Mechanical Properties of Fluids Chapter-Wise Test 5

\( S = \frac{F}{2 l} \).

\( F = 4.8 \times 10^{-2} \, \text{N} \), \( l = 0.48 \, \text{m} \).

\( S = \frac{4.8 \times 10^{-2}}{2 \times 0.48} = 0.05 \, \text{N/m} \).

The submarine’s rigid structure resists the external hydrostatic pressure (\( P = \rho g h \)), which increases with depth, preventing collapse by balancing the inward force with structural strength.

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

\( S = 0.0727 \, \text{N/m} \), \( \rho = 1000 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \), \( a = 0.05 \times 10^{-3} \, \text{m} \).

\( h = \frac{2 \times 0.0727 \times 1}{1000 \times 9.8 \times 0.05 \times 10^{-3}} = 0.2967 \, \text{m} = 29.67 \, \text{cm} \).

Excess pressure: \( \Delta P = \frac{2 S}{r} \).

\( S = 0.000239 \, \text{N/m} \), \( r = 1 \times 10^{-3} \, \text{m} \).

\( \Delta P = \frac{2 \times 0.000239}{1 \times 10^{-3}} = 0.478 \, \text{Pa} \).

When the viscous force (from Stokes’ law) plus buoyant force equals the gravitational force, the net force becomes zero, and the object reaches terminal velocity, moving at a constant speed, as explained in the chapter.

Continuity equation: \( A_1 v_1 = A_2 v_2 \).

\( A_1 = 0.15 \, \text{m}^2 \), \( v_1 = 1.8 \, \text{m/s} \), \( A_2 = 0.05 \, \text{m}^2 \).

\( v_2 = \frac{A_1 v_1}{A_2} = \frac{0.15 \times 1.8}{0.05} = 5.4 \, \text{m/s} \).

\( P_i = P_a + \rho g h + \frac{2 S}{r} \).

\( P_a = 1.01 \times 10^5 \, \text{Pa} \), \( \rho g h = 1000 \times 10 \times 0.25 = 2500 \, \text{Pa} \).

\( \frac{2 S}{r} = \frac{2 \times 0.0727}{6 \times 10^{-3}} = 24.23 \, \text{Pa} \).

\( P_i = 1.01 \times 10^5 + 2500 + 24.23 = 1.03524 \times 10^5 \, \text{Pa} \).

\( F_1 = \frac{A_1}{A_2} F_2 \), \( F_2 = 2000 \times 10 = 20000 \, \text{N} \).

\( A_1 = \pi (0.05)^2 \), \( A_2 = \pi (0.2)^2 \), \( \frac{A_1}{A_2} = \frac{0.0025}{0.04} = \frac{1}{16} \).

\( F_1 = \frac{20000}{16} = 1250 \, \text{N} \).

\( P = P_a + \rho g h \).

\( P_a = 1.01 \times 10^5 \, \text{Pa} \), \( \rho = 1.03 \times 10^3 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \), \( h = 250 \, \text{m} \).

\( P = 1.01 \times 10^5 + 1.03 \times 10^3 \times 9.8 \times 250 = 1.01 \times 10^5 + 2.5235 \times 10^6 = 2.6245 \times 10^6 \, \text{Pa} \).

Continuity equation: \( A_1 v_1 = A_2 v_2 \).

\( A_1 = 0.05 \, \text{m}^2 \), \( v_1 = 1.5 \, \text{m/s} \), \( A_2 = 0.015 \, \text{m}^2 \).

\( v_2 = \frac{A_1 v_1}{A_2} = \frac{0.05 \times 1.5}{0.015} = 5 \, \text{m/s} \).

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

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

\( F_1 = 180 \, \text{N} \), \( A_1 = 0.012 \, \text{m}^2 \), \( A_2 = 0.06 \, \text{m}^2 \).

\( F_2 = 180 \times \frac{0.06}{0.012} = 180 \times 5 = 900 \, \text{N} \).

The equation of continuity assumes the fluid is incompressible, meaning its density (\( \rho \)) remains constant, even as velocity increases in a narrowing pipe, ensuring mass conservation (\( A v = \text{constant} \)).

\( P = P_a + \rho g h \).

\( P_a = 1.01 \times 10^5 \, \text{Pa} \), \( \rho = 1.03 \times 10^3 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \), \( h = 800 \, \text{m} \).

\( P = 1.01 \times 10^5 + 1.03 \times 10^3 \times 9.8 \times 800 = 1.01 \times 10^5 + 8.0784 \times 10^6 = 8.1794 \times 10^6 \, \text{Pa} \).

Gauge pressure is \( P_g = \rho g h \).

\( \rho = 1000 \, \text{kg/m}^3 \), \( g = 10 \, \text{m/s}^2 \), \( h = 5 \, \text{m} \).

\( P_g = 1000 \times 10 \times 5 = 5 \times 10^4 \, \text{Pa} \).

At lower altitudes, the air column above is taller and heavier, increasing atmospheric pressure (\( P_a = \rho g h \)), which raises the height of the mercury column in the barometer to balance the greater external pressure.