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

Options

A. `5cm`
B. `9cm`
C. `4.5cm`
D. `5.5cm`

Answer

C

Explanation

`C_(1)` is the centre of mass of cut portion and `C_(2)` that of remaining portion. We have to find `x_(2)`
`x_(1)=14-10.5=3.5cm`

Mass will be proportional to area. So mass of the whole disc is
`M=kpi(14)^(2)`
Mass of cut portion `m_(1)=kpi(10.5)^(2)`
Mass of the remaining portion `m_(2)=M-m_(1)-kpi(14^(2)-10.5^(2))`
`=kpi(24.5)xx(3.6)`
Now `m_(1)x_(1)=m_(2)x_(2)`
`implies x_(2)=(m_(1)x_(1)/m_(2) = (kpi (10.5)^(2)xx3.5))/(kpi(24.5)xx3.5)=4.5cm`