A large number of water droplets each of radius ' \(t\) ' combine to form a large drop of Radius ' \(R\) '. If the surface tension of water is ' \(T\) ' \& mechanical equivalent of heat is ' \(\mathrm{J}\) ' then the rise in temperature due to this is

Radius of each droplet \(=\mathrm{r}\)
Radius of the drop \(=\mathrm{R}\)
As volume remains constant,
\(
\begin{aligned}
& \mathrm{n} \times \frac{4}{3} \pi \mathrm{r}^3=\frac{4}{3} \pi \mathrm{R}^3 \\
& \therefore \quad \mathrm{n}=\frac{\mathrm{R}^3}{\mathrm{r}^3}
\end{aligned}
\)
Decrease in surface area \(=4 \pi r^2 n-4 \pi R^2 n\)
\(
\begin{aligned}
\Delta \mathrm{A} & =4 \pi\left[\mathrm{nr}^2-\mathrm{R}^2\right] \\
& =4 \pi\left[\frac{\mathrm{R}^3}{\mathrm{r}^3} \mathrm{r}^2-\mathrm{R}^2\right] \\
& =4 \pi \mathrm{R}^3\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]
\end{aligned}
\)
Enery released \(\mathrm{W}=\mathrm{T} \times \Delta \mathrm{T}\)
\(
\begin{aligned}
& \text { Heat produced } \mathrm{Q}=\frac{\mathrm{W}}{\mathrm{J}} \\
& \text { H }
\end{aligned}
\)
Heat produced \(\mathrm{Q}=\frac{\mathrm{W}}{\mathrm{J}}\)
\(
\mathrm{Q}=\mathrm{m} \cdot \mathrm{s} \cdot \Delta \theta
\)
Put (i) and (iii) into (ii)
\(
\begin{aligned}
& \mathrm{m} \cdot \mathrm{S} \Delta \theta=\frac{4 \pi \mathrm{R}^3 \mathrm{~T}}{\mathrm{~J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right] \\
& \frac{4}{3} \pi \mathrm{R} \rho_{\text {water }} \mathrm{S}_{\text {water }} \Delta \theta=\frac{4 \pi \mathrm{R}^3 \mathrm{~T}}{\mathrm{~J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right] \\
& \Delta \mathrm{Q}=\frac{3 \mathrm{~T}}{\mathrm{~J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]
\end{aligned}
\)