A large number of water drops, each of radius r, combine to have a drop of radius R. If the surface tension is T and mechanical equivalent of heat is J, the rise in heat energy per unit volume will be:

1
\(\frac{{{\rm{2T}}}}{{\rm{J}}}\left( {\frac{{\rm{1}}}{{\rm{r}}} - \frac{{\rm{1}}}{{\rm{R}}}} \right) \)
2
\(\frac{{{\rm{3T}}}}{{\rm{J}}}\left( {\frac{{\rm{1}}}{{\rm{r}}} - \frac{{\rm{1}}}{{\rm{R}}}} \right)\)
3
\(\frac{{{\rm{3T}}}}{{{\rm{rJ}}}}\)
4
\(\frac{{{\rm{2T}}}}{{{\rm{rJ}}}}\)

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