Question:
A cylindrical metal box whose flat surface has an area of \(0.01\text{ m}^2\) rests on liquid of \(0.3\text{ mm}\) thickness. If upon applying a horizontal force of magnitude \(\frac{1}{3}\text{ N}\), the box slides with a constant speed of \(0.09\text{ m s}^{-1}\), the coefficient of viscosity of the liquid is nearly
A cylindrical metal box whose flat surface has an area of \(0.01\text{ m}^2\) rests on liquid of \(0.3\text{ mm}\) thickness. If upon applying a horizontal force of magnitude \(\frac{1}{3}\text{ N}\), the box slides with a constant speed of \(0.09\text{ m s}^{-1}\), the coefficient of viscosity of the liquid is nearly
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For a body moving over a thin liquid layer, viscous force is given by
\[
F=\eta A\frac{v}{d}.
\]
Always convert thickness from mm to m before substitution.
Updated On: Jun 25, 2026
- \(2.5\times 10^{-2}\text{ Pa.s}\)
- \(1.1\times 10^{-1}\text{ Pa.s}\)
- \(1.1\times 10^{-2}\text{ Pa.s}\)
- \(2.5\times 10^{-1}\text{ Pa.s}\)
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The Correct Option is B
Solution and Explanation
Step 1: Use Newton's law of viscosity.
For viscous force, \[ F=\eta A\frac{v}{d} \] where \[ F=\text{applied force}, \] \[ \eta=\text{coefficient of viscosity}, \] \[ A=\text{area of contact}, \] \[ v=\text{constant speed}, \] and \[ d=\text{thickness of liquid layer} \]
Step 2: Rearrange the formula.
From \[ F=\eta A\frac{v}{d}, \] we get \[ \eta=\frac{Fd}{Av} \]
Step 3: Substitute the given values.
Given: \[ F=\frac{1}{3}\text{ N} \] \[ d=0.3\text{ mm}=0.3\times 10^{-3}\text{ m}=3\times 10^{-4}\text{ m} \] \[ A=0.01\text{ m}^2 \] \[ v=0.09\text{ m s}^{-1} \] Now, \[ \eta=\frac{\left(\frac{1}{3}\right)(3\times 10^{-4})}{(0.01)(0.09)} \] \[ \eta=\frac{10^{-4}}{9\times 10^{-4}} \] \[ \eta=\frac{1}{9} \] \[ \eta=0.111\text{ Pa.s} \] \[ \eta\approx 1.1\times 10^{-1}\text{ Pa.s} \]
Step 4: Final conclusion.
Therefore, the coefficient of viscosity of the liquid is nearly \[ \boxed{1.1\times 10^{-1}\text{ Pa.s}} \]
For viscous force, \[ F=\eta A\frac{v}{d} \] where \[ F=\text{applied force}, \] \[ \eta=\text{coefficient of viscosity}, \] \[ A=\text{area of contact}, \] \[ v=\text{constant speed}, \] and \[ d=\text{thickness of liquid layer} \]
Step 2: Rearrange the formula.
From \[ F=\eta A\frac{v}{d}, \] we get \[ \eta=\frac{Fd}{Av} \]
Step 3: Substitute the given values.
Given: \[ F=\frac{1}{3}\text{ N} \] \[ d=0.3\text{ mm}=0.3\times 10^{-3}\text{ m}=3\times 10^{-4}\text{ m} \] \[ A=0.01\text{ m}^2 \] \[ v=0.09\text{ m s}^{-1} \] Now, \[ \eta=\frac{\left(\frac{1}{3}\right)(3\times 10^{-4})}{(0.01)(0.09)} \] \[ \eta=\frac{10^{-4}}{9\times 10^{-4}} \] \[ \eta=\frac{1}{9} \] \[ \eta=0.111\text{ Pa.s} \] \[ \eta\approx 1.1\times 10^{-1}\text{ Pa.s} \]
Step 4: Final conclusion.
Therefore, the coefficient of viscosity of the liquid is nearly \[ \boxed{1.1\times 10^{-1}\text{ Pa.s}} \]
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