JEE MAIN - Physics (2022 - 28th July Evening Shift - No. 19)
The distance of centre of mass from end A of a one dimensional rod (AB) having mass density $$\rho=\rho_{0}\left(1-\frac{x^{2}}{L^{2}}\right) \mathrm{kg} / \mathrm{m}$$ and length L (in meter) is $$\frac{3 L}{\alpha} \mathrm{m}$$. The value of $$\alpha$$ is ___________. (where x is the distance from end A)
Answer
8
Explanation
$$\rho = {\rho _0}\left( {1 - {{{x^2}} \over {{L^2}}}} \right)$$ kg/m
$${x_{cm}} = {{A\int\limits_0^L {{\rho _0}\left( {1 - {{{x^2}} \over {{L^2}}}} \right)x\,dx} } \over {A\int\limits_0^L {{\rho _0}\left( {1 - {{{x^2}} \over {{L^2}}}} \right)\,dx} }}$$
$${x_{cm}} = {{{{{L^2}} \over 2} - {{{L^2}} \over 4}} \over {L - {L \over 3}}} = {{{{{L^2}} \over 4}} \over {{{2L} \over 3}}} = {{3L} \over 8}$$
$$ \Rightarrow \alpha = 8$$
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