JEE Advanced 2022 Paper 1 · Q01 · Escape Velocity

Question

Two spherical stars A and B have densities ρ_A and ρ_B, respectively. A and B have the same radius, and their masses M_A and M_B are related by M_B = 2 M_A. Due to an interaction process, star A loses some of its mass, so that its radius is halved, while its spherical shape is retained, and its density remains ρ_A. The entire mass lost by A is deposited as a thick spherical shell on B with the density of the shell being ρ_A. If v_A and v_B are the escape velocities from A and B after the interaction process, the ratio v_B/v_A = √(10n/15^(1/3)). The value of n is ___.

SolveKar AI · Accepted Answer

Let R be the original radius and let ρ_B denote the original density of B. M_A = (4/3)πR³ρ_A and M_B = 2M_A = (4/3)πR³ρ_B ⇒ ρ_B = 2ρ_A. After interaction, star A has radius R/2 and density ρ_A, so its new mass M_A' = (4/3)π(R/2)³ρ_A = M_A/8. Mass lost by A = M_A − M_A/8 = 7M_A/8. This is deposited as a thick shell on B with density ρ_A. The new outer radius R' of B satisfies (4/3)π(R'³ − R³)ρ_A = 7M_A/8 = 7·(4/3)πR³ρ_A/8 ⇒ R'³ = R³(1 + 7/8) = 15R³/8 ⇒ R' = R·(15/8)^(1/3). New mass of B: M_B' = M_B + 7M_A/8 = 2M_A + 7M_A/8 = 23M_A/8. Escape velocity v = √(2GM/R). v_A from new A: v_A = √(2G·(M_A/8)/(R/2)) = √(GM_A/(2R))·√1 = √(GM_A/(2R))... evaluating carefully: v_A² = 2G(M_A/8)/(R/2) = GM_A/(2R). v_B² = 2G(23M_A/8)/(R·(15/8)^(1/3)) = (23/4)·GM_A/R / (15/8)^(1/3). Ratio v_B²/v_A² = [(23/4)/(15/8)^(1/3)] / (1/2) = (23/2)/(15/8)^(1/3) = (23/2)·(8/15)^(1/3) = (23/2)·2/(15)^(1/3) = 23/(15)^(1/3). So v_B/v_A = √(23/15^(1/3)) = √(10n/15^(1/3)) ⇒ 10n = 23 ⇒ n = 2.3.

JEE Advanced 2022 Paper 1 · Q01 · Escape Velocity

Solution