Question

A metal block of base area $0.10 \,m ^{2}$ is connected to a $0.010 \,kg$ mass via a string that passes over an ideal pulley (considered massless $\&$ frictionless). A liquid with a film thickness of $0.30\, mm$ is placed between the block and the table. When released the block moves to the right with a constant speed of $0.085\, m / s$. Find the coefficient of viscosity of the liquid. $\left( g =9.8\, m / s ^{2}\right)$

• A. $6 \times 10^{-2} Pa . s$
• B. $3.45 \times 10^{-5}$ Pa.s
• C. $3.45 \times 10^{-3} Pa . s$
• D. $3.45 \times 10^{-6}$ Pa.s

Answer 1

Answer: (C)

$\eta=\frac{\text { Shear stress }}{\text { Strain Rate }}$
$=\frac{ F / A }{ v / \ell}=\frac{ mg \ell}{ Av }$
$\ell$ is thickness of film
$\eta=\frac{0.01 \times 9.8 \times 0.3 \times 10^{-3}}{0.1 \times 0.085}$
$=3.45 \times 10^{-3} Pa - s$