JEE Main 2023PhysicsMathematics in PhysicsEasyMCQ

JEE Main 2023 — Mathematics in Physics Question with Solution

JEE Main 2023 (12 Apr Shift 1)

Question

Three forces F1=10 N, F2=8 N, F3=6 N are acting on a particle of mass 5 kg. The forces F2 and F3 are applied perpendicularly so that particle remains at rest. If the force F1 is removed, then the acceleration of the particle is

Choose an option

Show full solutionCorrect option: D
Correct answer
D2 m s–2

Step-by-step explanation

When the force F1 is removed, the particle starts accelerating because of the resultant force between F2 and F3.

The resultant force can be calculated as follows:

F=F22+F32=10 N

Hence, the acceleration of the particle is given by

a=10 N5 kg=2 m s–2

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Mathematics in Physics 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 2023, covering the Mathematics in Physics 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.