Physics · Fluid Mechanics

JEE Main 2024 — 1 February, Shift 1 — Question 50

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle $\theta$ with each other. When suspended in water the angle remains the same. If density of the material of the sphere is $1.5 \mathrm{~g} / \mathrm{cc}$ , the dielectric constant of water will be $\_\_\_\_\_\_\_\_\_\_\_$ (Take density of water $=1 \mathrm{~g} / \mathrm{cc})$

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

\text{Given:} \quad \rho_s = 1.5\,\text{g/cc}, \quad \rho_w = 1.0\,\text{g/cc}, \quad \theta_{\text{air}} = \theta_{\text{water}} = \theta, \quad K = \text{dielectric constant of water}.

\text{In air:} \quad F_e = \frac{kq^2}{r^2}, \qquad W = mg,

\tan\theta = \frac{F_e}{W} = \frac{kq^2}{mgr^2}.

\text{In water:} \quad F_e' = \frac{kq^2}{Kr^2}, \qquad F_b = \rho_w V g,

W' = mg - F_b = mg - \rho_w V g.

V = \frac{m}{\rho_s}, \qquad W' = mg\left(1 - \frac{\rho_w}{\rho_s}\right).

\tan\theta = \frac{F_e'}{W'} = \frac{kq^2/Kr^2}{mg\left(1 - \frac{\rho_w}{\rho_s}\right)} = \frac{kq^2}{Kmgr^2\left(1 - \frac{\rho_w}{\rho_s}\right)}.

\text{Since angles are equal:} \qquad \frac{kq^2}{mgr^2} = \frac{kq^2}{Kmgr^2\left(1 - \frac{\rho_w}{\rho_s}\right)}.

1 = \frac{1}{K\left(1 - \frac{\rho_w}{\rho_s}\right)} \quad\Rightarrow\quad K = \frac{1}{1 - \frac{\rho_w}{\rho_s}}.

K = \frac{1}{1 - \frac{1}{1.5}} = \frac{1}{1 - \frac{2}{3}} = \frac{1}{1/3} = 3.

\boxed{K = 3}

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 1 February, Shift 1
Subject: Physics
Chapter: Fluid Mechanics
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