Question:
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 :
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 :
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Heat produced equals the decrease in surface energy; divide by the volume \(\frac43\pi R^3\) and by \(J\) to express in calories.
Updated On: Nov 4, 2026
- \(\frac{2T}{J}\left(\frac1r-\frac1R\right)\)
- \(\frac{3T}{J}\left(\frac1r-\frac1R\right)\)
- \(\frac{3T}{rJ}\)
- \(\frac{2T}{rJ}\)
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The Correct Option is B
Solution and Explanation
Volume conservation: \(n\cdot\frac43\pi r^3=\frac43\pi R^3\Rightarrow n=\frac{R^3}{r^3}\).
Decrease in surface energy \(=T\cdot4\pi\left(nr^2-R^2\right)\).
Per unit volume: \(\frac{4\pi T(nr^2-R^2)}{\frac43\pi R^3}=3T\left(\frac{nr^2}{R^3}-\frac1R\right)=3T\left(\frac1r-\frac1R\right)\).
Heat (in calories) \(=\frac{3T}{J}\left(\frac1r-\frac1R\right)\).
Answer: (2) \(\frac{3T}{J}\left(\frac1r-\frac1R\right)\)
Decrease in surface energy \(=T\cdot4\pi\left(nr^2-R^2\right)\).
Per unit volume: \(\frac{4\pi T(nr^2-R^2)}{\frac43\pi R^3}=3T\left(\frac{nr^2}{R^3}-\frac1R\right)=3T\left(\frac1r-\frac1R\right)\).
Heat (in calories) \(=\frac{3T}{J}\left(\frac1r-\frac1R\right)\).
Answer: (2) \(\frac{3T}{J}\left(\frac1r-\frac1R\right)\)
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