Q96 / 100 · BITSAT 2023 (Memory Based Paper 2) · Physics
Eight equal drops of water are falling through air with a steady speed of $10 \mathrm{~cm} / \mathrm{s}$. If the drops coalesce, the new velocity is:
- A$10 \mathrm{~cm} / \mathrm{s}$
- Correct answerB$40 \mathrm{~cm} / \mathrm{s}$
- C$16 \mathrm{~cm} / \mathrm{s}$
- D$5 \mathrm{~cm} / \mathrm{s}$
Solution
Let $\mathrm{r}$ be the radius of small drops of water.
$\mathrm{R}=$ radius of big drop formed as volume remain same.
$\therefore 8 . \frac{4}{3} \pi \mathrm{r}^3-\frac{4}{3} \pi \mathrm{R}^3 \Rightarrow \mathrm{R}=2 \mathrm{r}$
Terminal velocity,
$\begin{aligned}<br />& \mathrm{v}_{\mathrm{T}}=\frac{2}{9 \eta}(\rho-\sigma) \mathrm{r}^2 \mathrm{~g} \\<br />& \therefore \mathrm{v}_{\mathrm{T}} \propto \mathrm{r}^2 \\<br />& \therefore \frac{\mathrm{v}_1}{\mathrm{v}_2}=\left(\frac{\mathrm{r}}{\mathrm{R}}\right)^2 \\<br />& \Rightarrow \frac{10}{\mathrm{v}_2}=\left(\frac{1}{2}\right)^2\left(\because \mathrm{v}_1=10 \mathrm{~cm} / \mathrm{s} \text { given }\right) \\<br />& \quad \Rightarrow \mathrm{v}_2=40 \mathrm{~cm} / \mathrm{s}<br />\end{aligned}$
$\mathrm{R}=$ radius of big drop formed as volume remain same.
$\therefore 8 . \frac{4}{3} \pi \mathrm{r}^3-\frac{4}{3} \pi \mathrm{R}^3 \Rightarrow \mathrm{R}=2 \mathrm{r}$
Terminal velocity,
$\begin{aligned}<br />& \mathrm{v}_{\mathrm{T}}=\frac{2}{9 \eta}(\rho-\sigma) \mathrm{r}^2 \mathrm{~g} \\<br />& \therefore \mathrm{v}_{\mathrm{T}} \propto \mathrm{r}^2 \\<br />& \therefore \frac{\mathrm{v}_1}{\mathrm{v}_2}=\left(\frac{\mathrm{r}}{\mathrm{R}}\right)^2 \\<br />& \Rightarrow \frac{10}{\mathrm{v}_2}=\left(\frac{1}{2}\right)^2\left(\because \mathrm{v}_1=10 \mathrm{~cm} / \mathrm{s} \text { given }\right) \\<br />& \quad \Rightarrow \mathrm{v}_2=40 \mathrm{~cm} / \mathrm{s}<br />\end{aligned}$
Hence, the correct answer is option: 'B'.