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NEET PHYSICSELECTROSTATIC POTENTIAL AND CAPACITANCEMedium

Question

Twenty seven drops of same size are charged at 220 V each. They combine to form a bigger drop. Calculate the potential of the bigger drop:

A

1520 V

B

1980 V

C

660 V

D

1320 V

Step-by-Step Solution

  1. Conservation of Volume: Let rr be the radius of a small drop and RR be the radius of the big drop. The total volume remains constant during the combination. 43πR3=27×43πr3  ⟹  R=(27)1/3r=3r\frac{4}{3}\pi R^3 = 27 \times \frac{4}{3}\pi r^3 \implies R = (27)^{1/3} r = 3r.
  2. Conservation of Charge: Let qq be the charge on a small drop. The total charge QQ on the big drop is the sum of the charges on the 27 small drops. Q=27qQ = 27q.
  3. Potential Relationship: The electric potential VV of a conducting sphere is given by V=kQRV = \frac{kQ}{R} (analogous to the gravitational potential V∝MRV \propto \frac{M}{R} discussed in Class 11 Physics, Chapter 8).
  • Potential of small drop: Vsmall=kqr=220 VV_{small} = \frac{kq}{r} = 220 \text{ V}.
  • Potential of big drop: Vbig=kQR=k(27q)3r=9(kqr)=9VsmallV_{big} = \frac{kQ}{R} = \frac{k(27q)}{3r} = 9 \left( \frac{kq}{r} \right) = 9 V_{small}.
  1. Calculation: Vbig=9×220 V=1980 VV_{big} = 9 \times 220 \text{ V} = 1980 \text{ V}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ELECTROSTATIC POTENTIAL AND CAPACITANCE. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

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