JEE Main 2015PhysicsRotational MotionHardMCQ

JEE Main 2015 — Rotational Motion Question with Solution

JEE Main 2015 (11 Apr Online)

Question

A particle of mass 2 kg is on a smooth horizontal table and moves in a circular path of radius 0.6 m. The height of the table from the ground is 0.8 m. If the angular speed of the particle is 12 rad s-1 , the magnitude of its angular momentum about a point on the ground right under the center of the circle is:

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Show full solutionCorrect option: A
Correct answer
A14.4 kg m2 s-1

Step-by-step explanation



p→=mv j^=2×v j^=2×0.6×12=7.2×2=14.4j^     v=rω

r→=(0.8 k^+0.6 i^)⇒L→=r→×p→

=0.6 i^+0.8 k^×14.4 j^ 

=0.6×14.4 k^-0.8×14.4 i^

=14.40.6 k^-0.8 i^

⇒|L→|=14.4 0.62+0.82

=14.40.36+0.64

=14.4×1

=14.4 kg m2 s-1

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About this question

This is a previous-year question from JEE Main 2015, covering the Rotational Motion chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.