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  4. Mechanical Properties Of Fluids Chapter-Wise Test 2

Mechanical Properties Of Fluids Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Why does the pressure in a fluid decrease when its speed increases in a horizontal pipe?

Bernoulli’s principle states that in steady flow, an increase in kinetic energy (\( \frac{1}{2} \rho v^2 \)) corresponds to a decrease in pressure energy, as total energy is conserved along a horizontal streamline.

Due to viscosity
Because of energy conservation
Due to increased density
Because of turbulence
2

Which of the following statements is correct about atmospheric pressure?

Atmospheric pressure decreases with altitude as the weight of the air column above reduces, a fundamental concept in the chapter’s discussion of pressure variation.

It increases with altitude
It is constant everywhere
It decreases with altitude
It depends only on temperature
3

A container has whole blood (\( \rho = 1.06 \times 10^3 \, \text{kg/m}^3 \)) up to \( 1.5 \, \text{m} \). What is the gauge pressure at the bottom? (Take \( g = 9.8 \, \text{m/s}^2 \))

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

\( \rho = 1.06 \times 10^3 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \), \( h = 1.5 \, \text{m} \).

\( P_g = 1.06 \times 10^3 \times 9.8 \times 1.5 = 15582 \, \text{Pa} \).

1.2 × 10⁴ Pa
1.56 × 10⁴ Pa
1.8 × 10⁴ Pa
2.0 × 10⁴ Pa
2

A \( 60 \, \text{kg} \) person stands on two bones, each with area \( 25 \, \text{cm}^2 \). What is the average pressure on the bones? (Take \( g = 10 \, \text{m/s}^2 \))

\( P_{av} = \frac{F}{A} \), \( F = mg = 60 \times 10 = 600 \, \text{N} \).

Total area \( A = 2 \times 25 \times 10^{-4} = 5.0 \times 10^{-3} \, \text{m}^2 \).

\( P_{av} = \frac{600}{5.0 \times 10^{-3}} = 1.2 \times 10^5 \, \text{Pa} \).

1.0 × 10⁵ Pa
1.1 × 10⁵ Pa
1.2 × 10⁵ Pa
1.3 × 10⁵ Pa
3

A spray tube (\( 10 \, \text{cm}^2 \)) has 50 holes of diameter \( 0.9 \, \text{mm} \). If the flow speed is \( 2.5 \, \text{m/min} \), what is the ejection speed?

\( A_1 v_1 = A_2 v_2 \), \( v_1 = 2.5 \, \text{m/min} = 0.04167 \, \text{m/s} \), \( A_1 = 10 \times 10^{-4} \, \text{m}^2 \).

Hole area: \( A_h = \pi (0.45 \times 10^{-3})^2 \), total \( A_2 = 50 \times \pi \times 2.025 \times 10^{-7} \approx 3.181 \times 10^{-5} \, \text{m}^2 \).

\( v_2 = \frac{10 \times 10^{-4} \times 0.04167}{3.181 \times 10^{-5}} \approx 1.31 \, \text{m/s} \).

1.0 m/s
1.3 m/s
1.5 m/s
1.8 m/s
2

What happens to the fluid velocity when a pipe narrows, assuming steady flow?

The equation of continuity (\( A v = \text{constant} \)) for an incompressible fluid in steady flow implies that as the cross-sectional area decreases, the velocity must increase to maintain a constant mass flow rate.

It decreases
It remains constant
It increases
It becomes zero
3

A mercury barometer shows a height of \( 75 \, \text{cm} \) at sea level. What is the atmospheric pressure? (\( \rho = 13.6 \times 10^3 \, \text{kg/m}^3 \), \( g = 10 \, \text{m/s}^2 \))

\( P_a = \rho g h \).

\( \rho = 13.6 \times 10^3 \, \text{kg/m}^3 \), \( g = 10 \, \text{m/s}^2 \), \( h = 0.75 \, \text{m} \).

\( P_a = 13.6 \times 10^3 \times 10 \times 0.75 = 1.02 \times 10^5 \, \text{Pa} \).

0.76 × 10⁵ Pa
1.02 × 10⁵ Pa
1.36 × 10⁵ Pa
1.52 × 10⁵ Pa
2

Water flows through a horizontal pipe at a speed of \( 3 \, \text{m/s} \) with a pressure of \( 1.5 \times 10^5 \, \text{Pa} \). If the speed increases to \( 6 \, \text{m/s} \), what is the new pressure? (\( \rho = 1000 \, \text{kg/m}^3 \))

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

\( P_1 = 1.5 \times 10^5 \, \text{Pa} \), \( v_1 = 3 \, \text{m/s} \), \( v_2 = 6 \, \text{m/s} \), \( \rho = 1000 \, \text{kg/m}^3 \).

\( P_2 = P_1 + \frac{1}{2} \rho (v_1^2 - v_2^2) = 1.5 \times 10^5 + \frac{1}{2} \times 1000 (9 - 36) \).

\( P_2 = 1.5 \times 10^5 - 13500 = 1.365 \times 10^5 \, \text{Pa} \).

1.365 × 10⁵ Pa
1.5 × 10⁵ Pa
1.635 × 10⁵ Pa
1.8 × 10⁵ Pa
1

What is the total pressure at a depth of \( 3.4 \, \text{m} \) in seawater (\( \rho = 1.03 \times 10^3 \, \text{kg/m}^3 \)) with atmospheric pressure \( 1.01 \times 10^5 \, \text{Pa} \)? (Take \( g = 9.8 \, \text{m/s}^2 \))

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

\( P_a = 1.01 \times 10^5 \, \text{Pa} \), \( \rho = 1.03 \times 10^3 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \), \( h = 3.4 \, \text{m} \).

\( P = 1.01 \times 10^5 + 1.03 \times 10^3 \times 9.8 \times 3.4 = 1.01 \times 10^5 + 34319.6 = 1.353196 \times 10^5 \, \text{Pa} \).

1.32 × 10⁵ Pa
1.34 × 10⁵ Pa
1.35 × 10⁵ Pa
1.36 × 10⁵ Pa
3

A person stands on two bones, each with area \( 22 \, \text{cm}^2 \), exerting a total force of \( 588 \, \text{N} \). What is the average pressure on the bones?

\( P_{av} = \frac{F}{A} \), \( F = 588 \, \text{N} \).

Total area \( A = 2 \times 22 \times 10^{-4} = 4.4 \times 10^{-3} \, \text{m}^2 \).

\( P_{av} = \frac{588}{4.4 \times 10^{-3}} = 1.33636 \times 10^5 \, \text{Pa} \).

1.2 × 10⁵ Pa
1.3 × 10⁵ Pa
1.34 × 10⁵ Pa
1.4 × 10⁵ Pa
3

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