250+ TOP MCQs on Pressure Distribution in a Fluid – 2 and Answers

Fluid Mechanics Questions and Answers for Experienced people 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
Answer: c
Clarification: 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
Answer: b
Clarification: Pressure increases in the vertically downward direction but remains constant in the horizontal direction. Thus,
P_B = P_A + ρgh
where P_B = Pressure at B, P_A = 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 60^o 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
Answer: c
Clarification: Total height of the water in the beaker = 2 + ¹⁄₂ * 10 cos 60^o 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 (S_w = 1) and half-filled with oil (S_o = 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
Answer: b
Clarification: Gauge pressure at the base of the beaker = S_o * 10³ * 0.05 * g + S_w * 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 = ¹⁄₂ * 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 = ¹⁄₂ * 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 (S_w = 1) up to a height of 10 cm. Now oil (S_o = 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
Answer: b
Clarification: Gauge pressure at the base of the beaker = S_o * 10³ * 0.2 * g + S_w * 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 = ¹⁄₃ * 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 = ¹⁄₃ * 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
Answer: a
Clarification: The change of pressure with the vertical direction y is given by
dP/dy = – ρg
dP = -ρg dy
If P_a and P_b 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