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JEE Mains · Physics · STD 11 - Surface tension
  • JEE Mains 2023
  • Medium
  • MCQ

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 \(\left(\text { Take } \pi=\frac{22}{7}\right)\)

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Answer & Solution

Correct Answer

(B) \(7.92 \times 10^{-4}\,J\)

Step-by-step Solution

Detailed explanation

\(1000 \times \frac{4 \pi}{3}(1)^3=\frac{4 \pi}{3} R ^3\)

\(R =10\,mm\)

\(T \times 1000 \times 4 \pi\left(10^{-3}\right)^2- T \times 4 \pi\left(10 \times 10^{-3}\right)^2=\Delta E\)

\(\Delta E =4 \times \pi \times 7 \times 10^{-2}[1000-100] \times 10^{-6}\)

\(\Delta E =7.92 \times 10^{-4}\,J\)

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