Mechanical Properties of Fluids Chapter-Wise Test 6

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the reason a liquid’s terminal velocity depends on its radius when falling through a viscous medium?

Stokes’ law (\( F = 6 \pi \eta a v \)) shows that viscous force is proportional to radius (\( a \)), while weight (\( \frac{4}{3} \pi a^3 \rho g \)) depends on \( a^3 \). At terminal velocity, these balance, making \( v \propto a^2 \).

Viscosity decreases with radius
Surface tension varies
Viscous force depends on radius
Pressure changes with depth
3

In streamline flow, why do streamlines never cross each other?

In streamline flow, each streamline represents the path of a fluid particle with a unique velocity at each point. If streamlines crossed, a particle at the intersection would have two different velocities simultaneously, which is impossible in steady flow.

Due to high viscosity
Because each streamline has a unique velocity
Due to turbulent flow
Because of pressure differences
2

What is the excess pressure inside an ethanol drop of radius \( 2.2 \, \text{mm} \) at \( 20^\circ \text{C} \)? (Surface tension = \( 0.0227 \, \text{N/m} \))

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

\( S = 0.0227 \, \text{N/m} \), \( r = 2.2 \times 10^{-3} \, \text{m} \).

\( \Delta P = \frac{2 \times 0.0227}{2.2 \times 10^{-3}} = 20.64 \, \text{Pa} \).

18 Pa
20 Pa
22 Pa
24 Pa
3

What is the excess pressure inside a water drop of radius \( 1.8 \, \text{mm} \) at \( 20^\circ \text{C} \)? (Surface tension = \( 0.0727 \, \text{N/m} \))

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

\( S = 0.0727 \, \text{N/m} \), \( r = 1.8 \times 10^{-3} \, \text{m} \).

\( \Delta P = \frac{2 \times 0.0727}{1.8 \times 10^{-3}} = 80.78 \, \text{Pa} \).

70 Pa
80 Pa
90 Pa
100 Pa
2

A soap film supports \( 4.0 \times 10^{-2} \, \text{N} \) over a \( 40 \, \text{cm} \) slider. What is the surface tension?

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

\( F = 4.0 \times 10^{-2} \, \text{N} \), \( l = 0.4 \, \text{m} \).

\( S = \frac{4.0 \times 10^{-2}}{2 \times 0.4} = 0.05 \, \text{N/m} \).

0.04 N/m
0.045 N/m
0.05 N/m
0.055 N/m
3

What is the primary source of resistance to motion in a viscous fluid?

Viscosity, the internal friction between fluid layers moving at different velocities, resists relative motion, causing a shear stress proportional to the velocity gradient, as seen in Stokes’ law.

Surface tension
Pressure gradient
Viscosity
Gravity
3

A pipe narrows from \( 0.12 \, \text{m}^2 \) to \( 0.04 \, \text{m}^2 \). If water flows at \( 1.6 \, \text{m/s} \) in the wider section, what is the speed in the narrower section?

Continuity equation: \( A_1 v_1 = A_2 v_2 \).

\( A_1 = 0.12 \, \text{m}^2 \), \( v_1 = 1.6 \, \text{m/s} \), \( A_2 = 0.04 \, \text{m}^2 \).

\( v_2 = \frac{A_1 v_1}{A_2} = \frac{0.12 \times 1.6}{0.04} = 4.8 \, \text{m/s} \).

4.0 m/s
4.4 m/s
4.8 m/s
5.2 m/s
3

A plate of area \( 0.08 \, \text{m}^2 \) moves at \( 0.4 \, \text{m/s} \) over a \( 0.25 \, \text{mm} \) thick oil film (\( \eta = 0.034 \, \text{Pa s} \)). What force is required?

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

\( \eta = 0.034 \, \text{Pa s} \), \( v = 0.4 \, \text{m/s} \), \( A = 0.08 \, \text{m}^2 \), \( l = 0.25 \times 10^{-3} \, \text{m} \).

\( F = 0.034 \times \frac{0.4 \times 0.08}{0.25 \times 10^{-3}} = 0.034 \times 128 = 4.352 \, \text{N} \).

