A large number of water drops, each of radius $\mathrm{r}$, combine to have a drop of radius $\mathrm{R}$. If the surface tension is $T$ and mechanical equivalent of heat is $\mathrm{J}$, the rise in heat energy per unit volume will be :
Correct Option: , 4
(4)
$\mathrm{R}$ is the radius of bigger drop. is the radius of $n$ water drops. Water drops are combined to make bigger drop. So,
Volume of $\mathrm{n}$ drops $=$ volume of bigger drop
$n\left(\frac{4}{3} \pi r^{3}\right)=\frac{4}{3} \pi R^{3}$
$R=r n^{1 / 3} \Rightarrow n=\left(\frac{R}{r}\right)^{3}$
$\Delta \mathrm{U}=\mathrm{T}$ (Change in surface area) $\Delta \mathrm{U}=\mathrm{T}\left(\mathrm{n} 4 \pi \mathrm{r}^{2}-4 \pi \mathrm{R}^{2}\right)$
$\Delta \mathrm{U}=4 \pi \mathrm{T}\left[\left(\frac{\mathrm{R}}{\mathrm{r}}\right)^{3} \mathrm{r}^{2}-\mathrm{R}^{2}\right] \Rightarrow \frac{4 \pi \mathrm{T}\left(\frac{\mathrm{R}^{3}}{\mathrm{r}}-\mathrm{R}^{2}\right)}{\mathrm{J}}$
$\frac{\Delta \mathrm{U}}{\mathrm{V}}=\frac{4 \pi \mathrm{T}\left(\frac{\mathrm{R}^{3}}{\mathrm{r}}-\mathrm{R}^{2}\right)}{\mathrm{J} \times \frac{4}{3} \pi \mathrm{R}^{3}}=\frac{3 \mathrm{~T}}{\mathrm{~J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]$
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