Physics - Gravitation Question with Solution | TestHub

We know, $\mathrm{g}=\frac{\mathrm{GM}}{{\mathrm{R}}^2}$ If the density is d, then $\Rightarrow M=\frac{4}{3}\pi R^{3}d\dots (\because M=V\times d)$ $\mathrm{g}⩽\frac{4}{3}\pi \mathrm{GRd}$ Given, $\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}=\frac{\mathrm{x}}{\mathrm{y}}$ and $\frac{{\mathrm{d}}_1}{{\mathrm{d}}_2}=\frac{\mathrm{m}}{\mathrm{n}}$ ${\mathrm{g}}_1=\frac{4}{3}\pi {\mathrm{GR}}_1{\mathrm{d}}_1$ ${\mathrm{g}}_2=\frac{4}{3}\pi {\mathrm{GR}}_2{\mathrm{d}}_2$ Therefore, $\frac{{\mathrm{g}}_1}{{\mathrm{g}}_2}=\frac{{\mathrm{R}}_{\mathrm{l}}^{\prime }}{{\mathrm{R}}_2^{\prime }}\times \frac{{\mathrm{d}}_1^{\prime }}{{\mathrm{d}}_2}$ $\therefore \frac{{\mathrm{g}}_1}{{\mathrm{g}}_2}=\frac{\mathrm{xm}}{\mathrm{yn}}$