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 \(\left(\text { Take } \pi=\frac{22}{7}\right)\)
Pick an option to check your answer.
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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