Mechanical Properties of Fluids Chapter-Wise Test 17

Pascal’s law: \( P = \frac{F_1}{A_1} = \frac{F_2}{A_2} \).

\( F_2 = F_1 \times \frac{A_2}{A_1} \).

\( F_1 = 250 \, \text{N} \), \( A_1 = 0.02 \, \text{m}^2 \), \( A_2 = 0.08 \, \text{m}^2 \).

\( F_2 = 250 \times \frac{0.08}{0.02} = 250 \times 4 = 1000 \, \text{N} \).

Bernoulli’s equation: \( P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \).

\( P_1 = 1.85 \times 10^5 \, \text{Pa} \), \( v_1 = 3.2 \, \text{m/s} \), \( v_2 = 5.8 \, \text{m/s} \), \( \rho = 1000 \, \text{kg/m}^3 \).

\( P_2 = 1.85 \times 10^5 + \frac{1}{2} \times 1000 (10.24 - 33.64) = 1.85 \times 10^5 - 11700 = 1.733 \times 10^5 \, \text{Pa} \).

Fluids cannot resist shear stress without flowing due to their lack of rigidity; solids, not fluids, resist shear stress with a fixed shape. The incorrect statement is that fluids resist shear stress like solids.

Pascal’s law: \( P = \frac{F_1}{A_1} = \frac{F_2}{A_2} \).

\( F_2 = F_1 \times \frac{A_2}{A_1} \).

\( F_1 = 350 \, \text{N} \), \( A_1 = 0.025 \, \text{m}^2 \), \( A_2 = 0.1 \, \text{m}^2 \).

\( F_2 = 350 \times \frac{0.1}{0.025} = 350 \times 4 = 1400 \, \text{N} \).

Pascal’s law assumes an incompressible fluid. If pressure exceeds the fluid’s incompressibility limit, the fluid compresses or the system fails (e.g., leaks or bursts), as the assumption of uniform pressure transmission breaks down.

Liquids and gases are fluids that can flow and take the shape of their container due to their inability to maintain a fixed shape, unlike solids, which have a definite shape due to strong intermolecular forces resisting deformation.

Capillary rise: \( h = \frac{2 S \cos \theta}{\rho g a} \).

\( S = 0.4355 \, \text{N/m} \), \( \rho = 13.6 \times 10^3 \, \text{kg/m}^3 \), \( g = 10 \, \text{m/s}^2 \), \( a = 0.2 \times 10^{-3} \, \text{m} \), \( \cos \theta = -0.5 \).

\( h = \frac{2 \times 0.4355 \times (-0.5)}{13.6 \times 10^3 \times 10 \times 0.2 \times 10^{-3}} = -0.016 \, \text{m} = -1.6 \, \text{cm} \).

Gauge pressure: \( P_g = \rho g h = 1.03 \times 10^3 \times 9.8 \times 200 = 2.0194 \times 10^6 \, \text{Pa} \).

\( F = P_g A = 2.0194 \times 10^6 \times 0.09 = 1.81746 \times 10^5 \, \text{N} \).

Capillary rise (\( h = 2S \cos \theta / \rho g r \)) is inversely proportional to radius, not directly proportional. The incorrect statement is that height increases with tube radius.

The barometer measures atmospheric pressure (\( P_a = \rho g h \)). During a storm, atmospheric pressure drops, reducing the force supporting the mercury column, causing its height to decrease, as noted in the chapter.