QJEEM22Ph4e45: JEE Main 2022 Physics - Class 11 JEEM22Ph4e45 Solution: A pulley of radius is rotated about its axis by a force applied …

$FR = I\alpha$

$\alpha = {{(12t - 3{t^2}) \times 1.5} \over {4.5}} = 4t - {t^2}$

$w = \int {\alpha \,dt = 2{t^2} - {{{t^3}} \over 3}}$

$w = 0$

$\Rightarrow {t^2}\left[ {2 - {t \over 3}} \right] = 0$

$t = 6$ sec

$\left. {\theta = \int\limits_0^6 {\left[ {2{t^2} - {{{t^3}} \over 3}} \right]dt = \left[ {{{2{t^3}} \over 3} - {{{t^4}} \over {12}}} \right]} } \right|_0^6$

$= \left[ {{2 \over 3} \times {6^3} - {{{6^4}} \over {12}}} \right] = 36$

$n = {{36} \over {2\pi }}$

$= {{18} \over \pi }$