The density of the core of 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. The ratio of denisty `rho_(1)//rho_(2)` will be

Options

A. `7/3`
B. `5/3`
C. `8/3`
D. `1/3`

Answer

A

Explanation

Let M and m be the mass of the planet and that of the core
`(GM)/(4R^(2))=(Gm)/(R^(2)), M/4=mimpliesM=4m`
`4/3piR^(3) rho_(1)+4/3pi(8R^(3)-R^(3))rho_(2)=4(4/3pi R^(3)rho_(1))`
`rho_(1)+7rho_(2)=4rho_(1)`
`(rho_(1))/(rho_(2))=7/3`