99 Percentile Qs Bank for JEE MainPhysicsCenter of Mass, Momentum and Collision
In the figure shown a semicircular area is removed from a uniform square plate of side l and mass (before removing) m . The x - coordinate of centre of mass of remaining portion is (The origin is at the centre of square)
Options
- A- π ( π - 2 ) ℓ 2 ( 8 - π )
- Bπ ( π - 2 ) ℓ 2 ( 8 - π )
- C- π π - 2 ℓ 8 - π
- DNone of these
Correct answer
D. None of these
Step-by-step solution
The centre of mass about x - axis is given as, x cm = m 1 x 1 + m 2 x 2   m 1 + m 2 Here, m 1 = mass of the square plate = m x 1 = centre of mass of the square plate = 0 (As the origin is at centre of the square) m 2 = mass of the removed part = mass density of square × area of the semicircular = - m l 2 π l 2 2 2 = - π 8 m x 2 = centre of mass of the removed part = l 2 - 4 3 π l 2 = l 2 1 - 4 3 π Now, the centre of mass of the system will be, x cm = - π m 8 × l 2 1 - 4 3