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Q.

A planet with mass equal to four times the mass of the Earth, has the average density half that of the Earth. Time period of a pendulum is 2 second on the surface of the Earth, The time period of the same pendulum on the surface of the planet is

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a

4 sec

b

2 sec

c

2 2 sec

d

2 sec

answer is B.

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Detailed Solution

M4M=43πR13ρ43πR23ρ2=2R13R23 18=R13R23⇒R1R2=12 g1g2=M4M2RR2=1 g=GMR2 as g are same time period will also be same as T=2πlg

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