Fluid Mechanics Questions and Answers – Pressure Distribution in a Fluid – 2

This set of Fluid Mechanics Questions and Answers for Experienced people focuses on “Pressure Distribution in a Fluid – 2”.

1. Three beakers 1, 2 and 3 of different shapes are kept on a horizontal table and filled with water up to a height h. If the pressure at the base of the beakers are P1, P2 and P3 respectively, which one of the following will be the relation connecting the three?
The pressure on the surface of the liquid in the beakers is the same P1 = P2 = P3
a) P1 > P2 > P3
b) P1 < P2 < P3
c) P1 = P2 = P3
d) P1 > P2 < P3
View Answer

Answer: c
Explanation: The pressure on the surface of the liquid in the beakers is the same. Pressure varies in the downward direction according to the formula P = ρgh, where ρ is the density of the liquid and h is the height of the liquid column from the top.
P1 = ρgh
P2 = ρgh
P3 = ρgh
Since all the beakers contain water up to to the same height, P1 = P2 = P3.

2. A beaker is filled with a liquid of specific gravity S = 1:2 as shown. What will be the pressure difference (in kN/m2) between the two points A and B, 30 cm below and 10 cm to the right of point A?
Find pressure difference between two points A & B, 30 cm below & 10 cm to point A
a) 2.5
b) 3.5
c) 4.5
d) 5.5
View Answer

Answer: b
Explanation: Pressure increases in the vertically downward direction but remains constant in the horizontal direction. Thus,
PB = PA + ρgh
where PB = Pressure at B, PA = Pressure at A, ρ = density of the liquid, g = acceleration due to gravity and h = vertical distance separating the two points.
PB – PA = 1:2 * 103 * 9.81 * 0.3 N/m2 = 3.53 kN/m2

3. The arm of a teapot is 10 cm long and inclined at an angle of 60o to the vertical. The center of the arm base is 2 cm above the base of the beaker. Water is poured into the beaker such that half the arm is filled with it. What will be the pressure at the base of the beaker if the atmospheric pressure is 101.3 kPa?
Find the pressure at the base of the beaker if the atmospheric pressure is 101.3 kPa
a) 101.3
b) 101.5
c) 101.7
d) 101.9
View Answer

Answer: c
Explanation: Total height of the water in the beaker = 2 + 12 * 10 cos 60o cm = 4:5 cm. Pressure at the base of the beaker = 101.3 + 103 * 9.81 * 0.045 Pa = 101.3 + 0.44 kPa = 101.74 kPa.
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4. A beaker of height 10 cm is half-filled with water (Sw = 1) and half-filled with oil (So = 1). At what distance (in cm) from the base will the pressure be half the pressure at the base of the beaker?
Find T what distance from base will pressure be half pressure at base of beaker
a) 4.375
b) 4.5
c) 5.5
d) 5.625
View Answer

Answer: b
Explanation: Gauge pressure at the base of the beaker = So * 103 * 0.05 * g + Sw * 103 * 0.05 * g = 882.9Pa. Let the required height be h m from the base.
If 0.05 ≤ h < 0.1,
800(0.1 – h)g = 12 * 882.9
Thus, h = 0.04375 (out of the range considered).
If 0 < h ≤ 0:05,
800 * 0.05 * g + 103 * (0.05 – h) * g = 12 * 882.9
Thus, h = 0.045 (in the range considered). Hence, the correct answer will be 45 cm.

5. A beaker of height 30 cm is filled with water (Sw = 1) up to a height of 10 cm. Now oil (So = 0:9) is poured into the beaker till it is completely filled. At what distance (in cm) from the base will the pressure be one-third the pressure at the base of the beaker?
Find distance from base will the pressure be one-third pressure at base of beaker
a) 27.33
b) 19.2
c) 10.8
d) 2.67
View Answer

Answer: b
Explanation: Gauge pressure at the base of the beaker = So * 103 * 0.2 * g + Sw * 103 * 0.1 * g = 2550.6Pa. Let the required height be h m from the base.
If 0.1 ≤ h < 0.3,
800(0.3 – h)g = 13 * 2550.6
Thus, h = 0.192 (in the range considered).
Even if there’s no need to check for the other range, it’s shown here for demonstration purpose.If
0 < h ≤ 0.1,
800 * 0.2 * g + 103 * (0.2 – h) * g = 13 * 2550.6
Thus, h = 0.2733 (out of the range considered). Hence, the correct answer will be 19.2 cm.

6. An oil tank of height 6 m is half-filled with oil and the air above it exerts a pressure of 200 kPa on the upper surface. The density of oil varies according to the given relation:
The density of oil for oil tank of height 6 m is half-filled with oil & air above it
What will be the percentage error in the calculation of the pressure at the base of the tank if the density is taken to be a constant equal to 800?
Find percentage error of pressure at base of tank if density is constant to 800
a) 0.01
b) 0.05
c) 0.10
d) 0.15
View Answer

Answer: a
Explanation: The change of pressure with the vertical direction y is given by
dP/dy = – ρg
dP = -ρg dy
If Pa and Pb be the pressures at the top and bottom surfaces of the tank,
The change of pressure with the vertical direction y is given in figure
Thus, Pb = 223.5746kPa. If the density is assumed to be constant,
Pb = 200 + 800 * 9.81 * 3 * 103 = 223.544 kPa. Hence, precentage error Pa & Pb be the pressures at the top & bottom surfaces of the tank

