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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 _______ \(\mathrm{kg}\  \mathrm{m} \mathrm{s}^{-1}\).

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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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