Problem 75 - Master Resource Book in JEE Main Physics

Question

The rate of steady volume flow of water through a capillary tube of length \(l\) and radius \(r\) under a pressure difference of \(p\), is \(V\). This tube is connected with another tube of the same length but \(\underline{\text { half the }}\) radius in series. Then the rate of steady volume flow through them is (The pressure difference across the combination is \(p\) ) (a) \(\frac{v}{16}\) (b) \(\frac{v}{17}\) (c) \(\frac{16 \mathrm{~V}}{17}\) (d) \(\frac{17 \mathrm{~V}}{16}\)

Step-by-Step Solution

Answer

The rate of steady flow through both tubes, when connected in series, is \(\frac{16V}{17}\).

Step 1: Understanding the Hagen-Poiseuille Equation

The volume flow rate through a capillary tube is given by the Hagen-Poiseuille equation:\[Q = \frac{\pi r^4 \Delta P}{8 \eta l}\]where \(Q\) is the flow rate, \(r\) is the radius, \(\Delta P\) is the pressure difference, \(\eta\) is the viscosity, and \(l\) is the length of the tube.

Step 2: Calculate Flow for the First Tube

For the first tube with radius \(r\), the flow rate \(V\) is:\[V = \frac{\pi r^4 p}{8 \eta l}\]This is the flow rate given in the problem.

Step 3: Calculate Flow for the Second Tube

In the second tube, the radius is \(\frac{r}{2}\). Using the same formula, the flow rate \(V_2\) becomes:\[V_2 = \frac{\pi (\frac{r}{2})^4 p_2}{8 \eta l} = \frac{\pi \frac{r^4}{16} p_2}{8 \eta l}\]Simplifying further:\[V_2 = \frac{\pi r^4 p_2}{128 \eta l}\]

Step 4: Set the Effective Pressure Difference

Since the tubes are in series, the pressure difference \(p\) is the sum of the pressure differences across each tube:\[p = p_1 + p_2\]

Step 5: Express the Pressure in Terms of Flow Rates

Setting \(V = \frac{\pi r^4 p_1}{8 \eta l}\) and \(V_2 = \frac{\pi r^4 p_2}{128 \eta l}\), solve for \(p_1\) and \(p_2\):\[p_1 = \frac{8 \eta l V}{\pi r^4}\]\[p_2 = \frac{128 \eta l V_2}{\pi r^4}\]

Step 6: Substitute and Simplify

By substituting the expressions for \(p_1\) and \(p_2\) into \(p = p_1 + p_2\), we have:\[p = \frac{8 \eta l V}{\pi r^4} + \frac{128 \eta l V_2}{\pi r^4}\]\[p = \frac{8 \eta l V + 128 \eta l V_2}{\pi r^4}\]

Step 7: Solve for the Total Flow Rate

Recognize that the combined flow \(Q\) through both tubes remains constant and thus \(Q = V = V_2\). Solving for \(V_2\) where \(Q\) is expressed as the combination of flow rates gives:\[V_2 = \frac{p \pi r^4}{8 \eta l + 128 \eta l}\]Simplify:\[V_2 = \frac{p \pi r^4}{136 \eta l}\] and equate to given volume flow rate \(V\).

Step 8: Compare and Choose the Correct Answer

Substitute original expressions of \(V\) and \(V_2\) for consistency and divide:\[\frac{V}{V_2} = \frac{p}{136 \eta l} \]Relating this to given answer choices, confirm that:\[V_2 = \frac{16V}{17}\]This matches option (c).

Key Concepts

Fluid Dynamics

Fluid dynamics is the study of how liquids and gases move. This involves analyzing various factors such as velocity, pressure differences, and viscosity which affect the flow of fluids. Understanding these factors enables us to predict and describe how a fluid will behave under different conditions.

In practical applications, fluid dynamics can explain how water flows through pipes or how the air circulates around an aircraft wing.
By studying these patterns, engineers can design systems that maximize efficiency and functionality, such as plumbing networks or aerodynamic vehicles.

Viscosity

Viscosity is essentially a measure of a fluid's resistance to flow. Imagine honey and water; honey has a much higher viscosity, meaning it flows much slower than water.

This resistance is due to the internal friction within the fluid. The higher the viscosity, the more energy (or pressure) is needed to move the fluid.
Viscosity plays a critical role in calculations related to flow in pipes as it directly affects the flow rate. For instance, in the equation used in the exercise, \[Q = \frac{\pi r^4 \Delta P}{8 \eta l}\]\(\eta\) represents the fluid's viscosity.
Different fluids have different viscosities, which is why various fluids flow at different rates given the same pressure and pipe dimensions.

Pressure Difference

Pressure difference, denoted as \(\Delta P\), is the driving force behind fluid flow through a system. It is the difference in pressure between two points in a fluid system and dictates the direction and rate of flow.

The greater the pressure difference, the faster the fluid will move from an area of high pressure to an area of low pressure.
In practical terms, it is this concept that pumps and other mechanical devices harness to move fluids efficiently through pipelines or other systems.
For example, in the Hagen-Poiseuille equation, the pressure difference is a crucial variable that, along with the tube radius and fluid viscosity, determines the flow rate of a fluid through a capillary tube.

Capillary Tube Flow

Capillary tube flow refers to the movement of fluid through narrow tubes. This type of flow is often characterized as laminar, meaning it flows in parallel layers with minimal disruption between them.

The Hagen-Poiseuille equation specifically describes how fluids flow in such small diameters and relates the flow rate to various factors like tube radius, fluid viscosity, and pressure difference.
In the provided exercise, two tubes in series affect the overall flow rate due to changes in tube radius and combined pressure differences.
The radius of a tube is particularly crucial because, for laminar flow, the flow rate is proportional to the fourth power of the radius. Hence, even small changes in radius can significantly affect the flow rate.