JEE Advanced Previous Year Questions with Solutions on Gravitation

Newton made a systematic study of the forces acting on different particles and formulated the results in a simple law. This is known as the Law of Gravitation. According to the Law of Gravitation, the force of attraction between any two particles in this system, is directly proportional to the product of the masses of the particles, and inversely proportional to the square of the distance between them. The force acts along the line joining the two particles.

Download Gravitation JEE Advanced Previous Year Solved Questions PDF

Question 1) If the angular momentum of a planet of mass m, moving around the Sun in a circular orbit is L, about the centre of the Sun, its areal velocity is

(a) L/m

(b) 4L/m

(c) L/2m

(d) 2L/m

Answer: (c) L/2m

Solution:

Area velocity, dA/dt

dA = (½)r²dθ

⇒ (dA/dt) = (½)r²(dθ/dt)

= (½)r²ω

Also, L = mvr =mr²ω

Therefore, (dA/dt) = (½)(L/m)

Question 2) A bullet is fired vertically upwards with velocity v from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to the planet’s gravity is (¼)th of its value of the surface of the planet. If the escape velocity from the planet is v_esc = v√N, then the value of N is (ignore energy loss due to atmosphere).

Answer: 2

Solution:

Let h be the height to which the bullet rises

With the height the acceleration due to gravity varies as,

⇒ h = R

We know the escape speed,

Now applying the conservation of energy principle

Loss of kinetic energy = gain in gravitational potential energy

(½) mv² = -GMm/2R – (- GMm/R)

Comparing equation (1) and (2) we get N = 2

Question 3) A box weighs 196 N on a spring balance at the north pole. Its weight recorded on the same balance if it is shifted to the equator is close is (Take g = 10ms^-2 at the north pole and the radius of the earth = 6400 km)

(a) 195.66 N

(b) 194.32 N

(c) 194.66 N

(d) 195.32 N

Answer: (d) 195.32 N

Solution:

Weight at the pole, w = mg = 196 N

⇒ m = 19.6 Kg

Weight at the equator, w’ = mg’ = m(g – ω²R)

= 19.6 (10 – 0.034) = 195.32 N

Question 4) If the radius of earth is R, then the height h at which the value of g becomes one-fourth,

1) R/8

2) 3R/8

3) 3R/4

4) R

Answer: (4) R

Solution:

We know , g = GM/R²

g_h = GM/(R+h)²

g_h = (1/4) g

h + R = 2R

h = R

Question 5) A body weighs 500 N on the surface of the earth. How much would it weigh halfway below the surface of the earth?

1) 125 N

2) 250 N

3) 500 N

4) 1000 N

Answer: 2) 250 N

Solution:

Weight (W) of a body given by the product of mass (m) and acceleration due to gravity (g)

W = mg

The value of g below the surface of the earth at a distance h is given by

g′ = g(1 − h/R)

Therefore weight halfway below the surface of the earth, w′ = w(1 − h/R)

Given, h = R/2,w = 500N

Therefore, w′ = 500(1−1/2)

= 500/2 = 250N

Question 6) If g is the acceleration due to gravity on the earth’s surface, the gain in the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth is

1) (1/2) mg R

2) 2 mg R

3) mg R

4) (1/4) mg R

Answer: 1) (1/2) mg R

Solution:

Initial potential energy ,

Final potential energy,

But g = GM/R²

Therefore, GM = gR² …..(2)

From (1) and (2) we get

Question 7)The escape velocity from the earth is 11 kms ^-1. The escape velocity from a planet having twice the radius and same mean density as that of the earth is

1) 5.5 kms ^-1

2) 11 kms ^-1

3) 22 kms ^-1

4) None of these

Answer: 3) 22 kms ^-1

Solution:

The escape velocity of earth is given by

Radius of planet is twice the radius of earth ie R_p=2R_e

ρ_p = ρ_e

Escape velocity

Therefore,

Question 8) A solid sphere of mass M and radius R has a spherical cavity of radius R/2 such that the centre of the cavity is at a distance of R / 2 from the centre of the sphere. A point mass m is placed inside the cavity at a distance of R / 4 from the centre of the sphere. The gravitational pull between the sphere and the point mass m is

Answer:

Solution:

Field strength is uniform inside the cavity. Let us find it at its centre.

E_T = E_R + E_C (T = total, R = Remaining, C- Cavity)

E_R = E_T – E_C

Question 9) A particle of mass 10 g is kept on the surface of a uniform sphere of mass 100 kg and radius 10 cm. Find the work to be done against the gravitational force between them. to take the particle far away from the sphere. (take, G = 6.67 × 10^-11 Nm²kg ^-2)

(1) 13.34 × 10^-10 J

(2) 3.33 × 10^-10 J

(3) 6.67 × 10^-9 J

(4) 6.67 × 10^-10 J

Answer: (4) 6.67 × 10^-10 J

Solution:

The initial potential energy of the particle, U_i = – GMm/r

= -6.67 x 10^-10 J

We know that Work done = Difference in potential energy

W = ΔU = U_f – U_i

W =- U_i (since U_f = 0)

= 6.67 x 10^-10 J

Question 10) The value of g on the earth surface is 980 cm/sec². Its value at a height of 64 km from the earth surface is …………… cms^–2.

(1) 960.40

(2) 984.90

(3) 982.45

(4) 977.55

Answer: (1) 960.40

Solution:

(g’/g) = [R/(R+h)]²

Given h = 64 km

g = 980 cm/s²

g’ = g [R/(R + h)]²

= 980 x [(6400)/(6400 + 64)]²

= 960. 4 cm/s²

Also Read:

Gravitation JEE Main Previous Year Questions With Solutions

Gravitation