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JEE Mains · Physics · STD 11 - 9.2 surface tension

A water drop of diameter \(cm\) is broken into \(64\) equal droplets. The surface tension of water is \(0.075\,N / m\). In this process the gain in surface energy will be ...........\(J\)

  1. A \(2.8 \times 10^{-4}\)
  2. B \(1.5 \times 10^{-3}\)
  3. C \(1.9 \times 10^{-4}\)
  4. D \(9.4 \times 10^{-5}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(2.8 \times 10^{-4}\)

Step-by-step Solution

Detailed explanation

\(d =2\,cm ; \quad r =1\,cm ; \quad T =0.075\) \(\Delta SE = T \Delta A\) \(=0.075\left( A _{ f }- A _{1}\right)\) \(A _{ i }=4 \pi r ^{2}\) \(A _{ t }=4 \pi r _{0}^{2} \times 64\) By volume conservation \(\frac{4}{3} \pi r ^{3}=64 \cdot \frac{4}{3} \pi r_{0}^{3}\)…
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