JEE Main · Physics ↓ Falling

Gravitation appeared 22 times across 3 years — 2.5% of Physics. This question is from Gravitational Potential Energy.

Year 2026 2025 2024 Total
Questions 5 9 8 22
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Three masses 200 kg, 300 kg and 400 kg are placed at the vertices of an equilateral triangle with sides 20 m. They are rearranged on the vertices of a bigger triangle of side 25 m and with the same centre. The work done in this process ____ J. (Gravitational constant G = 6.7 × 10⁻¹¹ N m² / kg²)

Solution & Explanation

Related Formula
U = -(Gm₁ m₂)/(r) Wₑₓₜ = Δ U = Uf - Uᵢ
Core Logic

Equilateral triangle mass configuration scaling
Equilateral triangle mass configuration scaling

Initial potential energy of the system (rᵢ = 20 m):

Uᵢ = -(G)/(rᵢ) (m₁ m₂ + m₂ m₃ + m₁ m₃) Uᵢ = -6.67 × 10⁻¹¹20 [200 × 300 + 300 × 400 + 200 × 400] Uᵢ = -6.67 × 10⁻¹¹20 [60000 + 120000 + 80000] Uᵢ = -6.67 × 10⁻¹¹20 × 260000 = -86.71 × 10⁻⁸ J

Equilateral triangle mass configuration scaling
Equilateral triangle mass configuration scaling

Final potential energy of the system (rf = 25 m):

Uf = -(G)/(rf) (260000) Uf = -6.67 × 10⁻¹¹25 × 260000 Uf = -69.36 × 10⁻⁸ J
Step 1: Calculate Work Done

The work done by an external agent is equal to the change in potential energy:

Δ U = Uf - Uᵢ Δ U = (-69.36 × 10⁻⁸) - (-86.71 × 10⁻⁸) Δ U = (86.71 - 69.36) × 10⁻⁸ = 17.35 × 10⁻⁸ J ≈ 1.74 × 10⁻⁷ J
Pattern Recognition

When an entire geometric configuration is isotropically scaled (e.g., side a to b), the total potential energy scales by the ratio of their reciprocals. Uf = Uᵢ (rᵢ/rf).

Chapter Mix

Class 11 Physics: Gravitation Class 11 Physics: Work, Energy and Power

Reference Study Guides

Gravitation

More Gravitation Previous-Year Questions

Q44 jee_main_2026_21_jan_morning Satellite Energy
Initially a satellite of 100 kg is in a circular orbit of radius 1.5RE. This satellite can be moved to a circular orbit of radius 3RE by supplying α × 10⁶J of energy. The value of α is ____. (Take Radius of Earth RE = 6 × 10⁶ m and g = 10 m/s²)
  • A. 150
  • B. 500
  • C. 100
  • D. 1000

Solution

Related Formula
E = (-GME m)/(2r) Δ E = Ef - Eᵢ g = (GME)/(RE²) GME = g RE²
Core Logic

Energy of a satellite in a circular orbit is given as E = (-GME m)/(2r) where r is the radius of the circular orbit. Required energy to be supplied Δ E = Ef - Eᵢ:

Δ E = ( (-GME m)/(2(3RE)) ) - ( (-GME m)/(2(1.5RE)) ) Δ E = (-GME m)/(6RE) + (GME m)/(3RE) = (GME m)/(6RE)
Step 1: Evaluate Delta E

Substitute GME = g RE² into the expression:

Δ E = ((g RE²) m)/(6RE) = (1)/(6) m g RE = (1)/(6) × 100 × 10 × (6 × 10⁶) = 1000 × 10⁶ J

Comparing with α × 10⁶ J, we get α = 1000.

Pattern Recognition

Orbital transition energy is always Δ E = (GMm)/(2) ((1)/(rᵢ) - (1)/(rf)). Never forget the factor of 2 in the denominator (which accounts for kinetic energy contribution in orbit).

