Question

An ideal fluid flows (laminar flow) through a pipe of non-uniform diameter. The maximum and minimum diameters of the pipe are 6.4 cm and 4.8 cm, respectively. The ratio of minimum and maximum velocities of fluid in this pipe is:
A. \[\dfrac{9}{{16}}\]
B. \[\dfrac{{81}}{{256}}\]
C. \[\dfrac{3}{4}\]
D. \[\dfrac{{\sqrt 3 }}{2}\]

Answer 1

Hint: Calculate the maximum and minimum area of the pipe using given diameters. The velocity of the flow is maximum when the area of cross-section is minimum. Use the continuity equation for the flow of fluid through a pipe.

Complete step by step answer:

Equation of continuity.
\[{A_{\max }}{v_{\min }} = {A_{\min }}{v_{\max }}\]

Here, \[{v_{\min }}\] is the minimum velocity and \[{v_{\max }}\] is the maximum velocity.

Complete step by step answer:

We know that the circular cross-section of pipe of radius r has area, \[A = \pi {r^2}\]. Therefore, the maximum area of the pipe is,
\[{A_{\max }} = \pi r_{\max }^2\]
\[ \Rightarrow {A_{\max }} = \pi {\left( {\dfrac{{{d_{\max }}}}{2}} \right)^2}\]
\[ \Rightarrow {A_{\max }} = \dfrac{{\pi d_{\max }^2}}{4}\] …… (1)

Also, the minimum area of the pipe is,
\[{A_{\min }} = \dfrac{{\pi d_{\min }^2}}{4}\] …… (2)

According to equation of continuity in the laminar flow, we have,
\[{A_{\max }}{v_{\min }} = {A_{\min }}{v_{\max }}\]

Here, \[{v_{\min }}\] is the minimum velocity and \[{v_{\max }}\] is the maximum velocity.

The above equation implies that the velocity of the flow is maximum through the minimum area of cross-section of the pipe.

We rearrange the above equation as follows,
\[\dfrac{{{v_{\min }}}}{{{v_{\max }}}} = \dfrac{{{A_{\min }}}}{{{A_{\max }}}}\]

Use equation (1) and (2) to rewrite the above equation as follows,
\[\dfrac{{{v_{\min }}}}{{{v_{\max }}}} = \dfrac{{\dfrac{{\pi d_{\min }^2}}{4}}}{{\dfrac{{\pi d_{\max }^2}}{4}}}\]
\[ \Rightarrow \dfrac{{{v_{\min }}}}{{{v_{\max }}}} = {\left( {\dfrac{{{d_{\min }}}}{{{d_{\max }}}}} \right)^2}\]

Substitute 4.8 cm for \[{d_{\min }}\] and 6.4cm for \[{d_{\max }}\] in the above equation.
\[\dfrac{{{v_{\min }}}}{{{v_{\max }}}} = {\left( {\dfrac{{4.8\,cm}}{{6.4\,cm}}} \right)^2}\]
\[ \Rightarrow \dfrac{{{v_{\min }}}}{{{v_{\max }}}} = {\left( {0.75} \right)^2}\]
\[ \Rightarrow \dfrac{{{v_{\min }}}}{{{v_{\max }}}} = \dfrac{9}{{16}}\]

Note:The equation of continuity can be applied to any flow on the condition the flow should cover the whole area of cross-section through which the liquid is flowing. The equation of continuity can be used to calculate the velocity of the water flowing through the lower opening of the water tank.