Mechanical Properties of Fluids Chapter-Wise Test 9

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Why does Bernoulli’s principle not apply to fluids with high viscosity?

Bernoulli’s principle assumes no energy loss due to friction, which is valid for non-viscous fluids. High viscosity causes frictional losses between fluid layers, converting kinetic energy into heat, thus violating the energy conservation assumption.

It applies only to turbulent flow
Viscosity causes energy loss due to friction
High viscosity increases pressure
It requires compressibility
2

Which of the following statements is correct about a liquid drop?

Surface tension minimizes surface area, making a sphere (with the least surface area for a given volume) the natural shape of a liquid drop in the absence of external forces like gravity.

It is flat due to pressure
It is spherical due to surface tension
It forms a cube due to viscosity
It spreads due to gravity
2

A hydraulic lift raises a \( 1800 \, \text{kg} \) load using a small piston of radius \( 3 \, \text{cm} \) and a large piston of radius \( 15 \, \text{cm} \). What force is applied on the small piston? (Take \( g = 9.8 \, \text{m/s}^2 \))

\( F_1 = \frac{A_1}{A_2} F_2 \), \( F_2 = 1800 \times 9.8 = 17640 \, \text{N} \).

\( A_1 = \pi (0.03)^2 \), \( A_2 = \pi (0.15)^2 \), \( \frac{A_1}{A_2} = \frac{0.0009}{0.0225} = \frac{1}{25} \).

\( F_1 = \frac{17640}{25} = 705.6 \, \text{N} \approx 706 \, \text{N} \).

600 N
706 N
800 N
900 N
2

Why does a spinning ball deviate from a straight path in air, as explained by the Magnus effect?

The Magnus effect occurs due to Bernoulli’s principle: a spinning ball drags air, creating a velocity difference (faster on one side, slower on the other), resulting in a pressure difference that generates a net force perpendicular to its path.

Due to gravitational pull
Due to pressure difference from spinning
Because of increased viscosity
Due to streamline crossing
2

A pipe’s cross-sectional area changes from \( 0.1 \, \text{m}^2 \) to \( 0.025 \, \text{m}^2 \). If water flows at \( 2.2 \, \text{m/s} \) in the larger section, what is the speed in the smaller section?

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

\( A_1 = 0.1 \, \text{m}^2 \), \( v_1 = 2.2 \, \text{m/s} \), \( A_2 = 0.025 \, \text{m}^2 \).

\( v_2 = \frac{A_1 v_1}{A_2} = \frac{0.1 \times 2.2}{0.025} = 8.8 \, \text{m/s} \).

6.6 m/s
7.7 m/s
8.8 m/s
9.9 m/s
3

A manometer with mercury (\( \rho = 13.6 \times 10^3 \, \text{kg/m}^3 \)) shows a height difference of \( 0.28 \, \text{m} \). What is the pressure difference? (Take \( g = 10 \, \text{m/s}^2 \))

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

\( \rho = 13.6 \times 10^3 \, \text{kg/m}^3 \), \( g = 10 \, \text{m/s}^2 \), \( h = 0.28 \, \text{m} \).

\( \Delta P = 13.6 \times 10^3 \times 10 \times 0.28 = 38080 \, \text{Pa} \).

3.6 × 10⁴ Pa
3.8 × 10⁴ Pa
4.0 × 10⁴ Pa
4.2 × 10⁴ Pa
2

A pipe narrows from \( 0.06 \, \text{m}^2 \) to \( 0.02 \, \text{m}^2 \). If water flows at \( 1.8 \, \text{m/s} \) in the wider section, what is the speed in the narrower section?

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

\( A_1 = 0.06 \, \text{m}^2 \), \( v_1 = 1.8 \, \text{m/s} \), \( A_2 = 0.02 \, \text{m}^2 \).

\( v_2 = \frac{A_1 v_1}{A_2} = \frac{0.06 \times 1.8}{0.02} = 5.4 \, \text{m/s} \).

3.6 m/s
4.5 m/s
5.4 m/s
6.0 m/s
3

Why does water rise in a capillary tube against gravity?

