Question

The shear stress at a point in a liquid is found to be $0.03{\text{N/m}}^2$. The velocity gradient for the fluid flow is $0.15{\text{s}}^{-1}$. What will be its viscosity (in $\text{Poise}$) ?

• A. $20$
• B. $2$
• C. $0.2$
• D. $0.5$

Answer 1

The correct option is B $2$

$F=\eta A\frac{dv}{dz}...\left(i\right)$
where $F=$viscous force,
$A=$ area on which viscous force is acting tangentially
$\frac{dv}{dz}=$ velocity gradient
$\eta =$ coeffcient of viscosity.

From Eq.$\left(i\right)$,
$\eta =\frac{\frac{F}{A}}{\frac{dv}{dz}}$
where $\frac{F}{A}$ is the shear stress.
Given, $\frac{F}{A}=0.03{\text{N/m}}^2$
& $\frac{dv}{dz}=0.15{\text{s}}^{-1}$
$\Rightarrow \eta =\frac{0.03}{0.15}{\text{N.s/m}}^2=0.2{\text{Ns/m}}^2$
$\therefore \eta =2\text{Poise}$
[since 1 Poise $=0.1{\text{N.s/m}}^2$]