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(C) Let $n$ be the number of small droplets of radius $r$ that combine to form a large drop of radius $R$. Since the total volume remains constant,we

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$\frac{3T}{J}\left(\frac{1}{r}-\frac{1}{R}\right)$

Vedclass · Answer

(C) Let $n$ be the number of small droplets of radius $r$ that combine to form a large drop of radius $R$. Since the total volume remains constant,we have: $n \cdot \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3 \implies n = \frac{R^3}{r^3}$. The decrease in surface area $\Delta A$ is given by: $\Delta A = n(4 \pi r^2) - 4 \pi R^2 = 4 \pi (n r^2 - R^2) = 4 \pi \left(\frac{R^3}{r^3} \cdot r^2 - R^2ight) = 4 \pi R^3 \left(\frac{1}{r} - \frac{1}{R}ight)$. The energy released $W$ due to the decrease in surface area is $W = T \cdot \Delta A = 4 \pi R^3 T \left(\frac{1}{r} - \frac{1}{R}ight)$. This energy is converted into heat $Q = \frac{W}{J} = \frac{4 \pi R^3 T}{J} \left(\frac{1}{r} - \frac{1}{R}ight)$. Also,the heat produced is $Q = m \cdot s \cdot \Delta 	heta$,where $m$ is the mass of the large drop,$s$ is the specific heat of water,and $\Delta 	heta$ is the rise in temperature. $m = ho \cdot V = ho \cdot \frac{4}{3} \pi R^3$. Equating the two expressions for $Q$: $ho \cdot \frac{4}{3} \pi R^3 \cdot s \cdot \Delta 	heta = \frac{4 \pi R^3 T}{J} \left(\frac{1}{r} - \frac{1}{R}ight)$. Solving for $\Delta 	heta$: $\Delta 	heta = \frac{3T}{ho s J} \left(\frac{1}{r} - \frac{1}{R}ight)$. Assuming the question implies the rise in temperature per unit density and specific heat (or that $ho s = 1$ in the given context),the correct expression matches option $C$.

$A$ large number of water droplets each of radius '$r$' combine to form a large drop of radius '$R$'. If the surface tension of water is '$T$' and the mechanical equivalent of heat is '$J$',then the rise in temperature due to this process is:

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