Mechanical Properties of Fluids Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Water (\( \rho = 1000 \, \text{kg/m}^3 \)) flows horizontally at \( 2.8 \, \text{m/s} \) with pressure \( 1.9 \times 10^5 \, \text{Pa} \). If the speed increases to \( 5.2 \, \text{m/s} \), what is the new pressure?

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

\( P_1 = 1.9 \times 10^5 \, \text{Pa} \), \( v_1 = 2.8 \, \text{m/s} \), \( v_2 = 5.2 \, \text{m/s} \), \( \rho = 1000 \, \text{kg/m}^3 \).

\( P_2 = 1.9 \times 10^5 + \frac{1}{2} \times 1000 (7.84 - 27.04) = 1.9 \times 10^5 - 9600 = 1.804 \times 10^5 \, \text{Pa} \).

1.75 × 10⁵ Pa
1.80 × 10⁵ Pa
1.85 × 10⁵ Pa
1.90 × 10⁵ Pa
2

What is the excess pressure inside an ethanol drop of radius \( 1.5 \, \text{mm} \) at \( 20^\circ \text{C} \)? (Surface tension = \( 0.0227 \, \text{N/m} \))

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

\( S = 0.0227 \, \text{N/m} \), \( r = 1.5 \times 10^{-3} \, \text{m} \).

\( \Delta P = \frac{2 \times 0.0227}{1.5 \times 10^{-3}} = 30.27 \, \text{Pa} \).

15 Pa
30 Pa
45 Pa
60 Pa
2

A soap film supports \( 3.6 \times 10^{-2} \, \text{N} \) over a \( 60 \, \text{cm} \) slider. What is the surface tension?

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

\( F = 3.6 \times 10^{-2} \, \text{N} \), \( l = 0.6 \, \text{m} \).

\( S = \frac{3.6 \times 10^{-2}}{2 \times 0.6} = 0.03 \, \text{N/m} \).

0.025 N/m
0.028 N/m
0.030 N/m
0.033 N/m
3

Which statement is incorrect about hydraulic systems?

Hydraulic systems rely on incompressible fluids to transmit pressure effectively (Pascal’s law). The incorrect statement is that they use highly compressible fluids, which would reduce efficiency.

They use incompressible fluids
Pressure is transmitted equally
They use highly compressible fluids
Force is amplified by area ratio
3

Water (\( \rho = 1000 \, \text{kg/m}^3 \)) flows horizontally at \( 4.5 \, \text{m/s} \) with pressure \( 2.0 \times 10^5 \, \text{Pa} \). If the speed increases to \( 7.0 \, \text{m/s} \), what is the new pressure?

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

\( P_1 = 2.0 \times 10^5 \, \text{Pa} \), \( v_1 = 4.5 \, \text{m/s} \), \( v_2 = 7.0 \, \text{m/s} \), \( \rho = 1000 \, \text{kg/m}^3 \).

\( P_2 = 2.0 \times 10^5 + \frac{1}{2} \times 1000 (20.25 - 49) = 2.0 \times 10^5 - 14375 = 1.85625 \times 10^5 \, \text{Pa} \).

1.85 × 10⁵ Pa
1.86 × 10⁵ Pa
1.87 × 10⁵ Pa
1.88 × 10⁵ Pa
2

Which of the following statements is correct about the equation of continuity?

The equation of continuity (\( A v = \text{constant} \)) assumes an incompressible fluid, ensuring mass flow rate conservation, not requiring viscosity or turbulence considerations.

It applies only to viscous fluids
It requires turbulent flow
It assumes incompressibility
It depends on pressure variation
3

What is the gauge pressure at a depth of \( 4.5 \, \text{m} \) in ethyl alcohol (\( \rho = 806 \, \text{kg/m}^3 \))? (Take \( g = 10 \, \text{m/s}^2 \))

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

\( \rho = 806 \, \text{kg/m}^3 \), \( g = 10 \, \text{m/s}^2 \), \( h = 4.5 \, \text{m} \).

\( P_g = 806 \times 10 \times 4.5 = 36270 \, \text{Pa} \).

3.2 × 10⁴ Pa
3.6 × 10⁴ Pa
4.0 × 10⁴ Pa
4.4 × 10⁴ Pa
2

A plate of area \( 0.06 \, \text{m}^2 \) moves at \( 0.6 \, \text{m/s} \) over a \( 0.4 \, \text{mm} \) thick glycerine film (\( \eta = 0.83 \, \text{Pa s} \)). What force is required?

