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+2 votes
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in Physics by (35.1k points)

The disc of mass M with uniform surface mass density σ is shown in the figure. The centre of mass of the quarter disc (the shaded area) is at the position \(\frac x3\frac a\pi,\frac x3\frac a\pi\) where x is...........(Round off to the Nearest Integer)

[a is an area as shown in the figure]

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1 Answer

+2 votes
by (34.7k points)

Answer is 4

C.O.M of quarter disc is at \(\frac {4a}{3\pi},\frac{4a}{3\pi}\)= 4

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