A thin bar of length L has a mass per unit length `lambda`, that increases linerarly with distance from one end. If its total mass is M and its mass per unit length at the lighter end is `lambda_(0)`, then the distance of the centre of mass from the lighter end is

Options

A. `(L)/(2)-(lambda_(0)L^(2))/(4M)`
B. `(L)/(2)+(lambda_(0)L^(2))/(4M)`
C. `(L)/(3)+(lambda_(0)L^(2))/(4M)`
D. `(2L)/(3)-(lambda_(0)L^(2))/(4M)`

Answer

D

Explanation

Mass per unit length, `lamda=lamda_(0)+Bx=m`
`M=int_(0)^(L)mdx=int_(0)^(L)(lamda_(0)+Bx)dx`
`M=lamda_(0)L+(BL^(2))/(2)`
`B=(2M)/(L^(2))-(lamda_(0))/(L)" "......(i)`
`X_(cm)=(intdm(r))/(intdm)`
`=(int(lamdadx)x)/(M)`
`=(int(lamda_(0)+betax)dx.x)/(M)`
`=(int_(0)^(L)(lamda_(0)x+betax^(2))dx)/(M)`
`X_(cm)=(lamda_(0)L+(BL^(2))/(2))/(M)`
From equation (i)
`X_(cm)=(2L)/(3)-(lamda_(0)L^(2))/(6M)`