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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
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Verified Answer
The correct answer is:
$\frac{3 \mathrm{~T}}{\mathrm{~J}}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$
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)
$$
\begin{aligned}
& \mathrm{m} \cdot \mathrm{S} \Delta \theta=\frac{4 \pi \mathrm{R}^3 \mathrm{~T}}{\mathrm{~J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right] \\
& \frac{4}{3} \pi \mathrm{R} \rho_{\text {water }} \mathrm{S}_{\text {water }} \Delta \theta=\frac{4 \pi \mathrm{R}^3 \mathrm{~T}}{\mathrm{~J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right] \\
& \Delta \mathrm{Q}=\frac{3 \mathrm{~T}}{\mathrm{~J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]
\end{aligned}
$$
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)
$$
\begin{aligned}
& \mathrm{m} \cdot \mathrm{S} \Delta \theta=\frac{4 \pi \mathrm{R}^3 \mathrm{~T}}{\mathrm{~J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right] \\
& \frac{4}{3} \pi \mathrm{R} \rho_{\text {water }} \mathrm{S}_{\text {water }} \Delta \theta=\frac{4 \pi \mathrm{R}^3 \mathrm{~T}}{\mathrm{~J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right] \\
& \Delta \mathrm{Q}=\frac{3 \mathrm{~T}}{\mathrm{~J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]
\end{aligned}
$$
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