MHT CET · Physics · Mechanical Properties of Fluids
' \(n\) ' small drops of same size fall through air with constant velocity \(5 \mathrm{~cm} / \mathrm{s}\). They coalesce to form a big drop. The terminal velocity of the big drop is
- A \(7 \mathrm{n}^{2 / 3} \mathrm{~cm} / \mathrm{s}\)
- B \(5 \mathrm{n}^{2 / 3} \mathrm{~cm} / \mathrm{s}\)
- C \(3 \mathrm{n}^{2 / 3} \mathrm{~cm} / \mathrm{s}\)
- D \(9 \mathrm{n}^{2 / 3} \mathrm{~cm} / \mathrm{s}\)
Answer & Solution
Correct Answer
(B) \(5 \mathrm{n}^{2 / 3} \mathrm{~cm} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \text { Volume }=\frac{4}{3} \pi r^3 n=\frac{4}{3} \pi R^3 \\
& \therefore \mathrm{R}=\mathrm{n}^{1 / 3} \mathrm{r} \therefore \frac{\mathrm{R}}{\mathrm{r}}=\mathrm{n}^{\frac{1}{3}}
\end{aligned}
\)
Terminal velocity, \(\mathrm{v} \propto \mathrm{r}^2\)
\(
\begin{aligned}
& \frac{\mathrm{v}_2}{\mathrm{v}_1}=\left(\frac{\mathrm{r}_2}{\mathrm{r}_1}\right)^2=\left(\frac{\mathrm{R}}{\mathrm{r}}\right)^2=\mathrm{n}^{2 / 3} \\
& \therefore \mathrm{v}_2=\mathrm{n}^{2 / 3} \mathrm{v}_1=5 \mathrm{n}^{2 / 3} \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
\begin{aligned}
& \text { Volume }=\frac{4}{3} \pi r^3 n=\frac{4}{3} \pi R^3 \\
& \therefore \mathrm{R}=\mathrm{n}^{1 / 3} \mathrm{r} \therefore \frac{\mathrm{R}}{\mathrm{r}}=\mathrm{n}^{\frac{1}{3}}
\end{aligned}
\)
Terminal velocity, \(\mathrm{v} \propto \mathrm{r}^2\)
\(
\begin{aligned}
& \frac{\mathrm{v}_2}{\mathrm{v}_1}=\left(\frac{\mathrm{r}_2}{\mathrm{r}_1}\right)^2=\left(\frac{\mathrm{R}}{\mathrm{r}}\right)^2=\mathrm{n}^{2 / 3} \\
& \therefore \mathrm{v}_2=\mathrm{n}^{2 / 3} \mathrm{v}_1=5 \mathrm{n}^{2 / 3} \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
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