If a number of small droplets of water each of radius ‘r’ coalesce to form a single drop of radius R, the rise of temperature is (s=specific heat capacity of water, d= density of water, T= surface Tension of water)

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$\frac{3 Td}{s} (\frac{1}{R} - \frac{1}{r})$

$\frac{3 s}{Td} (\frac{1}{r} - \frac{1}{R})$

$\frac{3 T}{ds} (\frac{1}{r} - \frac{1}{R})$

$\frac{3 Ts}{d} (\frac{1}{r} - \frac{1}{R})$

answer is C.

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Detailed Solution

$\begin{array}{l} E = msΔ θ \\ \Rightarrow 4 π ({nr}^{2} - R^{2}) T = msΔ θ \\ \Rightarrow Δ θ = \frac{4 πT}{ms} ({nr}^{2} - R^{2}) \\ \Rightarrow Δ θ = \frac{4 πT}{\frac{4}{3} πds} (\frac{{nr}^{2}}{{nr}^{3}} - \frac{R^{2}}{R^{3}}) \\ \Rightarrow Δ θ = \frac{3 T}{ds} (\frac{1}{r} - \frac{1}{R}) \end{array}$

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