Topic: Fluid Mechanics
A liquid of density $1200 kg/m^{3}$ is flowing steadily through a tube of varying cross section as shown below. The cross section at point A is $1 cm^{2}$ and at B is $20 mm^{2}$. The points A and B in the same horizontal plane. The speed of liquid at point A is 10cm/sec. The difference of pressure at A and B is
- A. 100 Pa
- B. 170 Pa
- C. 144 Pa
- D. 180 Pa
Answer: C. 144 Pa
Explanation: First, use the continuity equation $A_{1}v_{1} = A_{2}v_{2}$ to find $v_2$. $A_1 = 1\text{ cm}^2 = 10^{-4}\text{ m}^2$, $v_1 = 10\text{ cm/s} = 0.1\text{ m/s}$, $A_2 = 20\text{ mm}^2 = 20 \times 10^{-6}\text{ m}^2$. Thus, $v_2 = \frac{10^{-4}}{20 \times 10^{-6}} \times 0.1 = 0.5$ m/s. Using Bernoulli's equation for a horizontal pipe ($Z_1 = Z_2$), $\frac{P_{1}-P_{2}}{\rho g} = \frac{v_{2}^{2}-v_{1}^{2}}{2g}$. The pressure difference is $P_1 - P_2 = \frac{\rho(v_2^2 - v_1^2)}{2} = \frac{1200(0.5^2 - 0.1^2)}{2} = 600(0.24) = 144$ Pa.