A spherical planet of radius R has spherically symmetrical distribution of mass density, varying as square of the distance from the centre, from zero at centre to maximum value $ρ_{0}$ at its surface.

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The value of acceleration due to gravity ‘g’ varies inside the planet as cube of the distance from centre.

The value of escape velocity of a mass m at the surface of planet is $\sqrt{\frac{4 π G ρ_{0} R^{2}}{5}}$

The value of escape velocity from surface is same as the escape velocity from another planet of same total mass & radius but having uniform mass density.

The energy required to impart escape velocity to particles of masses ‘m’ & ‘2m’, at the surface of planet, will be in ratio 1:2.

answer is B, C, D.

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Detailed Solution

$ρ = k r^{2} = \frac{ρ_{0}}{R^{2}} r^{2} (A) v_{e s c} = \sqrt{\frac{2 G M}{R}} M = \int_{0}^{R} \frac{ρ_{0}}{R^{2}} r^{2} 4 π r^{2} d r = \frac{ρ_{0} 4 π}{R^{2}} \frac{R^{5}}{5} = \frac{ρ_{0} 4 π R^{3}}{5} s o, v_{e s c} = \sqrt{\frac{2 G ρ_{0} 4 π R^{3}}{5 R}} = \sqrt{\frac{8 π G ρ_{0} R^{2}}{5}}$

B) $g - \frac{G M}{r^{2}} = \frac{G}{r^{2}} \int_{0}^{r} \frac{ρ_{0}}{R^{2}} r^{2} 4 π r^{2} d r -$

$\frac{G ρ_{0} 4 π r^{3}}{5 R^{2}}$

C) as $v_{e s c} = \sqrt{\frac{2 G M}{R}}$ so will be same for other planet also.
D) Energy required $= \frac{1}{2} m v_{e s c}^{2} α m$
So $\frac{E_{1}}{E_{2}} = \frac{1}{2}$