NTA Abhyas JEE Main2020PhysicsCenter of Mass, Momentum and CollisionPractice
If linear density of a rod of length 3 m varies as λ =2+ x , then the position of the centre of gravity of the rod is
Options
- A7 3 m
- B12 7 m
- C10 7 m
- D9 7 m
Correct answer
B. 12 7 m
Step-by-step solution
Let rod is placed along x -axis. Mass of element P Q of length d x situated at x = x is d m = λ d x = 2 + x d x The CM of the element has coordinates ( x , 0, 0). Therefore, x -coordinates of CM of the rod will be x C M = ∫ 0 3 x d m ∫ 0 3 d m = ∫ 0 3 x 2 + x d x ∫ 0 3 2 + x d x = ∫ 0 3 2 x + x 2 d x ∫ 0 3 2 + x d x = 2 x 2 2 + x 3 3 0 3 2 x + x 2 2 0 3 = 3 2 + 3 3 3 2 × 3 + 3 2 2 = 9 + 9 6 + 9 / 2 = 18 × 2 21 = 12 7 m