Question

In a horizontal pipe of a non-uniform cross-section, water flows with a velocity of \[1m{s^{ - 1}}\] at a point where the diameter of the pipe is \[20\] cm. The velocity of water in \[\left( {m{s^{ - 1}}} \right)\] at a point where the diameter of the pipe is \[5\] cm is:
A) $\left( {\text{A}} \right){\text{ 64}}$
B) $\left( {\text{B}} \right){\text{ 24}}$
C) $\left( {\text{C}} \right){\text{ 8}}$
D) $\left( {\text{D}} \right){\text{ 32}}$
E) $\left( {\text{E}} \right){\text{ 16}}$

Answer 1

Hint:Viscosity of fluid: It is the property of the fluid by virtue of which an internal force of friction comes into action when the fluid is in motion and which opposes the relative motion between different layers of the fluid.

Streamline flow: It is the specific flow of fluid such that each particle of the liquid passing a given point moves along the same path and has the same velocity as its predecessor.

Equation of continuity: It defines that during a streamline flow of the non-viscous and incompressible fluid through a pipe of varying cross-section, the product of the area of cross-section of the pipe and the normal fluid velocity inside the pipe (av) remains constant throughout the flow.

Bernoulli’s principle: It states that the sum of pressure energy and potential energy per unit volume of an incompressible, non-viscous fluid in a streamlined irrotational flow remains constant along a streamline.

Formula used:
Equation of continuity, ${{\text{a}}_1}{{\text{v}}_1} = {{\text{a}}_2}{{\text{v}}_2}$
Here \[{a_1},{\text{ }}{a_2} = \] area of cross-section of the pipe, \[{v_1},{\text{ }}{v_2} = \] velocity of the fluid

Complete step-by-step answer:
It is given that in the question, Diameter of the pipe at the first point \[\left( {{d_1}} \right) = 20{\text{ }}cm = 0.2{\text{ }}m\],
Radius of the pipe at the first point \[\left( {{r_1}} \right) = \dfrac{{{{\text{d}}_1}}}{2}\]
$ = \dfrac{{0.2}}{2}$
$ = 0.1m$
Diameter of the pipe at the second point \[\left( {{d_2}} \right) = 5cm = 0.05m\]
 Radius of the pipe at the second point $\left( {{r_2}} \right) = \dfrac{{{{\text{d}}_2}}}{2}$
$ = \dfrac{{0.05}}{2}$
Let us divide the terms we get,
$ = 0.025m$
Velocity of water at the first point \[\left( {{v_1}} \right) = 1m{s^{ - 1}}\],
Now using the equation of continuity, ${{\text{a}}_1}{{\text{v}}_1} = {{\text{a}}_2}{{\text{v}}_2}$
$ \Rightarrow \pi {{\text{r}}_1}^2{{\text{v}}_1} = \pi {{\text{r}}_2}^2{{\text{v}}_2}$
Cancel the same terms and we get
$ \Rightarrow {{\text{r}}_1}^2{{\text{v}}_1} = {{\text{r}}_2}^2{{\text{v}}_2}$
Taking \[{v^2}\] as LHS and remaining terms as in the RHS we get,
$ \Rightarrow {{\text{v}}_2} = {\left( {\dfrac{{{{\text{r}}_1}}}{{{{\text{r}}_2}}}} \right)^2}{{\text{v}}_1}$
Substitute the values of the RHS as we have to find out in the previous steps,
$ \Rightarrow {{\text{v}}_2} = {\left( {\dfrac{{0.1}}{{0.025}}} \right)^2} \times {\text{1}}$
On some simplification we get,
$ \Rightarrow {{\text{v}}_2} = {4^2}$
On squaring the RHS we get,
$ \Rightarrow {{\text{v}}_2} = 16m{s^{ - 1}}$
$\therefore $The velocity of water in ${{\text{v}}_2} = 16m{s^{ - 1}}$ at a point.

Hence, the correct option is E

Note:Bernoulli’s equation is derived from the assumption that there is no loss of energy due to friction among fluid particles.
Some of the kinetic energy converted into heat energy due to the work done against the internal energy or friction or the viscous forces.
Angular momentum of the fluid is not taken into consideration in Bernoulli’s equation.