Graviational acceleration on the surface of plane fo `(sqrt6)/(11)g.` where g is the gracitational acceleration on the surface of the earth. The average mass density of the planet is `(2)/(3)` times that of the earth. If the escape speed on the surface of the earht is taken to be `11 kms^(-1)` the escape speed on teh surface of the planet in `kms^(-1)` will be

Options

A. `3`
B. `6`
C. `9`
D. `12`

Answer

A

Explanation

`g=(GM)/(R^(2))=(G(4/3piR^(3))rho)/(R^(2))`
`gproprhoR, Rpropg/rho`
Now escape velocity, `v_(e)=sqrt(2gR)`
`v_(e)propsqrt(gR), v_(e)propsqrt(gxxg/rho)propsqrt((g^(2))/rho)`
`(v_(e))_("planet")=(11kms^(-1))sqrt(6/121xx3/2)=3 km s^(-1)`