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(B) The linear mass density is given by $\lambda = \alpha + \beta x$.Consider a small element of length $dx$ at a distance $x$ from the origin. The ma

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$\frac{(3 \alpha + 2 \beta L) L}{3(2 \alpha + \beta L)}$

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(B) The linear mass density is given by $\lambda = \alpha + \beta x$.Consider a small element of length $dx$ at a distance $x$ from the origin. The mass of this element is $dm = \lambda dx = (\alpha + \beta x) dx$.The centre of mass $x_{cm}$ is given by the formula $x_{cm} = \frac{\int x dm}{\int dm}$.Substituting the values,we get $x_{cm} = \frac{\int_0^L x(\alpha + \beta x) dx}{\int_0^L (\alpha + \beta x) dx}$.Evaluating the numerator: $\int_0^L (\alpha x + \beta x^2) dx = [\frac{\alpha x^2}{2} + \frac{\beta x^3}{3}]_0^L = \frac{\alpha L^2}{2} + \frac{\beta L^3}{3} = \frac{L^2(3\alpha + 2\beta L)}{6}$.Evaluating the denominator: $\int_0^L (\alpha + \beta x) dx = [\alpha x + \frac{\beta x^2}{2}]_0^L = \alpha L + \frac{\beta L^2}{2} = \frac{L(2\alpha + \beta L)}{2}$.Dividing the two results: $x_{cm} = \frac{L^2(3\alpha + 2\beta L)}{6} 	imes \frac{2}{L(2\alpha + \beta L)} = \frac{L(3\alpha + 2\beta L)}{3(2\alpha + \beta L)}$.

The linear mass density $(\lambda)$ of a rod of length $L$ kept along the $x$-axis varies as $\lambda = \alpha + \beta x$,where $\alpha$ and $\beta$ are positive constants. The centre of mass of the rod is at ..........

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