Viscous force: \( F = \eta \frac{v A}{l} \).

\( \eta = 0.83 \, \text{Pa s} \), \( v = 0.6 \, \text{m/s} \), \( A = 0.06 \, \text{m}^2 \), \( l = 0.4 \times 10^{-3} \, \text{m} \).

\( F = 0.83 \times \frac{0.6 \times 0.06}{0.4 \times 10^{-3}} = 0.83 \times 90 = 74.7 \, \text{N} \).

60 N
70 N
74.7 N
80 N
3

A soap bubble of radius \( 15 \, \text{mm} \) has a surface tension of \( 0.023 \, \text{N/m} \). What is the excess pressure inside?

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

\( S = 0.023 \, \text{N/m} \), \( r = 15 \times 10^{-3} \, \text{m} \).

\( \Delta P = \frac{4 \times 0.023}{15 \times 10^{-3}} = 6.13 \, \text{Pa} \).

5 Pa
6 Pa
7 Pa
8 Pa
2

A hydraulic system applies \( 150 \, \text{N} \) on a small piston of area \( 0.015 \, \text{m}^2 \). What force is exerted by a large piston of area \( 0.075 \, \text{m}^2 \)?

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

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

\( F_1 = 150 \, \text{N} \), \( A_1 = 0.015 \, \text{m}^2 \), \( A_2 = 0.075 \, \text{m}^2 \).

\( F_2 = 150 \times \frac{0.075}{0.015} = 150 \times 5 = 750 \, \text{N} \).

600 N
700 N
750 N
800 N
3

What property of a fluid does the coefficient of viscosity measure?

The coefficient of viscosity (\( \eta \)) measures a fluid’s internal resistance to flow, defined as the ratio of shear stress to the strain rate (\( \eta = F l / v A \)), reflecting its “thickness” or friction between layers.

Density
Surface tension
Resistance to flow
Compressibility
3

A manometer with seawater (\( \rho = 1.03 \times 10^3 \, \text{kg/m}^3 \)) shows a height difference of \( 0.19 \, \text{m} \). What is the pressure difference? (Take \( g = 10 \, \text{m/s}^2 \))

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

\( \rho = 1.03 \times 10^3 \, \text{kg/m}^3 \), \( g = 10 \, \text{m/s}^2 \), \( h = 0.19 \, \text{m} \).

\( \Delta P = 1.03 \times 10^3 \times 10 \times 0.19 = 1957 \, \text{Pa} \).

1800 Pa
1900 Pa
1957 Pa
2000 Pa
3

An aircraft (\( m = 3.0 \times 10^5 \, \text{kg} \), wing area \( 450 \, \text{m}^2 \)) flies level. What is the pressure difference across the wings? (Take \( g = 10 \, \text{m/s}^2 \))

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

\( mg = 3.0 \times 10^5 \times 10 = 3.0 \times 10^6 \, \text{N} \), \( A = 450 \, \text{m}^2 \).

\( \Delta P = \frac{3.0 \times 10^6}{450} = 6666.67 \, \text{Pa} \approx 6667 \, \text{Pa} \).

6000 Pa
6500 Pa
6667 Pa
7000 Pa
3

A capillary tube of radius \( 0.55 \, \text{mm} \) 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 \( g = 10 \, \text{m/s}^2 \))

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

\( S = 0.0727 \, \text{N/m} \), \( \rho = 1000 \, \text{kg/m}^3 \), \( g = 10 \, \text{m/s}^2 \), \( a = 0.55 \times 10^{-3} \, \text{m} \).

\( h = \frac{2 \times 0.0727 \times 1}{1000 \times 10 \times 0.55 \times 10^{-3}} = 0.02645 \, \text{m} = 2.645 \, \text{cm} \).

2.5 cm
2.6 cm
2.7 cm
2.8 cm
3

What is the excess pressure inside an oxygen drop of radius \( 0.5 \, \text{mm} \) at \( -183^\circ \text{C} \)? (Surface tension = \( 0.0132 \, \text{N/m} \))

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

\( S = 0.0132 \, \text{N/m} \), \( r = 0.5 \times 10^{-3} \, \text{m} \).

\( \Delta P = \frac{2 \times 0.0132}{0.5 \times 10^{-3}} = 52.8 \, \text{Pa} \).

40 Pa
53 Pa
60 Pa
70 Pa
2

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