Questions
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
detailed solution
Correct option is B
M4M=43πR13ρ43πR23ρ2=2R13R23 18=R13R23⇒R1R2=12 g1g2=M4M2RR2=1 g=GMR2 as g are same time period will also be same as T=2πlgTalk to our academic expert!
Similar Questions
The diameters of two planets are in ratio 4 : 1. Their mean densities have ratio 1 : 2. The ratio of on the planets will be
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