A rod of length L has non-uniform linear mass density given by $ρ (x) = a + b {(\frac{c}{L})}^{2}$ , where a and b are constants and $0 \leq x \leq L$ . The value of x for the center of mass of the rod is at:

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$\frac{4}{3} (\frac{a + b}{2 a + 3 b}) L$

$\frac{3}{2} (\frac{2 a + b}{3 a + b}) L$

$\frac{3}{4} (\frac{2 a + b}{3 a + b}) L$

$\frac{3}{2} (\frac{a + b}{2 a + b}) L$

answer is B.

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Detailed Solution

Given, Linear mass density, $ρ (x) = a + b {(\frac{x}{L})}^{2}$

$X_{CM} = \frac{\int xdm}{\int dm}$

$\int dm = \int_{0}^{L} ρ (x) dx = \int_{0}^{L} [a + b {(\frac{x}{L})}^{2}] dx$

$= aL + \frac{bL}{3}$

$\int_{0}^{L} xdm = \int_{0}^{L} (ax + \frac{{bx}^{3}}{L^{2}}) dx = (\frac{{aL}^{2}}{2} + \frac{{bL}^{2}}{4})$

$∴ X_{CM} = \frac{(\frac{{aL}^{2}}{2} + \frac{{bL}^{2}}{4})}{aL + \frac{bL}{3}}$

$\Rightarrow X_{CM} = \frac{3 L}{4} (\frac{2 a + b}{3 a + b})$

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