The density of a linear rod of length 'L' varies as p = A+Bx where x is the distance from the left end. Then the position of the centre of mass from the left end is

$\frac{3 A L + 2 B L^{2}}{3 (2 A + B L)}$

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$\frac{A L + 3 B L^{2}}{(2 A + B L)}$

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$\frac{3 A L + 2 B L^{2}}{(2 A + B L)}$

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$\frac{3 A L + 2 B L^{2}}{3 (2 A + B L)}$

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Solution

The correct option is D
$\frac{3 A L + 2 B L^{2}}{3 (2 A + B L)}$

$x_{c m}$ = $\frac{\int_{0}^{L} d m x}{\int_{0}^{L} d m}$

where dm = $ρ d x$

dm = (A+Bx)dx

$\Rightarrow x_{c m} = \frac{\int_{0}^{L} [(A + B x) d x] x}{\int_{0}^{L} (A + B x) d x} = \frac{\int_{0}^{L} A x d x + \int_{0}^{L} B x^{2} d x}{\int_{0}^{L} A d x + \int_{0}^{L} B x d x}$

$\Rightarrow x_{c m} = \frac{3 A L + 2 B L^{2}}{3 (2 A + B L)}$

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