Problem 69 - Master Resource Book in JEE Main Physics

Question

Two water pipes \(P\) and \(Q\) having diameter \(2 \times 10^{-2} \mathrm{~m}\) and \(4 \times 10^{-2} \mathrm{~m}\) respectively are joined in series with the main supply line of water. The velocity of water flowing in pipe \(P\) is (a) 4 times that of \(Q\) (b) 2 times that of \(Q\) (c) \(1 / 2\) times that of 0 (d) \(1 / 4\) times that of 0

Step-by-Step Solution

Answer

The velocity of water in pipe P is 4 times that of Q (option a).

Step 1: Understand the Flow Relationship

When two pipes are joined in series, they carry the same volumetric flow rate. This implies that the flow rate in pipe \( P \) is equal to the flow rate in pipe \( Q \). The formula for flow rate \( Q \) in terms of velocity \( v \) and cross-sectional area \( A \) is given by \( Q = A \times v \). Since we have two pipes in series with constant flow, \( A_P \times v_P = A_Q \times v_Q \).

Step 2: Calculate Cross-sectional Areas

The cross-sectional area \( A \) of a pipe is calculated using the formula \( A = \pi \times r^2 \), where \( r \) is the radius. For pipe \( P \), the diameter is \( 2 \times 10^{-2} \mathrm{~m} \), so the radius is \( r_P = 1 \times 10^{-2} \mathrm{~m} \). The area is \( A_P = \pi \times (1 \times 10^{-2})^2 = \pi \times 1 \times 10^{-4} \ \mathrm{m^2} \). Similarly, for pipe \( Q \), the radius is \( r_Q = 2 \times 10^{-2} \mathrm{~m} \) and the area is \( A_Q = \pi \times (2 \times 10^{-2})^2 = \pi \times 4 \times 10^{-4} \ \mathrm{m^2} \).

Step 3: Relate Velocities Using Areas

Using the relationship \( A_P \times v_P = A_Q \times v_Q \), we substitute the areas: \( \pi \times 1 \times 10^{-4} \times v_P = \pi \times 4 \times 10^{-4} \times v_Q \). Simplifying, the \( \pi \) cancels out, leading to \( v_P = 4 \times v_Q \).

Step 4: Interpret and Confirm the Answer

The equation \( v_P = 4 \times v_Q \) shows that the velocity of water in pipe \( P \) is 4 times that in pipe \( Q \). Therefore, the correct answer is option (a).

Key Concepts

Continuity Equation

The Continuity Equation is a crucial principle in fluid mechanics, describing how fluid flows in a pipe. This equation assumes a constant flow rate through pipes connected in series. It is based on the conservation of mass in a closed system.

Imagine you're squeezing a tube of toothpaste. If you pinch it at one end and press down, toothpaste flows out at the other. The amount you pressed (or the mass of the toothpaste) remains constant as it moves through the tube. Similarly, the Continuity Equation states that the volumetric flow rate must remain constant across different sections of a pipe system. The equation is:

\[ A_1 \times v_1 = A_2 \times v_2 \]

Where:

• \(A_1\) and \(A_2\) are the cross-sectional areas of the two sections of the pipe.
• \(v_1\) and \(v_2\) are the fluid velocities at these sections.

Using this equation, we can analyze how velocity changes when it flows from a narrower pipe to a wider one, such as in the exercise with pipes \(P\) and \(Q\).

Volumetric Flow Rate

Volumetric Flow Rate is a key concept that helps us understand the amount of fluid moving through a pipe per unit time. It tells us how much of a fluid passes through a point in a system every second, which is crucial for designing and controlling fluid systems.

The formula for volumetric flow rate \(Q\) is:
\[ Q = A \times v \]

Where:

• \(Q\) is the volumetric flow rate.
• \(A\) is the cross-sectional area of the pipe.
• \(v\) is the velocity of the fluid flowing through the pipe.

In the example with pipes \(P\) and \(Q\), the flow rate must be identical for both pipes since they are in series. This concept allows us to solve for unknown velocities when the areas of pipes are different, as the exercise required.

By maintaining the same flow rate across pipes, we can manage and predict how changes to one section of a system can impact the rest of the system.

Cross-sectional Area

The Cross-sectional Area of a pipe is an essential parameter in determining how a fluid behaves as it flows through the pipe. It is a measure of the size of a cut made perpendicular to the pipe's length, which is effectively the pipe's 'opening size.'

Mathematically, for a circular pipe, the area \(A\) is calculated using the diameter \(D\) of the pipe with the formula:

\[ A = \pi \times \left(\frac{D}{2}\right)^2 \]

It's like cutting across a cylindrical fish tank to see its surface - that surface area is your cross-sectional area here.

In the textbook exercise, knowing the cross-sectional areas of pipes \(P\) and \(Q\) was crucial. With diameters of \(2 \times 10^{-2}\) meters and \(4 \times 10^{-2}\) meters respectively, calculating these areas helped determine how they impacted the velocity of water.

By understanding and calculating this area, you can predict changes in fluid velocity as it moves from one section of piping to another, demonstrating the physical concept of constraint in flow systems.