Fluid Mechanics Test

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Fluid Mechanics Test - 4

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Weekly Quiz Competition

Question 1
1 / -0

The pressure of 1 atm = ______ Pa

A
1.013 x 10^-5 Pa

B
1.013 x 10³ Pa

C
1.013 x 10⁵ Pa

D
1.013 x 10^-3 Pa

Solution

Pressure of 1 atm = 1.013 x 10⁵ Pa
Question 2
1 / -0

In a car lift, compressed air exerts a force F₁ on a small piston having a radius of 5cm. This pressure is transmitted to the second piston of a radius of 15cm. If the mass of the car to be lifted is 1350 kg. What is F₁?

A
14.7 * 10³N

B
2.47 * 10³N

C
1.47 * 10³N

D
24.7 * 10³N

Solution

From Pascal’s law: P₁= P₂
⇒ F₁/A₁= F₂/A₂
⇒ F₁/πr₁² = F₂/πr₂²
⇒ F₁ = (F₂* r₁² ) / r₂²= 1350 * 9.8 * (5 * 10^-2)² / (15 * 10^-2)² = 1470 N = 1.47 * 10³N
Question 3
1 / -0

Which of the following is not an application of Pascal’s Law?

A
Brahma Press

B
Submarine

C
Hydraulic Lift

D
Both A and C

Solution

Pascal's law states that the magnitude of pressure within the fluid is equal in all parts.
Option A: Brahma's press is a Hydraulic Press that works on the principle of Pascal's law.
Option B: Submarines don't work on the principle of Pascal's law.
Option C: Hydraulic lifts works on the principle of pascals law i.e. a force applied on a smaller cylinder is transmitted to lift heavy loads using larger cylinders.
Hence, option B is correct.
Question 4
1 / -0

The formula used to find the pressure on a swimmer h meters below the surface of a lake is: (where P_a is the atmospheric pressure)

A
P_a + hρ

B
hρg

C
P_a - hρg

D
P_a + hρg

Solution

We know that the pressure at some point inside the water can be represented by: P_a + ρhg
where,
ρ = Density of the liquid
P_a = Atmospheric pressure
H = Depth at which the body is present
g = Gravitational acceleration
Question 5
1 / -0

The pressure at the bottom of a tank containing a liquid does not depend on:

A
Area of the bottom surface.

B
Acceleration due to gravity.

C
Height of the liquid column.

D
Nature of the liquid.

Solution

Pressure at the bottom of a tank containing liquid is given as P = hρg which is independent of the surface area bottom of the tank.
Question 6
1 / -0

An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46 cm. What will be the length of the air column above mercury in the tube now? (Atmospheric pressure = 76 cm of Hg)

A
16

B
22

C
38

D
6

Solution

Since the system is accelerating horizontally such that no component of acceleration in the vertical direction. Hence, the pressure in the vertical direction will remain unaffected.

For air trapped in tube, p₁V₁ = p₂V₂
p₁ = p_atm = pg76
V₁= A.8 [ A = area of cross section]
p₂ = p_atm - ρg(54-x) = ρg(22+x)
V = A.x
ρg76 x 8A = ρg (22+x) (Ax)
x² + 22x - 78 x 8 by solving, x = 16.
Question 7
1 / -0

Find the density when a liquid 5 m high in a column exerts a pressure of 80 Pa.

A
625 kg/m³

B
1.6 kg/m³

C
0.625 kg/m³

D
2.625 kg/m³

Solution

► Pressure = Density x Gravity x Height = ρgh
⇒ ρ = P/(g*h) = 80 Pa / (9.8 m/s² x 5 m)
⇒ Density = 1.632 kg/m³
Question 8
1 / -0

Two vessels with equal base and unequal height have water filled to the same height. The force at the base of the vessels is:

A
Force doesn’t depend on such factors

B
Equal

C
Varies with time

D
Unequal

Solution

Force at the bases of two vessels will be equal as force depends on height and area. Since it is the same here, force is equal.
Question 9
1 / -0

Water is flowing continuously from a tap having an internal diameter 8 x 10^-3 m. The water velocity as it leaves the tap is 0.4 ms^-1. The diameter of the water stream at a distance 2 x 10^-1 m below the tap is close to:

A
7.5 x 10^-3 m

B
9.6 x 10^-3 m

C
3.6 x 10^-3 m

D
5.0 x 10^-3 m

Solution
Question 10
1 / -0

If the terminal speed of a sphere of gold (density = 19.5 kg/m3 ) is 0.2 m/s in a viscous liquid (density = 1.5 kg/m3 ) of the same size in the same liquid.

A
0.2 m/s

B
0.4 m/s

C
0.133 m/s

D
0.1m/s

Solution

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Fluid Mechanics Test - 4

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Question 1

1.013 x 10 -5 Pa

1.013 x 10 3 Pa

1.013 x 10 5 Pa

1.013 x 10 -3 Pa

Solution

Question 2

14.7 * 103 N

2.47 * 103 ​N

1.47 * 103 ​N

24.7 * 103 ​N

Solution

Question 3

Brahma Press

Submarine

Hydraulic Lift

Both A and C

Solution

Question 4

Pa + hρ

hρg

Pa - hρg

Pa + hρg

Solution

Question 5

Area of the bottom surface.

Acceleration due to gravity.

Height of the liquid column.

Nature of the liquid.

Solution

Question 6

16

22

38

6

Solution

Question 7

625 kg/m3

1.6 kg/m3

0.625 kg/m3

2.625 kg/m3

Solution

Question 8

Force doesn’t depend on such factors

Equal

Varies with time

Unequal

Solution

Question 9

7.5 x 10-3 m

9.6 x 10-3 m

3.6 x 10-3 m

5.0 x 10-3 m

Solution

Question 10

0.2 m/s

0.4 m/s

0.133 m/s

0.1m/s

Solution

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1.013 x 10^-5 Pa

1.013 x 10³ Pa

1.013 x 10⁵ Pa

1.013 x 10^-3 Pa

14.7 * 10³N

2.47 * 10³N

1.47 * 10³N

24.7 * 10³N

P_a + hρ

P_a - hρg

P_a + hρg

625 kg/m³

1.6 kg/m³

0.625 kg/m³

2.625 kg/m³

7.5 x 10^-3 m

9.6 x 10^-3 m

3.6 x 10^-3 m

5.0 x 10^-3 m