Q. A metal plate of area $10^{3}\ {cm}^{2}$ rests on a layer of oil 6 mm$Type\ equation\ here.$ thick. A tangential force of $10^{- 2\ }N$ is applied on it to move it with a constant velocity of ${6\ cms}^{- 1}$. The coefficient of viscosity of the liquid is
A) 0.1 poise
B) 0.5 poise
C) 0.7 poise
D) 0.9 poise
Explanation
(a) $ɳ = \frac{F}{A(dv/dy)}$
$\therefore\ ɳ = \frac{10^{- 2}}{\left( 10^{3} \times 10^{- 4} \right)\left( \frac{6 \times 10^{- 2}}{6 \times 10^{- 3}} \right)}$
$\ = \frac{10^{- 2} \times 6 \times 10^{- 3}}{10^{- 1} \times 6 \times 10^{- 2}}$
$= 10^{- 2}{Nsm}^{2} = 0.1\ poise$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.