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JEE Advanced Physics Electrostatics 2020 JEE Advanced 2020 (Paper 2)

JEE Advanced Physics Question (2020) — Solution

Question

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 . The spheres are now immersed in a dielectric liquid of density 800   kg   m - 3 and dielectric constant  21 . If the angle between the strings remains the same after the immersion, then

Options

  1. A. electric force between the spheres remains unchanged
  2. B. electric force between the spheres reduces
  3. C. mass density of the spheres is 840   kg   m - 3
  4. D. the tension in the strings holding the spheres remains unchanged

Answer

C. mass density of the spheres is 840   kg   m - 3

Step-by-step solution

T sin θ = F ; T cos θ = m g Divide, tan θ = F m g . . . 1 As force between two charged bodies doesn't depend upon medium. Hence, the force between them remains same because the distance spheres is same. When it is submerged in liquid then T ' sin θ = F ε r T ' cos θ = m g 1 - ρ ℓ ρ s Divide, tan θ = F mg 1 - ρ l ρ S ε r . . . 2 Equating (1) & (2) 1 - ρ ℓ ρ s = 1 ε r ; 1 - 800 ρ s = 1 21 ρ s = 840 kg m - 3 T ' = T ε r = T 21

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