Problem 90 - Master Resource Book in JEE Main Physics

Question

An ideal fluid flows through a pipe of circular cross-section made of two sections with diameters \(2.5\) \(\mathrm{cm}\) and \(3.75 \mathrm{~cm}\). the ratio of the velocities in the two pipes is (a) \(9: 4\) (b) \(3: 2\) (c) \(\sqrt{3} ; \sqrt{2}\) (d) \(\sqrt{2} ; \sqrt{3}\)

Step-by-Step Solution

Answer

The ratio of the velocities in the two pipe sections is 9:4.

Step 1: Understand the Problem

We are given an ideal fluid flowing through two sections of a pipe with different diameters. We need to find the ratio of velocities in these sections using the principle of continuity for incompressible fluids.

Step 2: Recall the Principle of Continuity

The principle of continuity states that for an incompressible fluid flowing through a pipe, the product of the cross-sectional area and velocity of the fluid is constant across different sections of the pipe. Mathematically, this is expressed as: \(A_1 \cdot v_1 = A_2 \cdot v_2\), where \(A\) is the cross-sectional area and \(v\) is the velocity.

Step 3: Calculate Cross-Sectional Areas

Calculate the cross-sectional areas of the two sections using the formula for the area of a circle, \(A = \pi r^2\). For the first section, with diameter \(2.5\, \mathrm{cm}\), the radius is \(1.25\, \mathrm{cm}\). For the second section, with diameter \(3.75\, \mathrm{cm}\), the radius is \(1.875\, \mathrm{cm}\). Thus, \(A_1 = \pi (1.25)^2\) and \(A_2 = \pi (1.875)^2\).

Step 4: Simplify the Area Ratio

Calculate the ratio \(\frac{A_2}{A_1}\):\[\frac{A_2}{A_1} = \frac{(1.875)^2}{(1.25)^2} = \frac{3.515625}{1.5625} = 2.25\]This implies that \(\frac{v_1}{v_2}\) is the reciprocal of \(\frac{A_2}{A_1}\) due to the continuity equation.

Step 5: Use the Continuity Equation to Find Velocity Ratios

Since \(A_1 \cdot v_1 = A_2 \cdot v_2\), we can rearrange to find \(\frac{v_1}{v_2} = \frac{A_2}{A_1} = 2.25\). This simplifies to \(\frac{v_2}{v_1} = \frac{1}{2.25} = \frac{4}{9}\).

Step 6: Identify the Correct Option

The calculated velocity ratio is \(v_1 : v_2 = 9:4\). Therefore, the correct option is (a) \(9:4\).

Key Concepts

Principle of Continuity

When dealing with fluid dynamics, especially with ideal incompressible fluids, one fundamental concept is the Principle of Continuity. This principle is very straightforward. It simply states that the flow rate of a fluid remains constant from one cross-section of a pipe to another.

In practical terms, if a pipe carrying fluid narrows or widens, the speed of the fluid adjusts in such a way that the volume of fluid passing through any section per unit time stays the same. Mathematically, this concept can be expressed as: \[ A_1 \cdot v_1 = A_2 \cdot v_2 \] where:

• \( A_1 \) and \( A_2 \) are the cross-sectional areas at two different points in the pipe,
• \( v_1 \) and \( v_2 \) are the fluid velocities at these points.

This equation confirms that as the cross-sectional area of the pipe changes, the velocity of the fluid must change in such a way to keep their product constant.

Incompressible Fluids

Incompressible fluids are a central concept in fluid dynamics. This term describes fluids that have a constant density regardless of the pressure applied to them.

This characteristic is crucial in many fluid dynamics problems and simplifies calculations considerably because the fluid's volume and density do not change as pressure changes within the system.

When we say a fluid is incompressible, we're making an idealization that enables easier computation, particularly in steady-flow scenarios like those involving pipes and channels. Water is often treated as an incompressible fluid in many engineering applications because its compressibility is extremely low under normal conditions.

Cross-Sectional Area

The cross-sectional area of a pipe or other conduit through which a fluid flows is a crucial factor in determining the behavior of the fluid flow. It is simply the area of the cut surface that the fluid flows across.

For circular pipes, which are very common in practical applications, the cross-sectional area \(A\) is calculated as:\[ A = \pi r^2 \]where \(r\) is the radius of the pipe.

In problems like the one given, finding the cross-sectional areas of different sections of the pipe helps us to understand how fluid velocities will change due to the changes in the pipe's diameter. Bigger diameters mean larger cross-sectional areas and typically result in slower velocities, while smaller diameters yield higher velocities, given a constant flow rate.

Velocity Ratios

Velocity ratios offer a way to compare the relative speeds of a fluid between two or more sections of pipe. These ratios are determined using the principle of continuity.

Once cross-sectional areas are known, velocity ratios can be discussed. For example, using the continuity equation \(A_1 \cdot v_1 = A_2 \cdot v_2\), the velocity ratio can be expressed as:\[ \frac{v_1}{v_2} = \frac{A_2}{A_1} \]In the exercise problem, after calculating cross-sectional areas, this ratio helps us see how adjusting the pipe’s size impacts fluid speed. Initially, when cross-sectional areas were calculated and compared, it led to the realization that the velocity in the smaller section is higher compared to the larger section, clarifying how fluid dynamics work in pipes.