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Acceleration due to gravity and variation with altitude and depth: NEET MCQs with answers

63 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. Which one of the following graphs best represents the variation of acceleration due to gravity (g') with distance r from the centre of the Earth (radius R)?

  1. g' decreases linearly with r for r < R, and is inversely proportional to r for r ≥ R
  2. g' increases linearly with r for r < R, and is inversely proportional to r² for r ≥ R
  3. g' is constant for r < R, and decreases as 1/r² for r ≥ R
  4. g' increases as r² for r < R, and decreases as 1/r for r ≥ R
Show answer and explanation

Answer: (B) g' increases linearly with r for r < R, and is inversely proportional to r² for r ≥ R

Inside the Earth (r < R), g' = (G M / R³) r, which means g' ∝ r (linear increase from zero at r = 0 to maximum at r = R). Outside the Earth (r ≥ R), g' = G M / r², which means g' ∝ 1/r².

medium

2. Two spheres of the same material have radii in the ratio 1 : 3. The ratio of the acceleration due to gravity on their surfaces is:

  1. 1 : 27
  2. 1 : 9
  3. 3 : 1
  4. 1 : 3
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Answer: (D) 1 : 3

For a sphere of uniform density ρ, g = G M / R² = G (4/3 π R³ ρ) / R² = (4/3) π G ρ R. Since density ρ is the same for both, g is directly proportional to R. Therefore, g₁ : g₂ = R₁ : R₂ = 1 : 3.

hard

3. At what fraction of the Earth's radius above the surface (h/R) is the acceleration due to gravity equal to the acceleration due to gravity at a depth equal to the same fraction of the Earth's radius below the surface (d/R = h/R = x)?

  1. √5 - 2
  2. (√3 - 1) / 2
  3. (√5 - 1) / 2
  4. (3 - √5) / 2
Show answer and explanation

Answer: (C) (√5 - 1) / 2

Let x = h/R = d/R. We are given g_h = g_d, so g / (1 + x)² = g(1 - x). This gives (1 - x)(1 + x)² = 1, which expands to (1 - x)(1 + 2x + x²) = 1 + x - x² - x³ = 1. Thus, x(1 - x - x²) = 0. Since x > 0, we have x² + x - 1 = 0. Solving the quadratic equation gives x = (-1 + √5) / 2 = (√5 - 1) / 2.

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4. A body is taken from the Earth's surface to a height h = R/2. Then it is brought back to the surface and taken into a mine to a depth d = R/2. The ratio of the acceleration due to gravity at height h to that at depth d is:

  1. 4 : 9
  2. 9 : 8
  3. 2 : 3
  4. 8 : 9
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Answer: (D) 8 : 9

At height h = R/2, g_h = g / (1 + h/R)² = g / (1 + 1/2)² = g / (3/2)² = 4g/9. At depth d = R/2, g_d = g(1 - d/R) = g(1 - 1/2) = g/2. The ratio of g_h to g_d is (4g/9) / (g/2) = 8/9.

easy

5. The acceleration due to gravity on the surface of the Earth is g. At what height h above the surface does the acceleration due to gravity become g/9, where R is the radius of the Earth?

  1. h = R
  2. h = 2R
  3. h = 3R
  4. h = 8R
Show answer and explanation

Answer: (B) h = 2R

The variation of acceleration due to gravity with height is given by g_h = g / (1 + h/R)². Setting g_h = g/9 gives (1 + h/R)² = 9, so 1 + h/R = 3, which yields h = 2R.

hard

6. A hypothetical planet of mass M and radius R has a non-uniform mass density given by ρ(r) = Cr, where r is the distance from the centre and C is a constant. The variation of acceleration due to gravity g(r) inside the planet (r ≤ R) with distance r is proportional to:

  1. r
  2. r²
  3. 1 / r
  4. r³
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Answer: (B) r²

The mass enclosed within radius r is M(r) = ∫₀ʳ 4π x² ρ(x) dx = 4π C ∫₀ʳ x³ dx = π C r⁴. The gravitational acceleration at distance r inside the planet is g(r) = G M(r) / r² = G (π C r⁴) / r² = π G C r². Therefore, g(r) ∝ r².

medium

7. At what depth d below the surface of the Earth does the acceleration due to gravity become equal to its value at a height h = R/2 above the surface?

  1. 5R/9
  2. 4R/9
  3. 2R/3
  4. 7R/9
Show answer and explanation

Answer: (A) 5R/9

At height h = R/2, g_h = g / (1 + 1/2)² = 4g/9. At depth d, g_d = g(1 - d/R). Setting g_d = g_h gives 1 - d/R = 4/9, which yields d/R = 5/9, so d = 5R/9.

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8. Given below are two statements: Statement I: The acceleration due to gravity is zero at the centre of the Earth. Statement II: The rate of decrease of g with depth is twice the rate of decrease of g with height near the surface of the Earth. In light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are true
  2. Both Statement I and Statement II are false
  3. Statement I is true but Statement II is false
  4. Statement I is false but Statement II is true
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Answer: (C) Statement I is true but Statement II is false

Statement I is true because at the centre d = R, so g' = g(1 - R/R) = 0. Statement II is false because near the surface (dg/dh) = -2g/R, whereas (dg/dd) = -g/R, meaning the rate of decrease with height is twice that with depth, not vice-versa.

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9. The acceleration due to gravity at a depth equal to R/3 (R = radius of Earth) is what percentage of its value at the surface?

  1. 133 %
  2. 66.7 %
  3. 33.3 %
  4. 56.3 %
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Answer: (B) 66.7 %

g_d = g(1 − d/R) = g(1 − 1/3) = 66.7% of g.

medium

10. A planet has radius 2R and mean density equal to half that of the Earth. If the acceleration due to gravity on the Earth's surface is g, what is the acceleration due to gravity on the surface of this planet?

  1. g
  2. 2g
  3. g/2
  4. 4g
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Answer: (A) g

The acceleration due to gravity in terms of density ρ and radius R is g = (4/3)πGρR. For the planet, g' = (4/3)πG(ρ/2)(2R) = (4/3)πGρR = g.

easy

11. A physical balance with two identical pans is used to measure the mass of an object. If the balance is taken from the Earth's surface to the bottom of a deep mine of depth d = R/2, the reading of the mass will:

  1. decrease by 50%
  2. increase by 50%
  3. decrease by 25%
  4. remain completely unchanged
Show answer and explanation

Answer: (D) remain completely unchanged

A physical balance works on the principle of moments and compares the gravitational force on the object with standard weights: m₁·g' = m₂·g', so m₁ = m₂ regardless of the local value of g'. Therefore, the measured mass remains completely unchanged.

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12. The acceleration due to gravity at a height h above the Earth's surface is 4% less than that at the surface. If h << R, the percentage decrease in the acceleration due to gravity at a depth equal to h below the surface will be:

  1. 8%
  2. 4%
  3. 1%
  4. 2%
Show answer and explanation

Answer: (D) 2%

For small heights (h << R), the fractional decrease at altitude is Δg/g = 2h/R = 4%. At depth d = h, the fractional decrease is Δg'/g = d/R = h/R. Since 2h/R = 4%, we have h/R = 2%.

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