QJEEM24Ph0890: JEE Main 2024 Physics - Class 11 JEEM24Ph0890 Solution: A body of mass moving with a uniform speed in plane along the…

$y-x-4=0$

$d_1$ is perpendicular distance of given line from origin.

$\mathrm{d}_1=\left|\frac{-4}{\sqrt{1^2+1^2}}\right| \Rightarrow 2 \sqrt{2} \mathrm{~m}$

So

$\begin{aligned}|\overrightarrow{\mathrm{L}}|=\mathrm{mvd}_1 & =5 \times 3 \sqrt{2} \times 2 \sqrt{2} \mathrm{~kg} \mathrm{~m}^2 / \mathrm{s} \\& =60 \mathrm{~kg} \mathrm{~m}^2 / \mathrm{s}\end{aligned}$