Question

125 small water drops of same size fall through air with constant velocity 4 cm/s. They coalesce to form a big drop. The terminal velocity of the big drop is ______.

• A. 0.5 m/s
• B. 1 m/s
• C. 1.5 m/s
• D. 2.5 m/s

Hint:

Shortcut for coalescing drops: The new terminal velocity is always $V_{new} = V_{old} \times (N^{2/3})$. Here, $N=125$, so $125^{2/3} = (5^3)^{2/3} = 5^2 = 25$. Multiply old speed by 25!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Terminal velocity ($v_t$) of a spherical drop is proportional to the square of its radius ($v_t \propto r^2$). When $n$ drops coalesce, the total volume is conserved.

Step 2: Formula Application:
Volume of big drop $V = n \times v \implies \frac{4}{3}\pi R^3 = n \times \frac{4}{3}\pi r^3$. So, $R = n^{1/3}r$. For $n=125$, $R = (125)^{1/3}r = 5r$.

Step 3: Explanation:
Since $v_t \propto r^2$, the new terminal velocity $V_T = v_t \times (\frac{R}{r})^2$. $V_T = 4 \text{ cm/s} \times (5)^2 = 4 \times 25 = 100$ cm/s. Converting to meters: $100$ cm/s $= 1$ m/s.

Step 4: Final Answer:
The terminal velocity of the big drop is 1 m/s.