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Universal law of gravitation: 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. Two point masses are separated by a fixed distance r. If both masses are doubled and the distance between them is halved, the magnitude of the gravitational attraction between them becomes:

  1. 4 times the initial value
  2. 8 times the initial value
  3. 32 times the initial value
  4. 16 times the initial value
Show answer and explanation

Answer: (D) 16 times the initial value

Initial force is F = G·m₁·m₂/r². When both masses are doubled (m' = 2m) and distance is halved (r' = r/2), the new force is F' = G(2m₁)(2m₂)/(r/2)² = 16·G·m₁·m₂/r² = 16F.

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2. Two identical stars, each of mass M, orbit around their common centre of mass in circular orbits of radius R under their mutual gravitational attraction. The orbital speed of each star is:

  1. 2 √(G M / R)
  2. √(G M / (2 R))
  3. √(G M / R)
  4. √(G M / (4 R))
Show answer and explanation

Answer: (D) √(G M / (4 R))

The distance between the two stars is 2R, so the gravitational attraction between them is F = G M² / (2R)² = G M² / (4 R²). This force provides the centripetal force for circular motion of radius R: M v² / R = G M² / (4 R²). Solving for v gives v = √(G M / (4 R)).

hard

3. A uniform thin hemispherical shell of mass M and radius R has its base on the x-y plane centered at the origin. A point particle of mass m is located at the origin. The magnitude of the gravitational attraction between the shell and the particle is:

  1. Zero
  2. G M m / R²
  3. G M m / (4 R²)
  4. G M m / (2 R²)
Show answer and explanation

Answer: (D) G M m / (2 R²)

Divide the hemispherical shell into rings of angular radius θ and width dθ. The ring has mass dM = M sin θ dθ. Each point on the ring is at distance R from the origin. The axial force is dF = (G m dM / R²) cos θ = (G M m / R²) sin θ cos θ dθ. Integrating from θ = 0 to π/2 gives F = (G M m / R²) [sin² θ / 2] from 0 to π/2 = G M m / (2 R²).

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4. Four identical particles, each of mass m, are kept at the four corners of a square of side a. The magnitude of the net gravitational force on any one particle due to the other three particles is:

  1. (G·m²/a²) · (2 + 1/√2)
  2. (G·m²/a²) · (2√2 + 1)
  3. (G·m²/a²) · (√2 + 1/2)
  4. (G·m²/a²) · (1 + 1/√2)
Show answer and explanation

Answer: (C) (G·m²/a²) · (√2 + 1/2)

The forces from the two adjacent masses each have magnitude F₁ = G·m²/a² at 90° to each other, giving a resultant of √2·F₁ directed along the diagonal. The diagonally opposite mass exerts F₂ = G·m²/(√2 a)² = F₁/2 along the same diagonal. The total force is F_net = √2·F₁ + F₁/2 = (√2 + 1/2)·(G·m²/a²).

easy

5. The value of the universal gravitational constant G is about:

  1. 6.02 × 10²³ N m² kg⁻²
  2. 6.67 × 10⁻¹¹ N m² kg⁻²
  3. 9.8 N m² kg⁻²
  4. 6.67 × 10⁻¹¹ N kg² m⁻²
Show answer and explanation

Answer: (B) 6.67 × 10⁻¹¹ N m² kg⁻²

In F = Gm1m2/r², G = 6.67 × 10⁻¹¹ N m² kg⁻². It was first measured experimentally by Cavendish.

NCERT Class 11

hard

6. Three identical stars, each of mass M, move in a common circular orbit of radius R under their mutual gravitational attraction, always remaining at the vertices of an equilateral triangle. The orbital speed of each star is:

  1. √(G·M / R)
  2. √(G·M / (√3·R))
  3. √(√3·G·M / R)
  4. √(2·G·M / R)
Show answer and explanation

Answer: (B) √(G·M / (√3·R))

The side length of the equilateral triangle inscribed in a circle of radius R is a = √3 R. The net inward gravitational force on any star is F_net = √3·G·M²/a² = √3·G·M²/(3R²) = G·M²/(√3·R²). This provides the necessary centripetal force: M·v²/R = G·M²/(√3·R²), which yields v = √(G·M / (√3·R)).

