TS EAMCET · Physics · Mechanical Properties of Fluids
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
- A \(2.5 \times 10^{-2} \frac{\mathrm{Ns}}{\mathrm{m}^2}\)
- B \(0.25 \times 10^{-2} \frac{\mathrm{Ns}}{\mathrm{m}^2}\)
- C \(5 \times 10^{-2} \frac{\mathrm{Ns}}{\mathrm{m}^2}\)
- D \(0.5 \times 10^{-2} \frac{\mathrm{Ns}}{\mathrm{m}^2}\)
Answer & Solution
Correct Answer
(A) \(2.5 \times 10^{-2} \frac{\mathrm{Ns}}{\mathrm{m}^2}\)
Step-by-step Solution
Detailed explanation
\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…
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