Mechanical Properties of Fluids Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the total pressure at a depth of \( 2.8 \, \text{m} \) in whole blood (\( \rho = 1.06 \times 10^3 \, \text{kg/m}^3 \)) with atmospheric pressure \( 1.01 \times 10^5 \, \text{Pa} \)? (Take \( g = 9.8 \, \text{m/s}^2 \))

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

\( P_a = 1.01 \times 10^5 \, \text{Pa} \), \( \rho = 1.06 \times 10^3 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \), \( h = 2.8 \, \text{m} \).

\( P = 1.01 \times 10^5 + 1.06 \times 10^3 \times 9.8 \times 2.8 = 1.01 \times 10^5 + 29094.4 = 1.300944 \times 10^5 \, \text{Pa} \).

1.28 × 10⁵ Pa
1.30 × 10⁵ Pa
1.32 × 10⁵ Pa
1.34 × 10⁵ Pa
2

A horizontal pipe carries water (\( \rho = 1000 \, \text{kg/m}^3 \)) at \( 2 \, \text{m/s} \) and \( 1.4 \times 10^5 \, \text{Pa} \). If the speed becomes \( 5 \, \text{m/s} \), what is the pressure?

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

\( P_1 = 1.4 \times 10^5 \, \text{Pa} \), \( v_1 = 2 \, \text{m/s} \), \( v_2 = 5 \, \text{m/s} \), \( \rho = 1000 \, \text{kg/m}^3 \).

\( P_2 = 1.4 \times 10^5 + \frac{1}{2} \times 1000 (4 - 25) = 1.4 \times 10^5 - 10500 = 1.295 \times 10^5 \, \text{Pa} \).

1.25 × 10⁵ Pa
1.295 × 10⁵ Pa
1.35 × 10⁵ Pa
1.40 × 10⁵ Pa
2

What is the fundamental difference between gauge pressure and absolute pressure?

Absolute pressure is the total pressure, including atmospheric pressure (\( P = P_a + \rho g h \)), while gauge pressure is the excess pressure above atmospheric pressure (\( P_g = \rho g h \)), as defined in the chapter.

Gauge pressure includes viscosity
Absolute pressure includes atmospheric pressure
Gauge pressure is always higher
Absolute pressure excludes depth
2

Water flows at \( 2.0 \, \text{m/s} \) at \( 1.5 \, \text{m} \) height with pressure \( 1.65 \times 10^5 \, \text{Pa} \). What is the pressure at \( 0.2 \, \text{m} \) height with speed \( 3.0 \, \text{m/s} \)? (\( \rho = 1000 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \))

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: \( 1.65 \times 10^5 + \frac{1}{2} \times 1000 \times 4 + 1000 \times 9.8 \times 1.5 = 1.65 \times 10^5 + 2000 + 14700 = 1.817 \times 10^5 \).

Right: \( P_2 + \frac{1}{2} \times 1000 \times 9 + 1000 \times 9.8 \times 0.2 = P_2 + 4500 + 1960 = P_2 + 6460 \).

\( 1.817 \times 10^5 = P_2 + 6460 \), \( P_2 = 1.7524 \times 10^5 \, \text{Pa} \).

1.70 × 10⁵ Pa
1.75 × 10⁵ Pa
1.80 × 10⁵ Pa
1.85 × 10⁵ Pa
2

Water flows at \( 3 \, \text{m/s} \) at \( 2 \, \text{m} \) height with pressure \( 1.3 \times 10^5 \, \text{Pa} \). What is the pressure at \( 0 \, \text{m} \) height with speed \( 4 \, \text{m/s} \)? (\( \rho = 1000 \, \text{kg/m}^3 \), \( g = 10 \, \text{m/s}^2 \))

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: \( 1.3 \times 10^5 + \frac{1}{2} \times 1000 \times 9 + 1000 \times 10 \times 2 = 1.3 \times 10^5 + 4500 + 20000 = 1.545 \times 10^5 \).

Right: \( P_2 + \frac{1}{2} \times 1000 \times 16 = P_2 + 8000 \).

\( 1.545 \times 10^5 = P_2 + 8000 \), \( P_2 = 1.465 \times 10^5 \, \text{Pa} \).

1.40 × 10⁵ Pa
1.465 × 10⁵ Pa
1.50 × 10⁵ Pa
1.55 × 10⁵ Pa
2

What prevents a fluid from flowing when the pressure gradient is zero?

A pressure gradient drives fluid flow by creating a net force. When the gradient is zero (uniform pressure), no net force exists to initiate motion, keeping the fluid at rest unless acted upon by other forces.

