Linear mass density of a bar is ρ=ρoLoL+ρo where ρo is constant and Lo is length of bar. Its centre of mass from one of the end is (Assume the other dimensions of the bar to be unity)
A
3Lo5
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B
4Lo7
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C
5Lo9
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D
2Lo5
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Solution
The correct option is C5Lo9 Let us take a small mass dm at distance L from one of the end of width dl
We know, C.O.M of continuous body, rC.O.M=∫Ldm∫dm =Lo∫0L(ρoLoL+ρo)dLLo∫0(ρoLoL+ρo)dL[∵ρ=dmdL=ρoLoL+ρo] rC.O.M=[ρoLo×L33+ρoL22]Lo0[ρoLo×L22+ρoL]Lo0 =ρoL2o3+ρoL2o2ρoLo2+ρoLo rC.O.M=59Lo Centre of mass from one of the end is 5Lo9.