Mechanical Properties of Fluids Chapter-Wise Test 16

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A spray tube (\( 9 \, \text{cm}^2 \)) has 45 holes of diameter \( 0.6 \, \text{mm} \). If the flow speed is \( 2.0 \, \text{m/min} \), what is the ejection speed?

\( A_1 v_1 = A_2 v_2 \), \( v_1 = 2.0 \, \text{m/min} = 0.0333 \, \text{m/s} \), \( A_1 = 9 \times 10^{-4} \, \text{m}^2 \).

Hole area: \( A_h = \pi (0.3 \times 10^{-3})^2 \), total \( A_2 = 45 \times \pi \times 9 \times 10^{-8} \approx 1.272 \times 10^{-5} \, \text{m}^2 \).

\( v_2 = \frac{9 \times 10^{-4} \times 0.0333}{1.272 \times 10^{-5}} \approx 2.36 \, \text{m/s} \).

1.8 m/s
2.0 m/s
2.4 m/s
2.8 m/s
3

Which of the following statements is correct about pressure in a fluid at rest?

Pressure in a fluid at rest acts equally in all directions (Pascal’s law) and increases with depth (\( P = P_a + \rho g h \)), not influenced by the container’s shape, ruling out options involving directionality or shape dependence.

It acts only downward
It depends on the shape of the container
It acts equally in all directions
It is independent of depth
3

A plate of area \( 0.09 \, \text{m}^2 \) moves at \( 0.5 \, \text{m/s} \) over a \( 0.6 \, \text{mm} \) thick machine oil film (\( \eta = 0.113 \, \text{Pa s} \)). What force is required?

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

\( \eta = 0.113 \, \text{Pa s} \), \( v = 0.5 \, \text{m/s} \), \( A = 0.09 \, \text{m}^2 \), \( l = 0.6 \times 10^{-3} \, \text{m} \).

\( F = 0.113 \times \frac{0.5 \times 0.09}{0.6 \times 10^{-3}} = 0.113 \times 75 = 8.475 \, \text{N} \).

7.5 N
8.5 N
9.5 N
10.5 N
2

A spray tube (\( 5 \, \text{cm}^2 \)) has 25 holes of diameter \( 1.3 \, \text{mm} \). If the flow speed is \( 2.7 \, \text{m/min} \), what is the ejection speed?

\( A_1 v_1 = A_2 v_2 \), \( v_1 = 2.7 \, \text{m/min} = 0.045 \, \text{m/s} \), \( A_1 = 5 \times 10^{-4} \, \text{m}^2 \).

Hole area: \( A_h = \pi (0.65 \times 10^{-3})^2 \), total \( A_2 = 25 \times \pi \times 4.225 \times 10^{-7} \approx 3.318 \times 10^{-5} \, \text{m}^2 \).

\( v_2 = \frac{5 \times 10^{-4} \times 0.045}{3.318 \times 10^{-5}} \approx 0.678 \, \text{m/s} \).

0.5 m/s
0.6 m/s
0.7 m/s
0.8 m/s
3

What is the gauge pressure at a depth of \( 2.7 \, \text{m} \) in mercury (\( \rho = 13.6 \times 10^3 \, \text{kg/m}^3 \))? (Take \( g = 9.8 \, \text{m/s}^2 \))

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

\( \rho = 13.6 \times 10^3 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \), \( h = 2.7 \, \text{m} \).

\( P_g = 13.6 \times 10^3 \times 9.8 \times 2.7 = 359856 \, \text{Pa} = 3.59856 \times 10^5 \, \text{Pa} \).

3.2 × 10⁵ Pa
3.4 × 10⁵ Pa
3.6 × 10⁵ Pa
3.8 × 10⁵ Pa
3

What is the excess pressure inside a water drop of radius \( 2.5 \, \text{mm} \) at \( 100^\circ \text{C} \)? (Surface tension = \( 0.3 \times 10^{-3} \, \text{N/m} \))

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

\( S = 0.3 \times 10^{-3} \, \text{N/m} \), \( r = 2.5 \times 10^{-3} \, \text{m} \).

\( \Delta P = \frac{2 \times 0.3 \times 10^{-3}}{2.5 \times 10^{-3}} = 0.24 \, \text{Pa} \).

0.20 Pa
0.24 Pa
0.28 Pa
0.32 Pa
2

A \( 45 \, \text{kg} \) person stands on two bones, each with area \( 18 \, \text{cm}^2 \). What is the average pressure on the bones? (Take \( g = 9.8 \, \text{m/s}^2 \))

\( P_{av} = \frac{F}{A} \), \( F = mg = 45 \times 9.8 = 441 \, \text{N} \).

Total area \( A = 2 \times 18 \times 10^{-4} = 3.6 \times 10^{-3} \, \text{m}^2 \).

\( P_{av} = \frac{441}{3.6 \times 10^{-3}} = 1.225 \times 10^5 \, \text{Pa} \).

1.0 × 10⁵ Pa
1.225 × 10⁵ Pa
1.5 × 10⁵ Pa
1.75 × 10⁵ Pa
2

A hydraulic lift raises a \( 2100 \, \text{kg} \) load using a small piston of radius \( 3.0 \, \text{cm} \) and a large piston of radius \( 12 \, \text{cm} \). What force is applied on the small piston? (Take \( g = 9.8 \, \text{m/s}^2 \))

\( F_1 = \frac{A_1}{A_2} F_2 \), \( F_2 = 2100 \times 9.8 = 20580 \, \text{N} \).

\( A_1 = \pi (0.03)^2 \), \( A_2 = \pi (0.12)^2 \), \( \frac{A_1}{A_2} = \frac{0.0009}{0.0144} = \frac{1}{16} \).

\( F_1 = \frac{20580}{16} = 1286.25 \, \text{N} \approx 1286 \, \text{N} \).

1200 N
1250 N
1286 N
1300 N
3

What is the total pressure at a depth of \( 3.2 \, \text{m} \) in water (\( \rho = 1000 \, \text{kg/m}^3 \)) with atmospheric pressure \( 1.01 \times 10^5 \, \text{Pa} \)? (Take \( g = 9.8 \, \text{m/s}^2 \))

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

\( P_a = 1.01 \times 10^5 \, \text{Pa} \), \( \rho = 1000 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \), \( h = 3.2 \, \text{m} \).

\( P = 1.01 \times 10^5 + 1000 \times 9.8 \times 3.2 = 1.01 \times 10^5 + 31360 = 1.3236 \times 10^5 \, \text{Pa} \).

1.30 × 10⁵ Pa
1.32 × 10⁵ Pa
1.35 × 10⁵ Pa
1.38 × 10⁵ Pa
2

A soap film supports \( 2.8 \times 10^{-2} \, \text{N} \) over a \( 56 \, \text{cm} \) slider. What is the surface tension?

\( S = \frac{F}{2 l} \).

\( F = 2.8 \times 10^{-2} \, \text{N} \), \( l = 0.56 \, \text{m} \).

\( S = \frac{2.8 \times 10^{-2}}{2 \times 0.56} = 0.025 \, \text{N/m} \).

0.020 N/m
0.022 N/m
0.025 N/m
0.028 N/m
3

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