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A certain planet completes one rotation ...

A certain planet completes one rotation about its axis in time T. The weight of an object placed at the equator on the planet's surface is a fraction f (f is close to unity) of its weight recorded at a latitude of `60^@`. The density of the planet (assumed to be uniform) is given

A

`(4-f)/(1-f) (3pi)/(4GT^2)`

B

`(4-f)/(1+f) (3pi)/(4GT^2)`

C

`(4-3f)/(1-f) (3pi)/(4GT^2)`

D

`(4-2f)/(1-f) (3pi)/(4GT^2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`v=sqrt((GM)/r)` also `T=(2pir)/v` or `T=2pisqrt((r^3)/(GM))` when v is replaced by `sqrt((GM)/r)`
now `g_(eff)=gomega^2R_ecos^2phi`
now `f=(g-omega^2R_ecos0^(@))/(g-omega^2R_ecos60^(@))`
solving we get `R_e=(4g(f-1))/(omega^2(f-4))` or
`(GM)/(R_e^2)(f-1)=(omega^2R_e)/4(f-4),R_e^3=(4GM)/(omega^2)((f-1))/((f-4))`
Now, `p=M/(4/3piR_e^3)=3/16(omega^2)/(piG)((f-4))/((f-1))`
also `T=(2pi)/(omega)thereforep=(3pi(f-4))/(4T^2G(f-1))`
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