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Question

The density of a linear rod of length L varies as ρ=A+Bx where x is the distance from the left end. Its centre of mass is at a distance (from the left end) of

A
3AL2BL23(2ABL)
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B
3AL+2BL23(2A+BL)
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C
3AL+2BL22A+BL
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D
3AL+2BL22ABL
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Solution

The correct option is B 3AL+2BL23(2A+BL)
In this case, the rod is linear. So, its centre of mass will lie along xaxis. For the non-uniform body, the xcoordinate of COM will be,
xcom=xdmdm
Let us take a very small element of length dx from the left end of the rod as shown in the figure.


xcom=L0xρdxL0ρdx
[ρ=dmdxdm=ρdx]
xcom=L0x(A+Bx)dxL0(A+Bx)dx
[ρ=A+Bx given]
xcom=L0(Ax+Bx2)dxL0(A+Bx)dx
xcom=[Ax22+Bx33]L0[Ax+Bx22]L0
xcom=[AL22+BL33(0+0)][AL+BL22(0+0)]
xcom=3AL2+2BL362AL+BL22
xcom=3AL+2BL23(2A+BL)
Hence centre of mass of rod is at a distance of 3AL+2BL23(2A+BL) from the left end.

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