Eight drops of a liquid of density 3 and each of radius a are falling through air with a constant velocity \(3.75 \mathrm{~cm} \mathrm{~S}^{1}\) when the eight drops coalesce to form a single drop the terminal velocity of the new drop will be (A) \(15 \times 10^{-2} \mathrm{~ms}^{-1}\) (B) \(2.4 \times 10^{-2} \mathrm{~m} / \mathrm{s}\) (C) \(0.75 \times 10^{-2} \mathrm{~ms}^{-1}\) (D) \(25 \times 10^{-2} \mathrm{~m} / \mathrm{s}\)

To find the volume, we will consider that each drop is a sphere with radius a. The formula for the volume of a sphere is given by: \(V = \frac{4}{3}\pi r^3\) Since there are eight smaller drops, each with volume V, the total volume of the coalesced drop \(V_{new}\) will be: \(V_{new} = 8V = 8(\frac{4}{3}\pi a^3)\)

Since the new drop is also a sphere, the relationship between its volume and radius is given by \(V_{new} = \frac{4}{3}\pi R^3\), where R is the radius of the new drop. We can solve for R: \(\frac{4}{3}\pi R^3 = 8(\frac{4}{3}\pi a^3)\) Dividing both sides by \(\frac{4}{3}\pi\), we get: \(R^3 = 8a^3\) Now we can find R by taking the cube root of both sides: \(R = 2a\)

For smaller drops with Reynold's number less than 1, the terminal velocity can be determined using Stokes's Law. The formula relating terminal velocity, radius, and force is: \(v = \frac{2}{9}\frac{g r^2}{\eta}\) where v is the terminal velocity, g is the gravitational acceleration, r is the radius of the drop, and η is the dynamic viscosity of air. Since the terminal velocity of the smaller drops is given as \(3.75 \times 10^{-2} \mathrm{m.s^{-1}}\), we can write the equation for the terminal velocity for these drops as: \(v_{small} = \frac{2}{9}\frac{g a^2}{\eta}\) Now, we can write a similar equation for the terminal velocity of the coalesced drop, which has a radius of 2a: \(v_{new} = \frac{2}{9}\frac{g (2a)^2}{\eta}\) Now, we can rearrange the initial formula for the terminal velocity of a smaller drop to find the value of \(\frac{ga^2}{\eta}\): \(\frac{ga^2}{\eta} = \frac{9 v_{small}}{2} = \frac{9 (3.75 \times 10^{-2})}{2}\) Plug this value back into the equation for the terminal velocity of the coalesced drop: \(v_{new} = 2 (\frac{9 (3.75 \times 10^{-2})}{2}) = 3.75 \times 10^{-2} \mathrm{m.s^{-1}}\) Comparing with the given options, the terminal velocity of the new drop is closest to: (C) \(0.75 \times 10^{-2} \mathrm{~ms}^{-1}\)