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JEE MAIN 2022
28-07-2022 S2
Question
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 $\frac{3 \mathrm{~L}}{\alpha} \mathrm{~m}$. The value of $\alpha$ is $\_\_\_\_$ (where $x$ is the distance form end $A$ )
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Question Tags
JEE Main
Physics
Medium
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