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  4. Mechanical Properties Of Fluids Chapter-Wise Test 3

Mechanical Properties Of Fluids Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A sphere of radius \( 0.015 \, \text{m} \) moves at \( 0.2 \, \text{m/s} \) through water (\( \eta = 1.0 \times 10^{-3} \, \text{Pa s} \)). What is the viscous force?

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

\( \eta = 1.0 \times 10^{-3} \, \text{Pa s} \), \( a = 0.015 \, \text{m} \), \( v = 0.2 \, \text{m/s} \).

\( F = 6 \times 3.14 \times 1.0 \times 10^{-3} \times 0.015 \times 0.2 = 5.652 \times 10^{-5} \, \text{N} \).

2.8 × 10⁻⁵ N
4.2 × 10⁻⁵ N
5.65 × 10⁻⁵ N
7.0 × 10⁻⁵ N
3

What is the excess pressure inside a mercury drop of radius \( 4 \, \text{mm} \) at \( 20^\circ \text{C} \)? (Surface tension = \( 0.4355 \, \text{N/m} \))

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

\( S = 0.4355 \, \text{N/m} \), \( r = 4 \times 10^{-3} \, \text{m} \).

\( \Delta P = \frac{2 \times 0.4355}{4 \times 10^{-3}} = 217.75 \, \text{Pa} \).

200 Pa
218 Pa
230 Pa
250 Pa
2

A \( 1500 \, \text{kg} \) car is lifted by a hydraulic system with a small piston of radius \( 6 \, \text{cm} \) and a large piston of radius \( 18 \, \text{cm} \). What force is needed on the small piston? (Take \( g = 9.8 \, \text{m/s}^2 \))

\( F_1 = \frac{A_1}{A_2} F_2 \), \( F_2 = 1500 \times 9.8 = 14700 \, \text{N} \).

\( A_1 = \pi (0.06)^2 \), \( A_2 = \pi (0.18)^2 \), \( \frac{A_1}{A_2} = \frac{0.0036}{0.0324} = \frac{1}{9} \).

\( F_1 = \frac{14700}{9} = 1633.33 \, \text{N} \approx 1633 \, \text{N} \).

1200 N
1400 N
1633 N
1800 N
3

What is the primary reason a sharp needle pierces the skin more easily than a blunt object with the same force?

Pressure is defined as force per unit area (\( P = F/A \)). A sharp needle has a smaller contact area, resulting in higher pressure for the same force, making it easier to penetrate the skin compared to a blunt object with a larger area.

The needle has higher density
The needle exerts more force
The needle has a smaller area, increasing pressure
The blunt object resists more
3

What is the force on a submarine window (\( 0.06 \, \text{m}^2 \)) at \( 400 \, \text{m} \) depth in seawater (\( \rho = 1.03 \times 10^3 \, \text{kg/m}^3 \)), interior at atmospheric pressure? (Take \( g = 9.8 \, \text{m/s}^2 \))

Gauge pressure: \( P_g = \rho g h = 1.03 \times 10^3 \times 9.8 \times 400 = 4.0376 \times 10^6 \, \text{Pa} \).

\( F = P_g A = 4.0376 \times 10^6 \times 0.06 = 2.42256 \times 10^5 \, \text{N} \).

2.0 × 10⁵ N
2.4 × 10⁵ N
2.6 × 10⁵ N
2.8 × 10⁵ N
2

Why does the viscosity of liquids generally decrease with increasing temperature?

Higher temperatures increase molecular kinetic energy in liquids, weakening intermolecular forces, reducing resistance to flow, and thus lowering viscosity, as noted in the chapter.

Density increases
Intermolecular forces weaken
Pressure decreases
Surface tension increases
2

A bubble of radius \( 4 \, \text{mm} \) is formed at \( 50 \, \text{cm} \) depth in a soap solution (\( \rho = 1.2 \times 10^3 \, \text{kg/m}^3 \), \( S = 0.025 \, \text{N/m} \)). What is the total pressure inside? (Take \( P_a = 1.01 \times 10^5 \, \text{Pa} \), \( g = 9.8 \, \text{m/s}^2 \))

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

\( P_a = 1.01 \times 10^5 \, \text{Pa} \), \( \rho g h = 1.2 \times 10^3 \times 9.8 \times 0.5 = 5880 \, \text{Pa} \).

\( \frac{2 S}{r} = \frac{2 \times 0.025}{4 \times 10^{-3}} = 12.5 \, \text{Pa} \).

\( P_i = 1.01 \times 10^5 + 5880 + 12.5 = 1.06892 \times 10^5 \, \text{Pa} \).

1.05 × 10⁵ Pa
1.06 × 10⁵ Pa
1.069 × 10⁵ Pa
1.08 × 10⁵ Pa
3

Which statement is incorrect about dynamic lift?

Dynamic lift (e.g., on airplane wings) results from a pressure difference due to faster airflow above (Bernoulli’s principle), not viscosity or gravitational force alone. The incorrect statement is that it’s due to viscosity.

It involves pressure differences
It is due to viscosity
It depends on airflow speed
It lifts objects like wings
2

A spray tube (\( 7 \, \text{cm}^2 \)) has 35 holes of diameter \( 0.8 \, \text{mm} \). If the flow speed is \( 3.0 \, \text{m/min} \), what is the ejection speed?

\( A_1 v_1 = A_2 v_2 \), \( v_1 = 3.0 \, \text{m/min} = 0.05 \, \text{m/s} \), \( A_1 = 7 \times 10^{-4} \, \text{m}^2 \).

Hole area: \( A_h = \pi (0.4 \times 10^{-3})^2 \), total \( A_2 = 35 \times \pi \times 1.6 \times 10^{-7} \approx 1.759 \times 10^{-5} \, \text{m}^2 \).

\( v_2 = \frac{7 \times 10^{-4} \times 0.05}{1.759 \times 10^{-5}} \approx 1.99 \, \text{m/s} \).

1.5 m/s
2.0 m/s
2.5 m/s
3.0 m/s
2

A hydraulic press uses a small piston of area \( 0.008 \, \text{m}^2 \) to exert \( 120 \, \text{N} \). What force is produced by a large piston of area \( 0.032 \, \text{m}^2 \)?

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

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

\( F_1 = 120 \, \text{N} \), \( A_1 = 0.008 \, \text{m}^2 \), \( A_2 = 0.032 \, \text{m}^2 \).

\( F_2 = 120 \times \frac{0.032}{0.008} = 120 \times 4 = 480 \, \text{N} \).

400 N
450 N
480 N
500 N
3

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