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
- A. electric force between the spheres remains unchanged
- B. electric force between the spheres reduces
- C. mass density of the spheres is 840   kg   m - 3
- 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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