A circular plate of uniform thickness has a diameter fo 56cm. A circular portion of diameter 42 cm is removed from one edge of the plate as shown in figure. Find the position of the centre of mass of the remaining portion.

Options

A. 9 cm
B. 7 cm
C. 5 cm
D. 4 cm

Answer

A

Explanation

Initial area of the plane `=pi(28)^(2)`
Area of the removed plate `(A_(1))=pi(21)^(2)`
Area of the remaining plate `(A_(2))=pi(28^(2)-21^(2))`
`=pi(49xx7)`
Let `d_(1) and d_(2)` be the distances of the centres of mass of the removed plate and remaining plate from the centre of the original plate.
`d_(1)=27-21=7cm`
Since the plate is of uniform thickness, weight of the plate is proportional to the area .
`[W-mg = v xx rhog =Ah rhog therefore W prop A]`
`therefore A_(1)d_(1)=A_(2)d_(2)`
`d_(2)=((A_(1))/(A_(2)))d_(1)=(pi(21)^(2))/(pi(49xx7))xx7=9cm`
`therefore` It is at a distance of 9 cm from the centre of the original plate.