The density of the core of a planet is $ρ_{1}$ and that of the outer shell is $ρ_{2}$ , the radii of the core and that of the planet are $R$ and $2 R$ respectively. The acceleration due to gravity at the surface of the planet is same as at a depth $R$ . Find the ratio of $\frac{ρ_{1}}{ρ_{2}}$ .

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$7 : 2$

$6 : 2$

$7 : 3$

$6 : 3$

answer is D.

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Detailed Solution

Let $m_{1}$ , be the mass of the core and $m_{2}$ the mass of outer shell.

$g_{A} = g_{B}$ (given)

Then, $\frac{G m_{1}}{R^{2}} = \frac{G (m_{1} + m_{2})}{(2 R)^{2}}$

$∴ 4 m_{1} = (m_{1} + m_{2})$

or $4 \{\frac{4}{3} π R^{3} ρ_{1}\} = \frac{4}{3} π R^{3} \cdot ρ_{1} + \{\frac{4}{3} π (2 R)^{3} - \frac{4}{3} π R^{3}\} ρ_{2}$

$4 ρ_{1} = ρ_{1} + 7 ρ_{2}$

$\frac{ρ_{1}}{ρ_{3}} = \frac{7}{3}$ Ans.

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