Home/
Competitive Exams/
JEE Main & Advanced/
Physics/
Laws of Motion and Friction/
Question 30282
JEE Physics · Hard

Laws of Motion and Friction: Advanced — Block between Walls MCQ

Solve this quality-checked JEE multiple-choice question, then review the correct answer and explanation.

Timer-based practiceDetailed answer reviewMobile-friendly flow
Advanced — Block between WallsHardQuestion 30282

Question

Two vertical walls are at distance d apart. A horizontal block (mass m) is wedged between them with normal forces N₁ and N₂ on each wall. μ between block and wall = 0.3. The maximum weight that can be supported without the block falling is:
  1. A
    0.3(N₁+N₂)
    Correct
  2. B
    0.3×2N₁
  3. C
    0.3mg
  4. D
    0.6N₁

Correct answer

0.3(N₁+N₂)

Explanation

Friction on each wall acts upward: f₁ = μN₁ and f₂ = μN₂. Vertical equilibrium: f₁+f₂ = mg → μ(N₁+N₂) = mg. Max weight = 0.3(N₁+N₂).

Review Laws of Motion and Friction notesStart timed chapter practice