MHT CET · Physics · Mechanical Properties of Fluids
The speed of a ball of radius \(2 \mathrm{~cm}\) in a viscous liquid is \(20 \mathrm{~cm} / \mathrm{s}\). What will be the speed of a ball of radius \(1 \mathrm{~cm}\) in same liquid?
- A \(10 \mathrm{~cm} / \mathrm{s}\)
- B \(4 \mathrm{~cm} / \mathrm{s}\)
- C \(5 \mathrm{~cm} / \mathrm{s}\)
- D \(8 \mathrm{~cm} / \mathrm{s}\)
Answer & Solution
Correct Answer
(C) \(5 \mathrm{~cm} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
Terminal velocity \(\mathrm{V} \propto \mathrm{r}^2\)
\(
\begin{aligned}
& \therefore \frac{\mathrm{V}_2}{\mathrm{~V}_1}=\left(\frac{\mathrm{r}_2}{\mathrm{r}_1}\right)^2=\left(\frac{1}{2}\right)^2=\frac{1}{4} \\
& \therefore \mathrm{V}_2=\frac{\mathrm{V}_1}{4}=\frac{20}{4}=5 \mathrm{~cm} / \mathrm{s}
\end{aligned}
\)
\(
\begin{aligned}
& \therefore \frac{\mathrm{V}_2}{\mathrm{~V}_1}=\left(\frac{\mathrm{r}_2}{\mathrm{r}_1}\right)^2=\left(\frac{1}{2}\right)^2=\frac{1}{4} \\
& \therefore \mathrm{V}_2=\frac{\mathrm{V}_1}{4}=\frac{20}{4}=5 \mathrm{~cm} / \mathrm{s}
\end{aligned}
\)
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