Mechanical Properties of Fluids

Question: If $\sigma$ be the surface tension, the work done in breaking a big drop of radius $R$ in $n$ drops of equal radius is

1) $R{{n}^{2/3}}\sigma $

2) $({{n}^{2/3}}-1)R\sigma $

3) $({{n}^{1/3}}-1)R\sigma $

4) $4\pi {{R}^{2}}({{n}^{1/3}}-1)\sigma $

Solution:

Explanation: No Explanation

Surface tension surface energy

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