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 as $\lambda = 2 + x$.For a small element of length $dx$ at a distance $x$,the mass is $dm = \la

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$\frac{12}{7} \ m$

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(B) The linear density of the rod varies with distance as $\lambda = 2 + x$.For a small element of length $dx$ at a distance $x$,the mass is $dm = \lambda \ dx = (2 + x) \ dx$.The position of the centre of mass $x_{cm}$ is given by:$x_{cm} = \frac{\int x \ dm}{\int dm} = \frac{\int_{0}^{3} x(2 + x) \ dx}{\int_{0}^{3} (2 + x) \ dx}$Calculating the numerator:$\int_{0}^{3} (2x + x^2) \ dx = [x^2 + \frac{x^3}{3}]_{0}^{3} = (3^2 + \frac{3^3}{3}) - 0 = 9 + 9 = 18$.Calculating the denominator (total mass $M$):$\int_{0}^{3} (2 + x) \ dx = [2x + \frac{x^2}{2}]_{0}^{3} = (2(3) + \frac{3^2}{2}) - 0 = 6 + 4.5 = 10.5 = \frac{21}{2}$.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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