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Mechanical Engineering Exam  >  Mechanical Engineering Test  >  Fluid Mechanics  >  Test: Pressure Distribution in a Fluid- 1 - Mechanical Engineering MCQ

Pressure Distribution in a Fluid- 1 - Free MCQ Practice Test with solutions,


MCQ Practice Test & Solutions: Test: Pressure Distribution in a Fluid- 1 (20 Questions)

You can prepare effectively for Mechanical Engineering Fluid Mechanics for Mechanical Engineering with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Pressure Distribution in a Fluid- 1". These 20 questions have been designed by the experts with the latest curriculum of Mechanical Engineering 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 30 minutes
  • - Number of Questions: 20

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Test: Pressure Distribution in a Fluid- 1 - Question 1
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Which one of the following is the unit of pressure?

Detailed Solution: Question 1

Explanation: Pressure is defined as the force per unit area acting normal to a surface. The SI unit of force is N and area is m2. Thus, the unit of pressure will be N / m2.

Test: Pressure Distribution in a Fluid- 1 - Question 2
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Which one of the following statements is true regarding pressure?

Detailed Solution: Question 2

Explanation:Pressure is defined as the force per unit area acting normal to a surface. Both force and area are vectors. but the division of one by the other leads to a scalar quantity.

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Test: Pressure Distribution in a Fluid- 1 - Question 3
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If the pressure at apoint is 1m of water, what will be it's value in terms of 'm' in oil?(Take the specific gravity of oil to be 0.8)

Detailed Solution: Question 3

  • Explanation: Pressure at a pointy P is equal to ρgh, where ρ is  the density and h is the height of the liquid column. Therefore, ρwater*1*g = ρoil*h*g, where h is the pressure in terms of m of oil.

Thus, h= ρwater/ρoil  = 1/0.8 = 1.25.

Test: Pressure Distribution in a Fluid- 1 - Question 4
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A beaker half-filled with water is exposed to the atmosphere. If the pressure at points A, B and C as shown are Pa, Pb and Pc respectively, which one of the following will be the relation connecting the three?
 

Detailed Solution: Question 4

Explanation: Since the beaker is exposed to the atmosphere, the pressure at point A will be atmospheric, Pa = 0. Pressure increases in the vertically downward direction, Pa < Pb and Pa < Pc.
Pressure remains constant in the horizontal direction, Pb = Pc. Therefore, Pa < Pb = Pc.

Test: Pressure Distribution in a Fluid- 1 - Question 5
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A beaker is filled with a liquid up to a height h. If A and B are two points, one on the free surface and one at the base as shown, such that the minimum distance between the two is l, what will be the pressure at point B?

 

Detailed Solution: Question 5

Explanation: For a constant density liquid, pressure varies linearly in the vertically downward 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. Since A is at the free surface, PA= 0, PB = ρgh.

Test: Pressure Distribution in a Fluid- 1 - Question 6
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A beaker of height h is filled with a liquid of density ρ up to a certain limit. The beaker is rotated by an angle θ such that further increase in the angle will result in over flow of the liquid. If the liquid surface is exposed to the atmosphere, what will be the gauge pressure at point B?

Detailed Solution: Question 6

Explanation: Vertical distance below the free surface at which the point B is located will be h cos θ.
Since the pressure at the free surface is atmospheric, the gauge pressure at B will be = 0 + ρgh cos θ.

Test: Pressure Distribution in a Fluid- 1 - Question 7
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An arm of a teapot is completely filled with tea (density=ρ) If the arm has a length of l and is inclined at 30o to the horizontal, what will be the pressure difference between the two points, C at the mouth and D at the base of the arm?

Detailed Solution: Question 7

Explanation: Vertical distance difference between the two points, C at the mouth and D at the base of the arm will be l sin θ = l sin 30o = l/2. Thus, pressure difference between C and D is = ρgl/2.

Test: Pressure Distribution in a Fluid- 1 - Question 8
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A beaker is filled with a liquid of density ρ1 up to a certain height. The pressure at the base of the beaker id Pb. If the liquid is replaced by an equal volume of another liquid of density ρ2, what will be the pressure at the base of the beaker now?

Detailed Solution: Question 8

Explanation: PB = ρ1gh, where h=height up to which the liquid is filled. Since equal volume of the second liquid is poured, it’ll also rise to a height of h. Thus, the pressure at the base will become

Test: Pressure Distribution in a Fluid- 1 - Question 9
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A mercury manometer is used to measure the static pressure at a point in a water pipe shown in figure. The level difference of mercury in the two limbs is 10 mm. The gauge pressure at that point is 

Detailed Solution: Question 9

Let pressure at that point is ‘p’ (absolute)

The pressure above the (1) – (1) will be equal. So


P(gauge) = 1236.06 Pa

Hence, the correct option is (a).

Test: Pressure Distribution in a Fluid- 1 - Question 10
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A beaker is filled with a liquid of density ρ1 up to a certain height. A is a point, h m downwards from the free surface of the liquid as shown. The liquid is replaced by equal volume of another liquid of density ρ2. If ρ1 > ρ2, how will the pressure at point A change?

Detailed Solution: Question 10

Explanation: P1= ρ1gh and P2 = ρ2gh, where P1 and P2 are the pressures at point A when liquids of density ρ1 and ρ2 are poured. If ρ1 > ρ2, P1 > P2. Thus the pressure at point A will decrease.

