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 :

3T \over J( 1 \over r - 1 \over R )

2T \over J( 1 \over r - 1 \over R )

Properties of Matter PYQ — JEE Main 2021 · 26 Feb · Shift 1 · Q47 with Solution

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Correct answer: B

For a liquid drop, surface energy is $E = T \times A$ where $T$ is surface tension and $A$ is surface area.

For a spherical drop of radius $a$ , $A = 4\pi a^2$ so surface energy of one drop is $E = 4\pi a^2 T$

Let the number of small drops be $n$ , each of radius $r$ .

Then initial total surface energy is $E_i = n(4\pi r^2 T)$

After combining, one large drop of radius $R$ is formed.

Its surface energy is $E_f = 4\pi R^2 T$

Since volume is conserved, $n\left(\frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi R^3$

Hence, $nr^3 = R^3$ $n = \frac{R^3}{r^3}$

The decrease in surface energy is converted into heat: $\Delta E = E_i - E_f$

Substitute: $\Delta E = 4\pi T(nr^2 - R^2)$

Using $n = R^3/r^3$ , $nr^2 = \frac{R^3}{r^3}r^2 = \frac{R^3}{r}$

So,

= 4\pi T R^2\left(\frac{R}{r}-1\right)$$ This released energy appears as heat. --- 6. **Heat energy per unit volume** Mechanical equivalent of heat is $J$, so heat produced is $$H = \frac{\Delta E}{J}$$ The final volume of the large drop is $$V = \frac{4}{3}\pi R^3$$ Therefore, heat energy per unit volume is $$\frac{H}{V} = \frac{\Delta E/J}{V} = \frac{4\pi T\left(\frac{R^3}{r} - R^2\right)}{J\cdot \frac{4}{3}\pi R^3}$$ Simplify: $$\frac{H}{V} = \frac{3T}{J} \cdot \frac{\left(\frac{R^3}{r} - R^2\right)}{R^3}$$ $$\frac{H}{V} = \frac{3T}{J}\left(\frac{1}{r} - \frac{1}{R}\right)$$ --- 7. **Match with the options** Thus, the rise in heat energy per unit volume is $$\boxed{\frac{3T}{J}\left(\frac{1}{r}-\frac{1}{R}\right)}$$ This matches **Option B**.