Question

Two drops of same radius are falling through air with steady velocity of $5~cms^{-1}$. If the two drops coalesce, the terminal velocity would be:

• A. $10~cms^{-1}$
• B. $5 \times 4^{1/3}~cms^{-1}$
• C. $5 \times 4^{2/3}~cms^{-1}$
• D. $2.5~cms^{-1}$

Hint:

Terminal velocity scales with the square of the radius; $v_t \propto r^2$.

Answer 1

The Correct Option is B

Step 1: Note the situation.
Two rain drops of the same radius fall at a steady terminal speed of 5 cm/s. They merge into one bigger drop, and we want the new terminal speed.

Step 2: Recall how terminal speed depends on size.
For a small sphere falling through air, the terminal velocity is proportional to the square of its radius: $v_t \propto r^{2}$. Bigger drops fall faster.

Step 3: Conserve volume when they merge.
The two drops join into one, so the total volume stays the same. Volume of a sphere is $\tfrac{4}{3}\pi r^{3}$, so $\tfrac{4}{3}\pi R^{3} = 2 \times \tfrac{4}{3}\pi r^{3}$.

Step 4: Find the new radius.
Cancelling gives $R^{3} = 2 r^{3}$, so $R = 2^{1/3} r$.

Step 5: Find the speed ratio.
Since speed goes as radius squared, $\dfrac{v_{new}}{v_{old}} = \left(\dfrac{R}{r}\right)^{2} = (2^{1/3})^{2} = 2^{2/3}$.

Step 6: Compute the new speed.
So $v_{new} = 5 \times 2^{2/3}$. Note $2^{2/3} = 4^{1/3}$, giving $v_{new} = 5 \times 4^{1/3}$ cm/s. \[ \boxed{5 \times 4^{1/3} \ \text{cm s}^{-1}} \]