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JEE Main 2021 (26 Feb, Shift 2) · Electrostatic Potential Energy

Electrostatic Potential Energy question from JEE Main 2021 (26 Feb, Shift 2)

Q46JEE Main 2021 · 26 FEB · SHIFT 2Medium

Electrostatic Potential Energy

27 similar drops of mercury are maintained at 10V each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is __________ times that of a smaller drop.

Numerical answer type. Work it out, then open the answer.

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Answer: 243

  1. Self-energy U=kQ22RU = \frac{kQ^2}{2R}, so U∝Q2RU \propto \frac{Q^2}{R}.
  2. 27 drops: R=3rR = 3r, Q=27qQ = 27q.
  3. UbigUsmall=2723=243\frac{U_{big}}{U_{small}} = \frac{27^2}{3} = 243
  4. So the answer is 243.
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