Answer
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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.
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.
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