Mechanical Properties of Fluids Chapter-Wise Test 4

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the absolute pressure at \( 450 \, \text{m} \) depth in seawater (\( \rho = 1.03 \times 10^3 \, \text{kg/m}^3 \)) with \( P_a = 1.01 \times 10^5 \, \text{Pa} \)? (Take \( g = 9.8 \, \text{m/s}^2 \))

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

\( P_a = 1.01 \times 10^5 \, \text{Pa} \), \( \rho = 1.03 \times 10^3 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \), \( h = 450 \, \text{m} \).

\( P = 1.01 \times 10^5 + 1.03 \times 10^3 \times 9.8 \times 450 = 1.01 \times 10^5 + 4.5411 \times 10^6 = 4.6421 \times 10^6 \, \text{Pa} \).

4.5 × 10⁶ Pa
4.6 × 10⁶ Pa
4.7 × 10⁶ Pa
4.8 × 10⁶ Pa
2

A hydraulic lift raises a \( 1700 \, \text{kg} \) load using a small piston of radius \( 3.5 \, \text{cm} \) and a large piston of radius \( 14 \, \text{cm} \). What force is applied on the small piston? (Take \( g = 10 \, \text{m/s}^2 \))

\( F_1 = \frac{A_1}{A_2} F_2 \), \( F_2 = 1700 \times 10 = 17000 \, \text{N} \).

\( A_1 = \pi (0.035)^2 \), \( A_2 = \pi (0.14)^2 \), \( \frac{A_1}{A_2} = \frac{0.001225}{0.0196} = \frac{1}{16} \).

\( F_1 = \frac{17000}{16} = 1062.5 \, \text{N} \approx 1063 \, \text{N} \).

900 N
1000 N
1063 N
1100 N
3

A sphere of radius \( 0.05 \, \text{m} \) moves at \( 0.11 \, \text{m/s} \) through blood (\( \eta = 2.7 \times 10^{-3} \, \text{Pa s} \)). What is the viscous drag force?

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

\( \eta = 2.7 \times 10^{-3} \, \text{Pa s} \), \( a = 0.05 \, \text{m} \), \( v = 0.11 \, \text{m/s} \).

\( F = 6 \times 3.14 \times 2.7 \times 10^{-3} \times 0.05 \times 0.11 = 0.0028 \, \text{N} \).

0.0025 N
0.0028 N
0.0030 N
0.0032 N
2

Water (\( \rho = 1000 \, \text{kg/m}^3 \)) flows horizontally at \( 4.2 \, \text{m/s} \) with pressure \( 1.95 \times 10^5 \, \text{Pa} \). If the speed increases to \( 6.5 \, \text{m/s} \), what is the new pressure?

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

\( P_1 = 1.95 \times 10^5 \, \text{Pa} \), \( v_1 = 4.2 \, \text{m/s} \), \( v_2 = 6.5 \, \text{m/s} \), \( \rho = 1000 \, \text{kg/m}^3 \).

\( P_2 = 1.95 \times 10^5 + \frac{1}{2} \times 1000 (17.64 - 42.25) = 1.95 \times 10^5 - 12305 = 1.82695 \times 10^5 \, \text{Pa} \).

1.80 × 10⁵ Pa
1.83 × 10⁵ Pa
1.85 × 10⁵ Pa
1.90 × 10⁵ Pa
2

Which statement is incorrect about Stokes’ law?

Stokes’ law (\( F = 6 \pi \eta a v \)) applies to small spherical objects in viscous, streamline flow, not turbulent flow, where inertial forces dominate. The incorrect statement is that it applies to turbulent flow.

It applies to spherical objects
It involves viscous force
It applies to turbulent flow
Velocity is constant at terminal speed
3

What is the reason a small insect can walk on water?

Surface tension creates a “skin-like” effect at the water’s surface, supporting the insect’s weight if the force per unit length (\( S \)) exceeds the downward force distributed over its contact area, due to its small size and light weight.

High viscosity
Low density of the insect
Surface tension supports its weight
Buoyant force alone
3

What is the excess pressure inside a water drop of radius \( 2.5 \, \text{mm} \) at \( 20^\circ \text{C} \)? (Surface tension = \( 0.0727 \, \text{N/m} \))

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

\( S = 0.0727 \, \text{N/m} \), \( r = 2.5 \times 10^{-3} \, \text{m} \).

\( \Delta P = \frac{2 \times 0.0727}{2.5 \times 10^{-3}} = 58.16 \, \text{Pa} \).

