JEE Advanced 2020 · Paper 2 · Physics Q11

← JEE Advanced 2020 · Paper 2

PhysicsMultiple correct+4 / -2Question 11 of 54

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

electric force between the spheres remains unchanged

electric force between the spheres reduces

mass density of the spheres is

840\ \mathrm{kg\,m^{-3}}

the tension in the strings holding the spheres remains unchanged

Answer and worked solution

Official answer: B, C

(A)/(B) Electric force. In a medium of dielectric constant 21 the Coulomb force at the same separation becomes $F/21$ . It reduces, so (B) is true and (A) is false.

(C) Density. In equilibrium $\tan\frac\alpha2=\dfrac{F}{W_{eff}}$ . For the same angle,
$\frac F{mg}=\frac{F/21}{mg\left(1-\frac{800}\rho\right)}\implies1-\frac{800}\rho=\frac1{21}\implies\rho=\frac{21\times800}{20}=840\ \mathrm{kg\,m^{-3}}.$
(C) is true.

(D) Tension. $T=\sqrt{F^2+W_{eff}^2}$ , and both terms fall by a factor of 21. So $T$ changes, and (D) is false.

(B), (C)

Topics:Coulomb's law and superposition Buoyancy and Archimedes' principle Dielectrics

← Previous question Next question →

JEE Advanced 2020 · Paper 2, Physics question 11