Q1: Two spherical stars A and B have densities ρA and ρB, respectively. A and B have the same radius, and their masses MA and MB are related by MB = 2MA. 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 ρA. The entire mass lost by A is deposited as a thick spherical shell on B with the density of the shell being ρA. If vA and vB are the escape velocities from A and B after the interaction process, the ratio . The value of n is __________ . [JEE Advanced 2022 Paper 1]
Ans: 2.2 to 2.4
Due to an interaction process, star A losses some of it's mass and radius becomes R/2. Let new mass of star A is M'A. Here in both cases density of star A remains same ρA.
Initially
Finally
Density remains same,
So, Lost mass by
This lost mass 7MA / 8 is attached on the star B and density of the attached mass stay ρA. So new radius of star B is R2.
Density of the removed part from star A is,
Density of the added part in star B stay's same as ρA,
Escape velocity from star A after interaction process,
And escape velocity from star B after interaction process,
Given,
Comparing equation (1) and (2), we get,
10n = 23
⇒ n = 2.3
Gravitational force = Centripetal force of the earth
(∵ M = total mass from 0 to r)
Differentiate on both sides, we get
(∵ volume = mass × density)
Now,
Thus, the correct mapping is P → 3.
Now, L = mvR
Thus, the correct mapping is Q → 2.
Now,
Thus, the correct mapping is R → R.
Now,
Thus, the correct mapping is S → 1.
Therefore, option (B) is correct.
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