MHT CET · Physics · Mechanical Properties of Fluids
A metal sphere of mass ' \(m\) ' and density ' \(\sigma_1\) ' falls with terminal velocity through a container containing liquid. The density of liquid is ' \(\sigma_2\) '. The viscous force acting on the sphere is
- A \(m g\left(1+\frac{\sigma_2}{\sigma_1}\right)\)
- B \(m g\left(1-\frac{\sigma_1}{\sigma_2}\right)\)
- C \(\operatorname{mg}\left(1-\frac{\sigma_2}{\sigma_1}\right)\)
- D \(\operatorname{mg}\left(1+\frac{\sigma_1}{\sigma_2}\right)\)
Answer & Solution
Correct Answer
(C) \(\operatorname{mg}\left(1-\frac{\sigma_2}{\sigma_1}\right)\)
Step-by-step Solution
Detailed explanation
Given: Mass of sphere \(=\mathrm{m}\), Density of sphere \(=\sigma_1\), Density of liquid \(=\sigma_2\).
\(\mathrm{At}=\mathrm{v}=\mathrm{v}_{\mathrm{t}}\),
Weight of sphere \((\mathrm{W})=\) Viscous Force \(\left(\mathrm{F}_{\mathrm{V}}\right)+\) Buoyant Force due to the medium \(\left(\mathrm{F}_{\mathrm{B}}\right)\)
\(\begin{aligned}
\Rightarrow \mathrm{W} & =\mathrm{F}_{\mathrm{V}}+\mathrm{F}_{\mathrm{B}} \\
\mathrm{Mg} & =\mathrm{F}_{\mathrm{V}}+\left(\sigma_2 \mathrm{~V}\right) \mathrm{g} \quad \ldots .(\because \mathrm{m}=\mathrm{D} . \mathrm{V}) \\
\therefore \quad \mathrm{F}_{\mathrm{V}} & =\mathrm{mg}-\left(\sigma_2 \mathrm{~V}\right) \mathrm{g} \\
& =\mathrm{mg}\left[1-\frac{\sigma_2 \mathrm{~V}}{\mathrm{~m}}\right] \\
& =\mathrm{mg}\left[1-\frac{\sigma_2 \mathrm{~V}}{\sigma_1 \mathrm{~V}}\right] \\
& =\mathrm{mg}\left[1-\frac{\sigma_2}{\sigma_1}\right]
\end{aligned}\)
\(\mathrm{At}=\mathrm{v}=\mathrm{v}_{\mathrm{t}}\),
Weight of sphere \((\mathrm{W})=\) Viscous Force \(\left(\mathrm{F}_{\mathrm{V}}\right)+\) Buoyant Force due to the medium \(\left(\mathrm{F}_{\mathrm{B}}\right)\)
\(\begin{aligned}
\Rightarrow \mathrm{W} & =\mathrm{F}_{\mathrm{V}}+\mathrm{F}_{\mathrm{B}} \\
\mathrm{Mg} & =\mathrm{F}_{\mathrm{V}}+\left(\sigma_2 \mathrm{~V}\right) \mathrm{g} \quad \ldots .(\because \mathrm{m}=\mathrm{D} . \mathrm{V}) \\
\therefore \quad \mathrm{F}_{\mathrm{V}} & =\mathrm{mg}-\left(\sigma_2 \mathrm{~V}\right) \mathrm{g} \\
& =\mathrm{mg}\left[1-\frac{\sigma_2 \mathrm{~V}}{\mathrm{~m}}\right] \\
& =\mathrm{mg}\left[1-\frac{\sigma_2 \mathrm{~V}}{\sigma_1 \mathrm{~V}}\right] \\
& =\mathrm{mg}\left[1-\frac{\sigma_2}{\sigma_1}\right]
\end{aligned}\)
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