(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)
(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)
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{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.
- 27 Pa
- 175 Pa
- 135 Pa
- 36 Pa
The Correct Option is B
Solution and Explanation
To determine the pressure difference \((P_A - P_B)\) between points A and B in the horizontal tube, we can apply the Bernoulli's Equation for incompressible and steady flow:
\[ P_A + \frac{1}{2} \rho v_A^2 = P_B + \frac{1}{2} \rho v_B^2 \]
Rearranging the equation to find \((P_A - P_B)\):
\[ P_A - P_B = \frac{1}{2} \rho (v_B^2 - v_A^2) \]
Step 1: Convert Units
First, convert all measurements to SI units.
- Cross-sectional areas:
\[ A_A = 1.5 \, \text{cm}^2 = 1.5 \times 10^{-4} \, \text{m}^2 \] \[ A_B = 25 \, \text{mm}^2 = 25 \times 10^{-6} \, \text{m}^2 \]
- Speed at B:
\[ v_B = 60 \, \text{cm/s} = 0.6 \, \text{m/s} \]
Step 2: Apply Continuity Equation
For steady, incompressible flow, the continuity equation states:
\[ A_A v_A = A_B v_B \]
Solving for \( v_A \):
\[ v_A = \frac{A_B}{A_A} \times v_B = \frac{25 \times 10^{-6}}{1.5 \times 10^{-4}} \times 0.6 = \frac{25}{150} \times 0.6 = \frac{1}{6} \times 0.6 = 0.1 \, \text{m/s} \]
Step 3: Calculate Pressure Difference
Given the density of the liquid \( \rho = 1000 \, \text{kg/m}^3 \), substitute the known values into the Bernoulli's equation:
\[ P_A - P_B = \frac{1}{2} \times 1000 \times (0.6^2 - 0.1^2) = 500 \times (0.36 - 0.01) = 500 \times 0.35 = 175 \, \text{Pa} \]
Therefore, the pressure difference \((P_A - P_B)\) is 175 Pa.
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