**A horizontal force of 1 N is required to move a metal plate of area 10 ^{-2} m^{2} 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.**Correct answer= 7.5 Ns/m ^{2}**

**Explaination::**

**Given **: F=1, A=10^{-2} m^{2, }v_{0 }=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/m ^{2}**