A cylindrical metal box whose flat surface has an area of $0.01 \mathrm{~m}^2$ rests on liquid of $0.3…
A cylindrical metal box whose flat surface has an area of $0.01 \mathrm{~m}^2$ rests on liquid of $0.3 \mathrm{~mm}$ thickness. If upon applying a horizontal force of magnitude $\frac{1}{3} \mathrm{~N}$, the box slides with a constant speed of $0.09 \mathrm{~ms}^{-1}$, the coefficient of viscosity of the liquid is nearly
$2.5 \times 10^{-2} \mathrm{Pa.s}$
$1.1 \times 10^{-1} \mathrm{Pa.s}$
$1.1 \times 10^{-2} \mathrm{Pa.s}$
$2.5 \times 10^{-1} \mathrm{~Pa} . \mathrm{s}$
Solution
We know that
$
\mathrm{F}_{\mathrm{drag}}=\eta \mathrm{A} \frac{\mathrm{dv}}{\mathrm{dx}}
$
For metal box to be moving with constant velocity
$
\begin{aligned}
& \mathrm{F}_{\mathrm{drag}}=\mathrm{F} \\
& \Rightarrow \eta \mathrm{A} \frac{\mathrm{dv}}{\mathrm{dx}}=\mathrm{F} \\
& \Rightarrow \eta=\mathrm{F} \frac{\mathrm{dx}}{\mathrm{Adx}}=\frac{\frac{1}{3} \times 0.3 \times 10^{-3}}{0.01 \times(0.09-0)} \\
& =1.1 \mathrm{PaS}
\end{aligned}
$