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:
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:
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Terminal velocity scales with the square of the radius; $v_t \propto r^2$.
Updated On: Jun 10, 2026
- $10~cms^{-1}$
- $5 \times 4^{1/3}~cms^{-1}$
- $5 \times 4^{2/3}~cms^{-1}$
- $2.5~cms^{-1}$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
Terminal velocity ($v_t$) $\propto r^2$.
Step 2: Analysis
If two drops of radius $r$ coalesce into one drop of radius $R$, then $\frac{4}{3}\pi R^3 = 2 \times \frac{4}{3}\pi r^3 \implies R = 2^{1/3}r$. The ratio of terminal velocities is $v_{new} / v_{old} = (R/r)^2 = (2^{1/3})^2 = 2^{2/3}$. $v_{new} = v_{old} \times 2^{2/3} = 5 \times 4^{1/3}$.
Step 3: Conclusion
The new terminal velocity is $5 \times 4^{1/3}~cms^{-1}$.
Final Answer: (B)
Terminal velocity ($v_t$) $\propto r^2$.
Step 2: Analysis
If two drops of radius $r$ coalesce into one drop of radius $R$, then $\frac{4}{3}\pi R^3 = 2 \times \frac{4}{3}\pi r^3 \implies R = 2^{1/3}r$. The ratio of terminal velocities is $v_{new} / v_{old} = (R/r)^2 = (2^{1/3})^2 = 2^{2/3}$. $v_{new} = v_{old} \times 2^{2/3} = 5 \times 4^{1/3}$.
Step 3: Conclusion
The new terminal velocity is $5 \times 4^{1/3}~cms^{-1}$.
Final Answer: (B)
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