Question

Luids of viscosities 1  =0.1 N.s/ms and 2  =0.15 N.s/m2 are contained between two plates (each plate 1

Answer 1

Answer:

a) 1 / 3 N

b) 5/3 m/s

Explanation:

For constant speed V at the interface of the two fluids the net force is zero.

Hence, F top = F bottom thus the shear stresses at top and bottom must be the same.

$t_{top} = t_{bottom}\\where\\t = u.\frac{dU}{dt} \\Hence\\(\frac{V - 1}{h_{2} })*u_{top} = (\frac{V}{h_{1} })*u_{bottom} \\\\0.3V = 0.5\\\\V = 5/3 m/s$

For force of top plate

$F_{top} = u_{top} * \frac{V-U}{h_{2} }*A=( 0.15)*(\frac{5/3 -1}{0.3 } )*(1)\\F_{top} = \frac{1}{3} N$