JEE Advanced Previous Year Questions (2018 - 2024): Gravitation | Physics for JEE Main & Advanced PDF Download

2024

Q1: A particle of mass m is under the influence of the gravitational field of a body of mass M (≫ m). The particle is moving in a circular orbit of radius r₀ with time period T₀ around the mass M. Then, the particle is subjected to an additional central force, corresponding to the potential energy V_c(r) = mα/r³, where α is a positive constant of suitable dimensions and r is the distance from the center of the orbit. If the particle moves in the same circular orbit of radius r₀ in the combined gravitational potential due to M and V_c(r), but with a new time period T₁, then (T₁² - T₀²) / T₁² is given by [G is the gravitational constant.]    [ JEE Advanced 2024 Paper 2]
(a) 3αGM r₀²
(b) α2GM r₀²
(c) αGM r₀²
(d) 2αGM r₀²

Ans: (a) 
Solution:
F₁ = GMmr₀²
F₂ = GMmr₀² - 3mαr₀⁴
ω₁²ω₀² = F₂F₁ = GMr₀² - 3αr₀⁴GMr₀²
T₁²T₀² = 1 - 3αGMr₀²
T₁² - T₀²T₁² = 3αGMr₀²

2023

(a) 3R /5
(b) R / 6
(c) 6R / 5
(d) 5R/5

Ans: (a)

Solution:  The formula to calculate the acceleration due to gravity at a height ( h ) from the surface of the Earth is expressed as :

To find the ratio of gravitational acceleration at two different heights ( h_P ) and ( h_Q ) above the Earth's surface, use the formula for each and form a ratio :

Given ℎ_P=R_e/3, the ratio simplifies to:

Solving for ℎ_Q in terms of R_e yields .

2022

Q1: Two spherical stars Aand B have densities ρAand ρB, respectively. Aand Bhave the same radius, and their masses M_Aand M_Bare related by M_B = 2M_A. Due to an interaction process, star Aloses 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 Ais deposited as a thick spherical shell on Bwith the density of the shell being ρA. If v_Aand v_Bare the escape velocities from Aand Bafter the interaction process, the ratio . The value of nis __________ . [JEE Advanced 2022 Paper 1]

Ans: 2.2 to 2.4

^Solution: 

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 7M_A / 8is 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

2021

Ans:9
Solution: For earth-sun system,

For binary system

Using Eqs. (i) and (ii), we get
T = 9T₀
So, n = 9 

2019

Q1: Consider a spherical gaseous cloud of mass density ρ(r) in free space where r is the radial distance from its center. The gaseous cloud is made of particles of equal mass m moving in circular orbits about the common center with the same kinetic energy K. The force acting on the particles is their mutual gravitational force. If ρ(r) is constant in time, the particle number density n(r) = ρ(r)/m is [G is universal gravitational constant]
(a)
(b)
(c)
(d) [JEE Advanced 2019 Paper 1]

Ans: (d)
^Solution:  

Gravitational force = Centripetal force of the earth

(∵ M = total mass from 0 to r)

Differentiate on both sides, we get

(∵ volume = mass × density)

2018

Q1: A planet of mass M, has two natural satellites with masses m₁and m₂.The radii of their circular orbits are R₁and R₂respectively, Ignore the gravitational force between the satellites. Define v₁,L₁,K₁and T₁to be , respectively, the orbital speed, angular momentum, kinetic energy and time period of revolution of satellite 1; and v₂,L₂,K₂,and T₂to be the corresponding quantities of satellite 2.Given m₁/m₂ = 2and R₁/R₂ = 1/4,match the ratios in List-Ito the numbers in List- II.

(a) P → 4, Q → 2, R → 1, S → 3
(b) P → 3, Q → 2, R → 4, S → 1
(c) P → 2, Q → 3, R → 1, S → 4
(d) P → 2, Q → 3, R → 4, S → 1                [JEE Advanced 2018 Paper 2]

Ans: (b)
Solution:  Given m1/m2 = 2 and R₁/R₂ = 1/4

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.