A circular plate of uniform thickness has a diameter of `28 cm`. A circular portion of diameter `21 cm` is removed from the plate as shown. `O`is the centre of mass of complete plate. The position of centre of mass of remaining portion will shift towards left from `'O'` by

Answer

4.5

Explanation

`C_1` is centre of mass of cut portion and `C_2` that of remaining portion. We have to find `x_2`.
`x_1 = 14 - 10.5 = 3.5 cm`
As mass is uniformly distributed hence the mass should be proportional to area. So mass of whole disc: `M = k pi (14)^2` .
k = proportionality constant. Mass of removed portion : `m_1 = k`

Mass of remaining portion : `m_2 = (M - m_1)`
`= k pi(14^2 - 10.5^2) " " = kpi(24.5) xx (3.5)`
Now : `m_1x_1 = m_2x_2 implies x_2 = (m_1x_1)/(m_2)= (k pi(10.5)^5 xx 3.5)/(k pi (24.5) xx 3.5) = 4.5 cm`