A particle of mass $15 \mathrm{~kg}$ is moving with a uniform speed $8 \mathrm{~ms}^{-1}$ in $x y$-plane along the line $3 y=4 x+10$, then the magnitude of its angular momentum about the origin in $\mathrm{kg} \cdot \mathrm{m}^2 \mathrm{~s}^{-1}$ is $\ldots\left(\sin 53^{\circ}=\frac{4}{5}\right)$

• Correct answer
  A

  240
• B

  80
• C

  120
• D

  280

Solution

Angular momentum $=m v r_1$ $$ =15 \times 8 \times\left(\frac{10}{\sqrt{3^2+4^2}}\right)=15 \times 8 \times 2=240 \mathrm{~kg}-\mathrm{m}^2 \mathrm{~s}^{-1} $$

Hence, the correct answer is option: 'A'.