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Question:
The disc of mass $M$ with uniform surface mass density $\sigma$ is shown in the figure. The centre of mass of the quarter disc (the shaded area) is at the position $\frac{\mathrm{x}}{3} \frac{\mathrm{a}}{\pi}, \frac{\mathrm{x}}{3} \frac{\mathrm{a}}{\pi}$ where $\mathrm{x}$ is___________
(Round off to the Nearest Integer)
[a is an area as shown in the figure]
Solution:
C.O.M of quarter disc is at $\frac{4 a}{3 \pi}, \frac{4 a}{3 \pi}$
$=4$
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