Question

A non-viscous fluid is flowing through a horizontal cylindrical tube of radius of cross-section $R$. Velocity of fluid at radial distance $r$ from the axis of tube is given by $V = {V_0}\left( {1 - \dfrac{{{r^2}}}{{{R^2}}}} \right)$. Coefficient of viscosity of the fluid is $\eta$. Force per unit area on the fluid layer parallel to velocity at radial distance $\dfrac{R}{2}$ from the axis is:

A) $\dfrac{{\eta {V_0}}}{R}$
B) $\dfrac{{2\eta {V_0}}}{R}$
C) $\dfrac{{\eta {V_0}}}{{2R}}$
D) $\dfrac{{\eta {V_0}}}{{4R}}$

Answer 1

To find the solution for the given question, we need to find the force per unit area on the fluid layer parallel to velocity at specific radial distance from the axis in which the velocity of the fluid is given. Before going to the solution we need to know the formula for the force of non-viscous fluid.

Formula used:
The force of non-viscous fluid,
$F = - \eta A\dfrac{{\partial v}}{{\partial r}}$
Where,
$\eta$ is the coefficient of viscosity of the fluid
$A$ is the area of the fluid
$v$ is the velocity of the fluid
$r$ is the radial distance of the fluid

Complete step by step answer:
The fluid which has no resistance or a zero resistance to the internal friction is called a non-viscous fluid. If the fluid has more resistance to flow then it is called a viscous fluid.
The given velocity of fluid at radial distance $r$ from the axis of tube is:
$\Rightarrow$ $V = {V_0}\left( {1 - \dfrac{{{r^2}}}{{{R^2}}}} \right)$
$\Rightarrow$ $F = - \eta A\dfrac{{\partial v}}{{\partial r}}$
$\Rightarrow$ $\dfrac{F}{A} = - \eta \dfrac{{\partial v}}{{\partial r}}$
$\therefore$ $\dfrac{F}{A} = - \eta \dfrac{\partial }{{\partial r}}\left( {{V_0}\left( {1 - \dfrac{{{r^2}}}{{{R^2}}}} \right)} \right)$
By integrating we get,
$\Rightarrow$ $\dfrac{F}{A} = - \eta {V_0}\left[ {\dfrac{{ - 2r}}{{{R^2}}}} \right]$
$\Rightarrow$ $\dfrac{F}{A} = \eta {V_0}\left( {\dfrac{{2r}}{{{R^2}}}} \right)$
$\therefore$ $\dfrac{F}{A} = \dfrac{{\eta {V_0}}}{R}$
For the value of $r = \dfrac{{R}}{2}$,
Therefore the force per unit area on the fluid layer parallel to velocity at radial distance $\dfrac{R}{2}$ from the axis is $\dfrac{{\eta {V_0}}}{R}$.

So therefore the option (A) is correct.

Note: The equation of continuity is in the form of the principle of conservation of mass. An incompressible non-viscous fluid then is called an ideal fluid. If the fluid is non viscous and there is no internal friction between the adjacent layers. The main advantage in viscosity is to maintain the performance of the machine and the automobiles by determining the thickness of the lubricating oil or a motor oil.