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Electrostatics

JEE Main level, numerical answer, medium

193 of 311

Question 193 of 311

$512$ identical drops of mercury are charged to a potential of $2V$ each. The drops are joined to form a single drop. The potential of this drop is $V$ in Volt.

Show answer and solution

Answer: 128

Q=512q

Volume{}_{i}=Volume{}_{f}

512\times \dfrac{4}{3}\pi {r}^{3}=\dfrac{4}{3}\pi {R}^{3}

{2}^{9}{r}^{3}={R}^{3}

R=8r

2=\dfrac{kq}{r}

V=\dfrac{kQ}{R}=\dfrac{k512q}{8r}

V=128V

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JEE Main 2025 Apr 2 Shift 1 question paper

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JEE Main Physics practice question on Electrostatics (Electric Potential and Potential Energy), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.

512512512 identical drops of mercury are charged to a potential of 2V2V2V each. The drops are joined to form a single drop. The potential of this drop is VVV in Volt.

JEE Main question paper PDF

More Physics practice

$512$ identical drops of mercury are charged to a potential of $2V$ each. The drops are joined to form a single drop. The potential of this drop is $V$ in Volt.