JEE Main · Physics ↓ Falling

Work, Energy and Power appeared 31 times across 3 years — 3.6% of Physics. This question is from Potential Energy of a Spring.

Year 2026 2025 2024 Total
Questions 8 15 8 31
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Two blocks with masses 100g and 200g are attached to the ends of springs A and B as shown in figure. The energy stored in A is E. The energy stored in B, when spring constants kA, kB of A and B, respectively satisfy the relation 4kA = 3kB, is :
Potential Energy of a Spring diagram for Q32 - JEE Main 2026 Morning
Two identical-looking springs vertically suspended with masses attached.
Potential Energy of a Spring diagram for Q32 - JEE Main 2026 Morning
Two identical-looking springs vertically suspended with masses attached.

Solution & Explanation

Related Formula

kx = mg

U = (1)/(2)kx² = (1)/(2) m²g²k
Core Logic

For a block suspended in equilibrium, the restoring force of the spring balances gravity. Thus, the extension is x = mg/k. Substituting this into the spring energy equation gives energy directly in terms of mass and spring constant.

Step 1: Compare Energy Ratios

Given energy proportional relation:

U ∝ m²k UAUB = ( mAmB)²( kBkA)
Step 2: Insert values

Given 4kA = 3kB kBkA = (4)/(3) Also mA = 100g, mB = 200g mAmB = (1)/(2)

UAUB = ((1)/(2))²((4)/(3)) = (1)/(4) × (4)/(3) = (1)/(3)
Step 3: Final Calculation

Since UA = E, we have:

EUB = (1)/(3) UB = 3E
Pattern Recognition

Sees: "energy stored" + "spring under gravity weight" → jump straight to U = ((mg)²)/(2k) instead of finding x first.

Chapter Mix

Class 11 Physics: Work, Energy and Power Class 11 Physics: Laws of Motion

Reference Study Guides

Work Energy And Power

More Work, Energy and Power Previous-Year Questions — Page 7

Q49 jee_main_2024_31_jan_morning Conservation Of Momentum
An artillery piece of mass M₁ fires a shell of mass M₂ horizontally. Instantaneously after the firing, the ratio of kinetic energy of the artillery and that of the shell is :
  • A. M₁ / (M₁ + M₂)
  • B. (M₂)/(M₁)
  • C. M₂ / (M₁ + M₂)
  • D. (M₁)/(M₂)

Solution

Related Formula
KE = (p²)/(2m)
Core Logic

By conservation of linear momentum (since no external horizontal force acts on the system):

0 = M₁ v₁ + M₂ v₂ | p₁| = | p₂| = p

Both the artillery and the shell acquire the exact same magnitude of momentum during firing.

Step 2: Kinetic Energy Ratio

The kinetic energy is related to momentum by KE = (p²)/(2m). Since p is identical for both bodies:

KE ∝ (1)/(m)

Therefore, the ratio of kinetic energy of the artillery (M₁) to the shell (M₂) is:

KE₁KE₂ = ((p²)/(2M₁))/((p²)/(2M₂)) = (M₂)/(M₁)
Chapter Mix

Class 11 Physics: Work, Energy And Power

More Work, Energy and Power Questions — jee_main_2026_23_january_morning

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Rankbit System
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%)