Question

A liquid flows in a tube from left to right as shown in figure \[{A_1}\]​ and \[{A_2}\] are the cross-section of the portion of the tube as shown. Then the ratio of speed \[{v_1}/{v_2}\] will be:

A. \[\dfrac{{{A_1}}}{{{A_2}}}\]
B. \[\dfrac{{{A_2}}}{{{A_1}}}\]
C. \[\sqrt {\dfrac{{{A_2}}}{{{A_1}}}} \]
D. \[\sqrt {\dfrac{{{A_1}}}{{{A_2}}}} \]

Answer 1

Hint: Using the conservation of mass, the mass of liquid inflow into the tube will be equal to the mass of liquid outflow. The volume of liquid flowing in a tube per unit time is called the flow rate of the liquid. The flow rate of liquid is equal to the product of the speed of flow and the area of cross-section of the tube.

Formula used:
\[Q = Av\]
where Q is the rate of flow, A is the area of cross-section of the tube and v is the speed of flow.

Complete step by step solution:

Image: The cross-section of tube

From the given image which describes the flow of the liquid in the tube, the area of cross-section at two points are \[{A_1}\] and \[{A_2}\]. The speed of flow of liquid corresponding to the area of cross-section \[{A_1}\] is \[{v_1}\] and the speed of flow of liquid corresponding to the area of cross-section \[{A_2}\] is \[{v_2}\]. The flow rate of the liquid at the first point is,
\[{Q_1} = {A_1}{v_1}\]
The flow rate of the liquid at the second point is,
\[{Q_2} = {A_2}{v_2}\]

According to the conservation of mass, the amount of liquid entering into the tube per unit time will be equal to the amount of liquid flowing out of the tube per unit time, i.e. the rate of flow at the first point is equal to the rate of flow of the liquid at the second point.
\[{Q_1} = {Q_2}\]
Putting the expression for the flow rate at respective points, we get
\[{A_1}{v_1} = {A_2}{v_2}\]
\[\therefore \dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{{A_2}}}{{{A_1}}}\]

Therefore, the correct option is B.

Note: The flow rate of the liquid is independent of the dimensional feature of the pipe through which it is flowing. The flow rate is constant at every point in the pipe.