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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.

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Correct answer= 7.5 Ns/m2

Explaination::

Given : F=1, A=10-2 m2, v=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

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