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Question

The centres of four spheres each of mass m and diameter 2a are at four corners of a square of side b. The moment of inertia of the system about the diagonal of square will be

A
25m[4a2+52b2]
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B
25m[5a2+4b2]
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C
25m[a2+b2]
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D
m[45a2+b2]
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Solution

The correct option is C 25m[4a2+52b2]
Two of the spheres rotate about their own axis thus there moment of inertia is given as 2×25ma2 and the moment of inertia of other two are found out using parallel axis theorem as 2×(25ma2+mb(2)2)
Thus we get total moment of inertia as m[85a2+b22]=25m[4a2+52b2]

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