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Physics · Rotational Dynamics

JEE Main 2024 — 31 January, Shift 2 — Question 52

A body of mass ' mm ' is projected with a speed ' uu ' making an angle of 45∘45^{\circ} with the ground. The angular momentum of the body about the point of projection, at the highest point is expressed as 2mu3Xg\frac{\sqrt{2} \mathrm{mu}^{3}}{\mathrm{Xg}}. The value of ' X ' is _______\_\_\_\_\_\_\_ .

Answer: 8

Numerical answer — enter this value.

Step-by-step solution

L=mucos⁡θu2sin⁡2θ2g\mathrm{L}=\mathrm{mu} \cos \theta \frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 g}

=mu3142g⇒x=8=m u^{3} \frac{1}{4 \sqrt{2} g} \Rightarrow x=8

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
31 January, Shift 2
Subject
Physics
Chapter
Rotational Dynamics
Topic
Angular Momentum and its Conservation

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