JEE MAIN - Physics (2024 - 30th January Morning Shift - No. 4)
A particle of mass $$\mathrm{m}$$ is projected with a velocity '$$\mathrm{u}$$' making an angle of $$30^{\circ}$$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height $$\mathrm{h}$$ is :
$$\frac{\mathrm{mu}^3}{\sqrt{2} \mathrm{~g}}$$
zero
$$\frac{\sqrt{3}}{2} \frac{\mathrm{mu}^2}{\mathrm{~g}}$$
$$\frac{\sqrt{3}}{16} \frac{\mathrm{mu}^3}{\mathrm{~g}}$$
Explanation
$$\begin{aligned}
& \mathrm{L}=m u \cos \theta H \\
& =m u \cos \theta \times \frac{u^2 \sin ^2 \theta}{2 g} \\
& =\frac{m u^3}{2 g} \times \frac{\sqrt{3}}{2} \times\left(\frac{1}{2}\right)^2=\frac{\sqrt{3} m u^3}{16 g}
\end{aligned}$$
Comments (0)
