Problem 70 - Master Resource Book in JEE Main Physics

Question

The rate of flow of liquid through a capillary tube of radius \(r\) is \(V\), when the pressure difference across the two ends of the capillary is \(p\). If pressure is increased by \(3 p\) and radius is reduced to \(r / 2\), then the rate of flow becomes (a) \(\sqrt{V / 9}\) (b) \(3 V / 8\) (c) \(V / 4\) (d) \(V / 3\)

Step-by-Step Solution

Answer

The rate of flow becomes \(3V/8\). This corresponds to option (b).

Step 1: Understand the Relationship

The rate of flow of liquid through a capillary tube follows Poiseuille’s law. The formula is given by: \[ V = \frac{\pi r^4 p}{8 \eta L} \]where \( V \) is the flow rate, \( r \) is the radius, \( p \) is the pressure difference, \( \eta \) is the viscosity of the liquid, and \( L \) is the length of the tube. This will help evaluate the changes in parameters.

Step 2: Calculate New Flow Rate Equation

If the pressure difference is increased to \(3p\) and the radius is decreased to \(\frac{r}{2}\), substitute these changes into the flow rate equation:\[ V' = \frac{\pi \left(\frac{r}{2}\right)^4 (3p)}{8 \eta L} \] Simplify this to find the new flow rate, \( V' \).

Step 3: Simplify the New Flow Equation

First, calculate \(\left(\frac{r}{2}\right)^4 = \frac{r^4}{16}\). Substituting into the formula:\[ V' = \frac{\pi \cdot \frac{r^4}{16} \cdot 3p}{8 \eta L} \]This simplifies to:\[ V' = \frac{3\pi r^4 p}{128 \eta L} \].

Step 4: Compare Original and New Flow Rates

Now, compare this new equation with the original flow representing \( V \):\[ V = \frac{\pi r^4 p}{8 \eta L} \]. By dividing the new rate \( V' \) by the original rate \( V \):\[ \frac{V'}{V} = \frac{\frac{3\pi r^4 p}{128 \eta L}}{\frac{\pi r^4 p}{8 \eta L}} \].

Step 5: Approximating Simplification

Simplify the ratio:\[ \frac{V'}{V} = \frac{3}{16} \times \frac{8}{1} = \frac{3}{16} \times 8 = \frac{3}{8} \]. Thus, the new flow \( V' = \frac{3V}{8} \).

Key Concepts

Flow Rate Equation

The flow rate equation is central to understanding Poiseuille's Law. This law describes how the flow rate of a liquid through a narrow tube is influenced by several variables. The fundamental formula is expressed as \( V = \frac{\pi r^4 p}{8 \eta L} \), where:

• \( V \) represents the flow rate of the liquid through the tube.
• \( r \) is the radius of the tube.
• \( p \) stands for the pressure difference across the tube's ends.
• \( \eta \) is the dynamic viscosity of the fluid.
• \( L \) denotes the length of the tube.

Here, the flow rate is directly proportional to \( r^4 \) and to the pressure difference \( p \), while inversely proportional to the fluid viscosity \( \eta \) and the length of the tube \( L \). Notably, because the radius is taken to the fourth power, even small changes in \( r \) can significantly impact the flow rate.

Pressure Difference

The pressure difference, symbolized by \( p \), is a critical factor in determining the flow rate of a fluid through a tube. In Poiseuille's Law, this difference in pressure between the two ends of the tube pushes the fluid forward. Here, the pressure difference is amplified from \( p \) to \( 3p \), tripling the force driving the liquid through the capillary.This change drastically impacts the resulting flow rate. As noted in the formula, the flow rate is directly proportional to the pressure difference. Thus, a threefold increase in pressure difference should, theoretically, lead to a direct increase in flow rate, assuming all other factors remain constant. However, this is also dependent on the radius and viscosity, as demonstrated in the calculations.

Viscosity

Viscosity, denoted by \( \eta \), is a measure of a fluid's resistance to flow. A fluid with high viscosity, like honey, flows more slowly compared to a fluid with low viscosity, such as water. In Poiseuille’s Law, viscosity plays an inverse role in determining flow rate. This means that as viscosity goes up, the flow rate tends to decrease, and vice versa.Understanding viscosity is crucial when assessing flow through capillary tubes. It acts as a frictional force within the fluid, resisting motion. In applications involving significant changes to factors like pressure and radius, viscosity can substantially influence the outcome. Despite higher pressure, if the viscosity is considerable, the expected increase in flow rate may be mitigated. In the exercise, although we increase the pressure, if viscosity remains constant, the changes in other factors like radius become vital to evaluating flow rate adjustments.