Mechanical Properties of Fluids Chapter-Wise Test 14

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A tank has a hole \( 1.5 \, \text{m} \) below the water surface, open to the atmosphere. What is the efflux speed? (Take \( g = 10 \, \text{m/s}^2 \))

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

\( g = 10 \, \text{m/s}^2 \), \( h = 1.5 \, \text{m} \).

\( v = \sqrt{2 \times 10 \times 1.5} = \sqrt{30} \approx 5.48 \, \text{m/s} \).

4.5 m/s
5.0 m/s
5.5 m/s
6.0 m/s
3

What is the gauge pressure at a depth of \( 1.9 \, \text{m} \) in ethyl alcohol (\( \rho = 806 \, \text{kg/m}^3 \))? (Take \( g = 10 \, \text{m/s}^2 \))

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

\( \rho = 806 \, \text{kg/m}^3 \), \( g = 10 \, \text{m/s}^2 \), \( h = 1.9 \, \text{m} \).

\( P_g = 806 \times 10 \times 1.9 = 15314 \, \text{Pa} \).

1.4 × 10⁴ Pa
1.5 × 10⁴ Pa
1.6 × 10⁴ Pa
1.7 × 10⁴ Pa
2

Water (\( \rho = 1000 \, \text{kg/m}^3 \)) flows horizontally at \( 5.0 \, \text{m/s} \) with pressure \( 2.1 \times 10^5 \, \text{Pa} \). If the speed increases to \( 7.5 \, \text{m/s} \), what is the new pressure?

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

\( P_1 = 2.1 \times 10^5 \, \text{Pa} \), \( v_1 = 5.0 \, \text{m/s} \), \( v_2 = 7.5 \, \text{m/s} \), \( \rho = 1000 \, \text{kg/m}^3 \).

\( P_2 = 2.1 \times 10^5 + \frac{1}{2} \times 1000 (25 - 56.25) = 2.1 \times 10^5 - 15625 = 1.94375 \times 10^5 \, \text{Pa} \).

1.92 × 10⁵ Pa
1.94 × 10⁵ Pa
1.96 × 10⁵ Pa
1.98 × 10⁵ Pa
2

Why does the pressure exerted by a fluid increase with depth?

Pressure increases with depth due to the weight of the fluid column above, as given by \( P = \rho g h \). The greater the depth, the more fluid mass contributes to the pressure at that point.

Due to viscosity
Because of surface tension
Due to the weight of the fluid above
Because of fluid velocity
3

A bubble of radius \( 2.5 \, \text{mm} \) is blown at \( 20 \, \text{cm} \) depth in water (\( \rho = 1000 \, \text{kg/m}^3 \), \( S = 0.0727 \, \text{N/m} \)). What is the total pressure inside? (Take \( P_a = 1.01 \times 10^5 \, \text{Pa} \), \( g = 9.8 \, \text{m/s}^2 \))

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

\( P_a = 1.01 \times 10^5 \, \text{Pa} \), \( \rho g h = 1000 \times 9.8 \times 0.2 = 1960 \, \text{Pa} \).

\( \frac{2 S}{r} = \frac{2 \times 0.0727}{2.5 \times 10^{-3}} = 58.16 \, \text{Pa} \).

\( P_i = 1.01 \times 10^5 + 1960 + 58.16 = 1.02958 \times 10^5 \, \text{Pa} \).

1.02 × 10⁵ Pa
1.03 × 10⁵ Pa
1.04 × 10⁵ Pa
1.05 × 10⁵ Pa
2

Why does a mercury barometer use mercury instead of water?

Mercury’s high density (\( 13.6 \times 10^3 \, \text{kg/m}^3 \)) results in a shorter column height (\( h = P_a / \rho g \)) to balance atmospheric pressure, making it more practical than water, which requires a much taller column due to lower density.

Higher viscosity
Lower surface tension
Higher density
Greater compressibility
3

What is the excess pressure inside a mercury drop of radius \( 1.5 \, \text{mm} \) at \( 20^\circ \text{C} \)? (Surface tension = \( 0.4355 \, \text{N/m} \))

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

\( S = 0.4355 \, \text{N/m} \), \( r = 1.5 \times 10^{-3} \, \text{m} \).

\( \Delta P = \frac{2 \times 0.4355}{1.5 \times 10^{-3}} = 580.67 \, \text{Pa} \).

550 Pa
580 Pa
600 Pa
620 Pa
2

Why does a liquid’s surface behave like a stretched membrane?

Surface tension arises from cohesive forces between liquid molecules, creating a contractile force at the surface, resembling a stretched membrane that resists expansion, as molecules at the surface have higher energy than those inside.

Due to viscosity
Because of pressure differences
Due to cohesive forces at the surface
Because of gravitational pull
3

Water flows at \( 2.3 \, \text{m/s} \) at \( 3.5 \, \text{m} \) height with pressure \( 1.6 \times 10^5 \, \text{Pa} \). What is the pressure at \( 1.5 \, \text{m} \) height with speed \( 3.2 \, \text{m/s} \)? (\( \rho = 1000 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \))

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: \( 1.6 \times 10^5 + \frac{1}{2} \times 1000 \times 5.29 + 1000 \times 9.8 \times 3.5 = 1.6 \times 10^5 + 2645 + 34300 = 1.96945 \times 10^5 \).

Right: \( P_2 + \frac{1}{2} \times 1000 \times 10.24 + 1000 \times 9.8 \times 1.5 = P_2 + 5120 + 14700 = P_2 + 19820 \).

\( 1.96945 \times 10^5 = P_2 + 19820 \), \( P_2 = 1.77125 \times 10^5 \, \text{Pa} \).

1.75 × 10⁵ Pa
1.77 × 10⁵ Pa
1.80 × 10⁵ Pa
1.85 × 10⁵ Pa
2

What is the total pressure at a depth of \( 4 \, \text{m} \) in seawater (\( \rho = 1.03 \times 10^3 \, \text{kg/m}^3 \)) if atmospheric pressure is \( 1.01 \times 10^5 \, \text{Pa} \)? (Take \( g = 9.8 \, \text{m/s}^2 \))

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

\( P_a = 1.01 \times 10^5 \, \text{Pa} \), \( \rho = 1.03 \times 10^3 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \), \( h = 4 \, \text{m} \).

\( P = 1.01 \times 10^5 + 1.03 \times 10^3 \times 9.8 \times 4 = 1.01 \times 10^5 + 40376 = 1.41376 \times 10^5 \, \text{Pa} \).

1.41 × 10⁵ Pa
1.35 × 10⁵ Pa
1.45 × 10⁵ Pa
1.50 × 10⁵ Pa
1

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