Mechanical Properties of Fluids Chapter-Wise Test 12

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the excess pressure inside a mercury drop of radius \( 2 \, \text{mm} \) at \( 20^\circ \text{C} \)? (Surface tension of mercury = \( 0.465 \, \text{N/m} \))

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

\( S = 0.465 \, \text{N/m} \), \( r = 2 \times 10^{-3} \, \text{m} \).

\( \Delta P = \frac{2 \times 0.465}{2 \times 10^{-3}} = 465 \, \text{Pa} \).

232.5 Pa
465 Pa
930 Pa
1860 Pa
2

Which statement is incorrect about surface tension?

Surface tension arises from cohesive forces at the liquid’s surface, decreasing with temperature due to weaker molecular interactions, not increasing. The incorrect statement is that it increases with temperature.

It acts like a stretched membrane
It decreases with temperature
It increases with temperature
It causes capillary action
3

Why does adding a wetting agent like soap reduce the angle of contact of water on a surface?

Wetting agents reduce surface tension and increase adhesion between the liquid and solid, lowering the angle of contact (\( \theta \)), as the liquid spreads more easily, enhancing wetting, as explained in the chapter.

It increases viscosity
It reduces surface tension
It increases pressure
It decreases density
2

A tank has a hole \( 1.3 \, \text{m} \) below the water surface, open to the atmosphere. What is the efflux speed? (Take \( g = 9.8 \, \text{m/s}^2 \))

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

\( g = 9.8 \, \text{m/s}^2 \), \( h = 1.3 \, \text{m} \).

\( v = \sqrt{2 \times 9.8 \times 1.3} = \sqrt{25.48} \approx 5.05 \, \text{m/s} \).

4.5 m/s
5.0 m/s
5.5 m/s
6.0 m/s
2

What is the total pressure at a depth of \( 1.8 \, \text{m} \) in oxygen liquid (\( \rho = 1.43 \, \text{kg/m}^3 \)) with atmospheric pressure \( 1.01 \times 10^5 \, \text{Pa} \)? (Take \( g = 9.8 \, \text{m/s}^2 \))

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

\( P_a = 1.01 \times 10^5 \, \text{Pa} \), \( \rho = 1.43 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \), \( h = 1.8 \, \text{m} \).

\( P = 1.01 \times 10^5 + 1.43 \times 9.8 \times 1.8 = 1.01 \times 10^5 + 25.21 = 1.0102521 \times 10^5 \, \text{Pa} \).

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

What is the gauge pressure at \( 2.5 \, \text{m} \) depth in a tank of mercury (\( \rho = 13.6 \times 10^3 \, \text{kg/m}^3 \))? (Take \( g = 10 \, \text{m/s}^2 \))

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

\( \rho = 13.6 \times 10^3 \, \text{kg/m}^3 \), \( g = 10 \, \text{m/s}^2 \), \( h = 2.5 \, \text{m} \).

\( P_g = 13.6 \times 10^3 \times 10 \times 2.5 = 3.4 \times 10^5 \, \text{Pa} \).

2.8 × 10⁵ Pa
3.0 × 10⁵ Pa
3.4 × 10⁵ Pa
3.6 × 10⁵ Pa
3

An aircraft (\( m = 2.8 \times 10^5 \, \text{kg} \), wing area \( 480 \, \text{m}^2 \)) flies level. What is the pressure difference across the wings? (Take \( g = 9.8 \, \text{m/s}^2 \))

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

\( mg = 2.8 \times 10^5 \times 9.8 = 2.744 \times 10^6 \, \text{N} \), \( A = 480 \, \text{m}^2 \).

\( \Delta P = \frac{2.744 \times 10^6}{480} = 5716.67 \, \text{Pa} \approx 5717 \, \text{Pa} \).

5500 Pa
5717 Pa
6000 Pa
6200 Pa
2

A plate of area \( 0.05 \, \text{m}^2 \) moves at \( 0.7 \, \text{m/s} \) over a \( 0.8 \, \text{mm} \) thick water film (\( \eta = 1.0 \times 10^{-3} \, \text{Pa s} \)). What force is required?

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

\( \eta = 1.0 \times 10^{-3} \, \text{Pa s} \), \( v = 0.7 \, \text{m/s} \), \( A = 0.05 \, \text{m}^2 \), \( l = 0.8 \times 10^{-3} \, \text{m} \).

\( F = 1.0 \times 10^{-3} \times \frac{0.7 \times 0.05}{0.8 \times 10^{-3}} = 1.0 \times 10^{-3} \times 43.75 = 0.04375 \, \text{N} \).

0.03 N
0.04 N
0.044 N
0.05 N
3

Which of the following statements is correct about the angle of contact?

The angle of contact depends on the relative strength of adhesive (liquid-solid) and cohesive (liquid-liquid) forces, determining wetting behavior (acute for wetting, obtuse for non-wetting).

It is always 90°
It depends on adhesive and cohesive forces
It is independent of the liquid
It increases with viscosity
2

What is the gauge pressure at a depth of \( 2.4 \, \text{m} \) in whole blood (\( \rho = 1.06 \times 10^3 \, \text{kg/m}^3 \))? (Take \( g = 10 \, \text{m/s}^2 \))

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

\( \rho = 1.06 \times 10^3 \, \text{kg/m}^3 \), \( g = 10 \, \text{m/s}^2 \), \( h = 2.4 \, \text{m} \).

\( P_g = 1.06 \times 10^3 \times 10 \times 2.4 = 25440 \, \text{Pa} \).

2.4 × 10⁴ Pa
2.5 × 10⁴ Pa
2.6 × 10⁴ Pa
2.7 × 10⁴ Pa
2

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