JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2024
- Hard
- Numerical
A body of mass \(5 \mathrm{~kg}\) moving with a uniform speed \(3 \sqrt{2} \mathrm{~ms}^{-1}\) in \(\mathrm{X}-\mathrm{Y}\) plane along the line \(\mathrm{y}=\mathrm{x}+4\).The angular momentum of the particle about the origin will be _______ \(\mathrm{kg}\ \mathrm{m} \mathrm{s}^{-1}\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(60\)
Step-by-step Solution
Detailed 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 \(|\overrightarrow{\mathrm{L}}|=\operatorname{mvd}_1 =5 \times 3 \sqrt{2} \times 2 \sqrt{2} \mathrm{~kg} \mathrm{~m}^2 / \mathrm{s} \)
\( =60 \mathrm{~kg} \mathrm{~m}^2 / \mathrm{s}\)
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