A body of mass ' m ' is projected with a speed ' u ' making an angle of 45^{circ} wi

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6.System of Particles and Rotational Motion

hard

A body of mass ' $m$ ' is projected with a speed ' $u$ ' making an angle of $45^{\circ}$ with the ground. The angular momentum of the body about the point of projection, at the highest point is expressed as $\frac{\sqrt{2} \mathrm{mu}^3}{\mathrm{Xg}}$. The value of ' $\mathrm{X}$ ' is

A

$8$

B

$9$

C

$10$

D

$11$

(JEE MAIN-2024)

Solution

$\mathrm{L}=\mathrm{mu} \cos \theta \frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}$

$=m \mathrm{u}^3 \frac{1}{4 \sqrt{2} \mathrm{~g}} \Rightarrow \mathrm{x}=8$

Standard 11

Physics

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