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Question

# A uniform square sheet has a side length of 2R. A circular sheet of maximum possible area is removed from one of the quadrants of the square sheet. The centre of mass of the remaining portion from the centre of the original sheet is at a distance of :

A
πR2[16π]
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B
π[16π]
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C
Rπ[16π]
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D
Rπ16π
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Solution

## The correct option is A πR√2[16−π]Let m1 be the mass of circular sheet of radius R and m2 be the mass of the remaining portion of square sheet after removing the circular sheet.Side of square =2RRadius of circle= R2Let x1 and x2 be the x components of centre of masses of m1 and m2 respectively.Let y1 and y2 be the y components of centre of masses of m1 and m2 respectively.x1 = R2y1 = R2x2=xy2=yLet the centre of square sheet be at origin.Let m be the mass per unit area.m1=mπR222m2=m(4R2−πR222)Centre of mass of m1 and m2 is at origin.0=πR222R2+(4R2−πR222)xx =-πR2(16−π)Similarly y=−πR2(16−π)Centre of mass of m2 is at (πR2(16−π),πR2(16−π))Distance of it from origin is πR√2×(16−π)

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