A large number of water drops,each of radius | vedclass

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(D) Let $n$ be the number of small drops of radius $r$ that combine to form a large drop of radius $R$.By conservation of volume: $n 	imes \frac{4}{3

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $n$ be the number of small drops of radius $r$ that combine to form a large drop of radius $R$.By conservation of volume: $n 	imes \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$,which implies $n = \frac{R^3}{r^3}$.The change in surface area is $\Delta A = n(4 \pi r^2) - 4 \pi R^2$.Substituting $n = \frac{R^3}{r^3}$,we get $\Delta A = \frac{R^3}{r^3}(4 \pi r^2) - 4 \pi R^2 = 4 \pi R^2 (\frac{R}{r} - 1)$ (This is the decrease in area).The energy released is $\Delta U = T 	imes \Delta A = T(n 4 \pi r^2 - 4 \pi R^2) = 4 \pi T (n r^2 - R^2)$.The heat energy produced is $Q = \frac{\Delta U}{J} = \frac{4 \pi T (n r^2 - R^2)}{J}$.The volume of the large drop is $V = \frac{4}{3} \pi R^3$.The rise in heat energy per unit volume is $\frac{Q}{V} = \frac{4 \pi T (n r^2 - R^2) / J}{4/3 \pi R^3} = \frac{3T}{J} (\frac{n r^2}{R^3} - \frac{R^2}{R^3})$.Since $n r^3 = R^3$,we have $\frac{n r^2}{R^3} = \frac{1}{r}$.Thus,the rise in heat energy per unit volume is $\frac{3T}{J} (\frac{1}{r} - \frac{1}{R})$.

$A$ large number of water drops,each of radius $r$,combine to form a single drop of radius $R$. If the surface tension is $T$ and the mechanical equivalent of heat is $J$,the rise in heat energy per unit volume will be:

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