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JEE Main 202127 JulyMorning Shift
Exam
JEE Main
Year
2021
Date
27 July
Shift
Morning Shift
Subject
Physics
Chapter
Oscillations

JEE Main 2021 · 27 July · Morning Shift · Physics · Oscillations

Two identical tennis balls each having mass m and charge q are...

Two identical tennis balls each having mass mm and charge qq are suspended from a fixed point by threads of length l.l . What is the equilibrium separation when each thread makes a small angle θ\theta with the vertical?Two identical tennis balls each having mass m and charge q are suspended from a fixed point by threads of length l .

Options

  1. Ax = (q2l2πε0mg)12\left(\dfrac{{q}^{2}l}{2\pi{ε}_{0}mg}\right)\dfrac{1}{2}
  2. Bx = (q2l2πε0mg)13\left(\dfrac{{q}^{2}l}{2\pi{ε}_{0}mg}\right)\dfrac{1}{3}
  3. Cx = (q2l22πε0m2g)13\left(\dfrac{{q}^{2}{l}^{2}}{2\pi{ε}_{0}{m}^{2}g}\right)\dfrac{1}{3}
  4. Dx = (q2l22πε0m2g2)13\left(\dfrac{{q}^{2}{l}^{2}}{2{\pi ε}_{0}{m}^{2}{g}^{2}}\right)\dfrac{1}{3}

Correct answer

B. x = (q2l2πε0mg)13\left(\dfrac{{q}^{2}l}{2\pi{ε}_{0}mg}\right)\dfrac{1}{3}

Step-by-step solution

Two identical tennis balls each having mass m and charge q are suspended from a fixed point by threads of length l .

Tcos\theta = mg<br/>Tsinθ=kq2x2<br />Tsin\theta = \dfrac{k{q}^{2}}{{x}^{2}}
tanθ=kq2x2mgtan\theta = \dfrac{k{q}^{2}}{{x}^{2}mg}
as tanθ≈sinθ≈x2Ltan\theta \approx sin\theta \approx \dfrac{x}{2L}
x2L=Kq2x2mg\dfrac{x}{2L} = \dfrac{K{q}^{2}}{{x}^{2}mg}
x=(q2L2πε0mg)13x = \left(\dfrac{{q}^{2}L}{2\pi{ε}_{0}mg}\right)\dfrac{1}{3}

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