Gravitation — Class 11 MCQs with Answers

Gravitation — important questions & MCQs with answers (Class 11 Physics)

13 solved questions from this chapter's difficulty levels, each with its answer and explanation. The other 53 are timed and scored when you sign in.

1. Q1Easy

   Newton's law of gravitation: F =

   A.Gm₁m₂r

   B.Gm₁m₂/r

   C.Gm₁m₂/r²✓ Correct

   D.G/(m₁m₂r²)

   Answer: C. Gm₁m₂/r²

   Explanation: Newton's law of gravitation is inverse-square: F = Gm₁m₂/r². Writing Gm₁m₂r or dropping the square on r is the classic slip — check that r sits squared in the denominator, not the numerator.

2. Q2Easy

   Newton's law of gravitation: F =

   A.m₁ + m₂

   B.G m₁m₂/r

   C.G/(m₁m₂)

   D.G m₁m₂/r²✓ Correct

   Answer: D. G m₁m₂/r²

   Explanation: Newton's law of gravitation: F = Gm₁m₂/r², the inverse-square law. G/(m₁m₂) inverts the masses and drops r² — force must scale directly with each mass, not inversely.

3. Q3Easy

   g near Earth's surface ≈

   A.9.8 km/s

   B.9.8 m/s²✓ Correct

   C.10 km/s

   D.Variable

   Answer: B. 9.8 m/s²

   Explanation: g near Earth's surface ≈ 9.8 m/s², not 9.8 km/s — mixing up m/s² and km/s is the common unit slip here, since 9.8 km/s would be close to escape velocity, far too large for surface gravity.

4. Q4Easy

   U(r) = ?

   A.−G m₁m₂/r✓ Correct

   B.G m₁m₂/r

   C.G m₁m₂/r²

   D.Zero

   Answer: A. −G m₁m₂/r

   Explanation: Gravitational potential energy U(r) = −Gm₁m₂/r is negative because the zero reference is set at infinite separation. Writing +Gm₁m₂/r drops this sign convention that marks the force as attractive.

5. Q5Easy

   Escape velocity from Earth's surface = ?

   A.gR

   B.√(GM/R)

   C.√(2GM/R) = √(2gR)✓ Correct

   D.Mass

   Answer: C. √(2GM/R) = √(2gR)

   Explanation: Escape velocity comes from setting KE equal to |PE| at the surface: ½mv² = GMm/R, giving v = √(2GM/R) = √(2gR). Writing just gR under the root omits the factor of 2 from that energy balance.

6. Q6Medium

   Force between two 1 kg masses 1 m apart:

   A.1 N

   B.9.8 N

   C.6.67 × 10⁻¹¹ N✓ Correct

   D.Cannot compute

   Answer: C. 6.67 × 10⁻¹¹ N

   Explanation: F = Gm₁m₂/r² with m₁ = m₂ = 1 kg and r = 1 m gives F = G = 6.67 × 10⁻¹¹ N. 9.8 N wrongly substitutes g for G — the gravitational constant G and surface gravity g are easy to mix up but numerically very different.

7. Q7Medium

   g at height 6400 km (= R) above Earth (R = 6400 km):

   A.g/16

   B.g/2

   C.g

   D.g/4✓ Correct

   Answer: D. g/4

   Explanation: At height h = R (total distance 2R from center), g_h = gR²/(R+h)² = gR²/(2R)² = g/4. Writing g/2 mistakes the squared denominator for a linear one — always square the (R+h) term first.

8. Q8Medium

   Escape velocity from planet of mass M and radius R:

   A.√GMr

   B.√(GM/R)

   C.√(2GM/R)✓ Correct

   D.GMR

   Answer: C. √(2GM/R)

   Explanation: Escape velocity is v = √(2GM/R), from setting ½mv² = GMm/R. Omitting the 2, as in √(GM/R), instead gives the orbital velocity at the surface — a factor-of-2 slip that is the most common trap in this topic.

9. Q9Medium

   At distance 2R from Earth's center, orbital speed (in terms of g, R):

   A.√(gR)

   B.√(gR/2)✓ Correct

   C.√(2gR)

   D.gR

   Answer: B. √(gR/2)

   Explanation: At r = 2R, orbital speed v = √(GM/r) = √(GM/2R). Since g = GM/R², this becomes √(gR/2). Writing √(2gR) mistakes this for the escape-velocity form — check whether the 2 belongs inside or outside the radius term.

10. Q10Medium

    Two planets at distances r and 4r from sun. Ratio of periods:

    A.1 : 16

    B.1 : 2

    C.1 : 4

    D.1 : 8✓ Correct

    Answer: D. 1 : 8

    Explanation: By Kepler's third law, T² ∝ r³, so T₁/T₂ = (r₁/r₂)^1.5 = (1/4)^1.5 = 1/8, a period ratio of 1:8. Writing 1:16 mistakenly uses the square of the distance ratio instead of the 3/2 power.

11. Q11Hard

    Acceleration due to gravity at height h above Earth's surface (h<<R):

    A.g·R

    B.g(1 + h/R)

    C.g·h

    D.g(1 − 2h/R)✓ Correct

    Answer: D. g(1 − 2h/R)

    Explanation: For h ≪ R, expanding g_h = gR²/(R+h)² gives g_h ≈ g(1 − 2h/R). Writing g(1 + h/R) borrows the depth formula's form with the wrong sign — gravity weakens with height, so the term must subtract.

12. Q12Hard

    Force on body of mass m placed at center of uniform spherical shell:

    A.Zero✓ Correct

    B.mg

    C.GMm/R²

    D.Maximum

    Answer: A. Zero

    Explanation: By Newton's shell theorem, a uniform spherical shell exerts zero net gravitational force anywhere inside it, since pulls from every part of the shell cancel exactly. Picking GMm/R² wrongly applies the surface formula to an interior point.

13. Q13Hard

    Earth's mass = 6 × 10²⁴ kg, radius = 6.4 × 10⁶ m. g at surface ≈

    A.9.8 m/s²✓ Correct

    B.19.6 m/s²

    C.4.9 m/s²

    D.Cannot determine

    Answer: A. 9.8 m/s²

    Explanation: g = GM/R² = (6.67 × 10⁻¹¹ × 6 × 10²⁴)/(6.4 × 10⁶)² ≈ 9.8 m/s². Doubling this to 19.6 m/s² comes from mishandling the squared radius in the denominator — always square R before dividing, not after.

Gravitation — FAQs

More Class 11 Physics chapters

• 01Physical World
• 02Units and Measurements
• 03Motion in a Straight Line
• 04Motion in a Plane
• 05Laws of Motion
• 06Work, Energy and Power

Class 11 — other subjects

• Class 11 Biology chapters
• Class 11 Chemistry chapters
• Class 11 Computer Science chapters
• Class 11 English chapters
• Class 11 Mathematics chapters