
Water is flowing in a pipe of diameter 4 cm with a velocity \[3\,m{s^{ - 1}}\]. The water then enters into a tube of diameter 2 cm. Find the velocity of water in the other pipe.
A. \[3\,m{s^{ - 1}}\]
B. \[6\,m{s^{ - 1}}\]
C. \[12\,m{s^{ - 1}}\]
D. \[8\,m{s^{ - 1}}\]
Answer
162.3k+ views
Hint: Before solving this question one should know about gravitational force, pressure, and surface tension and we need to use the equation of continuity that is, \[{A_1}{v_1} = {A_2}{v_2}\]. Here A is the area of cross-section and V is the velocity, using this formula we are going to find the solution to the problem.
Formula Used:
From the principle of continuity, the formula is,
\[{A_1}{v_1} = {A_2}{v_2}\]
Where, \[{A_1},{A_2}\] is the area and \[{v_1},{v_2}\] is the speed of water.
Complete step by step solution:
Consider water flowing in a pipe of diameter 4cm with a velocity \[3\,m{s^{ - 1}}\]. The water then enters into a tube of diameter 2cm. We need to find the velocity of water in the other pipe. Using the equation of continuity in the pipe at the entry and exit we have,
\[{A_1}{v_1} = {A_2}{v_2} \\ \]
\[\Rightarrow \dfrac{{{v_2}}}{{{v_1}}} = \dfrac{{{A_1}}}{{{A_2}}} = {\left( {\dfrac{{\pi {r_1}}}{{\pi {r_2}}}} \right)^2} \\ \]
\[\Rightarrow \dfrac{{{v_2}}}{{{v_1}}} = {\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right)^2} \\ \]
\[\Rightarrow {v_2} = {v_1}{\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right)^2}\]
Given that, \[{v_1} = 3\,m{s^{ - 1}}\], \[{d_1} = 4\,cm\] and \[{d_2} = 2\,cm\]
Therefore, \[{r_1} = 2\,cm\] and \[{r_2} = 1\,cm\]
Substitute the value on above equation, we get,
\[{v_2} = 3 \times {\left( {\dfrac{2}{1}} \right)^2}\]
\[\therefore {v_2} = 12\,m{s^{ - 1}}\]
Therefore, the velocity of water in the other pipe is \[12\,m{s^{ - 1}}\].
Hence, option C is the correct answer.
Note: The concept of continuity used in the above solution is applied to a flowing fluid in terms of mass conservation, that is, assume that a certain mass of fluid ‘m’ enters a pipe (domain) from one end and exits at the other end. According to the equation of continuity, the mass (m) entering the system should match the mass exiting the system.
Formula Used:
From the principle of continuity, the formula is,
\[{A_1}{v_1} = {A_2}{v_2}\]
Where, \[{A_1},{A_2}\] is the area and \[{v_1},{v_2}\] is the speed of water.
Complete step by step solution:
Consider water flowing in a pipe of diameter 4cm with a velocity \[3\,m{s^{ - 1}}\]. The water then enters into a tube of diameter 2cm. We need to find the velocity of water in the other pipe. Using the equation of continuity in the pipe at the entry and exit we have,
\[{A_1}{v_1} = {A_2}{v_2} \\ \]
\[\Rightarrow \dfrac{{{v_2}}}{{{v_1}}} = \dfrac{{{A_1}}}{{{A_2}}} = {\left( {\dfrac{{\pi {r_1}}}{{\pi {r_2}}}} \right)^2} \\ \]
\[\Rightarrow \dfrac{{{v_2}}}{{{v_1}}} = {\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right)^2} \\ \]
\[\Rightarrow {v_2} = {v_1}{\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right)^2}\]
Given that, \[{v_1} = 3\,m{s^{ - 1}}\], \[{d_1} = 4\,cm\] and \[{d_2} = 2\,cm\]
Therefore, \[{r_1} = 2\,cm\] and \[{r_2} = 1\,cm\]
Substitute the value on above equation, we get,
\[{v_2} = 3 \times {\left( {\dfrac{2}{1}} \right)^2}\]
\[\therefore {v_2} = 12\,m{s^{ - 1}}\]
Therefore, the velocity of water in the other pipe is \[12\,m{s^{ - 1}}\].
Hence, option C is the correct answer.
Note: The concept of continuity used in the above solution is applied to a flowing fluid in terms of mass conservation, that is, assume that a certain mass of fluid ‘m’ enters a pipe (domain) from one end and exits at the other end. According to the equation of continuity, the mass (m) entering the system should match the mass exiting the system.
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