JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
\(64\) identical drops each charged upto potential of \(10\,mV\) are combined to form a bigger dorp. The potential of the bigger drop will be \(..........\,mV\)
- A \(150\)
- B \(140\)
- C \(130\)
- D \(160\)
Answer & Solution
Correct Answer
(D) \(160\)
Step-by-step Solution
Detailed explanation
Let \(q =\) charge on each drop \(V =\frac{ Kq }{ r } \text {-.- (1) }\) Now for combination of 64 drop \(64 \times \frac{4}{3} \pi r^3=\frac{4}{3} \pi R^3\) \(R=4 r\) And \(Q =64 q\) Potential of bigger drop \(=\frac{ KQ }{ R }=\frac{ K 64 q }{4 r }=16 \frac{ Kq }{ r }\)…
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