The density of the core a planet is `rho_(1)` and that of the outer shell is `rho_(2)`. The radii of the core and that of the planet are `R` and `2R` respectively. The acceleration due to gravity at the surface of the planet is same as at a depth `R`. Find the radio of `(rho_(1))/(rho_(2))`

Answer

C

Explanation

Let `m_(1)` be the mass of the core and `m_(2)` the mass of outer shell.

`g_(A) = g_(B)` (given)
then `(Gm_(1))/(R^(2)) = G(m_(1) + m_(2))/((2R)^(2))`
`:. 4 m_(1) = (m_(1) + m_(2))`
or `4{(4)/(3) pi R^(3) rho_(1) } = (4)/(3) pi R^(3).rho_(1)`
` + { (4)/(3) pi (2R)^(3) - (4)/(3) pi R^(3)} rho_(2)`
`:. 4 rho_(1) = rho_(1) + 7 rho_(2)`
`:. (rho_(1))/(rho_(2)) = (7)/(3)`