Mechanical Properties of Fluids Chapter-Wise Test 15

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A soap film on a wire supports \( 1.2 \times 10^{-2} \, \text{N} \) over a \( 25 \, \text{cm} \) slider. What is the surface tension?

\( S = \frac{F}{2 l} \), \( F = 1.2 \times 10^{-2} \, \text{N} \), \( l = 0.25 \, \text{m} \).

\( S = \frac{1.2 \times 10^{-2}}{2 \times 0.25} = 0.024 \, \text{N/m} \).

0.012 N/m
0.024 N/m
0.036 N/m
0.048 N/m
2

A soap bubble of radius \( 8 \, \text{mm} \) has a surface tension of \( 0.027 \, \text{N/m} \). What is the excess pressure inside?

Excess pressure in a bubble: \( \Delta P = \frac{4 S}{r} \).

\( S = 0.027 \, \text{N/m} \), \( r = 8 \times 10^{-3} \, \text{m} \).

\( \Delta P = \frac{4 \times 0.027}{8 \times 10^{-3}} = 13.5 \, \text{Pa} \).

10 Pa
13.5 Pa
15 Pa
20 Pa
2

A bubble of radius \( 3.5 \, \text{mm} \) is blown at \( 15 \, \text{cm} \) depth in water (\( \rho = 1000 \, \text{kg/m}^3 \), \( S = 0.0727 \, \text{N/m} \)). What is the total pressure inside? (Take \( P_a = 1.01 \times 10^5 \, \text{Pa} \), \( g = 9.8 \, \text{m/s}^2 \))

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

\( P_a = 1.01 \times 10^5 \, \text{Pa} \), \( \rho g h = 1000 \times 9.8 \times 0.15 = 1470 \, \text{Pa} \).

\( \frac{2 S}{r} = \frac{2 \times 0.0727}{3.5 \times 10^{-3}} = 41.54 \, \text{Pa} \).

\( P_i = 1.01 \times 10^5 + 1470 + 41.54 = 1.02471 \times 10^5 \, \text{Pa} \).

1.01 × 10⁵ Pa
1.02 × 10⁵ Pa
1.025 × 10⁵ Pa
1.03 × 10⁵ Pa
3

The pressure at a depth of \( 3 \, \text{m} \) in ethyl alcohol (\( \rho = 806 \, \text{kg/m}^3 \)) is measured with atmospheric pressure as \( 1.01 \times 10^5 \, \text{Pa} \). What is the total pressure? (Take \( g = 10 \, \text{m/s}^2 \))

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

\( P_a = 1.01 \times 10^5 \, \text{Pa} \), \( \rho = 806 \, \text{kg/m}^3 \), \( g = 10 \, \text{m/s}^2 \), \( h = 3 \, \text{m} \).

\( P = 1.01 \times 10^5 + 806 \times 10 \times 3 = 1.01 \times 10^5 + 24180 = 1.2518 \times 10^5 \, \text{Pa} \).

1.15 × 10⁵ Pa
1.20 × 10⁵ Pa
1.25 × 10⁵ Pa
1.30 × 10⁵ Pa
3

A pipe’s cross-sectional area decreases from \( 0.04 \, \text{m}^2 \) to \( 0.01 \, \text{m}^2 \). If the speed at the larger end is \( 2.5 \, \text{m/s} \), what is the speed at the smaller end?

Continuity equation: \( A_1 v_1 = A_2 v_2 \).

\( A_1 = 0.04 \, \text{m}^2 \), \( v_1 = 2.5 \, \text{m/s} \), \( A_2 = 0.01 \, \text{m}^2 \).

\( v_2 = \frac{A_1 v_1}{A_2} = \frac{0.04 \times 2.5}{0.01} = 10 \, \text{m/s} \).

5 m/s
7.5 m/s
10 m/s
12.5 m/s
3

A metal plate of area \( 0.05 \, \text{m}^2 \) moves over a liquid film (\( \eta = 0.02 \, \text{Pa s} \)) of thickness \( 0.2 \, \text{mm} \) at \( 0.5 \, \text{m/s} \). What is the force required?

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

\( \eta = 0.02 \, \text{Pa s} \), \( v = 0.5 \, \text{m/s} \), \( A = 0.05 \, \text{m}^2 \), \( l = 0.2 \times 10^{-3} \, \text{m} \).

\( F = 0.02 \times \frac{0.5 \times 0.05}{0.2 \times 10^{-3}} = 0.02 \times 125 = 2.5 \, \text{N} \).

1.5 N
2.5 N
3.5 N
4.5 N
2

An aircraft (\( m = 4.0 \times 10^5 \, \text{kg} \), wing area \( 600 \, \text{m}^2 \)) flies level. What is the pressure difference across the wings? (Take \( g = 9.8 \, \text{m/s}^2 \))

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

\( mg = 4.0 \times 10^5 \times 9.8 = 3.92 \times 10^6 \, \text{N} \), \( A = 600 \, \text{m}^2 \).

\( \Delta P = \frac{3.92 \times 10^6}{600} = 6533.33 \, \text{Pa} \approx 6533 \, \text{Pa} \).

6000 Pa
6200 Pa
6533 Pa
6800 Pa
3

Water flows at \( 1.5 \, \text{m/s} \) at \( 2.5 \, \text{m} \) height with pressure \( 1.8 \times 10^5 \, \text{Pa} \). What is the pressure at \( 0.5 \, \text{m} \) height with speed \( 2.5 \, \text{m/s} \)? (\( \rho = 1000 \, \text{kg/m}^3 \), \( g = 10 \, \text{m/s}^2 \))

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: \( 1.8 \times 10^5 + \frac{1}{2} \times 1000 \times 2.25 + 1000 \times 10 \times 2.5 = 1.8 \times 10^5 + 1125 + 25000 = 2.06125 \times 10^5 \).

Right: \( P_2 + \frac{1}{2} \times 1000 \times 6.25 + 1000 \times 10 \times 0.5 = P_2 + 3125 + 5000 = P_2 + 8125 \).

\( 2.06125 \times 10^5 = P_2 + 8125 \), \( P_2 = 1.98 \times 10^5 \, \text{Pa} \).

1.95 × 10⁵ Pa
1.98 × 10⁵ Pa
2.00 × 10⁵ Pa
2.05 × 10⁵ Pa
2

Which statement is incorrect about hydrostatic pressure?

Hydrostatic pressure (\( P = \rho g h \)) depends on depth, not the container’s shape or fluid velocity (which applies to dynamic flow). The incorrect statement is that it depends on container shape.

It increases with depth
It depends on fluid density
It depends on container shape
It is independent of velocity
3

A \( 70 \, \text{kg} \) person stands on two bones, each with area \( 16 \, \text{cm}^2 \). What is the average pressure on the bones? (Take \( g = 9.8 \, \text{m/s}^2 \))

\( P_{av} = \frac{F}{A} \), \( F = mg = 70 \times 9.8 = 686 \, \text{N} \).

Total area \( A = 2 \times 16 \times 10^{-4} = 3.2 \times 10^{-3} \, \text{m}^2 \).

\( P_{av} = \frac{686}{3.2 \times 10^{-3}} = 2.14375 \times 10^5 \, \text{Pa} \).

1.9 × 10⁵ Pa
2.0 × 10⁵ Pa
2.14 × 10⁵ Pa
2.3 × 10⁵ Pa
3

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0