of pendulum is given as