NTA Abhyas JEE Main2020PhysicsCenter of Mass, Momentum and CollisionPractice
A thin bar of length L has a mass per unit length λ , that increases linearly with distance from one end. If its total mass is M and its mass per unit length at the lighter end is λ 0 , then the distance of the centre of mass from the lighter end is
Options
- AL 3 + λ 0 L 2 8 M
- BL 3 + λ 0 L 2 4 M
- CL 2 - λ 0 L 2 4 M
- D2 L 3 - λ 0 L 2 6 M
Correct answer
D. 2 L 3 - λ 0 L 2 6 M
Step-by-step solution
λ ∝ x λ = Kx + λ 0 dm = λ dx ∴ X cm = ∫ 0 L dm · x ∫ 0 L dm = ∫ 0 L Kx + λ 0 dx.x ∫ 0 L Kx + λ 0 dx = K · x 3 3 0 L + λ 0 x 2 2 0 L K x 2 2 0 L + λ 0 L = KL 3 3 + λ 0 L 2 2 KL 2 2 + λ 0 L = KL 3 + λ 0 2 L 2 KL 2 2 + λ 0 L Now, M = ∫ 0 L dm = KL 2 2 + λ 0 L 2 M = KL 2 + 2 λ 0 L K = 2 M - λ 0 L L 2 Putting the value of K. X cm = 2 M - λ 0 L 3 L + λ 0 2 L 2 M = 2 L 2 M - λ 0 L 3 LM + λ 0 L 2 2 M = 2 ML 3 M - 2 λ 0 L 2 3 M + λ 0 L 2 2 M = 2 L 3 - λ 0 L 2 6 M Hence, 2 L 3 - λ 0 L 2 6 M