MCQ
Eight identical small drops of water are falling down vertically through a medium, each with terminal velocity ' $V$ '. If they combine to form a single drop, then its terminal velocity will be
- A$6 V$
- ✓$4 V$
- C$3 V$
- D$5 V$
✓
Answer
Correct option: B.
$4 V$
(b) : Let radius of each drop $=r$
Terminal velocity, $V=\frac{2}{g} \frac{r^2}{\eta}(\rho-\sigma) g$........(i)
Let ' $R$ ' be the radius of big drop.
Volume of big drop = volume of 8 small drops
$
\Rightarrow \frac{4}{3} \pi R^3=8 \times \frac{4}{3} \pi r^3 \Rightarrow R=2 r
$
Let $V^{\prime \prime}$ be the terminal velocity of bigger drop,
$
\begin{aligned}
& V^{\prime}=\frac{2 R^2(\rho-\sigma)}{9 \eta}......(ii) \\
& \Rightarrow \frac{V^{\prime}}{V}=\frac{R^2}{r^2}=\frac{(2 r)^2}{r^2}=4 \Rightarrow V^{\prime}=4 V
\end{aligned}
$
Terminal velocity, $V=\frac{2}{g} \frac{r^2}{\eta}(\rho-\sigma) g$........(i)
Let ' $R$ ' be the radius of big drop.
Volume of big drop = volume of 8 small drops
$
\Rightarrow \frac{4}{3} \pi R^3=8 \times \frac{4}{3} \pi r^3 \Rightarrow R=2 r
$
Let $V^{\prime \prime}$ be the terminal velocity of bigger drop,
$
\begin{aligned}
& V^{\prime}=\frac{2 R^2(\rho-\sigma)}{9 \eta}......(ii) \\
& \Rightarrow \frac{V^{\prime}}{V}=\frac{R^2}{r^2}=\frac{(2 r)^2}{r^2}=4 \Rightarrow V^{\prime}=4 V
\end{aligned}
$
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