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Physics-

General

Easy

Question

A thin movable plate is separated from two fixed plates $P subscript 1$ and $P subscript 2$ by two highly viscous liquids of coefficients of viscosity $n subscript 1$ and $n subscript 2$ as shown, where $n subscript 2 end subscript equals 9 n subscript 1 end subscript.$ Area of contact of movable plate with each fluid is same. If the distance between two fixed plates is ‘h’, then the distance ‘ $h subscript 1 end subscript$ ’ of movable plate from upper plate such that movable plate can be moved with a finite velocity by applying the minimum possible force on movable plate is ( assume only linear velocity distribution in each liquid).

$\frac{h}{4}$
$\frac{h}{2}$
$\frac{h}{6}$
$\frac{h}{3}$

The correct answer is: $fraction numerator h over denominator 4 end fraction$

Viscous force due to upper liquid $equals n subscript 1 end subscript A open parentheses fraction numerator v minus o over denominator h subscript 1 end subscript end fraction close parentheses$
Viscous force due to lower liquid = $n subscript 2 end subscript A open parentheses fraction numerator v minus o over denominator h minus h subscript 1 end subscript end fraction close parentheses$
If total force is minimum
$fraction numerator d over denominator d h subscript 1 end subscript end fraction open square brackets fraction numerator n subscript 1 end subscript over denominator h subscript 1 end subscript end fraction plus fraction numerator n subscript 2 end subscript over denominator h minus h subscript 1 end subscript end fraction close square brackets equals 0$

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