IIT-JEE 2012 · Paper 2 · Physics Q19

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PhysicsMultiple correct+4Question 19 of 60

Two spherical planets $P$ and $Q$ have the same uniform density $\rho$ , masses $M_P$ and $M_Q$ , and surface areas $A$ and $4A$ , respectively. A spherical planet $R$ also has uniform density $\rho$ 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

V_Q>V_R>V_P

V_R>V_Q>V_P

V_R/V_P=3

V_P/V_Q=\dfrac12

Answer and worked solution

Official answer: B, D

Sizes. Area $4A$ means Q has twice P's radius. With equal density, $M\propto r^3$ : $M_Q=8M_P$ , and R has mass $9M_P$ , so $r_R=9^{1/3}r_P$ .

Escape speed. $v=\sqrt{\dfrac{2GM}r}\propto\sqrt{\rho r^2}\propto r$ at equal density. So
$V_P:V_Q:V_R=1:2:9^{1/3}\approx1:2:2.08.$

(A) False. (B) $V_R>V_Q>V_P$ . True.
(C) $V_R/V_P\approx2.08$ , not 3. False.
(D) $V_P/V_Q=\frac12$ . True.

(B), (D)

Topics:Escape velocity

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IIT-JEE 2012 · Paper 2, Physics question 19