Question

Sixty four rain drops of radius $1\,\text{mm}$ each falling down with a terminal velocity of $10\,\text{cm/s}$ coalesce to form a bigger drop. The terminal velocity of the bigger drop is ___ cm/s.

Hint:

Under Stokes’ law, terminal velocity of a falling spherical body is proportional to the square of its radius.

Answer 1

Correct Answer: 80

To solve the problem, we need to determine the terminal velocity of a larger drop formed by the coalescence of 64 smaller droplets, each with a terminal velocity of 10 cm/s. 
1. **Volume Conservation**: The volume of the larger drop must equal the total volume of the smaller drops combined. The volume \(V\) of a sphere is given by the formula: \(V = \frac{4}{3}\pi r^3\). For a single small drop with radius \(r = 1 \, \text{mm} = 0.1 \, \text{cm}\), the volume is:
\(V_{\text{small}} = \frac{4}{3}\pi (0.1)^3 = \frac{4}{3000}\pi \, \text{cm}^3\).
The total volume of 64 small drops is:
\(V_{\text{total small}} = 64 \times \frac{4}{3000}\pi = \frac{256}{3000}\pi \, \text{cm}^3\).
For the larger drop with radius \(R\), the volume is:
\(V_{\text{large}} = \frac{4}{3}\pi R^3\).
By volume conservation, equate the two volumes:
\(\frac{4}{3}\pi R^3 = \frac{256}{3000}\pi\).
Simplifying, we find:
\(R^3 = \frac{256}{4000} = \frac{64}{1000}\).
Thus, \(R = \left(\frac{64}{1000}\right)^{\frac{1}{3}} = \frac{4}{10} = 0.4 \, \text{cm}\).
2. **Terminal Velocity**: Terminal velocity \(v\) of a drop is proportional to the square of its radius (for small Reynolds numbers, terminal velocity \(v \propto R^2\)). So, \(v_{\text{large}}\) is calculated as follows:
\(\frac{v_{\text{large}}}{v_{\text{small}}} = \left(\frac{R}{r}\right)^2\), where \(v_{\text{small}} = 10 \, \text{cm/s}\).
\(v_{\text{large}} = v_{\text{small}} \left(\frac{R}{r}\right)^2 = 10 \left(\frac{0.4}{0.1}\right)^2 = 10 \times 16 = 160 \, \text{cm/s}\).
3. **Validation**: The computed terminal velocity \(160 \, \text{cm/s}\) is outside the expected range of 80 to 80 cm/s mentioned in the problem. This suggests the given range might be incorrect or imply something further not covered in the solution process.
Therefore, the calculated terminal velocity of the bigger drop is 160 cm/s.