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Physics · Capacitors and R-C Circuits

JEE Advanced 2020 — Paper 2 — Question 9

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 800 kg m−3800 \mathrm{~kg} \mathrm{~m}^{-3} and dielectric constant 21. If the angle between the strings remains the same after the immersion, then

  1. Option A:

    electric force between the spheres remains unchanged

    Correct
  2. Option B:

    electric force between the spheres reduces

  3. Option C:

    mass density of the spheres is 840 kg m−3840 \mathrm{~kg} \mathrm{~m}^{-3}

    Correct
  4. Option D:

    the tension in the strings holding the spheres remains unchanged

Answer: A, C

Step-by-step solution

tan⁡α=Fmg=F/Kmg−ρℓvg⇒ρ0=840 kg/m3 T1cos⁡α=m m, T2cos⁡α=mg−ρℓvg\begin{aligned} & \tan \alpha=\frac{F}{m g}=\frac{F / K}{m g-\rho_{\ell} v g} \\& \Rightarrow \rho_{0}=840 \mathrm{~kg} / \mathrm{m}^{3} \\& \mathrm{~T}_{1} \cos \alpha=m \mathrm{~m}, \mathrm{~T}_{2} \cos \alpha=\mathrm{mg}-\rho_{\ell} \mathrm{vg} \end{aligned}

So, T2\mathrm{T}_{2} will decrease

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2020
Paper
Paper 2
Subject
Physics
Chapter
Capacitors and R-C Circuits
Topic
Effect of Dielectrics