Question

64 rain drops of the same radius are falling through air with a steady velocity of 0.5 cm s$^{-1}$. If the drops coalesce, the terminal velocity would be

• A. 1.25 cm s\(^{-1}\)
• B. 0.08 m s\(^{-1}\)
• C. 0.8 m s\(^{-1}\)
• D. 1.25 m s\(^{-1}\)

Hint:

When dealing with coalescing drops, remember that the volume adds up and the terminal velocity depends on the square of the radius. Use this to calculate the new terminal velocity after coalescence.

Answer 1

The Correct Option is B

When rain drops coalesce, their volume adds up, meaning the volume of the new drop will be the sum of the volumes of all the smaller drops. The volume of a sphere is given by: \[ V = \frac{4}{3} \pi r^3 \] Thus, if 64 drops coalesce, the volume of the new drop will be: \[ V_{\text{new}} = 64 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (64 r)^3 = 64 V \] The terminal velocity is proportional to the square of the radius of the drop, i.e., \[ v \propto r^2 \] Since the radius increases by a factor of \( \sqrt[3]{64} = 4 \), the new terminal velocity will be: \[ v_{\text{new}} = 4^2 \times v = 16 \times 0.5 \, \text{cm/s} = 8 \, \text{cm/s} \] Converting to m/s: \[ v_{\text{new}} = 8 \, \text{cm/s} = 0.08 \, \text{m/s} \] Thus, the terminal velocity of the coalesced drop is 0.08 m/s.