JEE Mains · Physics · STD 11 - 9.2 surface tension
A water drop of radius \(1\,cm\) is broken into \(729\) equal droplets. If surface tension of water is \(75\,dyne / cm\), then the gain in surface energy upto first decimal place will be\(...\times 10^{-4}\,J\)
- A \(8.5\)
- B \(8.2\)
- C \(7.5\)
- D \(5.3\)
Answer & Solution
Correct Answer
(C) \(7.5\)
Step-by-step Solution
Detailed explanation
Initial surface energy \(= TA\) Where \(T\) is surface tension and \(A\) is surface area \(U _{ i }=\left(\frac{75 \times 10^{-5}}{10^{-2}} \frac{ N }{ m }\right) \times\left[4 \pi\left(1 \times 10^{-2}\right)^{2}\right]\)…
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