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. 1
B. 2
C. 3
D. 4

Answer

C

Explanation

Escape velocity is given as : `v = sqrt(2gR)`
`(v_(p))/(v_(e)) = (sqrt(2g_(p)R_(p))/(2g_(e)R_(e)))`
`rArr (v_(p))/(v_(e)) = sqrt((g_(p))/(g_(e))) xx sqrt((R_(p))/(R_(c )))` …(i)
It is given that
`g_(p) = (sqrt(6))/(11) g_(e)`
`(g_(p))/(g_(e)) = (sqrt(6))/(11)` ...(ii)
`(rho_(p))/(rho_(e)) = (2)/(3)` ...(iii)
Further we know that : `g = (GM)/(R^(2))`
`rArr g = (G rho(4)/(3) pi R^(3))/(R^(2)) = (4G p rho R)/(3)`
`rArr (g_(p))/(g_(e)) = (rho_(p))/(rho_(e)) xx (R_(p))/(R_(e))`
Using equations (ii) and (iii) in above equation, we get the following :
`rArr (sqrt(6))/(11) = (2)/(3) xx (R_(p))/(R_(e))`
`rArr (R_(p))/(R_(e)) = (3sqrt(6))/(22)` ...(iv)
Now substituting equations (ii) and (iv) in equation (i), we get the following :
`rArr (v_(p))/(v_(e)) = sqrt((sqrt(6))/(11) xx (3sqrt(6))/(22)) = (3)/(11)`
`rArr v_(p) = (3)/(11) v_(e) = (3)/(11) xx 11 = 3` km/s
Hence option (c ) is correct.