Two spherical planets P and Q have the same uniform density , masses and and surface areas A and 4A respectively. A spherical planet R also has uniform density and its mass is . The escape velocities from the planets P, Q and R are , and , respectivel

Question

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 [2012]

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(2, 4)

Here planets P and Q have the same uniform density $M_{P}$ and surface areas $A$ and $4 A$ respectively. Let the mass of P, $(M_{P} + M_{Q})$ = m.

$Then m = ρ \times \frac{4}{3} π r^{3} = ρ \times \frac{4}{3} π {[\frac{A}{4 π}]}^{3 / 2}$

$The mass of M_{Q} = ρ \times \frac{4}{3} π {[\frac{4 A}{4 π}]}^{3 / 2} = 8 m$

$∴$ The mass of planet $R = 8 m + m = 9 m$

If the radius of $P = r,$ then the radius of $Q = 2 r$ $[∵ r_{Q} = {(\frac{4 A}{4 π})}^{3 / 2} = 2 {(\frac{A}{4 π})}^{3 / 2}]$

$and radius of R = 9^{1 / 3} r$ $[∵ M_{R} = M_{P} + M_{Q}, r_{R}^{3} = r^{3} + {(2 r)}^{3} = 9 r^{3}]$

As we know, escape velocity from the planet

$V_{e} = \sqrt{\frac{2 G M}{R}}$ $∴$ $v_{P} = \sqrt{\frac{2 G M_{P}}{R_{P}}} = \sqrt{\frac{2 G m}{r}}$

$v_{Q} = \sqrt{\frac{2 G M_{Q}}{R_{Q}}} = \sqrt{\frac{2 G (8 m)}{2 r}} = 2 v_{P}$

$v_{R} = \sqrt{\frac{2 G (9 m)}{9^{1 / 3} r}} = 9^{1 / 3} v_{P}$

Solution

1 $V_{Q} > V_{R} > V_{P}$

2 $V_{R} > V_{Q} > V_{P}$

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

4 $\frac{V_{P}}{V_{Q}} = \frac{1}{2}$

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Solution

Q. Two spherical planets P and Q have the same uniform density ρ, masses MP and MQ and surface areas A and 4A respectively. A spherical planet R also has uniform density ρ and its mass is (MP+MQ). The escape velocities from the planets P, Q and R are VP, VQ and VR, respectively. Then [2012]

1 VQ>VR>VP

2 VR>VQ>VP

3 VRVP=3

4 VPVQ=12

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1 $V_{Q} > V_{R} > V_{P}$

2 $V_{R} > V_{Q} > V_{P}$

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

4 $\frac{V_{P}}{V_{Q}} = \frac{1}{2}$