Several spherical drops of a liquid of radius r coalesce to form a single drop of radius R. If T is surface tension and V is volume under consideration, then the release of energy is

The correct option is 3VT(1/r - 1/R).

Step-by-step Derivation:

1. Let $n$ be the number of small spherical drops of radius $r$ that coalesce to form a single large drop of radius $R$. Since the total volume of the liquid remains conserved during this process, the volume of the single large drop is equal to the sum of the volumes of the $n$ small drops.
Let $V$ be the volume under consideration, which is the total volume of the liquid:
$V=n·\left(\frac{4}{3}\pi r^3\right)=\frac{4}{3}\pi R^3$

2. From the volume conservation equation, we can write the number of small drops $n$ in terms of the radii $R$ and $r$:
$n=\frac{R^3}{r^3}$

3. Next, let us calculate the change in the surface area.
Initial surface area of $n$ small drops:
$A_i=n·\left(4\pi r^2\right)=\left(\frac{R^3}{r^3}\right)·4\pi r^2=\frac{4\pi R^3}{r}$
Final surface area of the single large drop:
$A_f=4\pi R^2$

4. The decrease in surface area ($\Delta A$) is given by:
$\Delta A=A_i-A_f=\frac{4\pi R^3}{r}-4\pi R^2$
Taking out $4\pi R^3$ as a common factor:
$\Delta A=4\pi R^3(\frac{1}{r}-\frac{1}{R})$

5. We can express this in terms of the total volume $V$. Since $V=\frac{4}{3}\pi R^3$, we have:
$4\pi R^3=3V$
Substituting this back into the expression for $\Delta A$:
$\Delta A=3V(\frac{1}{r}-\frac{1}{R})$

6. The energy released during coalescing is equal to the product of surface tension $T$ and the decrease in surface area:
$E=T·\Delta A=3VT(\frac{1}{r}-\frac{1}{R})$