JEE Advanced 2023 · Paper 2 · numerical answer

JEE Advanced 2023 Physics question on Rotational Motion

A thin circular coin of mass 5 gm and radius 4/3 cm is initially in a horizontal xyxy-plane. The coin is tossed vertically up (+z+z direction) by applying an impulse of π2×10−2\sqrt{\dfrac{\pi}{2}} \times 10^{-2} N-s at a distance 2/3 cm from its center. The coin spins about its diameter and moves along the +z+z direction. By the time the coin reaches back to its initial position, it completes nn rotations. The value of nn is ____.

[Given: The acceleration due to gravity g=10g = 10 m s−2^{-2}]
Show answer

Answer: 30

[JEE Advanced 2023 · Paper 2]

Practise Rotational Motion the way it is asked

A timed set of real JEE Advanced Rotational Motion questions, graded as the exam grades. No sign-up.

Start a Rotational Motion test

More: all JEE Advanced Rotational Motion questions · JEE Advanced 2023 · Physics chapter list