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

• A. $\frac{\text{VR}^2}{\text{r}^2}$
• B. $\frac{\text{Vr}^2}{\text{R}^2}$
• C. $\frac{VR}{r}$
• D. $\frac{\text{Vr}}{\text{R}}$

Hint:

Since volume is conserved, $R = n^{1/3}r$. Terminal velocity becomes $n^{2/3}V$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
When drops fall with constant velocity, it implies they have reached their terminal velocity.
Terminal velocity of a spherical drop depends on the square of its radius.
Step 2: Key Formula or Approach:
According to Stokes' Law, terminal velocity $v_t = \frac{2}{9}\frac{r^2(\rho-\sigma)g}{\eta}$.
Therefore, $v_t \propto \text{radius}^2$.
We use this proportionality to relate the velocities of the small drop and the big drop.
Step 3: Detailed Explanation:
For a small drop of radius $r$, terminal velocity is $V$: \[ V \propto r^2 \implies V = kr^2 \] For the big coalesced drop of radius $R$, let its terminal velocity be $V'$: \[ V' \propto R^2 \implies V' = kR^2 \] Taking the ratio of the two equations: \[ \frac{V'}{V} = \frac{kR^2}{kr^2} \] \[ \frac{V'}{V} = \frac{R^2}{r^2} \] Solving for the new terminal velocity $V'$: \[ V' = V \left(\frac{R^2}{r^2}\right) = \frac{VR^2}{r^2} \] Step 4: Final Answer:
The terminal velocity of the big drop is $\frac{\text{VR}^2}{\text{r}^2}$.