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Gravitational potential and potential energy: NEET MCQs with answers

49 questions on this topic of Gravitation (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. The ratio of gravitational potential at the centre of a uniform solid sphere of radius R to that at its surface (taking potential at infinity to be zero) is:

  1. 1 : 2
  2. 3 : 2
  3. 2 : 3
  4. 4 : 3
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Answer: (B) 3 : 2

For a uniform solid sphere, the potential at the centre is V_c = -(3/2)(GM/R) and at the surface is V_s = -GM/R. The ratio V_c / V_s is (3/2) : 1 = 3 : 2.

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2. Three particles, each of mass m, are situated at the vertices of an equilateral triangle of side L. The work done by an external agent to increase the side of the triangle to 2L slowly is:

  1. 3Gm² / L
  2. -3Gm² / (2L)
  3. 3Gm² / (2L)
  4. Gm² / (2L)
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Answer: (C) 3Gm² / (2L)

Initial potential energy U_i = 3 × (-Gm²/L) = -3Gm²/L. Final potential energy U_f = 3 × (-Gm²/(2L)) = -3Gm²/(2L). The work done by the external agent equals ΔU = U_f - U_i = -3Gm²/(2L) - (-3Gm²/L) = +3Gm²/(2L).

hard

3. A uniform thin spherical shell of mass M and radius R has a small hole. A particle of mass m is released from rest at a point on the line passing through the centre at a distance 2R from the centre. The speed with which it reaches the centre of the shell through the hole is:

  1. √(GM / (2R))
  2. √(2GM / R)
  3. √(GM / R)
  4. Zero
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Answer: (C) √(GM / R)

Initial potential energy at distance 2R is U_initial = -GMm / (2R). The gravitational potential anywhere inside the shell (including at the centre) is uniform and equals -GM/R, so U_final = -GMm / R. By conservation of energy: U_initial + 0 = U_final + (1/2)mv², which gives (1/2)mv² = -GMm/(2R) - (-GMm/R) = GMm / (2R). Hence, v = √(GM / R).

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4. The minimum work needed to move a body of mass m from the earth's surface to infinity is:

  1. GMm/R
  2. GMm/R²
  3. mgR²
  4. Zero
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Answer: (A) GMm/R

Work needed equals the increase in potential energy: 0 - (-GMm/R) = GMm/R. This is also why escape velocity is √(2GM/R).

NCERT Class 11

easy

5. A uniform ring of mass M and radius R is placed in the xy-plane with its centre at the origin. What is the gravitational potential at a point on its axis at a distance x = √3 R from the centre?

  1. - GM / (4R)
  2. - GM / (2R)
  3. - √3 GM / (2R)
  4. - GM / R
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Answer: (B) - GM / (2R)

Every element of the ring is at an equal distance r = √(R² + x²) from the point on the axis. For x = √3 R, the distance is r = √(R² + 3R²) = 2R. Thus, the gravitational potential is V = - GM / r = - GM / (2R).

hard

6. Two spherical bodies of masses M and 2M and radii R and 2R respectively are held with their centres separated by a distance 9R in free space. If they are released from rest and attract each other purely by gravity, their relative speed of approach just before collision is:

  1. √[4GM / (3R)]
  2. √[2GM / (3R)]
  3. √[GM / R]
  4. √[8GM / (9R)]
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Answer: (A) √[4GM / (3R)]

Initially, distance between centres is r₁ = 9R. At collision, the distance between centres is r₂ = R + 2R = 3R. Reduced mass μ = (M × 2M)/(M + 2M) = 2M/3. By energy conservation, (1/2) μ v_rel² = GM(2M) [1/(3R) - 1/(9R)] = 2GM² [2/(9R)] = 4GM²/(9R). Thus, (1/2)(2M/3) v_rel² = 4GM²/(9R) ⇒ (1/3) v_rel² = 4GM/(9R) ⇒ v_rel = √[4GM / (3R)].

