Question

A planet is made of two materials of density ${\rho }_1$ and ${\rho }_2$ as shown in figure.
If the acceleration due to gravity at the surface of planet is same as at depth $R$, the ratio $\frac{{\rho }_1}{{\rho }_2}$ is equal to $\frac{x}{3}$. Then the value of $x$ is . Write upto two digits after the decimal point.

Answer 1

Let the total mass of the planet is $M$ and mass of the part with density ${\rho }_1$ be $M^{\prime }$.
From given diagram, $M=\frac{4}{3}\pi R^3{\rho }_1+\frac{4}{3}\pi \left(\right(2R{)}^3-R^3){\rho }_2$ and $M^{\prime }=\frac{4}{3}\pi R^3{\rho }_1$.

From given condition,
$\Rightarrow \frac{GM}{(2R{)}^2}=\frac{G{M}^{\prime }}{R^2}\Rightarrow \frac{M}{4}=M^{\prime }\Rightarrow M=4M^{\prime }\Rightarrow \frac{4}{3}\pi R^3{\rho }_1+\frac{4}{3}\pi (8R^3-R^3){\rho }_2=4(\frac{4}{3}.\pi R^3.{\rho }_1)\Rightarrow {\rho }_1+7{\rho }_2=4{\rho }_1\Rightarrow \frac{{\rho }_1}{{\rho }_2}=\frac{7}{3}.$
Therefore the value of $x$ is 7.