JEE PYQ: Mechanical Properties Of Fluids Question 8
Question 8 - 2021 (26 Feb 2021 Shift 1)
A large number of water drops, each of radius $r$, combine to have a drop of radius $R$. If the surface tension is $T$ and mechanical equivalent of heat is $J$, the rise in heat energy per unit volume will be:
(1) $\frac{2T}{rJ}$
(2) $\frac{3T}{rJ}$
(3) $\frac{2T}{J}\left(\frac{1}{r} - \frac{1}{R}\right)$
(4) $\frac{3T}{J}\left(\frac{1}{r} - \frac{1}{R}\right)$
Show Answer
Answer: (4) $\frac{3T}{J}\left(\frac{1}{r} - \frac{1}{R}\right)$
Solution
Volume of $n$ drops = volume of bigger drop $n\left(\frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi R^3$ $R = rn^{1/3} \Rightarrow n = \left(\frac{R}{r}\right)^3$ $\Delta U = T$ (Change in surface area) $= T(n \cdot 4\pi r^2 - 4\pi R^2) \Rightarrow \frac{4\pi T(\frac{R^3}{r} - R^2)}{J}$ $\frac{\Delta U}{V} = \frac{4\pi T(\frac{R^3}{r} - R^2)}{J \times \frac{4}{3}\pi R^3} = \frac{3T}{J}\left(\frac{1}{r} - \frac{1}{R}\right)$
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