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Question

If the radius of earth shrinks by 0.2% without change in its mass, the time period of oscillation of a simple pendulum:

A
Increases by 0.2%
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B
Decreases by 0.2%
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C
Increases by 0.1%
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D
Decreases by 0.1%
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Solution

The correct option is B Decreases by 0.2%
Time period of oscillation of simple pendulum T=2πlg
Thus percentage change in time period ΔTT×100=12Δgg×100
Also using g=GMR2
We get percentage change in acceleration due to gravity Δgg×100=2ΔRR×100
ΔTT×100=ΔRR×100
We get ΔTT×100=0.2
Thus time period of simple pendulum decreases by 0.2%.

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