Capillary rise occurs due to surface tension, where the adhesive forces between water and the tube wall exceed cohesive forces within water, creating an upward force balanced by the pressure difference across the curved meniscus.

Due to viscosity
Because of surface tension
Due to atmospheric pressure alone
Because of fluid density
2

What is the total pressure at a depth of \( 3.6 \, \text{m} \) in mercury (\( \rho = 13.6 \times 10^3 \, \text{kg/m}^3 \)) with atmospheric pressure \( 1.01 \times 10^5 \, \text{Pa} \)? (Take \( g = 9.8 \, \text{m/s}^2 \))

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

\( P_a = 1.01 \times 10^5 \, \text{Pa} \), \( \rho = 13.6 \times 10^3 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \), \( h = 3.6 \, \text{m} \).

\( P = 1.01 \times 10^5 + 13.6 \times 10^3 \times 9.8 \times 3.6 = 1.01 \times 10^5 + 479808 = 5.80808 \times 10^5 \, \text{Pa} \).

5.7 × 10⁵ Pa
5.8 × 10⁵ Pa
5.9 × 10⁵ Pa
6.0 × 10⁵ Pa
2

What is the significance of the term \( \rho g h \) in the pressure equation for a fluid at rest?

\( \rho g h \) represents the hydrostatic pressure due to the weight of the fluid column above a point, increasing with depth, as it accounts for gravitational potential energy per unit volume.

Kinetic energy
Surface tension pressure
Hydrostatic pressure
Viscous resistance
3

What is the gauge pressure at a depth of \( 3.5 \, \text{m} \) in seawater (\( \rho = 1.03 \times 10^3 \, \text{kg/m}^3 \))? (Take \( g = 10 \, \text{m/s}^2 \))

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

\( \rho = 1.03 \times 10^3 \, \text{kg/m}^3 \), \( g = 10 \, \text{m/s}^2 \), \( h = 3.5 \, \text{m} \).

\( P_g = 1.03 \times 10^3 \times 10 \times 3.5 = 3.605 \times 10^4 \, \text{Pa} \).

3.0 × 10⁴ Pa
3.6 × 10⁴ Pa
4.0 × 10⁴ Pa
4.5 × 10⁴ Pa
2

Why does atmospheric pressure decrease with altitude?

Atmospheric pressure is the weight of the air column above a point. As altitude increases, the height and mass of the air column decrease, reducing the pressure exerted at higher elevations.

Increased viscosity
Higher temperature
Less air column weight
Reduced turbulence
3

A manometer with seawater (\( \rho = 1.03 \times 10^3 \, \text{kg/m}^3 \)) shows a height difference of \( 0.22 \, \text{m} \). What is the pressure difference? (Take \( g = 9.8 \, \text{m/s}^2 \))

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

\( \rho = 1.03 \times 10^3 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \), \( h = 0.22 \, \text{m} \).

\( \Delta P = 1.03 \times 10^3 \times 9.8 \times 0.22 = 2220.68 \, \text{Pa} \).

2000 Pa
2200 Pa
2400 Pa
2600 Pa
2

A plate of area \( 0.07 \, \text{m}^2 \) moves at \( 0.6 \, \text{m/s} \) over a \( 0.9 \, \text{mm} \) thick machine oil film (\( \eta = 0.113 \, \text{Pa s} \)). What force is required?

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

\( \eta = 0.113 \, \text{Pa s} \), \( v = 0.6 \, \text{m/s} \), \( A = 0.07 \, \text{m}^2 \), \( l = 0.9 \times 10^{-3} \, \text{m} \).

\( F = 0.113 \times \frac{0.6 \times 0.07}{0.9 \times 10^{-3}} = 0.113 \times 46.67 = 5.27 \, \text{N} \).

4.8 N
5.0 N
5.3 N
5.5 N
3

Which of the following statements is correct about viscosity?

Viscosity measures a fluid’s resistance to flow due to internal friction between layers, decreasing with temperature in liquids as cohesive forces weaken, not increasing.

It increases with temperature in liquids
It measures resistance to flow
It is independent of shear stress
It decreases with density
2

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