High viscosity
Surface tension
No net force
Turbulence
3

A hydraulic lift raises a \( 1900 \, \text{kg} \) load using a small piston of radius \( 4.5 \, \text{cm} \) and a large piston of radius \( 18 \, \text{cm} \). What force is applied on the small piston? (Take \( g = 9.8 \, \text{m/s}^2 \))

\( F_1 = \frac{A_1}{A_2} F_2 \), \( F_2 = 1900 \times 9.8 = 18620 \, \text{N} \).

\( A_1 = \pi (0.045)^2 \), \( A_2 = \pi (0.18)^2 \), \( \frac{A_1}{A_2} = \frac{0.002026}{0.0324} = \frac{1}{16} \).

\( F_1 = \frac{18620}{16} = 1163.75 \, \text{N} \approx 1164 \, \text{N} \).

1000 N
1100 N
1164 N
1200 N
3

An aircraft (\( m = 2.5 \times 10^5 \, \text{kg} \), wing area \( 500 \, \text{m}^2 \)) flies level. What is the pressure difference across the wings? (Take \( g = 9.8 \, \text{m/s}^2 \))

\( \Delta P \cdot A = mg \).

\( mg = 2.5 \times 10^5 \times 9.8 = 2.45 \times 10^6 \, \text{N} \), \( A = 500 \, \text{m}^2 \).

\( \Delta P = \frac{2.45 \times 10^6}{500} = 4900 \, \text{Pa} \).

4700 Pa
4800 Pa
4900 Pa
5000 Pa
3

A soap bubble of radius \( 4 \, \text{mm} \) has a surface tension of \( 0.025 \, \text{N/m} \). What is the excess pressure inside the bubble?

Excess pressure in a bubble: \( \Delta P = \frac{4 S}{r} \).

\( S = 0.025 \, \text{N/m} \), \( r = 4 \times 10^{-3} \, \text{m} \).

\( \Delta P = \frac{4 \times 0.025}{4 \times 10^{-3}} = 25 \, \text{Pa} \).

12.5 Pa
25 Pa
50 Pa
100 Pa
2

A sphere of radius \( 0.07 \, \text{m} \) moves at \( 0.14 \, \text{m/s} \) through honey (\( \eta = 0.2 \, \text{Pa s} \)). What is the viscous drag force?

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

\( \eta = 0.2 \, \text{Pa s} \), \( a = 0.07 \, \text{m} \), \( v = 0.14 \, \text{m/s} \).

\( F = 6 \times 3.14 \times 0.2 \times 0.07 \times 0.14 = 0.0369 \, \text{N} \).

0.03 N
0.036 N
0.04 N
0.045 N
2

What is the primary reason gases are more compressible than liquids?

Gases have weaker intermolecular forces and larger intermolecular spaces compared to liquids, allowing them to compress significantly under pressure, while liquids’ strong forces resist volume changes.

Higher viscosity
Lower density
Weaker intermolecular forces
Higher surface tension
3

A soap bubble of radius \( 12 \, \text{mm} \) has a surface tension of \( 0.024 \, \text{N/m} \). What is the excess pressure inside?

Excess pressure in a bubble: \( \Delta P = \frac{4 S}{r} \).

\( S = 0.024 \, \text{N/m} \), \( r = 12 \times 10^{-3} \, \text{m} \).

\( \Delta P = \frac{4 \times 0.024}{12 \times 10^{-3}} = 8 \, \text{Pa} \).

6 Pa
8 Pa
10 Pa
12 Pa
2

A soap bubble of radius \( 9 \, \text{mm} \) has a surface tension of \( 0.026 \, \text{N/m} \). What is the excess pressure inside?

Excess pressure in a bubble: \( \Delta P = \frac{4 S}{r} \).

\( S = 0.026 \, \text{N/m} \), \( r = 9 \times 10^{-3} \, \text{m} \).

\( \Delta P = \frac{4 \times 0.026}{9 \times 10^{-3}} = 11.56 \, \text{Pa} \).

8 Pa
11.5 Pa
14 Pa
16 Pa
2

Why does the velocity of a fluid layer in contact with a stationary surface become zero?

In viscous flow, the no-slip condition states that the fluid layer in contact with a stationary surface adheres to it due to frictional forces, resulting in zero velocity relative to the surface.

Due to high pressure
Because of the no-slip condition
Due to streamline flow
Because of low viscosity
2

A hydraulic press applies \( 200 \, \text{N} \) on a small piston of area \( 0.01 \, \text{m}^2 \). What force is exerted by a large piston of area \( 0.04 \, \text{m}^2 \)?

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

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

\( F_1 = 200 \, \text{N} \), \( A_1 = 0.01 \, \text{m}^2 \), \( A_2 = 0.04 \, \text{m}^2 \).

\( F_2 = 200 \times \frac{0.04}{0.01} = 200 \times 4 = 800 \, \text{N} \).

600 N
700 N
800 N
900 N
3

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