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Question

Four solid spheres each of diameter 5cm and mass 0.5kg are placed with their centers at the corners of a square of side 4cm. The moment of inertia of the system about the diagonal of the square is N×104kgm2, then N is:

A
6
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B
7
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C
8
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D
9
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Solution

The correct option is C 9
MI of the system I=I1+I2+I3+I4 where Ii is the MI of the ith sphere about the diagonal XY where i goes from 1 to 4.
I1=I3 by symmetry
I2=I4 by symmetry
I1=25mR2+m(2a2)2 using Parallel Axis Theorem since 25mR2 is the MI of the sphere about the Center of Mass of the sphere. 2a2 is the distance between the center of the square and sphere.
I2=25mR2
I=2I1+2I2
=2(25mR2+m(2a2)2)+2(25mR2) =85mR2+ma2
=850.5(52)2104+0.516102=9104kgm2
comparing I with N×104kgm2, we get N=9

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