QJEEM23Pha613: JEE Main 2023 Physics - Class 11 JEEM23Pha613 Solution: A car is moving on a circular path of radius 600 m such that…

${{dv} \over {dt}} = {{{v^2}} \over R} \Rightarrow {{{v^2}} \over R} = v{{dv} \over {ds}}$

$\Rightarrow {{dv} \over v} = {{ds} \over R} \Rightarrow \left. {\ln v} \right|_{15}^v = {s \over R}$

$\Rightarrow v = 15{e^{\Delta /R}} = {{ds} \over {dt}} \Rightarrow dt = {1 \over {15}}{e^{ - \Delta /R}}ds$

$\Delta t = {R \over {15}}[1 - {e^{ - \Delta /R}}]$

$= 40[1 - {e^{ - \pi /2}}]$ seconds

$\Rightarrow t = 40$