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

Answer

C

Explanation

We know that `v = sqrt2GR`
`:. (v_p)/(v) = sqrt((g_p)/(g)xx(R_P)/(R )) ….(i)`
`Given `(g_p)/(g_e) = (sqrt6)/(11) …..(ii)`
Also `g = (4)/(3)pi G_rhoR :.(g_p)/(g) = (rho_p)/(rho)xx(R_p)/(R )`
`:.(sqrt6)/(11) = (2)/(3)xx(R_p)/(R ) [:.(rho_p)/(rho) = (2)/(3) (given)]`
`:. (R_p)/(R ) = (3sqrt6)/(22) ...(iii)`
From (i), (ii) & (iii) `(v_p)/(v) = sqrt((sqrt6)/(11)xx(3sqrt6)/(22)) = sqrt((3xx6)/(11xx22) = (3)/(11)`
`:. v_p = (3)/(11)xxv = (3)/(11)xx11km/s =3km/s`