JEE Mains · Physics · STD 11 - 9.2 surface tension
Eight mercury drops, each of radius \(r\), coalesce to form a bigger drop. The surface energy released in this process is _______. (\(S\) is the surface tension of mercury).
- A \(8\pi r^2 S\)
- B \(16\pi r^2 S\)
- C \(64\pi r^2 S\)
- D \(4\pi r^2 S\)
Answer & Solution
Correct Answer
(B) \(16\pi r^2 S\)
Step-by-step Solution
Detailed explanation
Let the radius of the bigger drop be \(R\). Since the total volume remains conserved during the process: \(8 \times \dfrac{4}{3} \pi r^3 = \dfrac{4}{3} \pi R^3\) \(R^3 = 8r^3 \Rightarrow R = 2r\) Initial surface energy of the \(8\) drops is:…
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