# 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?

a) P1 > P2 > P3
b) P1 < P2 < P3
c) P1 = P2 = P3
d) P1 > P2 < P3

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?

a) 2.5
b) 3.5
c) 4.5
d) 5.5

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?

a) 101.3
b) 101.5
c) 101.7
d) 101.9

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.

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?

a) 4.375
b) 4.5
c) 5.5
d) 5.625

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?

a) 27.33
b) 19.2
c) 10.8
d) 2.67

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:

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?

a) 0.01
b) 0.05
c) 0.10
d) 0.15

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,

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

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)

a) 4:75% higher
b) 4:75% lower
c) 6:75% higher
d) 6:75% lower

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

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?

a) 0%
b) 5%higher
c) 5%lower
d) 10%higher

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?

a) 588.6
b) 637.65
c) 735.75
d) 833.85

Explanation: PA = 0.8 * 103 * 9.81 * 0.05 + 1 * 103 * 9.81 * 0.025 = 637.65 kPa.

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

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.

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?

a) 23.5
b) 24.5
c) 26.5
d) 27.5

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?

a) 30o
b) 50o
c) 60o
d) 70o

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?

a) 12o
b) 15o
c) 17o
d) 20o

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:

a) 766
b) 776
c) 786
d) 796

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?

a) 36
b) 40
c) 76
d) 116

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

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