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 . The spheres are now immersed in a dielectric liquid of density and dielectric constant . If the angle between the strings remains the same after the immersion, then
Options
- Aelectric force between the spheres remains unchanged
- Belectric force between the spheres reduces
- Cmass density of the spheres is
- Dthe tension in the strings holding the spheres remains unchanged
Correct answer
A. electric force between the spheres remains unchanged
Step-by-step solution
Tsin\theta = F; Tcos\theta = mgtan\theta = \dfrac{F}{mg} . . . \left(1\right){T}^{'}sin\theta = \dfrac{F}{{ε}_{r}}{T}^{'}cos\theta = mg\left(1 - \dfrac{{ρ}_{ℓ}}{{ρ}_{s}}\right)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}{ρ}_{s} = 840kg{m}^{- 3}{T}^{'} = \dfrac{T}{{ε}_{r}} = \dfrac{T}{21}