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Two spherical planets P and Q have the same uniform density $ρ,$ masses $M_{P}$ and $M_{Q}$ and surface areas A and 4A respectively, A spherical planet R also has uniform density $ρ$ and its mass is $(M_{P} + M_{Q}) .$ The escape velocities from the planets P, Q and R are $v_{P}, v_{Q}$ and $v_{R}$ respectively. Then

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$v_{R} > v_{Q} > v_{P}$

$\frac{v_{R}}{v_{P}} = 3$

$v_{Q} > v_{R} > v_{P}$

$\frac{v_{P}}{v_{Q}} = \frac{1}{2}$

answer is B, D.

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Detailed Solution

$V_{e s c} = \sqrt{\frac{2 G M}{R}} = \sqrt{\frac{8}{3} G ρ π R^{2}}$

Given that, $4 (4 π R_{P}^{3}) = (4 π R_{Q}^{2}), ∴ R_{Q} = 2 R_{P}$

Mass of R is $M_{R} = M_{P} + M_{Q}$

$R_{R}^{3} = R_{P}^{3} + R_{Q}^{3}$ So, $R_{R} = 9^{1 / 3} R_{P}$

So, $R_{R} > R_{Q} > R_{P} \Rightarrow v_{R} > v_{Q} > v_{P} A l s o, \frac{v_{R}}{v_{P}} = 9^{1 / 3} a n d \frac{v_{P}}{v_{Q}} = \frac{1}{2}$

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