JEE Main 2019PhysicsWork Power EnergyMediumMCQ

JEE Main 2019 — Work Power Energy Question with Solution

JEE Main 2019 (12 Jan Shift 2)

Question

A particle of mass 20 g is released with an initial velocity 5 m s-1 along the curve from the point A, as shown in the figure. The point A is at height h from point B. The particle slides along the frictionless surface. When the particle reaches point B, its angular momentum about O will be: (Take g=10 m s-2)

Choose an option

Show full solutionCorrect option: C
Correct answer
C6 kg m2 s-1

Step-by-step explanation

Applying conservation of energy,

mgh=12mvB2-12mvA2
⇒vB=2gh+vA2
⇒vB=2×10×10+25
⇒vB=15 m s-1
Angular momentum about O, 

LO=mvBh+a
⇒LO=20×10-3×15×20 kg m2 s-1
⇒LO=6 kg m2 s-1

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Work Power Energy chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2019, covering the Work Power Energy 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.