If the linear density of a rod of length 3 m | vedclass

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(B) The linear density of the rod varies with distance $x$ as $\lambda = \frac{dm}{dx} = 2 + x$.The position of the centre of mass $x_{cm}$ is given b

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$12/7 \ m$

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(B) The linear density of the rod varies with distance $x$ as $\lambda = \frac{dm}{dx} = 2 + x$.The position of the centre of mass $x_{cm}$ is given by the formula:$x_{cm} = \frac{\int x \, dm}{\int dm} = \frac{\int_{0}^{3} x \lambda \, dx}{\int_{0}^{3} \lambda \, dx}$Substituting $\lambda = 2 + x$:$x_{cm} = \frac{\int_{0}^{3} x(2 + x) \, dx}{\int_{0}^{3} (2 + x) \, dx} = \frac{\int_{0}^{3} (2x + x^2) \, dx}{\int_{0}^{3} (2 + x) \, dx}$Evaluating the integrals:Numerator: $\int_{0}^{3} (2x + x^2) \, dx = [x^2 + \frac{x^3}{3}]_{0}^{3} = (3^2 + \frac{3^3}{3}) - 0 = 9 + 9 = 18$Denominator: $\int_{0}^{3} (2 + x) \, dx = [2x + \frac{x^2}{2}]_{0}^{3} = (2(3) + \frac{3^2}{2}) - 0 = 6 + 4.5 = 10.5 = \frac{21}{2}$Therefore,$x_{cm} = \frac{18}{21/2} = \frac{18 	imes 2}{21} = \frac{36}{21} = \frac{12}{7} \ m$.

If the linear density of a rod of length $3 \ m$ varies as $\lambda = 2 + x$,then the position of the centre of mass of the rod is:

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