JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
Sixty four conducting drops each of radius \(0.02 m\) and each carrying a charge of \(5 \,\mu C\) are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be ............
- A \(1: 4\)
- B \(4: 1\)
- C \(1: 8\)
- D \(8: 1\)
Answer & Solution
Correct Answer
(B) \(4: 1\)
Step-by-step Solution
Detailed explanation
Let \(R=\) radius of combined drop \(r=\) radius of smaller drop Volume will remain same \(\frac{4}{3} \pi R ^{3}=64 \times \frac{4}{3} \pi r ^{3}\) \(R =4 r\) \(Q =64 q ;\) \(q\) : charge of smaller drop \(Q\) : Charge of combined drop…
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