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 center of mass of complete plate. The position of center of mass of remaining portion will shift towards left from O by (in cm)

C₁ is the center of mass of cut portion and C₂ that of remaining portion. We have to find x₂.

$x_1=14-10.5=3.5\mathrm{cm}$

Mass will be proportional to area. So mass of the whole disc is $M=k\pi {\left(14\right)}^2$

Mass of cut portion $m_1=k\pi (10.5{)}^2$

Mass of the remaining portion $m_2=M-m_1=k\pi \left({14}^2-{10.5}^2\right)=k\pi \left(24.5\right) \times \left(3.5\right)$

$\text{ Now, }{m}_1{x}_1=m_2{x}_2$

$\Rightarrow x_2=\frac{m_1{x}_1}{m_2}=\frac{k\pi (10.5{)}^2\times 3.5}{k\pi \left(24.5\right)\times 3.5}=4.5\mathrm{cm}$