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JEE Advanced Previous Year Questions (2018 - 2023): Gravitation | Physics for JEE Main & Advanced PDF Download
2023
Q1: Two satellites P and Q are moving in different circular orbits around the Earth (radius R ). The heights of P and Q from the Earth surface are ℎP and ℎQ, respectively, where ℎP = R/3. The accelerations of P and Q due to Earth's gravity are gP and gQ, respectively. If gP/gQ = 36/25, what is the value of ℎQ ?
[JEE Advanced 2023 Paper 1]
(a) 3R /5
(b) R / 6
(c) 6R / 5
(d) 5R/5
Ans: (a)
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 ( hP ) and ( hQ ) above the Earth's surface, use the formula for each and form a ratio :
Given ℎP=Re/3, the ratio simplifies to:
Solving for ℎQ in terms of Re yields
.2022
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
2021
Q1: The distance between two stars of masses 3MS and 6MS is 9R. Here R is the mean distance between the centers of the Earth and the Sun, and MS is the mass of the Sun. The two stars orbit around their common center of mass in circular orbits with period nT, where T is the period of Earth's revolution around the Sun. The value of n is __________. [JEE Advanced 2021 Paper 2]
Ans: 9
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)
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 m1 and m2. The radii of their circular orbits are R1 and R2 respectively, Ignore the gravitational force between the satellites. Define v1,L1,K1 and T1 to be , respectively, the orbital speed, angular momentum, kinetic energy and time period of revolution of satellite 1; and v2,L2,K2, and T2 to be the corresponding quantities of satellite 2. Given m1/m2 = 2 and R1/R2 = 1/4, match the ratios in List-I to 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)
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.
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FAQs on JEE Advanced Previous Year Questions (2018 - 2023): Gravitation - Physics for JEE Main & Advanced
| 1. What is the law of gravitation? |  |
Ans. The law of gravitation states that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
| 2. How does the force of gravity vary with distance? | |
Ans. The force of gravity follows an inverse square law with distance. This means that as the distance between two objects increases, the force of gravity decreases. Specifically, the force of gravity is inversely proportional to the square of the distance between the objects.
| 3. What is the difference between mass and weight? | |
Ans. Mass is a measure of the amount of matter in an object and is constant regardless of the location. Weight, on the other hand, is a measure of the force of gravity acting on an object and depends on the gravitational field strength. Weight can vary depending on the location, but mass remains the same.
| 4. How does the value of 'g' vary on different planets? | |
Ans. The value of 'g', which represents the acceleration due to gravity, varies on different planets. It depends on the mass and radius of the planet. Generally, planets with larger masses and smaller radii will have a higher value of 'g', resulting in a stronger gravitational pull.
| 5. How does the gravitational force between two objects change if their masses are doubled? | |
Ans. If the masses of two objects are doubled, the gravitational force between them will also double. This is because the force of gravity is directly proportional to the product of the masses. Therefore, increasing the masses by a factor of two will result in a corresponding increase in the gravitational force.
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