JEE Main 2013PhysicsOscillationsHardMCQ

JEE Main 2013 — Oscillations Question with Solution

JEE Main 2013 (07 Apr)

Question

Two charges, each equal to q , are kept at x = - a and x  =  a on the x-axis. A particle of mass m and charge q0=-q2  is placed at the origin. If charge q 0 is given a small displacement (y << a) along the y-axis, the net force acting on the particle is proportional to :

Choose an option

Show full solutionCorrect option: C
Correct answer
C-y

Step-by-step explanation




⇒ F net = 2Fcos θ

Fnet=2kqq2y2+a22yy2+a2

Fnet=2kqq2yy2+a23/2≈kq2a3y⇒F∝-y

Net force F is opposite the direction of displacement y. Hence, negative sign is used.

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About this question

This is a previous-year question from JEE Main 2013, covering the Oscillations 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.