Chapter Mix

Class 11 Physics: Gravitation

Q30 jee_main_2026_22_january_morning Escape Velocity
The escape velocity from a spherical planet A is 10 km/s. The escape velocity from another planet B whose density and radius are 10% of those of planet A, is \_\_\_\_ m/s.
  • A. 1000
  • B. 200√(5)
  • C. 100√(10)
  • D. 1000√(2)

Solution

Related Formula
Vₑ = √((2GM)/(R)) = √((2G ρ (4π R³)/(3))/(R)) Vₑ ∝ √(ρ) × R
Core Logic

Given density and radius of B are 10% (0.1) of planet A:

((Vₑ)B)/((Vₑ)A) = √((ρB)/(ρA)) × (RB)/(RA) = √(0.1) × 0.1 = 110√(10) (Vₑ)B = 10 × 100010√(10) = 100√(10) m/sec
Pattern Recognition

Sees: Escape velocity scaling with density and radius. Shortcut: Express escape velocity in terms of density ρ and radius R, then take ratios. Check: Matches option (3). ✓

Chapter Mix

Class 11 Physics: Gravitation

Q35 jee_main_2026_22_january_morning Gravitational Force and Superposition
Net gravitational force at the centre of a square is found to be F₁ when four particles having mass M, 2M, 3M and 4M are placed at the four corners of the square as shown in figure and it is F₂ when the positions of 3M and 4M are interchanged. The ratio F₁F₂ is α√(5). The value of α is \_\_\_\_.
Gravitation diagram for Q35 - JEE Main 2026 January Morning
Four point masses placed at the corners of a square.
  • A. 2
  • B. 3
  • C. 1
  • D. 2√(5)

Solution

Related Formula
F = (G m₁ m₂)/(r²)
Core Logic

Solution diagram for Q35 - JEE Main 2026 Morning
Four point masses placed at the corners of a square.

Initial configuration:

F₁ = 2√(2) (Gmm₀)/(r²)$|

New configuration (after interchanging 3M and 4M):

F_2 = \sqrt{10} \frac{Gmm_0}{r^2}Ratio:\frac{F_1}{F_2} = \frac{2\sqrt{2}}{\sqrt{10}} = \frac{2}{\sqrt{5}} \implies \alpha = 2$$
Pattern Recognition

Sees: Gravitational force vector sum at center of square. Shortcut: Compute net vector resultants for both mass configurations and take ratio. Check: Matches option (1). ✓

Chapter Mix

Class 11 Physics: Gravitation

Q39 jee_main_2026_22_january_evening Satellite Motion and Time Period
Given below are two statements : Statement I : A satellite is moving around earth in the orbit very close to the earth surface. The time period of revolution of satellite depends upon the density of earth. Statement II: The time period of revolution of the satellite is T = 2π Rₑg (for satellite very close to the earth surface), where Rₑ radius of earth and g acceleration due to gravity. In the light of the above statements, choose the correct answer from the options given below :
  • A. Both Statement I and Statement II are false
  • B. Both Statement I and Statement II are true
  • C. Statement I is true but Statement II is false
  • D. Statement I is false but Statement II is true

Solution

Related Formula
T = 2π √((R³)/(GM)) M = ρ · (4)/(3)π R³ g = (GM)/(R²)
Core Logic

Evaluating Statement I: Substituting total mass M = ρ · (4)/(3)π R³ into time period formula:

T = 2π √((R³)/(G (ρ · (4)/(3)π R³))) = √((3π)/(G ρ))

Hence, time period depends inversely on the square root of Earth's density ρ. Statement I is true.

Evaluating Statement II: Substituting g = (GM)/(Rₑ²) into T = 2π √((Rₑ³)/(GM)):

T = 2π √((Rₑ)/(g))

Statement II is also true.

Step 1: Final Conclusion

Both Statement I and Statement II are true.

Pattern Recognition

Near-surface satellite shortcut: T = √(3π / (Gρ)) ≈ 84.6 min. Shows direct dependence on Earth's density ρ and formula T = 2π √(Rₑ / g).

Chapter Mix

Class 11 Physics: Gravitation

More Gravitation Questions — jee_main_2026_24_january_morning

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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)