QJEEM22Phcbd6: JEE Main 2022 Physics - Class 11 JEEM22Phcbd6 Solution: The distance of centre of mass from end A of a one dimensional…

$\rho = {\rho _0}\left( {1 - {{{x^2}} \over {{L^2}}}} \right)$ kg/m

${x_{cm}} = {{A\int\limits_0^L {{\rho _0}\left( {1 - {{{x^2}} \over {{L^2}}}} \right)x\,dx} } \over {A\int\limits_0^L {{\rho _0}\left( {1 - {{{x^2}} \over {{L^2}}}} \right)\,dx} }}$

${x_{cm}} = {{{{{L^2}} \over 2} - {{{L^2}} \over 4}} \over {L - {L \over 3}}} = {{{{{L^2}} \over 4}} \over {{{2L} \over 3}}} = {{3L} \over 8}$

$\Rightarrow \alpha = 8$