7. If a gas X be confined inside a bulb as shown, by what percent will the pressure of the gas be higher or lower than the atmospheric pressure? (Take the atmospheric pressure equal to 101.3 kPa)
Find percent of pressure of gas be higher or lower than atmospheric pressure
a) 4:75% higher
b) 4:75% lower
c) 6:75% higher
d) 6:75% lower
View Answer

Answer: a
Explanation: Pa = Patm = 101.3
Pb = Pa + 0.9 * 9.81 * 0.03 = 101.56
Pc = Pb + 13.6 * 9.81 * 0.04 = 106.9
Pd = Pc – 1 * 9.81 * 0.05 = 106.41
Pe = Pd – 0.9 * 9.81 * 0.04 = 106.1
PX = Pe = 106.1
Since, PX > Patm, the percentage by which the pressure of the gas is higher than the atmospheric pressure will be Percent of pressure of gas be higher or lower than atmospheric pressure is 4:75% higher
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8. A tank of height 3 m is completely filled with water. Now two-third of the liquid is taken out and an equal amount of two other immiscible liquids of specific gravities 0.8 and 1.2 are poured into the tank. By what percent will the pressure at the base of the tank change?
Percent for the pressure at the base of the tank change is 0%
a) 0%
b) 5%higher
c) 5%lower
d) 10%higher
View Answer

Answer: a
Explanation: Pressure at the base initially = 1 * 9.81 * 3 = 29.43 kPa; Pressure at the base after adding the other two liquids= 0.8 * 9.81 * 1 + 1 * 9.81 * 1 + 1.2 * 9.81 * 1 kPa; Thus the pressure at the base remains the same.

9. A beaker of height 15 cm is completely filled with water. Now two-third of the liquid is taken out and an equal amount of two other immiscible liquids of specific gravities 0.8 and 1.2 are poured into the tank. What will be the pressure (in kPa) at a point situated at a height, half the height of the beaker?
Pressure at point situated at height half the height of the beaker is 637.65
a) 588.6
b) 637.65
c) 735.75
d) 833.85
View Answer

Answer: b
Explanation: PA = 0.8 * 103 * 9.81 * 0.05 + 1 * 103 * 9.81 * 0.025 = 637.65 kPa.
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10. A beaker of height h is completely filled with water. Now two-third of the liquid is replaced by another liquid. If the pressure at the base of the beaker doubled, what is the specific gravity of the liquid poured?
a) 0.5
b) 1
c) 2
d) 2.5
View Answer

Answer: d
Explanation: Pressure at the base initially = Sw * h3 * g; Pressure at the base after pouring the second liquid = Sw * h3 * g + Sl * 2h3 * g, where Sw and Sl are the specific gravities of water and the second liquid.
The specific gravity of the liquid poured is 2.5 if pressure at base of beaker is doubled

11. A beaker, partially filled with a liquid is rotated by an angle 30o as shown. If the pressure at point B becomes 12 bar, what will be the height (in cm) of the beaker?
The height of the beaker is 24.5 for the pressure at point B becomes 12 bar
a) 23.5
b) 24.5
c) 26.5
d) 27.5
View Answer

Answer: b
Explanation: If the height of the beaker is h, the pressure at point B = 103 * g * h * cos 30o = 12 * 103kPa; h = 24.5 cm.

12. A beaker of height 15 cm is partially filled with a liquid and is rotated by an angle θ as shown.
If the pressure at point B becomes 5 bar, what will be the value of θ?
The value of θ is 70o for beaker of height 15 cm is partially filled with a liquid
a) 30o
b) 50o
c) 60o
d) 70o
View Answer

Answer: d
Explanation: If the angle of inclination is taken to be θ, the pressure at point B = 103 * g * 0.15 * cos θ = 5 * 103 kPa; θ = 70.12o.

13. A beaker of height 30 cm is partially filled with a liquid and is rotated by an angle θ as shown.
At this point, the pressure at point B is found to be 5 bar. By what angle should θ be increased such that the pressure at B gets halved?
The angle θ increased such that pressure at B gets halved is 15o in given diagram
a) 12o
b) 15o
c) 17o
d) 20o
View Answer

Answer: b
Explanation: Let θ1 and θ2 be the angles at which the beaker is inclided for the two cases mentioned.
103 * 9.81 * 0.15 * cos θ1 = 5 * 100; θ1 = 70.12o

103 * 9.81 * 0.3 * cos θ2 = 12 * 5 * 100; θ1 = 85.12o

θ2 – θ1 = 15o

14. A closed tank (each side of 5 m) is partially filled with fluid as shown. If the pressure of the air above the fluid is 2 bar, find the pressure at the bottom of the tank. Assume the density ρ of the fluid to vary according to the given relation:
Find pressure at bottom of tank if pressure of the air above the fluid is 2 bar
a) 766
b) 776
c) 786
d) 796
View Answer

Answer: c
Explanation:
PA = Patm = 760
PB = PA + 30
PC = PB – 50 / 13.6 = 786.32
PX = PC = 786.3.

15. For what height of the mercury column will the pressure inside the gas be 40 cm Hg vacuum?
Find the height of the mercury column if the pressure inside the gas be 40 cm Hg vacuum
a) 36
b) 40
c) 76
d) 116
View Answer

Answer: b
Explanation:
Pgas = Patm – ρgH
Taking gauge pressure in terms of cm of Hg,
-40 = 0 – H; H = 40.

Sanfoundry Global Education & Learning Series – Fluid Mechanics.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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