Test: Pressure Distribution in a Fluid- 1 - Question 11
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A hinged gate of length 5 m, inclined at 30° with the horizontal and with water mass on its left, is shown in the figure. Density of water is 1000 kg/m3. The minimum mass of the gate in kg per unit width (perpendicular to the plane of paper), required to keep it closed is 

Detailed Solution: Question 11

Density of water = ρw = 1000 kg/m3

Height of center of gravity of hinged gate

w = weight of the gate = mg

A = area of the gate for unit width

= 1 × 5 = 5 m2

Height of center of pressure from water level = h*

= 1.667 m

Force on the gate

Take the moment about hinged support

m = 9623 kg

Hence, the correct option is (d).

Test: Pressure Distribution in a Fluid- 1 - Question 12
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In Figure below, both fluids are at 20°C. If surface tension effects are negligible, what is the density of the oil, in kg/m3?

Detailed Solution: Question 12

Explanation: Move around the U-tube from left atmosphere to right atmosphere:

Pa= (9790N/m3)(0.06m) - γoil(0.08m)= Pa

solve for γoil≈ 7343N/m3,

or: ρoil= 7343/9,81≈ 748kg/m3

*Answer can only contain numeric values
Test: Pressure Distribution in a Fluid- 1 - Question 13
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The arrangement shown in the figure measures the velocity V of a gas of density 1 kg/m3 flowing through a pipe. The acceleration due to gravity is 9.81 m/s2. If the manometric fluid is water (density 1000 kg/m3) and the velocity V is 20 m/s, the differential head h (in mm) between the two arms of the manometer is ______.


Detailed Solution: Question 13

Hence, the correct answer is (20.408).

Assume the velocity at point ‘2’ is zero

Applying Bernoulli’s equation between ‘1’ and ‘2’.

Also, 

h(in mm) = 20.408 mm

Hence, the correct answer is (20.408).

Test: Pressure Distribution in a Fluid- 1 - Question 14
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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?

Detailed Solution: Question 14

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

Test: Pressure Distribution in a Fluid- 1 - Question 15
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 A beaker is filled with a liquid of density ρ up to a height h. If half the liquid is replaced by equal volume of another liquid of twice the density, what will be the change in the base pressure?

Detailed Solution: Question 15

Explanation: Base pressure when the beaker is filled with a liquid of density ρ up to a height h = ρgh
Base pressure when half the liquid is replaced by equal volume of another liquid of twice the density
= ρg h⁄2 + 2ρg h⁄2 = 3⁄2 ρgh
Thus the change in base pressure is = ρgh / 2. Since, P2 > P1, there will be an increase in pressure.

Test: Pressure Distribution in a Fluid- 1 - Question 16
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A cuboidal container (each side of 30 cm0) is completely filled with water. A is a point, 25 cm above the base such that the pressure at point A is P. At what height (in cm) from the base will the pressure be 2P? 

Detailed Solution: Question 16

Explanation: Pressure at a point P is equal to ρgh, where ρ is the density and h is the height of the liquid column from the top. Thus, ρ * g *(30 – h) = 2 * ρ * g *( 30-25), where h from the base where the pressure will be 2P. Thus, h = 30 – 2(30 – 25) = 20.

Test: Pressure Distribution in a Fluid- 1 - Question 17
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 A closed tank (of height 5 m) is PArtially filled with a liquid 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 of the liquid to vary according to the following relation:

where y is the height from the base

 

Detailed Solution: Question 17

Explanation: The change of pressure P with vertical direction y is given by


Test: Pressure Distribution in a Fluid- 1 - Question 18
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 The pressure gauges G1 and G2 installed on the system shown pressures of Pg1 = 5.00 bar and PG2 = 1.00 bar. The value of unknown pressure P is

Atmospheric Pressure is 1.01 bar

Detailed Solution: Question 18

PG1 = 5 bar, PG2 = 1 bar, Patm = 1.01 bar

Pressure gauge measure the pressure relative to its atmosphere.

Absolute pressure in chamber (B) = PG2 + Patm PB = 2.01 bar

Absolute pressure in chamber (A) = PG1 + PB = 7.01 bar

Hence, the correct option is (d).

Test: Pressure Distribution in a Fluid- 1 - Question 19
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The pressure gauges 1, 2 and 3 are installed on the system as shown. If the readings of the gauges be P1 = 1 bar, P2 = 2bar and P3 = 3 bar, what will be the value of P? (Take Patm = 1.01 bar)

Detailed Solution: Question 19

Explanation: PA = PAtm + P1
PB = PA + P2
PC = PB + P3
P = PC = PAtm + P1 + P2 + P3 = 1.01 + 1 + 2 + 3 = 7.01.

Test: Pressure Distribution in a Fluid- 1 - Question 20
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For a completely submerged body with centre of gravity ‘G’ and centre of buoyancy ‘B’, the condition of stability will be 

Detailed Solution: Question 20

For submerged bodies, equilibrium prevails when center of gravity (G) lies below the center of Buoyancy (B)

Hence, the correct option is (a).

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This test contains 20 MCQs on Pressure Distribution in a Fluid- 1 with detailed solutions for each question. It is part of the Fluid Mechanics for Mechanical Engineering course on EduRev, designed to align with the current Mechanical Engineering syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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