A circular plate of uniform thickness has a diameter of 56cm. A circular portion of diameter 42cm is removed from one edge as shown in the figure. The centre of mass of the remaining portion of from the centre of plate will be ____cm

If we consider the mass removed as negative mass

${\mathrm{x}}_{\mathrm{com}}=\frac{{\mathrm{m}}_1{\mathrm{x}}_1-{\mathrm{m}}_2{\mathrm{x}}_2}{{ \mathrm{m}}_1-{\mathrm{m}}_2}$

m₁ is the mass of original disc

m₂ is the mass of removed disc

x₁ is the center of mass of the original disc

x₂ is the center of mass of the removed disc

$\sigma$ is the mass per unit area of the disc

${\mathrm{x}}_{\mathrm{com}}=\frac{\mathrm{\sigma }\times \mathrm{\pi }\times {\left(28\times {10}^{-2}\right)}^2\times 0-\mathrm{\sigma }\times \mathrm{\pi }\times {\left(21\times {10}^{-2}\right)}^2\times \left(7\times {10}^{-2}\right)}{\mathrm{\sigma }\times \mathrm{\pi }\times {\left(28\times {10}^{-2}\right)}^2-\mathrm{\sigma }\times \mathrm{\pi }\times {\left(21\times {10}^{-2}\right)}^2}$

$\mathrm{on} \mathrm{solving}, {\mathrm{x}}_{\mathrm{com}}==-0.09 \mathrm{m}$