NTA Abhyas JEE Main2020PhysicsLaws of MotionPractice
When a block moves down a smooth inclined plane of inclination θ , its velocity on reaching the bottom is v . When it slides down a rough inclined plane of same inclination, its velocity on reaching the bottom is v / n , where n is a number greater than 1 . The coefficient of friction between the block and the rough surface is
Options
- Aμ = 1 - 1 n 2 tan θ
- Bμ = 1 - 1 n 2 cot ⁡ θ
- Cμ = 1 - 1 n 2 1 2 tan θ
- Dμ = 1 - 1 n 2 1 2 cot θ
Correct answer
A. μ = 1 - 1 n 2 tan θ
Step-by-step solution
For smooth incline, v 2 = 2 g sin θ × s       … … ( 1 ) For rough incline, v n 2 = 2 g sin θ - μ cos θ × s       … … ( 2 ) On solving (1) and (2), we obtain, μ = 1 - 1 n 2 tan θ