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JEE Main 202524 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Morning Shift
Subject
Physics
Chapter
Rotational Motion

JEE Main 2025 · 24 January · Morning Shift · Physics · Rotational Motion

An object of mass ' m ' is projected from origin in a vertical x y...

An object of mass ' m ' is projected from origin in a vertical xyx y plane at an angle 45∘45^{\circ} with the x axis with an initial velocity v0\mathrm{v}_0. The magnitude and direction of the angular momentum of the object with respect to origin, when it reaches at the maximum height, will be [ g is acceleration due to gravity]An object of mass ' m ' is projected from origin in a vertical x y plane at an angle 45^ with the x axis with an

Options

  1. Amvo322g\frac{m v_o^3}{2 \sqrt{2} g} along negative zz-axis
  2. Bmvo342g\frac{m v_o^3}{4 \sqrt{2} g} along positive zz-axis
  3. Cmvo342g\frac{m v_o^3}{4 \sqrt{2} g} along negative zz-axis
  4. Dmvo322g\frac{m v_o^3}{2 \sqrt{2} g} along positive zz-axis

Correct answer

C. mvo342g\frac{m v_o^3}{4 \sqrt{2} g} along negative zz-axis

Step-by-step solution

An object of mass ' m ' is projected from origin in a vertical x y plane at an angle 45^ with the x axis with an
H=(v02)22 g=v024 g L=mvh L=mv02v024 g\begin{aligned} & \mathrm{H}=\frac{\left(\frac{\mathrm{v}_0}{\sqrt{2}}\right)^2}{2 \mathrm{~g}}=\frac{\mathrm{v}_0^2}{4 \mathrm{~g}} \\ & \mathrm{~L}=\mathrm{mvh} \\ & \mathrm{~L}=\mathrm{m} \frac{\mathrm{v}_0}{\sqrt{2}} \frac{\mathrm{v}_0^2}{4 \mathrm{~g}} \end{aligned}
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