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JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion

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 _______.

  1. A \(8\)
  2. B \(9\)
  3. C \(10\)
  4. D \(11\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(8\)

Step-by-step Solution

Detailed explanation

\(\mathrm{L}=\mathrm{mu} \cos \theta \frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}\) \(=m \mathrm{u}^3 \frac{1}{4 \sqrt{2} \mathrm{~g}} \Rightarrow \mathrm{x}=8\)
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