Center of Mass, Momentum & Collision

JEE Main level, numerical answer, medium

52 of 130

Question 52 of 130

The disc of mass MM with uniform surface mass density σ\sigma is shown in the figure. The center of mass of the quarter disc (the shaded area) is at the position (xa3π,xa3π)\left(\dfrac{xa}{3\pi },\dfrac{xa}{3\pi }\right) where xx is _______ . (Round off to the Nearest Integer) [aa is an area as shown in the figure]

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Answer: 4

C.O.M of quarter disc is at 4a3π,4a3π\dfrac{4a}{3\pi },\dfrac{4a}{3\pi }
=4=4

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