JEE Mains · Physics · STD 11 - 9.2 surface tension
If \(1000\) droplets of water of surface tension \(0.07\,N / m\). having same radius \(1\,mm\) each, combine to from a single drop. In the process the released surface energy is \(\left(\text { Take } \pi=\frac{22}{7}\right)\)
- A \(7.92 \times 10^{-6}\,J\)
- B \(7.92 \times 10^{-4}\,J\)
- C \(9.68 \times 10^{-4}\,J\)
- D \(8.8 \times 10^{-5}\,J\)
Answer & Solution
Correct Answer
(B) \(7.92 \times 10^{-4}\,J\)
Step-by-step Solution
Detailed explanation
\(1000 \times \frac{4 \pi}{3}(1)^3=\frac{4 \pi}{3} R ^3\) \(R =10\,mm\) \(T \times 1000 \times 4 \pi\left(10^{-3}\right)^2- T \times 4 \pi\left(10 \times 10^{-3}\right)^2=\Delta E\) \(\Delta E =4 \times \pi \times 7 \times 10^{-2}[1000-100] \times 10^{-6}\)…
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