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 an axis passing through one corner of the square and perpendicular to its plane will be
A
45m[2a2+5b2]
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B
54m[a2+2b2]
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C
25m[3a2+4b2]
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D
34m[2a2+4b2]
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Solution
The correct option is A45m[2a2+5b2] One of the sphere rotate about its own axis, thus its moment of inertia is given as 25ma2 and the moment of inertia of other three are found out using parallel axis theorem. Two of them are at a distance of b from the rotational axis the sum of their moment of inertia is 2×(25ma2+mb2) and the third one is at a diagonal distance of √2b, thus its moment of inertia is 25ma2+2mb2 Thus we get total moment of inertia as 45m[2a2+5b2]