Question:
A tiny water droplet falls through the air at a uniform terminal speed of \(1\) cm/s. If \(27\) such identical droplets coalesce to form a single larger drop, what will be the new terminal velocity of the resulting drop, assuming the air remains the same?
A tiny water droplet falls through the air at a uniform terminal speed of \(1\) cm/s. If \(27\) such identical droplets coalesce to form a single larger drop, what will be the new terminal velocity of the resulting drop, assuming the air remains the same?
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Terminal speed is proportional to the square of the radius, and the coalesced drop has three times the radius of one droplet.
Updated On: Nov 12, 2026
- \(3\) cm/s
- \(27\) cm/s
- \(81\) cm/s
- \(9\) cm/s
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
For a small sphere in a viscous fluid, the terminal velocity is \(v_{T}=\frac{2r^{2}(\rho-\sigma)g}{9\eta}\), so \(v_{T}\propto r^{2}\).
Step 2: Radius of the big drop
Volume is conserved: \(27\cdot\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi R^{3}\), so \(R=3r\).
Step 3: Calculation
\(\frac{v_{T}'}{v_{T}}=\frac{R^{2}}{r^{2}}=9\).
Step 4: New speed
\(v_{T}'=9\times 1=9\) cm/s, which is option (D).
Final Answer:
The new terminal velocity is 9 cm/s. \[ \boxed{\text{(D) }9\text{ cm/s}} \]
For a small sphere in a viscous fluid, the terminal velocity is \(v_{T}=\frac{2r^{2}(\rho-\sigma)g}{9\eta}\), so \(v_{T}\propto r^{2}\).
Step 2: Radius of the big drop
Volume is conserved: \(27\cdot\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi R^{3}\), so \(R=3r\).
Step 3: Calculation
\(\frac{v_{T}'}{v_{T}}=\frac{R^{2}}{r^{2}}=9\).
Step 4: New speed
\(v_{T}'=9\times 1=9\) cm/s, which is option (D).
Final Answer:
The new terminal velocity is 9 cm/s. \[ \boxed{\text{(D) }9\text{ cm/s}} \]
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