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7. A satellite of mass m is orbiting the Earth of mass M and radius R in a circular orbit of radius 2R. The additional energy required to transfer the satellite into another stable circular orbit of radius 3R is:

  1. GMm / (4R)
  2. GMm / (6R)
  3. GMm / (24R)
  4. GMm / (12R)
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Answer: (D) GMm / (12R)

The total mechanical energy of a satellite in a circular orbit of radius r is E = -GMm / (2r). For r₁ = 2R, E₁ = -GMm / (4R). For r₂ = 3R, E₂ = -GMm / (6R). The required additional energy is ΔE = E₂ - E₁ = -GMm/(6R) - (-GMm/(4R)) = GMm/R × (1/4 - 1/6) = GMm / (12R).

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8. Three identical particles, each of mass m, are held at the vertices of an equilateral triangle of side a. If the separation between each pair of particles is doubled to 2a, the work done by the gravitational force during this process is:

  1. + 3 Gm² / (2a)
  2. - 3 Gm² / (2a)
  3. - 3 Gm² / a
  4. + 3 Gm² / a
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Answer: (B) - 3 Gm² / (2a)

Initial potential energy is U_i = - 3 Gm² / a. Final potential energy is U_f = - 3 Gm² / (2a). The work done by the gravitational force is W_g = - ΔU = - (U_f - U_i) = - [- 3 Gm² / (2a) - (- 3 Gm² / a)] = - 3 Gm² / (2a).

easy

9. The gravitational potential at a distance r from the centre of the earth (r ≥ R) is:

  1. GM/r
  2. -GMr
  3. -GM/r
  4. -GM/r²
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Answer: (C) -GM/r

Gravitational potential is the potential energy per unit mass, V = -GM/r, taking zero at infinity. It is negative because gravity is attractive.

NCERT Class 11

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10. Two point masses of 8 kg and 2 kg are fixed at a distance of 9 m apart in free space. The gravitational potential at the point on the line segment joining them where the net gravitational field intensity is zero is:

  1. -(4/3) G J kg⁻¹
  2. -4 G J kg⁻¹
  3. -(8/3) G J kg⁻¹
  4. -2 G J kg⁻¹
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Answer: (D) -2 G J kg⁻¹

Let the neutral point be at distance x from the 8 kg mass. Then G(8)/x² = G(2)/(9 - x)², which gives 2/x = 1/(9 - x), so 18 - 2x = x, yielding x = 6 m. The distance from the 2 kg mass is 9 - 6 = 3 m. The gravitational potential at this point is V = -G(8/6 + 2/3) = -G(4/3 + 2/3) = -2G J kg⁻¹.

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11. A thin uniform spherical shell of mass M and radius R has a concentric smaller thin uniform spherical shell of mass m and radius r (r < R). The gravitational potential at a point located at a distance x from the common centre, such that r < x < R, is:

  1. -G [ M/R + m/r ]
  2. -G [ M/x + m/x ]
  3. -G [ M/R + m/x ]
  4. -G [ M/x + m/r ]
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Answer: (C) -G [ M/R + m/x ]

The point lies inside the outer shell of radius R and outside the inner shell of radius r. The potential due to the outer shell at any internal point is constant and equals -GM/R. The potential due to the inner shell at a distance x > r is -Gm/x. By superposition, V = -G(M/R + m/x).

easy

12. An infinity of bodies, each of mass 2 kg, are placed along the x-axis at distances 1 m, 2 m, 4 m, 8 m, ... from the origin. The total gravitational potential at the origin due to this system is:

  1. - 2 G J kg⁻¹
  2. - 4 G J kg⁻¹
  3. - 8 G J kg⁻¹
  4. - G J kg⁻¹
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Answer: (B) - 4 G J kg⁻¹

The gravitational potential at the origin is V = - G ∑ (m / r_n) = - G (2) [1/1 + 1/2 + 1/4 + 1/8 + ...]. The series inside is an infinite geometric progression with first term a = 1 and common ratio r = 1/2, whose sum is S = 1 / (1 - 1/2) = 2. Therefore, V = - 2G(2) = - 4 G J kg⁻¹.

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