JEE Main20208 Jan 2020Evening ShiftPhysicsCenter of Mass, Momentum and CollisionActual
As shown in figure. When a spherical cavity (centred at O ) of radius 1 is cut out of a uniform sphere of radius R (centred at C ), the centre of mass of remaining (shaded part of sphere is at G , i.e., on the surface of the cavity. R can be determined by the equation:
Options
- AR 2 + R + 1 2 - R = 1
- BR 2 - R - 1 2 - R = 1
- CR 2 - R + 1 2 - R = 1
- DR 2 + R - 1 2 - R = 1
Correct answer
A. R 2 + R + 1 2 - R = 1
Step-by-step solution
M 1 = 4 3 π R 3 ρ M 2 = 4 3 π 1 3 - ρ X c o m = M 1 X 1 + M 2 X 2 M 1 + M 2 ⇒ 4 3 π R 3 ρ 0 + 4 3 π 1 3 - ρ R - 1 4 3 π R 3 ρ + 4 3 π 1 3 - ρ - 2 - R ⇒ R - 1 ( R 3 - 1 ) = 2 - R R ≠ 1 R - 1 R - 1 R 2 + R + 1 = 2 - 1 R 2 + R + 1 2 - R = 1 Alternative: M r e m a i n i n g 2 - R = M c a v i t y 1 - R ⇒ R 3 - 1 3 2 - R = 1 3 R - 1 ⇒ R 2 + R + 1 2 - R = 1