JEE Physics • Official PYQs

Work, Power & Energy

Solve official JEE Main Work, Power & Energy chapter-wise PYQs online. Work-energy theorem, spring potential energy, collisions & detailed solutions.

📝 15 Solved PYQs
📅 Exam Years: 2019–2024
📊 Weightage: 6%–8% (2 Questions per shift)
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🎯 High-Yield Syllabus Breakdown

These high-frequency concepts account for over 90% of all questions asked in Work, Power & Energy:

⚡ Essential Formulas & Speed Shortcuts

Formulas and time-saving shortcuts to solve Work & Energy questions in under 45 seconds:

Speed Trick #1:

Spring Potential Energy: U = ½ k x².

Speed Trick #2:

Stable Equilibrium: dU/dx = 0 and d²U/dx² > 0 (Potential energy minimum).

Speed Trick #3:

1D Elastic Head-on Collision with equal masses (m₁ = m₂): Velocities simply swap.

📖 Sample Solved Exam Questions

Preview authentic questions from this chapter with full step-by-step solutions:

Question 1 of 3 JEE Main 2024 (Online) 27th Jan Shift-1
A particle of mass 2 kg moves under a force F = (3x² - 2x + 1) N. The work done by the force in moving the particle from x = 0 to x = 2 m is:
(A) 6 J
(B) 4 J
(C) 8 J
(D) 10 J
💡 View Step-by-Step Solution & Answer

Correct Answer: 6 J

  1. Step 1: Work done by a variable force is W = ∫ F dx.
  2. Step 2: W = ∫₀² (3x² - 2x + 1) dx.
  3. Step 3: Integrate term by term: [x³ - x² + x]₀².
  4. Step 4: Substitute limits: (2³ - 2² + 2) - 0 = 8 - 4 + 2 = 6 J.

⚡ Speed Tip: Variable Work Integration: [x³ - x² + x] from 0 to 2 = 8 - 4 + 2 = 6 J.

Question 2 of 3 JEE Main 2023 (Online) 24th Jan Shift-1
If the kinetic energy of a body increases by 300%, then the percentage increase in its linear momentum is:
(A) 100%
(B) 200%
(C) 50%
(D) 150%
💡 View Step-by-Step Solution & Answer

Correct Answer: 100%

  1. Step 1: Relation between kinetic energy and momentum: KE = p² / (2m) ⇒ p = √(2m · KE).
  2. Step 2: A 300% increase means the new kinetic energy is KE' = KE + 3(KE) = 4 KE.
  3. Step 3: New momentum is p' = √(2m · 4KE) = 2 · √(2m · KE) = 2p.
  4. Step 4: Percentage increase in momentum = ((p' - p) / p) × 100% = ((2p - p) / p) × 100% = 100%.

⚡ Speed Tip: Square Root Scaling: Momentum scales as √KE. If KE becomes 4×, momentum becomes 2× (+100%).

Question 3 of 3 JEE Main 2024 (Online) 29th Jan Shift-2
An engine pumps water continuously through a hose with a speed v. If mass per unit length of water inside the hose is m, the rate at which kinetic energy is imparted to water is:
(A) ½ m v³
(B) m v²
(C) ½ m v²
(D) m v³
💡 View Step-by-Step Solution & Answer

Correct Answer: ½ m v³

  1. Step 1: Mass of water flowing out per unit time: dm/dt = m · v (mass per unit length × velocity).
  2. Step 2: Kinetic energy imparted per unit time is Power = d(KE)/dt = ½ (dm/dt) v².
  3. Step 3: Substitute dm/dt = m v: Power = ½ (m v) v².
  4. Step 4: Power = ½ m v³.

⚡ Speed Tip: Continuous Flow Power: P = ½ (dm/dt) v² = ½ (mv) v² = ½ m v³.

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❓ Frequently Asked Questions (FAQs)

What is the condition for complete vertical circle looping?

Minimum velocity at lowest point for a string must be v_min = √(5gr), and at highest point v_top = √(gr).

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