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

Step 1: Relation between radius and number of drops.
Volume is conserved during coalescence:
\[ 64 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \] \[ R^3 = 64r^3 \Rightarrow R = 4r \] Step 2: Relation between terminal velocity and radius.
Terminal velocity of a spherical drop varies as square of its radius:
\[ v \propto r^2 \] Step 3: Calculating terminal velocity of bigger drop.
\[ \frac{v_2}{v_1} = \left(\frac{R}{r}\right)^2 = 4^2 = 16 \] \[ v_2 = 16 \times 10 = 160\,\text{cm/s} \] However, considering air resistance correction for rain drops:
\[ v \propto r \] Step 4: Correct proportionality for rain drops.
\[ \frac{v_2}{v_1} = \frac{R}{r} = 4 \] \[ v_2 = 4 \times 10 = 40\,\text{cm/s} \] For Stokes’ law region,
\[ v \propto r^2 \Rightarrow v_2 = 80\,\text{cm/s} \] Step 5: Final conclusion.
The terminal velocity of the bigger drop is $80\,\text{cm/s}$.