Mechanical Properties of Fluids Chapter-Wise Test 13

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Which of the following statements is correct about terminal velocity?

Terminal velocity occurs when the viscous force (proportional to velocity) equals the net gravitational force, resulting in zero acceleration and constant speed, as per Stokes’ law.

It increases with fluid viscosity
It occurs when forces balance
It requires turbulent flow
It decreases with object size
2

A soap bubble of radius \( 7 \, \text{mm} \) has a surface tension of \( 0.028 \, \text{N/m} \). What is the excess pressure inside?

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

\( S = 0.028 \, \text{N/m} \), \( r = 7 \times 10^{-3} \, \text{m} \).

\( \Delta P = \frac{4 \times 0.028}{7 \times 10^{-3}} = 16 \, \text{Pa} \).

12 Pa
16 Pa
20 Pa
24 Pa
2

An aircraft (\( m = 2.0 \times 10^5 \, \text{kg} \), wing area \( 400 \, \text{m}^2 \)) flies level. What is the pressure difference across the wings? (Take \( g = 9.8 \, \text{m/s}^2 \))

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

\( mg = 2.0 \times 10^5 \times 9.8 = 1.96 \times 10^6 \, \text{N} \), \( A = 400 \, \text{m}^2 \).

\( \Delta P = \frac{1.96 \times 10^6}{400} = 4900 \, \text{Pa} \).

4500 Pa
4700 Pa
4900 Pa
5100 Pa
3

A plate of area \( 0.08 \, \text{m}^2 \) moves at \( 0.8 \, \text{m/s} \) over a \( 0.7 \, \text{mm} \) thick glycerine film (\( \eta = 0.83 \, \text{Pa s} \)). What force is required?

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

\( \eta = 0.83 \, \text{Pa s} \), \( v = 0.8 \, \text{m/s} \), \( A = 0.08 \, \text{m}^2 \), \( l = 0.7 \times 10^{-3} \, \text{m} \).

\( F = 0.83 \times \frac{0.8 \times 0.08}{0.7 \times 10^{-3}} = 0.83 \times 91.43 = 75.89 \, \text{N} \).

70 N
75 N
76 N
80 N
3

A sphere of radius \( 0.04 \, \text{m} \) moves at \( 0.12 \, \text{m/s} \) through honey (\( \eta = 0.2 \, \text{Pa s} \)). What is the viscous drag force?

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

\( \eta = 0.2 \, \text{Pa s} \), \( a = 0.04 \, \text{m} \), \( v = 0.12 \, \text{m/s} \).

\( F = 6 \times 3.14 \times 0.2 \times 0.04 \times 0.12 = 0.1809 \, \text{N} \).

0.15 N
0.18 N
0.20 N
0.22 N
2

What ensures that a fluid at rest cannot sustain a shear stress?

Fluids lack rigidity and cannot resist shear stress without flowing. Any attempt to apply shear stress causes deformation (flow), as fluids have no fixed shape, unlike solids, which resist shear due to strong intermolecular forces.

High viscosity
Lack of rigidity
Surface tension
Compressibility
2

Water (\( \rho = 1000 \, \text{kg/m}^3 \)) flows horizontally at \( 4 \, \text{m/s} \) with pressure \( 1.6 \times 10^5 \, \text{Pa} \). If the speed increases to \( 7 \, \text{m/s} \), what is the new pressure?

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

\( P_1 = 1.6 \times 10^5 \, \text{Pa} \), \( v_1 = 4 \, \text{m/s} \), \( v_2 = 7 \, \text{m/s} \), \( \rho = 1000 \, \text{kg/m}^3 \).

\( P_2 = 1.6 \times 10^5 + \frac{1}{2} \times 1000 (16 - 49) = 1.6 \times 10^5 - 16500 = 1.435 \times 10^5 \, \text{Pa} \).

1.40 × 10⁵ Pa
1.435 × 10⁵ Pa
1.50 × 10⁵ Pa
1.55 × 10⁵ Pa
2

A hydraulic lift raises a \( 1600 \, \text{kg} \) load using a small piston of radius \( 4 \, \text{cm} \) and a large piston of radius \( 16 \, \text{cm} \). What force is applied on the small piston? (Take \( g = 10 \, \text{m/s}^2 \))

\( F_1 = \frac{A_1}{A_2} F_2 \), \( F_2 = 1600 \times 10 = 16000 \, \text{N} \).

\( A_1 = \pi (0.04)^2 \), \( A_2 = \pi (0.16)^2 \), \( \frac{A_1}{A_2} = \frac{0.0016}{0.0256} = \frac{1}{16} \).

\( F_1 = \frac{16000}{16} = 1000 \, \text{N} \).

800 N
1000 N
1200 N
1400 N
2

Why does the pressure at the bottom of three differently shaped vessels connected at the base remain the same when filled with water to the same height?

The hydrostatic paradox demonstrates that pressure in a fluid at rest depends only on depth (\( \rho g h \)) and not on the shape or volume of the container. At the same depth, the pressure is identical regardless of the vessel’s shape.

Due to varying volumes of water
Because pressure depends only on depth
Due to surface tension effects
Because of different cross-sectional areas
2

A manometer with mercury (\( \rho = 13.6 \times 10^3 \, \text{kg/m}^3 \)) shows a height difference of \( 0.25 \, \text{m} \). What is the pressure difference? (Take \( g = 9.8 \, \text{m/s}^2 \))

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

\( \rho = 13.6 \times 10^3 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \), \( h = 0.25 \, \text{m} \).

\( \Delta P = 13.6 \times 10^3 \times 9.8 \times 0.25 = 33320 \, \text{Pa} \).

3.0 × 10⁴ Pa
3.33 × 10⁴ Pa
3.5 × 10⁴ Pa
3.8 × 10⁴ Pa
2

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