Mechanical Properties Of Fluids Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A soap bubble of radius \( 6 \, \text{mm} \) has a surface tension of \( 0.03 \, \text{N/m} \). What is the excess pressure inside?

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

\( S = 0.03 \, \text{N/m} \), \( r = 6 \times 10^{-3} \, \text{m} \).

\( \Delta P = \frac{4 \times 0.03}{6 \times 10^{-3}} = 20 \, \text{Pa} \).

10 Pa
20 Pa
30 Pa
40 Pa
2

Water (\( \rho = 1000 \, \text{kg/m}^3 \)) flows horizontally at \( 3.5 \, \text{m/s} \) with pressure \( 1.7 \times 10^5 \, \text{Pa} \). If the speed rises to \( 6 \, \text{m/s} \), what is the new pressure?

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

\( P_1 = 1.7 \times 10^5 \, \text{Pa} \), \( v_1 = 3.5 \, \text{m/s} \), \( v_2 = 6 \, \text{m/s} \), \( \rho = 1000 \, \text{kg/m}^3 \).

\( P_2 = 1.7 \times 10^5 + \frac{1}{2} \times 1000 (12.25 - 36) = 1.7 \times 10^5 - 11875 = 1.58125 \times 10^5 \, \text{Pa} \).

1.55 × 10⁵ Pa
1.58 × 10⁵ Pa
1.60 × 10⁵ Pa
1.65 × 10⁵ Pa
2

A manometer with mercury (\( \rho = 13.6 \times 10^3 \, \text{kg/m}^3 \)) shows a height difference of \( 20 \, \text{cm} \). What is the pressure difference? (Take \( g = 9.8 \, \text{m/s}^2 \))

\( \Delta P = \rho g h \).

\( \rho = 13.6 \times 10^3 \, \text{kg/m}^3 \), \( g = 9.8 \, \text{m/s}^2 \), \( h = 0.2 \, \text{m} \).

\( \Delta P = 13.6 \times 10^3 \times 9.8 \times 0.2 = 26656 \, \text{Pa} \).

2.0 × 10⁴ Pa
2.67 × 10⁴ Pa
3.0 × 10⁴ Pa
3.5 × 10⁴ Pa
2

What is the force on a submarine window (\( 0.08 \, \text{m}^2 \)) at \( 300 \, \text{m} \) depth in seawater (\( \rho = 1.03 \times 10^3 \, \text{kg/m}^3 \)), interior at atmospheric pressure? (Take \( g = 10 \, \text{m/s}^2 \))

Gauge pressure: \( P_g = \rho g h = 1.03 \times 10^3 \times 10 \times 300 = 3.09 \times 10^6 \, \text{Pa} \).

\( F = P_g A = 3.09 \times 10^6 \times 0.08 = 2.472 \times 10^5 \, \text{N} \).

2.4 × 10⁵ N
2.47 × 10⁵ N
2.5 × 10⁵ N
2.6 × 10⁵ N
2

What is the physical significance of the term \( \frac{1}{2} \rho v^2 \) in Bernoulli’s equation?

In Bernoulli’s equation, \( \frac{1}{2} \rho v^2 \) represents the kinetic energy per unit volume of the fluid, reflecting the energy associated with its motion, balanced against pressure and potential energy terms.

Potential energy per unit volume
Pressure energy
Kinetic energy per unit volume
Viscous energy
3

Water flows at \( 2.9 \, \text{m/s} \) at \( 3.2 \, \text{m} \) height with pressure \( 1.9 \times 10^5 \, \text{Pa} \). What is the pressure at \( 1.8 \, \text{m} \) height with speed \( 4.0 \, \text{m/s} \)? (\( \rho = 1000 \, \text{kg/m}^3 \), \( g = 10 \, \text{m/s}^2 \))

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: \( 1.9 \times 10^5 + \frac{1}{2} \times 1000 \times 8.41 + 1000 \times 10 \times 3.2 = 1.9 \times 10^5 + 4205 + 32000 = 2.26205 \times 10^5 \).

Right: \( P_2 + \frac{1}{2} \times 1000 \times 16 + 1000 \times 10 \times 1.8 = P_2 + 8000 + 18000 = P_2 + 26000 \).

\( 2.26205 \times 10^5 = P_2 + 26000 \), \( P_2 = 2.00205 \times 10^5 \, \text{Pa} \).

1.95 × 10⁵ Pa
2.00 × 10⁵ Pa
2.05 × 10⁵ Pa
2.10 × 10⁵ Pa
2

Why does a hydraulic lift amplify force when lifting a heavy load?

Pascal’s law ensures pressure is transmitted equally. A smaller piston area (\( A_1 \)) with force \( F_1 \) creates pressure (\( P = F_1/A_1 \)), which, applied to a larger area (\( A_2 \)), results in a larger force (\( F_2 = P \cdot A_2 \)), amplifying the input force.

Due to viscosity
Because of larger piston area
Due to fluid compression
Because of streamline flow
2

A plate of area \( 0.07 \, \text{m}^2 \) moves at \( 0.3 \, \text{m/s} \) over a \( 0.5 \, \text{mm} \) thick water film (\( \eta = 1.0 \times 10^{-3} \, \text{Pa s} \)). What force is required?

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

\( \eta = 1.0 \times 10^{-3} \, \text{Pa s} \), \( v = 0.3 \, \text{m/s} \), \( A = 0.07 \, \text{m}^2 \), \( l = 0.5 \times 10^{-3} \, \text{m} \).

\( F = 1.0 \times 10^{-3} \times \frac{0.3 \times 0.07}{0.5 \times 10^{-3}} = 1.0 \times 10^{-3} \times 42 = 0.042 \, \text{N} \).

0.03 N
0.042 N
0.05 N
0.06 N
2

A spray tube (\( 7 \, \text{cm}^2 \)) has 35 holes of diameter \( 1.2 \, \text{mm} \). If the flow speed is \( 3.0 \, \text{m/min} \), what is the ejection speed?

\( A_1 v_1 = A_2 v_2 \), \( v_1 = 3.0 \, \text{m/min} = 0.05 \, \text{m/s} \), \( A_1 = 7 \times 10^{-4} \, \text{m}^2 \).

Hole area: \( A_h = \pi (0.6 \times 10^{-3})^2 \), total \( A_2 = 35 \times \pi \times 3.6 \times 10^{-7} \approx 3.958 \times 10^{-5} \, \text{m}^2 \).

\( v_2 = \frac{7 \times 10^{-4} \times 0.05}{3.958 \times 10^{-5}} \approx 0.884 \, \text{m/s} \).

0.7 m/s
0.8 m/s
0.9 m/s
1.0 m/s
3

A manometer with whole blood (\( \rho = 1.06 \times 10^3 \, \text{kg/m}^3 \)) shows a height difference of \( 0.18 \, \text{m} \). What is the pressure difference? (Take \( g = 10 \, \text{m/s}^2 \))

\( \Delta P = \rho g h \).

\( \rho = 1.06 \times 10^3 \, \text{kg/m}^3 \), \( g = 10 \, \text{m/s}^2 \), \( h = 0.18 \, \text{m} \).

\( \Delta P = 1.06 \times 10^3 \times 10 \times 0.18 = 1908 \, \text{Pa} \).

1600 Pa
1908 Pa
2100 Pa
2400 Pa
2

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