MHT CET · Physics · Mechanical Properties of Fluids
A spherical metal ball of radius ' \(r\) ' falls through viscous liquid with velocity ' \(\mathrm{V}\) '. Another metal ball of same material but of radius \(\left(\frac{\mathrm{r}}{3}\right)\) falls through same liquid, then its terminal velocity will be
- A \(\frac{\mathrm{V}}{3}\)
- B \(\frac{\mathrm{V}}{4}\)
- C \(\frac{\mathrm{V}}{6}\)
- D \(\frac{\mathrm{V}}{9}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{V}}{9}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{v}=\frac{2 \mathrm{r}^2(\rho-\sigma) \mathrm{g}}{9 \Delta}\)
\(\rho, \sigma\) and \(\eta\) are constant
\(\Rightarrow \mathrm{v} \propto \mathrm{r}^2\)
The ratio of terminal velocities is
\(
\begin{aligned}
\frac{\mathrm{v}_1}{\mathrm{v}_2} & =\frac{\mathrm{r}_1^2}{\mathrm{r}_2^2} \\
\frac{\mathrm{v}_1}{\mathrm{v}_2} & =\frac{\mathrm{r}^2}{\left(\frac{\mathrm{r}}{3}\right)^2} \\
\frac{\mathrm{v}_1}{\mathrm{v}_2} & =9 \\
\therefore \quad \mathrm{v}_2 & =\frac{\mathrm{v}}{9}
\end{aligned}
\)
\(\rho, \sigma\) and \(\eta\) are constant
\(\Rightarrow \mathrm{v} \propto \mathrm{r}^2\)
The ratio of terminal velocities is
\(
\begin{aligned}
\frac{\mathrm{v}_1}{\mathrm{v}_2} & =\frac{\mathrm{r}_1^2}{\mathrm{r}_2^2} \\
\frac{\mathrm{v}_1}{\mathrm{v}_2} & =\frac{\mathrm{r}^2}{\left(\frac{\mathrm{r}}{3}\right)^2} \\
\frac{\mathrm{v}_1}{\mathrm{v}_2} & =9 \\
\therefore \quad \mathrm{v}_2 & =\frac{\mathrm{v}}{9}
\end{aligned}
\)
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