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. 2
C. 9
D. 12

Answer

A

Explanation

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