JEE MAIN - Physics (2024 - 29th January Evening Shift - No. 24)
A body of mass $$5 \mathrm{~kg}$$ moving with a uniform speed $$3 \sqrt{2} \mathrm{~ms}^{-1}$$ in $$X-Y$$ plane along the line $$y=x+4$$. The angular momentum of the particle about the origin will be _________ $$\mathrm{kg} \mathrm{~m}^2 \mathrm{~s}^{-1}$$.
Answer
60
Explanation
$$y-x-4=0$$
$$d_1$$ 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
$$\begin{aligned} |\overrightarrow{\mathrm{L}}|=\mathrm{mvd}_1 & =5 \times 3 \sqrt{2} \times 2 \sqrt{2} \mathrm{~kg} \mathrm{~m}^2 / \mathrm{s} \\ & =60 \mathrm{~kg} \mathrm{~m}^2 / \mathrm{s} \end{aligned}$$
Comments (0)
