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. \(2.5\times 10^{-2}\text{ Pa.s}\)
• B. \(1.1\times 10^{-1}\text{ Pa.s}\)
• C. \(1.1\times 10^{-2}\text{ Pa.s}\)
• D. \(2.5\times 10^{-1}\text{ Pa.s}\)

Hint:

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.

Answer 1

The Correct Option is B

Step 1: Recall Newton's law of viscosity.
When a flat object slides at constant velocity over a thin liquid layer: $F = \eta A \frac{v}{d}$, where $\eta$ is viscosity, $A$ is area, $v$ is speed, $d$ is layer thickness. At constant velocity, applied force equals viscous drag.
Step 2: Rearrange to isolate $\eta$.
\[ \eta = \frac{Fd}{Av} \]
Step 3: Convert all values to SI units.
$F = \frac{1}{3}$ N, $d = 3 \times 10^{-4}$ m, $A = 0.01$ m$^2$, $v = 0.09$ ms$^{-1}$.
Step 4: Substitute and simplify.
\[ \eta = \frac{\frac{1}{3} \times 3 \times 10^{-4}}{0.01 \times 0.09} = \frac{10^{-4}}{9 \times 10^{-4}} = \frac{1}{9} \approx 0.111 \text{ Pa.s} \]
Step 5: Express in scientific notation.
$\eta \approx 1.1 \times 10^{-1}$ Pa.s, about 100 times more viscous than water ($10^{-3}$ Pa.s).
Step 6: State the final answer.
\[ \boxed{\eta \approx 1.1 \times 10^{-1} \text{ Pa.s}} \]