40 Pa
50 Pa
58 Pa
65 Pa
3

What is the total pressure at a depth of \( 2.2 \, \text{m} \) in hydrogen liquid (\( \rho = 9.0 \times 10^{-2} \, \text{kg/m}^3 \)) with atmospheric pressure \( 1.01 \times 10^5 \, \text{Pa} \)? (Take \( g = 9.8 \, \text{m/s}^2 \))

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

\( P_a = 1.01 \times 10^5 \, \text{Pa} \), \( \rho = 9.0 \times 10^{-2} \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \), \( h = 2.2 \, \text{m} \).

\( P = 1.01 \times 10^5 + 9.0 \times 10^{-2} \times 9.8 \times 2.2 = 1.01 \times 10^5 + 1.94 = 1.0100194 \times 10^5 \, \text{Pa} \).

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

A tank has a hole \( 1.4 \, \text{m} \) below the water surface, open to the atmosphere. What is the efflux speed? (Take \( g = 9.8 \, \text{m/s}^2 \))

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

\( g = 9.8 \, \text{m/s}^2 \), \( h = 1.4 \, \text{m} \).

\( v = \sqrt{2 \times 9.8 \times 1.4} = \sqrt{27.44} \approx 5.24 \, \text{m/s} \).

4.8 m/s
5.0 m/s
5.2 m/s
5.5 m/s
3

A bubble of radius \( 5.5 \, \text{mm} \) is blown at \( 45 \, \text{cm} \) depth in water (\( \rho = 1000 \, \text{kg/m}^3 \), \( S = 0.0727 \, \text{N/m} \)). What is the total pressure inside? (Take \( P_a = 1.01 \times 10^5 \, \text{Pa} \), \( g = 10 \, \text{m/s}^2 \))

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

\( P_a = 1.01 \times 10^5 \, \text{Pa} \), \( \rho g h = 1000 \times 10 \times 0.45 = 4500 \, \text{Pa} \).

\( \frac{2 S}{r} = \frac{2 \times 0.0727}{5.5 \times 10^{-3}} = 26.44 \, \text{Pa} \).

\( P_i = 1.01 \times 10^5 + 4500 + 26.44 = 1.05526 \times 10^5 \, \text{Pa} \).

1.04 × 10⁵ Pa
1.05 × 10⁵ Pa
1.055 × 10⁵ Pa
1.06 × 10⁵ Pa
3

A manometer with water (\( \rho = 1000 \, \text{kg/m}^3 \)) shows a height difference of \( 0.15 \, \text{m} \). What is the pressure difference? (Take \( g = 9.8 \, \text{m/s}^2 \))

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

\( \rho = 1000 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \), \( h = 0.15 \, \text{m} \).

\( \Delta P = 1000 \times 9.8 \times 0.15 = 1470 \, \text{Pa} \).

1200 Pa
1470 Pa
1600 Pa
1800 Pa
2

Why does the pressure inside a liquid drop exceed the external pressure?

Surface tension creates a contractile force at the drop’s surface, requiring higher internal pressure (\( \Delta P = 2S/r \)) to balance this inward force, as the curved surface tends to reduce its area.

Due to viscosity
Because of surface tension
Due to fluid flow
Because of density variation
2

A soap film supports \( 2.2 \times 10^{-2} \, \text{N} \) over a \( 44 \, \text{cm} \) slider. What is the surface tension?

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

\( F = 2.2 \times 10^{-2} \, \text{N} \), \( l = 0.44 \, \text{m} \).

\( S = \frac{2.2 \times 10^{-2}}{2 \times 0.44} = 0.025 \, \text{N/m} \).

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

Why does a fluid at rest exert the same pressure at all points at the same horizontal level?

Pascal’s law states that pressure in a fluid at rest is transmitted equally in all directions. At the same horizontal level, the depth (\( h \)) is constant, so \( P = P_a + \rho g h \) is uniform, assuming no horizontal flow.

Due to viscosity
Because of constant depth
Due to surface tension
Because of turbulence
2

A soap film supports \( 1.8 \times 10^{-2} \, \text{N} \) over a \( 35 \, \text{cm} \) slider. What is the surface tension?

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

\( F = 1.8 \times 10^{-2} \, \text{N} \), \( l = 0.35 \, \text{m} \).

\( S = \frac{1.8 \times 10^{-2}}{2 \times 0.35} = 0.0257 \, \text{N/m} \).

0.020 N/m
0.0257 N/m
0.030 N/m
0.035 N/m
2

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