Question:

'$n$' small water drops of same size (radius $r$) fall through air with constant velocity V. They coalesce to form a big drop of radius R. The terminal velocity of the big drop is

Show Hint

Since volume is conserved, $R = n^{1/3}r$. Terminal velocity becomes $n^{2/3}V$.
Updated On: May 12, 2026
  • $\frac{\text{VR}^2}{\text{r}^2}$
  • $\frac{\text{Vr}^2}{\text{R}^2}$
  • $\frac{VR}{r}$
  • $\frac{\text{Vr}}{\text{R}}$
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The Correct Option is A

Solution and Explanation


Step 1: Concept

Terminal velocity $v_t$ is proportional to the square of the radius ($v_t \propto r^2$).

Step 2: Meaning

For small drops, $V = k r^2$. For the big drop, $V_{big} = k R^2$.

Step 3: Analysis

Taking the ratio: $V_{big} / V = R^2 / r^2$.

Step 4: Conclusion

$V_{big} = \frac{VR^2}{r^2}$. Final Answer: (A)
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