The radius of planet is four times that of the moon. The acceleration due to gravity $\left({\mathrm{g}}_{\mathrm{m}}\right)$ on the moon is $\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$5,$}\right.$ where $\mathrm{g}$ is acceleration due to gravity on the planet. What will be escape velocity from the planet surface, if its value from moon surface is $2.5\mathrm{km}{s}^{-1}?$$\sqrt{5}=2.236?$

The gravitational and electrical force between two electron $10 \mathrm{cm}$ apart are ${\mathrm{F}}_{\mathrm{g}}$ and ${\mathrm{F}}_{\mathrm{e}}$ respectively. The ratio ${\mathrm{F}}_{\mathrm{g}}/{\mathrm{F}}_{\mathrm{e}}$ is of the order of :

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The magnitude of the gravitational field at distance ${\mathrm{r}}_1 \mathrm{and} {\mathrm{r}}_2$ from the centre of a uniform sphere of radius $\mathrm{R}$ and mass $\mathrm{M} \mathrm{are} {\mathrm{F}}_1 \mathrm{and} {\mathrm{F}}_2$ respectively. Then

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         A solid sphere of uniform density and radius $4$units is located with its centre at the origin $\mathrm{O}$ of coordinates. Two spheres of equal radii $1$ unit, with their centre at $A \left(-2,0,0\right) and B (2,0,0)$ respectively, are taken out of the solid leaving behind spherical cavities as shown. Then :                                                                                       

The average density of the earth :