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Question

Particles of masses 1g,2g,3g...,100g are kept at the marks 1cm,2cm.3cm,...,100cm respectively on a metre scale. The moment of inertia of the system of particles about a perpendicular bisector of the metre scale is about x×106gmcm2. Find x.

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Solution

I=50(0)2+(49+51)(1)2+(48+52)(2)2+........+(1+99)(49)2
=100[(1)2+(2)2+.....+(49)2] where we have used 12+22+32+...............+n2=n(n+1)(2n+1)6
4.3×106gmcm2=0.402kgm2

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