Mechanical Properties of Fluids Chapter-Wise Test 7

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A cylindrical vessel is filled with water (\( \rho = 1000 \, \text{kg/m}^3 \)) to a height of \( 2 \, \text{m} \). What is the pressure at the bottom of the vessel if the atmospheric pressure is \( 1.01 \times 10^5 \, \text{Pa} \)? (Take \( g = 10 \, \text{m/s}^2 \))

Using the pressure variation with depth: \( P = P_a + \rho g h \).

Given: \( P_a = 1.01 \times 10^5 \, \text{Pa} \), \( \rho = 1000 \, \text{kg/m}^3 \), \( g = 10 \, \text{m/s}^2 \), \( h = 2 \, \text{m} \).

\( P = 1.01 \times 10^5 + 1000 \times 10 \times 2 \).

\( P = 1.01 \times 10^5 + 20000 = 1.21 \times 10^5 \, \text{Pa} \).

1.11 × 10⁵ Pa
1.21 × 10⁵ Pa
1.31 × 10⁵ Pa
1.41 × 10⁵ Pa
2

What is the key assumption behind the equation of continuity for fluid flow?

The equation of continuity (\( A v = \text{constant} \)) assumes the fluid is incompressible, meaning its density remains constant, ensuring mass conservation in steady flow through varying cross-sections.

Fluid is viscous
Flow is turbulent
Fluid is incompressible
Pressure is constant
3

What is the primary condition for streamline flow to occur in a fluid?

Streamline flow occurs when viscous forces dominate inertial forces, keeping fluid motion steady and layered. This requires a low velocity, below the critical speed where turbulence begins.

High velocity
Low velocity
High density
Low pressure
2

What is the force on a submarine window (\( 0.07 \, \text{m}^2 \)) at \( 350 \, \text{m} \) depth in seawater (\( \rho = 1.03 \times 10^3 \, \text{kg/m}^3 \)), interior at atmospheric pressure? (Take \( g = 10 \, \text{m/s}^2 \))

Gauge pressure: \( P_g = \rho g h = 1.03 \times 10^3 \times 10 \times 350 = 3.605 \times 10^6 \, \text{Pa} \).

\( F = P_g A = 3.605 \times 10^6 \times 0.07 = 2.5235 \times 10^5 \, \text{N} \).

2.3 × 10⁵ N
2.5 × 10⁵ N
2.7 × 10⁵ N
2.9 × 10⁵ N
2

What is the total pressure at a depth of \( 1.6 \, \text{m} \) in air (\( \rho = 1.29 \, \text{kg/m}^3 \)) with atmospheric pressure \( 1.01 \times 10^5 \, \text{Pa} \)? (Take \( g = 9.8 \, \text{m/s}^2 \))

Total pressure: \( P = P_a + \rho g h \).

\( P_a = 1.01 \times 10^5 \, \text{Pa} \), \( \rho = 1.29 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \), \( h = 1.6 \, \text{m} \).

\( P = 1.01 \times 10^5 + 1.29 \times 9.8 \times 1.6 = 1.01 \times 10^5 + 20.23 = 1.0102023 \times 10^5 \, \text{Pa} \).

1.01 × 10⁵ Pa
1.02 × 10⁵ Pa
1.03 × 10⁵ Pa
1.04 × 10⁵ Pa
1

A tank has a hole \( 0.7 \, \text{m} \) below the water surface, open to the atmosphere. What is the efflux speed? (Take \( g = 9.8 \, \text{m/s}^2 \))

Torricelli’s law: \( v = \sqrt{2 g h} \).

\( g = 9.8 \, \text{m/s}^2 \), \( h = 0.7 \, \text{m} \).

\( v = \sqrt{2 \times 9.8 \times 0.7} = \sqrt{13.72} \approx 3.7 \, \text{m/s} \).

3.0 m/s
3.7 m/s
4.2 m/s
4.8 m/s
2

What is the main reason a plane’s wings generate lift during flight?

The aerofoil shape causes air to move faster over the top than underneath, reducing pressure above (Bernoulli’s principle) and creating a pressure difference that lifts the plane, known as dynamic lift.

Viscosity of air
Pressure difference due to airflow speed
Gravitational force
Surface tension of the wings
2

A soap film supports \( 2.4 \times 10^{-2} \, \text{N} \) over a \( 48 \, \text{cm} \) slider. What is the surface tension?

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

\( F = 2.4 \times 10^{-2} \, \text{N} \), \( l = 0.48 \, \text{m} \).

\( S = \frac{2.4 \times 10^{-2}}{2 \times 0.48} = 0.025 \, \text{N/m} \).

0.020 N/m
0.025 N/m
0.030 N/m
0.035 N/m
2

An aircraft (\( m = 2.5 \times 10^5 \, \text{kg} \), wing area \( 400 \, \text{m}^2 \)) flies at \( 900 \, \text{km/h} \). What is the pressure difference across the wings? (Take \( g = 10 \, \text{m/s}^2 \), \( \rho_{\text{air}} = 1.2 \, \text{kg/m}^3 \))

\( \Delta P \cdot A = mg \), \( mg = 2.5 \times 10^5 \times 10 = 2.5 \times 10^6 \, \text{N} \).

