Question

Water is moving with a speed of \[5.18\,m{s^{ - 1}}\] through a pipe with a cross-section area of \[4.2\,c{m^2}\]. The water gradually descends \[9.66\,m\] as the cross-sectional area of pipe is increased to \[7.60\,c{m^2}\]. Find the speed of the flow of water at a lower level.
A. \[2.86\,m{s^{ - 1}}\]
B. \[3.0\,m{s^{ - 1}}\]
C. \[3.82\,m{s^{ - 1}}\]
D. \[5.7\,m{s^{ - 1}}\]

Answer 1

Hint: Before going to solve this question let us understand the streamlined flow. The streamlined flow is also referred to as laminar flow. When there is no turbulence and fluctuations in velocity then it is called streamlined flow. It is a path traced by a particle in a flow.

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 a water that is moving with a speed of \[{v_1} = 5.18\,m{s^{ - 1}}\]through a pipe with a cross-section area of \[{A_1} = 4.2\,c{m^2}\]. The water gradually descends \[9.66\,m\] as the cross-sectional area of pipe is increased to \[{A_2} = 7.60\,c{m^2}\]. We need to find the speed of the flow of water at a lower level.

By the principle of continuity, the formula is,
\[{A_1}{v_1} = {A_2}{v_2}\]
\[\Rightarrow {v_2} = \dfrac{{{A_1}{v_1}}}{{{A_2}}}\]
Substitute the value of area and speed of flow of water in above equation, we obtain as follows,
\[{v_2} = \dfrac{{4.2 \times 5.18}}{{7.60}} \\ \]
\[\therefore {v_2} = 2.86\,m{s^{ - 1}}\]
Therefore, the speed of the flow of water at lower level is \[2.86\,m{s^{ - 1}}\].

Hence, option A is the correct answer.

Note: In this problem it is important to remember the equation for the continuity of motion. Since the water is flowing in a pipe moving at a certain speed, we are going to find the flow of water at the lower level.