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7. A straight wire of length 2L has a uniform linear mass density λ. A point mass m is placed on the perpendicular bisector of the wire at a distance d from its midpoint. The magnitude of the gravitational attraction between the wire and the point mass is:

  1. (G m λ) / √(L² + d²)
  2. (G m λ L) / (d² + L²)
  3. (2 G m λ) / d
  4. (2 G m λ L) / (d √(L² + d²))
Show answer and explanation

Answer: (D) (2 G m λ L) / (d √(L² + d²))

By dividing the wire into elements dx at position x, the force component along the perpendicular is dF = (G m λ dx / (x² + d²)) · (d / √(x² + d²)). Integrating from x = -L to L yields F = (2 G m λ L) / (d √(L² + d²)).

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8. A uniform wire of total mass M and length L is bent into a semicircle of radius R. A point mass m is placed at the centre of curvature of the semicircle. The magnitude of the gravitational attraction between the semicircular wire and the point mass is:

  1. 2 π G M m / R²
  2. G M m / (π R²)
  3. 4 G M m / (π² R²)
  4. 2 G M m / (π R²)
Show answer and explanation

Answer: (D) 2 G M m / (π R²)

The linear mass density is λ = M / (π R). Consider an element dθ of the semicircle. The force on mass m due to element dm = λ R dθ is dF = G m dm / R² along the radial direction. By symmetry, components perpendicular to the axis of symmetry cancel, and components along the axis add: F = ∫ (G m λ R dθ / R²) cos θ from -π/2 to π/2 = (G m λ / R) [sin θ] from -π/2 to π/2 = 2 G m λ / R. Substituting λ = M / (π R) gives F = 2 G M m / (π R²).

easy

9. Two point masses m and 9m are separated by a fixed distance L. A third point mass M is to be placed along the line joining them such that the net gravitational force on M is zero. The position of mass M from the mass m is:

  1. L/4
  2. L/3
  3. L/2
  4. 3L/4
Show answer and explanation

Answer: (A) L/4

Let the distance from m be x. The condition for equilibrium is G·m·M/x² = G·(9m)·M/(L - x)². Taking the square root gives 1/x = 3/(L - x), which simplifies to L - x = 3x, yielding x = L/4.

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10. The dimensional formula of G is:

  1. [M L³ T⁻²]
  2. [M⁻¹ L² T⁻²]
  3. [M L² T⁻²]
  4. [M⁻¹ L³ T⁻²]
Show answer and explanation

Answer: (D) [M⁻¹ L³ T⁻²]

G = Fr²/(m1m2), so its dimensions are [M L T⁻²][L²]/[M²] = [M⁻¹ L³ T⁻²].

NCERT Class 11

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11. Two bodies of masses 4 kg and 9 kg are separated by a distance of 10 m. At what distance from the 4 kg mass along the line joining them is the net gravitational force on a third mass equal to zero?

  1. 4.0 m
  2. 3.0 m
  3. 5.0 m
  4. 6.0 m
Show answer and explanation

Answer: (A) 4.0 m

Let the point be at distance x from the 4 kg mass. For the net force to be zero, G(4)m/x² = G(9)m/(10 - x)². Taking the square root on both sides yields 2/x = 3/(10 - x), which simplifies to 20 - 2x = 3x, giving 5x = 20, or x = 4.0 m.

easy

12. Two spherical bodies of masses 2 kg and 8 kg are fixed at a distance of 1.2 m apart. Where should a third point mass of 1 kg be placed on the line connecting their centres so that it experiences zero net gravitational force?

  1. 0.6 m from the 2 kg mass
  2. 0.3 m from the 2 kg mass
  3. 0.4 m from the 2 kg mass
  4. 0.8 m from the 2 kg mass
Show answer and explanation

Answer: (C) 0.4 m from the 2 kg mass

Let the point mass be placed at distance x from the 2 kg mass. For zero net force: G(2)(1)/x² = G(8)(1)/(1.2 - x)². Taking the square root on both sides: √2/x = √8/(1.2 - x) = 2√2/(1.2 - x), which simplifies to 1/x = 2/(1.2 - x). Thus, 1.2 - x = 2x, giving 3x = 1.2 m, or x = 0.4 m from the 2 kg mass.

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