A large number of liquid drops each of radius r coalesce to form a single drop of the radius R. The energy released in the process is converted into kinetic energy of the big drop so formed. The speed of the big drop is (given surface tension of the liquid T, density $$\rho$$)

We are given that a large number of liquid drops, each of radius $$r$$, coalesce to form a single big drop of radius $$R$$. The energy released during coalescence is converted into the kinetic energy of the big drop. We need to find the speed of the big drop, given the surface tension $$T$$ and density $$\rho$$.

First, recall that the surface energy of a liquid drop is given by the surface tension multiplied by the surface area. For a single small drop of radius $$r$$, the surface area is $$4\pi r^2$$, so its surface energy is $$T \times 4\pi r^2$$.

Let the number of small drops be $$n$$. The volume of one small drop is $$\frac{4}{3}\pi r^3$$, and the volume of the big drop is $$\frac{4}{3}\pi R^3$$. Since volume is conserved during coalescence:

$$n \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$$

Canceling $$\frac{4}{3}\pi$$ from both sides:

$$n r^3 = R^3$$

Solving for $$n$$:

$$n = \frac{R^3}{r^3}$$

The total surface energy before coalescence is the sum of the surface energies of all small drops:

$$\text{Initial surface energy} = n \times T \times 4\pi r^2 = \frac{R^3}{r^3} \times T \times 4\pi r^2 = T \times 4\pi \frac{R^3}{r}$$

The surface energy after coalescence is that of the big drop:

$$\text{Final surface energy} = T \times 4\pi R^2$$

The energy released is the difference between the initial and final surface energies:

$$E_{\text{released}} = \text{Initial surface energy} - \text{Final surface energy} = T \times 4\pi \frac{R^3}{r} - T \times 4\pi R^2$$

Factor out $$T \times 4\pi$$:

$$E_{\text{released}} = T \times 4\pi \left( \frac{R^3}{r} - R^2 \right)$$

Factor $$R^2$$ from the expression inside the parentheses:

$$E_{\text{released}} = T \times 4\pi R^2 \left( \frac{R}{r} - 1 \right)$$

Note that $$\frac{R}{r} - 1 = \frac{R - r}{r}$$. Also, observe that $$\frac{1}{r} - \frac{1}{R} = \frac{R - r}{R r}$$. Therefore:

$$R - r = R r \left( \frac{1}{r} - \frac{1}{R} \right)$$

Substitute this into the expression:

$$E_{\text{released}} = T \times 4\pi R^2 \times \frac{R - r}{r} = T \times 4\pi R^2 \times \frac{R r \left( \frac{1}{r} - \frac{1}{R} \right)}{r}$$

Simplify by canceling $$r$$ in the numerator and denominator:

$$E_{\text{released}} = T \times 4\pi R^2 \times R \left( \frac{1}{r} - \frac{1}{R} \right) = T \times 4\pi R^3 \left( \frac{1}{r} - \frac{1}{R} \right)$$

This energy is converted into the kinetic energy of the big drop. The kinetic energy is given by $$\frac{1}{2} M v^2$$, where $$M$$ is the mass of the big drop and $$v$$ is its speed.

The mass $$M$$ is the density times volume:

$$M = \rho \times \frac{4}{3}\pi R^3$$

Set the energy released equal to the kinetic energy:

$$T \times 4\pi R^3 \left( \frac{1}{r} - \frac{1}{R} \right) = \frac{1}{2} M v^2$$

Substitute $$M$$:

$$T \times 4\pi R^3 \left( \frac{1}{r} - \frac{1}{R} \right) = \frac{1}{2} \times \rho \times \frac{4}{3}\pi R^3 \times v^2$$

Divide both sides by $$\pi R^3$$ (assuming $$R \neq 0$$):

$$T \times 4 \left( \frac{1}{r} - \frac{1}{R} \right) = \frac{1}{2} \times \rho \times \frac{4}{3} \times v^2$$

Simplify the right side:

$$\frac{1}{2} \times \frac{4}{3} = \frac{4}{6} = \frac{2}{3}$$

So:

$$4T \left( \frac{1}{r} - \frac{1}{R} \right) = \frac{2}{3} \rho v^2$$

Solve for $$v^2$$. Multiply both sides by $$\frac{3}{2}$$:

$$v^2 = 4T \left( \frac{1}{r} - \frac{1}{R} \right) \times \frac{3}{2} \times \frac{1}{\rho}$$

Simplify:

$$v^2 = \frac{4 \times 3}{2} \times \frac{T}{\rho} \left( \frac{1}{r} - \frac{1}{R} \right) = \frac{12}{2} \times \frac{T}{\rho} \left( \frac{1}{r} - \frac{1}{R} \right) = 6 \times \frac{T}{\rho} \left( \frac{1}{r} - \frac{1}{R} \right)$$

Therefore:

$$v^2 = \frac{6T}{\rho} \left( \frac{1}{r} - \frac{1}{R} \right)$$

Take the square root of both sides:

$$v = \sqrt{ \frac{6T}{\rho} \left( \frac{1}{r} - \frac{1}{R} \right) }$$

Comparing with the options, this matches option B.

Hence, the correct answer is Option B.