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The mass and the diameter of a planet a...

The mass and the diameter of a planet are three times the repective value for the Earth. The period of oscillation of a simple pendulum on the Earth is 2 s. The period of oscillation of the same pendulum on the planet would be:

A

`(3)/(2)s`

B

`(2)/(sqrt(3))s`

C

`(sqrt(3))/(2)s`

D

`2sqrt(3)s`

Text Solution

Verified by Experts

The correct Answer is:
D
`g=(GM)/(R^(2)),(g_(p))/(g_(e))=(M_(p))/(M_(e))xx((R_(e))/(R_(p)))^(2)=3xx(1)/(9)=(1)/(3)`
`T=2pisqrt((l)/(g)),(T_(p))/(T_(e))=sqrt((g_(e))/(g_(p)))=sqrt(3)`
`T_(p)=sqrt(3)T_(e)`
Time period at Earth for second.s pendulum = 2sec
`therefore T_(p)=2sqrt(3)` sec
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