Water is flowing on a horizontal fixed surface, such that its flow velocity varies with y (vertical direction) as v=k ( 2 y^ 2 a^ 2 - y^ 3 a^ 3 ). If coefficient of viscosity for water is , what will be shear stress

Correct answer: A. ( k)/(a) Newton's law of viscosity, F= A (d v)/(d y) array l Stress =(F)/(A)= ((d v)/(d y))= ( 4 y a^ 2 - 3 y^ 2 a^ 3 ) At y=a, stress = k((4)/(a)-(3)/(a))=( k)/(a) array

BITSAT 2015

Exam: BITSAT
Year: 2015
Subject: Physics
Chapter: Mechanical Properties of Fluids

BITSAT 2015 · Physics · Mechanical Properties of Fluids

Water is flowing on a horizontal fixed surface, such that its flow...

Water is flowing on a horizontal fixed surface, such that its flow velocity varies with

y

(vertical direction)

as v=k

\left(\frac{2 y^{2}}{a^{2}}-\frac{y^{3}}{a^{3}}\right).

If coefficient of viscosity for water is

\eta

, what will be shear stress between layers of water at

y=a.

Options

A $\frac{\eta k}{a}$
B $\frac{\eta}{k a}$
C $\frac{\eta a}{k}$
DNone of these

Correct answer

A. $\frac{\eta k}{a}$

Step-by-step solution

Newton's law of viscosity,

F=\eta A \frac{d v}{d y}

\begin{array}{l} \text{ Stress }=\frac{F}{A}=\eta \left(\frac{d v}{d y}\right)=\etak\left(\frac{4 y}{a^{2}}-\frac{3 y^{2}}{a^{3}}\right)\\ \text{ At } y=a, \text{ stress }=\eta k\left(\frac{4}{a}-\frac{3}{a}\right)=\frac{\eta k}{a} \end{array}

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