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JEE Advanced 2020Paper 2
Exam
JEE Advanced
Year
2020
Paper
Paper 2
Subject
Physics
Chapter
Electrostatics

JEE Advanced 2020 · Paper 2 · Physics · Electrostatics

Two identical non-conducting solid spheres of same mass and charge...

Two identical non-conducting solid spheres of same mass and charge are suspended in air from a common point by two non-conducting, massless strings of same length. At equilibrium, the angle between the strings is α\alpha. The spheres are now immersed in a dielectric liquid of density 800kgm−3800kg{m}^{- 3} and dielectric constant 2121. If the angle between the strings remains the same after the immersion, thenTwo identical non-conducting solid spheres of same mass and charge are suspended in air from a common point by two

Options

  1. Aelectric force between the spheres remains unchanged
  2. Belectric force between the spheres reduces
  3. Cmass density of the spheres is 840kgm−3840kg{m}^{- 3}
  4. Dthe tension in the strings holding the spheres remains unchanged

Correct answer

A. electric force between the spheres remains unchanged

Step-by-step solution

Two identical non-conducting solid spheres of same mass and charge are suspended in air from a common point by two

Tsin\theta = F; Tcos\theta = mg</p><p>Divide,</p><p></p><p>Divide,</p><p>tan\theta = \dfrac{F}{mg} . . . \left(1\right)</p><p>Asforcebetweentwochargedbodiesdoesn′tdepend uponmedium.Hence,theforcebetweenthemremainssamebecausethedistancespheres is same.</p><p>Whenitissubmergedinliquidthen</p><p>As force between two charged bodies doesn't depend upon medium. Hence, the force between them remains same because the distance spheres is same.</p><p>When it is submerged in liquid then{T}^{'}sin\theta = \dfrac{F}{{ε}_{r}}</p><p></p><p>{T}^{'}cos\theta = mg\left(1 - \dfrac{{ρ}_{ℓ}}{{ρ}_{s}}\right)</p><p>Divide,</p><p></p><p>Divide,</p><p>tan\theta = \dfrac{F}{mg \left(1 - \dfrac{{ρ}_{l}}{{ρ}_{S}}\right)}{ε}_{r} . . . \left(2\right)</p><p>Equating (1) & (2)</p><p>1 - \dfrac{{ρ}_{ℓ}}{{ρ}_{s}} = \dfrac{1}{{ε}_{r}}; 1 - \dfrac{800}{{ρ}_{s}} = \dfrac{1}{21}</p><p></p><p>{ρ}_{s} = 840kg{m}^{- 3}</p><p></p><p>{T}^{'} = \dfrac{T}{{ε}_{r}} = \dfrac{T}{21}

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