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A metal cube of side $10 \mathrm{~cm}$ rests on a film of a liquid of thickness $0.2 \mathrm{~mm}$. If upon applying a horizontal force $\overrightarrow{\mathrm{F}}$ of magnitude $0.1 \mathrm{~N}$ the cube slides with a constant speed of $0.08 \mathrm{~m} / \mathrm{s}$, then the coefficient of viscosity is
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$2.5 \times 10^{-2} \frac{\mathrm{Ns}}{\mathrm{m}^2}$
$\begin{aligned} & A s \mathrm{v}=\text { constant } \\ & \text { So, } \mathrm{F}=\mathrm{F}_{\mathrm{drag}} \\ & \Rightarrow \quad 0.1=\eta \mathrm{A} \frac{\mathrm{dv}}{\mathrm{dx}} \\ & \Rightarrow \quad 0.1=\eta \times(0.1)^2 \times\left(\frac{0.08-0}{0.2 \times 10^{-3}}\right) \\ & \Rightarrow \quad \eta=2.5 \times 10^{-2} \frac{\mathrm{Ns}}{\mathrm{m}^2}\end{aligned}$
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