Question

A horizontal force of 1 N is required to move a metal plate of area 10^-2 m² with a velocity of 2 ×10^-2 m/s, when it rests on a layer of oil 1.5 ×10^-3 m thick. Find the coefficient of viscosity of oil.

Answer 1

Correct answer= 7.5 Ns/m²

Explaination::

Given : F=1, A=10^-2 m^2, v₀ =2 × 10^-2 m/s, y=1.5 × 10^-3 m

Solution:

Velocity gradient, \(\frac{dv}{dy} = \frac{2 × 10^{-2}}{  1.5 × 10-3 }= \frac{40}{3} s^{-1}\)

Viscous Force, F= η\(\frac{dv}{dy}\)

Therefore coefficient of viscosity is

η =\( \frac{F}{A(dv/dy)}=\frac{1}{10^{−2}(40/3)}\) 

=\(\frac{30}{4}\)

= 7.5 Ns/m²