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(B) The center of mass $x_{cm}$ of a rod with variable linear mass density $\mu(x)$ is given by the formula:$x_{cm} = \frac{\int_{0}^{L} x \mu(x) dx}{

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$2a = b$

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(B) The center of mass $x_{cm}$ of a rod with variable linear mass density $\mu(x)$ is given by the formula:$x_{cm} = \frac{\int_{0}^{L} x \mu(x) dx}{\int_{0}^{L} \mu(x) dx}$Substituting $\mu(x) = a + \frac{bx}{L}$:$x_{cm} = \frac{\int_{0}^{L} x(a + \frac{bx}{L}) dx}{\int_{0}^{L} (a + \frac{bx}{L}) dx} = \frac{\int_{0}^{L} (ax + \frac{bx^2}{L}) dx}{\int_{0}^{L} (a + \frac{bx}{L}) dx}$Evaluating the integrals:Numerator: $[\frac{ax^2}{2} + \frac{bx^3}{3L}]_{0}^{L} = \frac{aL^2}{2} + \frac{bL^2}{3} = L^2(\frac{a}{2} + \frac{b}{3})$Denominator: $[ax + \frac{bx^2}{2L}]_{0}^{L} = aL + \frac{bL}{2} = L(a + \frac{b}{2})$Thus,$x_{cm} = \frac{L^2(\frac{a}{2} + \frac{b}{3})}{L(a + \frac{b}{2})} = L \frac{(\frac{3a + 2b}{6})}{(\frac{2a + b}{2})} = L \frac{3a + 2b}{3(2a + b)}$Given $x_{cm} = \frac{7}{12}L$,we have:$\frac{3a + 2b}{3(2a + b)} = \frac{7}{12}$$\frac{3a + 2b}{2a + b} = \frac{7}{4}$$4(3a + 2b) = 7(2a + b)$$12a + 8b = 14a + 7b$$b = 2a$.

$A$ uniform thin rod $AB$ of length $L$ has linear mass density $\mu(x) = a + \frac{bx}{L}$,where $x$ is measured from $A$. If the $CM$ of the rod lies at a distance of $\frac{7}{12}L$ from $A$,then $a$ and $b$ are related as

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