Question

An ideal fluid is flowing in a tube of varying cross-section. At some point the radius of the tube is $r$ and the velocity of the flow is $v$ . The velocity of flow at another point, where radius is $\dfrac{r}{2}$, is:
(A) $\dfrac{v}{4}$
(B) $\dfrac{v}{2}$
(C) $2v$
(D) $4v$

Answer 1

First of all, we apply the technique that for an incompressible liquid, the volume of fluid passing through any given point per unit time inside the tube is equal. After that we will equate for the two given cases. We will manipulate accordingly and obtain the result.

Complete step by step solution:
In the given problem, we are supplied the following data:
The fluid which is flowing through a tube has varying cross-sectional area which means the radius of the tube is not uniform at the points.
We are given two points, one of whose radii is $r$ and the velocity of the flow is $v$ .
The other point has a radius of $\dfrac{r}{2}$ .
We are asked to find the velocity of the flow of the fluid through this point.
To begin with, we know that in fluid mechanics, the equation of continuity states that the rate at which mass enters a system is proportional to the rate at which the system leaves mass plus the mass accumulation within the system.
Let us proceed to solve the problem:
We know, when a fluid flows through a tube, we can apply the equation of continuity, which is given below:
${A_1}{v_1} = {A_2}{v_2}$ …… (1)
Where,
${A_1}$ indicates the area of the region where the radius is $r$ .
${v_1}$ indicates the velocity of the fluid through the region whose radius is $r$ .
${A_2}$ indicates the area of the region where the radius is $\dfrac{r}{2}$ .
${v_2}$ indicates the velocity of the fluid through the region whose radius is $\dfrac{r}{2}$.
Again, the area of the region where the radius is $r$ :
${A_1} = \pi {r^2}$
The area of the region where the radius is $\dfrac{r}{2}$ :
{A_2} = \pi {\left( {\dfrac{r}{2}} \right)^2} \\\ \Rightarrow {A_2} = \dfrac{{\pi {r^2}}}{4} \\\
Substituting the required values in the equation (1) we get:
{A_1}{v_1} = {A_2}{v_2} \\\ \Rightarrow \pi {r^2} \times v = \dfrac{{\pi {r^2}}}{4} \times {v_2} \\\ \Rightarrow v = \dfrac{1}{4} \times {v_2} \\\ \therefore {v_2} = 4v \\\
Hence, the velocity of the flow of the fluid through this point whose radius is $\dfrac{r}{2}$ found out to be $4v$ .

The correct option is (D).

Note: While solving the problem we should remember that the volume of the fluid passing through a given cross-section per unit time is always the same. Greater the radius of the cross-section, lower is the velocity of the fluid and vice-versa. The important point is that it only holds good, if the fluid is incompressible in nature.