2.5 N
3.5 N
4.35 N
5.0 N
3

A tank has a hole \( 1.1 \, \text{m} \) below the water surface, open to the atmosphere. What is the efflux speed? (Take \( g = 10 \, \text{m/s}^2 \))

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

\( g = 10 \, \text{m/s}^2 \), \( h = 1.1 \, \text{m} \).

\( v = \sqrt{2 \times 10 \times 1.1} = \sqrt{22} \approx 4.69 \, \text{m/s} \).

4.0 m/s
4.7 m/s
5.0 m/s
5.5 m/s
2

An aircraft (\( m = 3.5 \times 10^5 \, \text{kg} \), wing area \( 550 \, \text{m}^2 \)) flies level. What is the pressure difference across the wings? (Take \( g = 10 \, \text{m/s}^2 \))

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

\( mg = 3.5 \times 10^5 \times 10 = 3.5 \times 10^6 \, \text{N} \), \( A = 550 \, \text{m}^2 \).

\( \Delta P = \frac{3.5 \times 10^6}{550} = 6363.64 \, \text{Pa} \approx 6364 \, \text{Pa} \).

6000 Pa
6200 Pa
6364 Pa
6500 Pa
3

Water flows at \( 2.6 \, \text{m/s} \) at \( 4.0 \, \text{m} \) height with pressure \( 1.7 \times 10^5 \, \text{Pa} \). What is the pressure at \( 2.0 \, \text{m} \) height with speed \( 3.8 \, \text{m/s} \)? (\( \rho = 1000 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \))

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: \( 1.7 \times 10^5 + \frac{1}{2} \times 1000 \times 6.76 + 1000 \times 9.8 \times 4.0 = 1.7 \times 10^5 + 3380 + 39200 = 2.1258 \times 10^5 \).

Right: \( P_2 + \frac{1}{2} \times 1000 \times 14.44 + 1000 \times 9.8 \times 2.0 = P_2 + 7220 + 19600 = P_2 + 26820 \).

\( 2.1258 \times 10^5 = P_2 + 26820 \), \( P_2 = 1.8578 \times 10^5 \, \text{Pa} \).

1.80 × 10⁵ Pa
1.85 × 10⁵ Pa
1.90 × 10⁵ Pa
1.95 × 10⁵ Pa
2

A sphere of radius \( 0.06 \, \text{m} \) moves at \( 0.13 \, \text{m/s} \) through glycerine (\( \eta = 0.83 \, \text{Pa s} \)). What is the viscous drag force?

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

\( \eta = 0.83 \, \text{Pa s} \), \( a = 0.06 \, \text{m} \), \( v = 0.13 \, \text{m/s} \).

\( F = 6 \times 3.14 \times 0.83 \times 0.06 \times 0.13 = 0.1222 \, \text{N} \).

0.10 N
0.12 N
0.14 N
0.16 N
2

What is the primary reason mercury does not wet glass?

Mercury has an obtuse angle of contact with glass because its cohesive forces (between mercury molecules) are stronger than its adhesive forces with glass, causing it to form droplets rather than spread.

High viscosity
Low density
Strong cohesive forces
Weak surface tension
3

A sphere of radius \( 0.02 \, \text{m} \) falls through glycerine (\( \eta = 0.83 \, \text{Pa s} \)) at a terminal velocity of \( 0.1 \, \text{m/s} \). What is the viscous drag force?

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

\( \eta = 0.83 \, \text{Pa s} \), \( a = 0.02 \, \text{m} \), \( v = 0.1 \, \text{m/s} \).

\( F = 6 \times 3.14 \times 0.83 \times 0.02 \times 0.1 \approx 0.0313 \, \text{N} \).

0.015 N
0.031 N
0.045 N
0.062 N
2

What causes the speed of efflux from a hole in a tank to equal the speed of a freely falling body from the same height?

Torricelli’s law (\( v = \sqrt{2gh} \)) equates the kinetic energy of the efflux (\( \frac{1}{2}mv^2 \)) to the potential energy lost (\( mgh \)) as fluid falls from the surface to the hole, mirroring the free fall equation (\( v = \sqrt{2gh} \)), assuming no energy losses.

Surface tension
Conversion of potential energy to kinetic energy
Viscous drag
Atmospheric pressure variation
2

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