Two drops of the same radius are falling through air with a steady velocity for \(5 \mathrm{~cm}\) per sec. If the two drops coakesce the terminal velocity would be (A) \(10 \mathrm{~cm}\) per sec (B) \(2.5 \mathrm{~cm}\) per sec (C) \(5 \times(4)^{(1 / 3)} \mathrm{cm}\) per sec (D) \(5 \times \sqrt{2} \mathrm{~cm}\) per sec

We're given that two drops have the same radius and are falling with a terminal velocity of \(5 \mathrm{cm/s}\). Let these individual terminal velocities be denoted by \(v_1\) and \(v_2\). After merging into a single larger droplet, the terminal velocity becomes \(v\). Our initial step is to find the relationship between these terminal velocities. From the given information, we can write: \(v_1 = v_2 = 5 \mathrm{cm/s}\)

Terminal velocity depends upon the mass and the radius of the droplets involved. From the given information, we know that the two falling drops have the same radius and, therefore, have the same mass. Consider the situation where the two drops merge to form a single larger droplet. According to the volume conservation principle, the volume of the bigger droplet is equal to the sum of the initial droplets' volumes: \(V = V_1 + V_2\) Also, since the droplets are spherical, we can write their volumes in terms of their radii: \(\frac{4}{3}\pi R^3 = \frac{4}{3}\pi r^3 + \frac{4}{3}\pi r^3\) Cancel out the common factors and find the relationship between the radii of the combined droplet and initial droplets: \(R^3=2r^3\) \(R = (2)^{1/3}r\)

The terminal velocity formula is: \(v = kR^2\) Where \(v\) is the terminal velocity, \(k\) is a constant (that depends on factors like the viscosity of the fluid and density of the droplets) and \(R\) is the droplet's radius. Since the droplets have the same radius, we can apply the formula to both the individual droplets and the combined droplet: \(v_1 = k_1 r^2\) \(v = k_2 R^2\) Assuming the constant \(k\) is the same for both the individual droplets and the combined droplet: \(v_1 = kr^2\) \(v = k(2)^{2/3}r^2\) Divide the two equations to find the relationship between the terminal velocities: \(\frac{v}{v_1} = (2)^{2/3}\)

Since \(v_1 = 5 \mathrm{cm/s}\), we can now find the terminal velocity of the combined droplet using the relationship we found in the previous step: \(v = v_1 (2)^{2/3}\) \(v = 5 \times (2)^{2/3} \mathrm{cm/s}\) Converting this to a value, we get: \(v = 5 \times (4)^{1/3} \mathrm{cm/s}\) Hence, the terminal velocity of the combined droplet is: (C) \(5 \times (4)^{1/3} \mathrm{cm/s}\)