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 kg m$$^{−3}$$ and dielectric constant 21. If the angle between the strings remains the same after the immersion, then

Let each sphere have radius $$r$$, volume $$V = \frac{4}{3}\pi r^{3}$$, density $$\rho_s$$ and hence mass $$m = \rho_s V$$. Each string makes an angle $$\theta = \frac{\alpha}{2}$$ with the vertical.

1. Equilibrium in air
For one sphere the forces are:
  • weight downward: $$mg = \rho_s V g$$
  • tension $$T$$ along the string
  • horizontal electrostatic repulsion $$F_e$$ from the other sphere

Resolving along horizontal and vertical directions in air,

$$T\sin\theta = F_e \qquad -(1)$$
$$T\cos\theta = mg \qquad -(2)$$

Dividing $$(1)$$ by $$(2)$$ gives

$$\tan\theta = \frac{F_e}{mg} = \frac{F_e}{\rho_s V g} \qquad -(3)$$

2. Equilibrium in the dielectric liquid
Density of liquid $$\rho_l = 800\ \text{kg m}^{-3}$$, dielectric constant $$K = 21$$.

The forces now are:
  • effective weight (weight − buoyant force): $$m g - \rho_l V g = (\rho_s - \rho_l)Vg$$ downward
  • new tension $$T'$$ along the string
  • electrostatic force reduced inside the dielectric:
    $$F'_e = \frac{F_e}{K}$$ (Coulomb force is inversely proportional to the medium’s dielectric constant)

Resolving again,

$$T'\sin\theta = F'_e \qquad -(4)$$
$$T'\cos\theta = (\rho_s-\rho_l)Vg \qquad -(5)$$

Hence

$$\tan\theta = \frac{F'_e}{(\rho_s-\rho_l)Vg} = \frac{F_e/K}{(\rho_s-\rho_l)Vg} \qquad -(6)$$

3. Condition for unchanged angle
Given that the angle (and therefore $$\tan\theta$$) is the same in air and in the liquid, equate $$(3)$$ and $$(6)$$:

$$\frac{F_e}{\rho_s V g} \;=\; \frac{F_e/K}{(\rho_s-\rho_l)Vg}$$

Simplifying (the factors $$F_e$$, $$V$$, $$g$$ cancel):

$$\frac{1}{\rho_s} = \frac{1}{K(\rho_s-\rho_l)}$$

Cross-multiplying:

$$K(\rho_s-\rho_l) = \rho_s$$
$$K\rho_s - K\rho_l = \rho_s$$
$$(K-1)\rho_s = K\rho_l$$

Therefore

$$\rho_s = \frac{K\rho_l}{K-1} = \frac{21 \times 800}{21 - 1} = \frac{16800}{20} = 840\ \text{kg m}^{-3}$$

4. Checking the listed statements
A. “Electric force remains unchanged” - FALSE because $$F'_e = F_e/K$$ is smaller.
B. “Electric force reduces” - TRUE as shown above.
C. “Mass density of the spheres is 840 kg m$$^{-3}$$” - TRUE from the calculation.
D. “Tension in the strings remains unchanged” - FALSE since $$T' = \dfrac{(\rho_s-\rho_l)Vg}{\cos\theta}$$ is less than $$T = \dfrac{\rho_s V g}{\cos\theta}$$.

Final result: Option B and Option C are correct.