\( A = 400 \, \text{m}^2 \).

\( \Delta P = \frac{2.5 \times 10^6}{400} = 6250 \, \text{Pa} \).

5000 Pa
6250 Pa
7500 Pa
8750 Pa
2

A manometer with ethyl alcohol (\( \rho = 806 \, \text{kg/m}^3 \)) shows a height difference of \( 0.3 \, \text{m} \). What is the pressure difference? (Take \( g = 10 \, \text{m/s}^2 \))

\( \Delta P = \rho g h \).

\( \rho = 806 \, \text{kg/m}^3 \), \( g = 10 \, \text{m/s}^2 \), \( h = 0.3 \, \text{m} \).

\( \Delta P = 806 \times 10 \times 0.3 = 2418 \, \text{Pa} \).

2000 Pa
2418 Pa
2600 Pa
3000 Pa
2

A capillary tube of radius \( 0.65 \, \text{mm} \) is dipped in water (\( S = 0.0727 \, \text{N/m} \), \( \rho = 1000 \, \text{kg/m}^3 \), \( \cos \theta = 1 \)). What is the capillary rise? (Take \( g = 9.8 \, \text{m/s}^2 \))

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

\( S = 0.0727 \, \text{N/m} \), \( \rho = 1000 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \), \( a = 0.65 \times 10^{-3} \, \text{m} \).

\( h = \frac{2 \times 0.0727 \times 1}{1000 \times 9.8 \times 0.65 \times 10^{-3}} = 0.02285 \, \text{m} = 2.285 \, \text{cm} \).

2.0 cm
2.2 cm
2.3 cm
2.5 cm
3

A sphere of radius \( 0.045 \, \text{m} \) moves at \( 0.15 \, \text{m/s} \) through machine oil (\( \eta = 0.034 \, \text{Pa s} \)). What is the viscous drag force?

Stokes’ law: \( F = 6 \pi \eta a v \).

\( \eta = 0.034 \, \text{Pa s} \), \( a = 0.045 \, \text{m} \), \( v = 0.15 \, \text{m/s} \).

\( F = 6 \times 3.14 \times 0.034 \times 0.045 \times 0.15 = 0.0433 \, \text{N} \).

0.03 N
0.04 N
0.043 N
0.05 N
3

Water flows at \( 2.5 \, \text{m/s} \) at \( 3 \, \text{m} \) height with pressure \( 1.5 \times 10^5 \, \text{Pa} \). What is the pressure at \( 1 \, \text{m} \) height with speed \( 3.5 \, \text{m/s} \)? (\( \rho = 1000 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \))

Bernoulli’s: \( P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2 \).

Left: \( 1.5 \times 10^5 + \frac{1}{2} \times 1000 \times 6.25 + 1000 \times 9.8 \times 3 = 1.5 \times 10^5 + 3125 + 29400 = 1.82525 \times 10^5 \).

Right: \( P_2 + \frac{1}{2} \times 1000 \times 12.25 + 1000 \times 9.8 \times 1 = P_2 + 6125 + 9800 = P_2 + 15925 \).

\( 1.82525 \times 10^5 = P_2 + 15925 \), \( P_2 = 1.666 \times 10^5 \, \text{Pa} \).

1.60 × 10⁵ Pa
1.666 × 10⁵ Pa
1.70 × 10⁵ Pa
1.75 × 10⁵ Pa
2

A hydraulic lift has a small piston of area \( 0.01 \, \text{m}^2 \) and a large piston of area \( 0.09 \, \text{m}^2 \). If a force of \( 100 \, \text{N} \) is applied on the small piston, what is the force exerted by the large piston?

By Pascal’s law, pressure is transmitted equally: \( P = \frac{F_1}{A_1} = \frac{F_2}{A_2} \).

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

\( F_1 = 100 \, \text{N} \), \( A_1 = 0.01 \, \text{m}^2 \), \( A_2 = 0.09 \, \text{m}^2 \).

\( F_2 = 100 \times \frac{0.09}{0.01} = 100 \times 9 = 900 \, \text{N} \).

600 N
900 N
1200 N
1500 N
2

A sphere of radius \( 0.035 \, \text{m} \) moves at \( 0.09 \, \text{m/s} \) through glycerine (\( \eta = 0.83 \, \text{Pa s} \)). What is the viscous drag force?

Stokes’ law: \( F = 6 \pi \eta a v \).

\( \eta = 0.83 \, \text{Pa s} \), \( a = 0.035 \, \text{m} \), \( v = 0.09 \, \text{m/s} \).

\( F = 6 \times 3.14 \times 0.83 \times 0.035 \times 0.09 = 0.0493 \, \text{N} \).

0.04 N
0.05 N
0.06 N
0.07 N
2

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