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If the linear density of the rod of length L varies as λ=A+Bx , then its centre of mass is given by

a
Xcm=L(2A+BL)3(3A+2BL)
b
Xcm=L(3A+2BL)3(2A+BL)
c
Xcm=L(3A+2BL)3
d
Xcm=L(2A+3BL)3

detailed solution

Correct option is B

As the rod is along x -axis, for all points on it y  and z  will be zero,so,  YCM  = 0 and  ZCM   = 0i.e., the centre of mass will lie on the rod.  Now, consider an element of rod of length dx  at a distance x  from the origin, thendm  = λ   dx  = (A+Bx) dxSo,   XCM  = ∫0Lx dm∫0Ldm   =   ∫0Lx(A+Bx)dx∫0L(A+Bx)dxAL22   +  BL33AL​​  +   BL22    =   L(3A+2BL)3(2A+BL)

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