JEE Mains · Physics · STD 11 - 9.2 surface tension
A liquid drop of diameter \(2\) mm breaks into \(512\) droplets. The change in surface energy is \(\alpha \times 10^{-6}\) J. The value of \(\alpha\) is _______. (Take surface tension of liquid \(= 0.08\) N/m)
- A \(10\)
- B \(7\)
- C \(8\)
- D \(11\)
Answer & Solution
Correct Answer
(B) \(7\)
Step-by-step Solution
Detailed explanation
Let the radius of the initial drop be \(R\) and the radius of each small droplet be \(r\). Given diameter \(D = 2\) mm, so \(R = 1\) mm \(= 10^{-3}\) m. By conservation of volume: \(\dfrac{4}{3}\pi R^3 = 512 \times \dfrac{4}{3}\pi r^3\)…
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