Problem 81 - Master Resource Book in JEE Main Physics

Question

An incompressible fluid flows steadily through a cylindrical pipe which has radius \(2 R\) at a point \(A\) and radius \(R\) at a point \(B\). Further along the flow of direction if the velocity at point \(A\) is \(v\), its velocity at point \(B\) will be (a) \(v / 4\) (b) \(2 v\) (c) \(4 \underline{v}\) (d) \(-\frac{v}{2}\)

Step-by-Step Solution

Answer

(c) The velocity at point B is \( 4v \).

Step 1: Understand Continuity Equation

In incompressible fluid flow, the volume flow rate must be constant throughout the pipe. This is given by the continuity equation: \( A_1v_1 = A_2v_2 \), where \( A \) is the cross-sectional area and \( v \) is the velocity at points \( A \) and \( B \), respectively.

Step 2: Calculate Area at Points A and B

The cross-sectional area \( A \) of a circle is given by \( A = \pi r^2 \). Thus, the area at point \( A \) is \( \pi (2R)^2 = 4\pi R^2 \) and at point \( B \) is \( \pi (R)^2 = \pi R^2 \).

Step 3: Apply Continuity Equation

Using the continuity equation \( A_1v_1 = A_2v_2 \), we substitute the areas: \( 4\pi R^2 \cdot v = \pi R^2 \cdot v_2 \).

Step 4: Solve for Velocity at Point B

Cancel \( \pi R^2 \) from both sides of the equation, resulting in: \( 4v = v_2 \). Solving for \( v_2 \), we find that \( v_2 = 4v \).

Key Concepts

Incompressible Fluid Flow

Incompressible fluid flow is a fundamental concept in fluid mechanics, where the fluid density remains constant regardless of changes in pressure or velocity. This property is particularly relevant for fluids like water under typical conditions. When dealing with incompressible fluids, the amount of fluid entering a pipe system at one end should equal the amount exiting at the other unless there's a leakage or water storage involved.

The principle governing incompressible fluid flow is often expressed by the continuity equation. This important equation states that the product of the cross-sectional area and the velocity of the fluid must remain constant as it moves through a non-leaking, frictionless tube. This ensures that no fluid is lost or gained as it travels through the pipe. Understanding this concept helps in predicting how fluids behave in different pipe sections as they maintain a constant flow rate.

Cross-Sectional Area Calculation

Calculating the cross-sectional area is critical for understanding fluid flow in pipes. The cross-section of a pipe can often be represented as a circle. Hence, its area is determined using the formula for the area of a circle: \[ A = \pi r^2 \]where \( r \) is the radius of the circle.

In this specific exercise, we have two different pipe sections with different radii. At point A, the radius is \(2R\), making the area \(4\pi R^2\). At point B, where the radius is \(R\), the area is just \(\pi R^2\). Calculating these areas is vital because they allow us to apply the continuity equation effortlessly for determining how velocity changes from one point to another depending on the size of the pipe section.

Velocity in Pipes

Velocity in pipes is a key factor determined by the changes in the cross-sectional area while maintaining the same flow rate of an incompressible fluid. According to the continuity equation:\[ A_1v_1 = A_2v_2 \]Given that the cross-sectional areas are \(4\pi R^2\) at point A and \(\pi R^2\) at point B, and knowing the velocity at A is \(v\), we set up the equation:\[ 4\pi R^2 \cdot v = \pi R^2 \cdot v_2 \]
By canceling \(\pi R^2\) from both sides, we simplify to:\[ 4v = v_2 \]
This shows that the velocity at point B, \(v_2\), is four times the velocity at point A. Such understanding confirms how different pipe sections and their areas affect the speed at which fluids move, an essential insight for designing efficient fluid systems.