Question

The figure shows a liquid of given density flowing steadily in horizontal tube of varying cross - section. Cross sectional areas at A is 1.5 cm², and B is is 25 mm², if the speed of liquid at B is 60 cm/s then \((P_A-P_B)\) is: 
(Given \(P_A\) and \(P_B\) are liquid pressures at A and B points.
density \(ρ = 1000 \space kg \space m^{-3}\)
A and B are on the axis of tube)

• A. 27 Pa
• B. 175 Pa
• C. 135 Pa
• D. 36 Pa

Hint:

{Bernoulli's Principle} states that for an incompressible, frictionless fluid, the following quantity is constant along a streamline: \[ P + \frac{1}{2} \rho v^2 + \rho g h = \text{constant} \] In horizontal flow (\( h \) is constant), the pressure and velocity are inversely related: as the velocity increases, the pressure decreases, and vice versa.

Answer 1

The Correct Option is B

Sure! Here is the detailed solution to the problem using the Bernoulli's principle: 

To find the pressure difference \((P_A - P_B)\) in the liquid flowing through the tube, we will use Bernoulli's equation. It states that for an incompressible, non-viscous fluid flowing in a streamline, the following relation holds:

\(P + \frac{1}{2} \rho v^2 + \rho gh = \text{constant}\)

Since the tube is horizontal, the height term \(h\) is constant and cancels out. The equation simplifies to:

\(P_A + \frac{1}{2} \rho v_A^2 = P_B + \frac{1}{2} \rho v_B^2\)

Rearranging for \((P_A - P_B)\), we get:

\(P_A - P_B = \frac{1}{2} \rho (v_B^2 - v_A^2)\)

First, we need to find \(v_A\) using the continuity equation, which states:

\(A_A \cdot v_A = A_B \cdot v_B\)

Given:

• \(A_A = 1.5 \, \text{cm}^2 = 1.5 \times 10^{-4} \, \text{m}^2\)
• \(A_B = 25 \, \text{mm}^2 = 25 \times 10^{-6} \, \text{m}^2\)
• \(v_B = 60 \, \text{cm/s} = 0.6 \, \text{m/s}\)
• \(\rho = 1000 \, \text{kg/m}^3\)

Using the continuity equation:

\((1.5 \times 10^{-4}) \cdot v_A = (25 \times 10^{-6}) \cdot 0.6\)

Solving for \(v_A\):

\(v_A = \frac{25 \times 10^{-6} \cdot 0.6}{1.5 \times 10^{-4}}\)\(v_A = 0.1 \, \text{m/s}\)

Now substitute \(v_A\) and \(v_B\) back into the Bernoulli's equation:

\(P_A - P_B = \frac{1}{2} \times 1000 \times (0.6^2 - 0.1^2)\)\(P_A - P_B = 500 \times (0.36 - 0.01)\)\(P_A - P_B = 500 \times 0.35 = 175 \, \text{Pa}\)

Thus, the pressure difference \((P_A - P_B)\) is 175 Pa.

This detailed explanation helps understand the application of Bernoulli's principle in fluid dynamics.