JEE Main20209 Jan 2020Evening ShiftPhysicsCenter of Mass, Momentum and CollisionActual
A rod of length l has non-uniform linear mass density given by ρ x = a + b x l 2 , where a and b are constants and 0 ≤ x ≤ l The value of x for the centre of mass of the rod is at:
Options
- A3 2 a + b 2 a + b L
- B3 4 2 a + b 3 a + b L
- C4 3 a + b 2 a + 3 b L
- D3 2 2 a + b 3 a + b L
Correct answer
B. 3 4 2 a + b 3 a + b L
Step-by-step solution
X c m = 1 M ∫ 0 ι x . d M d M = ρ . d x = a + b x ı 2 . d x x c m = ∫ x d M ∫ d m = ∫ x ρ d x ∫ ρ d x = ∫ 0 ι x a + b x 2 ι 2 d x ∫ 0 ι a + b x 2 ι 2 d x = a x 2 2 0 ι + b ι 2 x 4 4 0 ι a x 0 ι + b ι 2 x 3 3 0 ι = a ι 2 2 + b ι 2 4 a ι + b ι 3 = 2 a + b 3 a + b ι 4 × 3 = 3 ι 4 2 a + b 3 a + b