MHT CET · Physics · Mechanical Properties of Fluids
Two rain drops of same radius \(r\) falling with terminal velocity \(V\) merge and form bigger drop with radius \(R\), terminal velocity is
- A \(\frac{V R^2}{r^2}\)
- B \(\frac{V R}{r}\)
- C \(\frac{V r^2}{R^2}\)
- D \(\frac{2 V R}{r}\)
Answer & Solution
Correct Answer
(A) \(\frac{V R^2}{r^2}\)
Step-by-step Solution
Detailed explanation
Expression for terminal velocity can be obtained by balancing weight of the droplet with viscous retardation force and the buoyancy force.
Terminal velocity is given by \(v=\frac{2}{9}\left(\frac{r^2(\rho-\sigma)}{\eta}\right)\)
\(\therefore v \propto r^2\)
Now, let \(V^{\prime}\) be the terminal velocity of the merged droplet
\(\begin{aligned}
& \therefore \frac{V}{V^{\prime}}=\frac{r^2}{R^2} \\
& \Rightarrow V^{\prime}=\frac{V R^2}{r^2}
\end{aligned}\)
Terminal velocity is given by \(v=\frac{2}{9}\left(\frac{r^2(\rho-\sigma)}{\eta}\right)\)
\(\therefore v \propto r^2\)
Now, let \(V^{\prime}\) be the terminal velocity of the merged droplet
\(\begin{aligned}
& \therefore \frac{V}{V^{\prime}}=\frac{r^2}{R^2} \\
& \Rightarrow V^{\prime}=\frac{V R^2}{r^2}
\end{aligned}\)
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