# The density of core of a planet is rho_1 and that of outer shell is rho_2. The radius of core is R and that of planet is 2R. Gravitational field at outer surface of planet is same as at the surface of core what is the ratio rho/rho_2. ?

## (1)3/4 (2)5/3 (3)7/3 (4)3/5

Mar 15, 2018

$3$

#### Explanation:

Suppose, mass of the core of the planet is $m$ and that of the outer shell is $m '$

So,field on the surface of core is $\frac{G m}{R} ^ 2$

And,on the surface of the shell it will be $\frac{G \left(m + m '\right)}{2 R} ^ 2$

Given,both are equal,

so, $\frac{G m}{R} ^ 2 = \frac{G \left(m + m '\right)}{2 R} ^ 2$

or, $4 m = m + m '$

or, $m ' = 3 m$

Now,$m = \frac{4}{3} \pi {R}^{3} {\rho}_{1}$ (mass=volume $\cdot$ density)

and, $m ' = \frac{4}{3} \pi \left({\left(2 R\right)}^{3} - {R}^{3}\right) {\rho}_{2} = \frac{4}{3} \pi 7 {R}^{3} {\rho}_{2}$

Hence,$3 m = 3 \left(\frac{4}{3} \pi {R}^{3} {\rho}_{1}\right) = m ' = \frac{4}{3} \pi 7 {R}^{3} {\rho}_{2}$

So,${\rho}_{1} = \frac{7}{3} {\rho}_{2}$

or, $\frac{{\rho}_{1}}{{\rho}_{2}} = \frac{7}{3}$