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Mechanical Engineering Exam  >  Mechanical Engineering Tests  >  Test: Hydrostatic Force on Plane Area - 2 - Mechanical Engineering MCQ

Test: Hydrostatic Force on Plane Area - 2 - Mechanical Engineering MCQ


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10 Questions MCQ Test - Test: Hydrostatic Force on Plane Area - 2

Test: Hydrostatic Force on Plane Area - 2 for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Test: Hydrostatic Force on Plane Area - 2 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Test: Hydrostatic Force on Plane Area - 2 MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Hydrostatic Force on Plane Area - 2 below.
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Test: Hydrostatic Force on Plane Area - 2 - Question 1
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The greatest and the least depth of a circular plate of 4 m diameter from the free surface of water are 3m and 1 m respectively as shown. What will be the total pressure in (kN) on the plate?

Detailed Solution for Test: Hydrostatic Force on Plane Area - 2 - Question 1

Explanation: Total liquid pressure on the lamina = F = γA, where γ = specific weight of the liquid,  = depth of centroid of the lamina from the free surface, A= area of the centroid. Now, γ = 9.81 * 103 N / m3;  = 1 + 3 – 1 / 2 – 2m, A = π ⁄ 4 * 42 = 4π m2. Hence, F = 246.55 kN.

Test: Hydrostatic Force on Plane Area - 2 - Question 2
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 The greatest and the least depth of a circular plate of 4 m diameter from the free surface of water are 3m and 1 m respectively as shown. What will be the depth (in m) of it’s centre of pressure?

Detailed Solution for Test: Hydrostatic Force on Plane Area - 2 - Question 2

Explanation: The depth of the centroid y and the centre of pressure yCP are related by:

where I= the moment of inertia and A = area and θ = the angle of inclination of the lamina to the horizontal. Now,
y = 1 + 3 – 1 / 2 = 2, I = π ⁄ 64 * 42 = 4π, A = π ⁄ 4 * 42 = 4π, sin θ = 1 ⁄ 2 Thus, yCP = 2.125m.

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Test: Hydrostatic Force on Plane Area - 2 - Question 3
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The highest and lowest vertices of a diagonal of a square lamina (each side equal to 4m) are 1 m and 3 m respectively as shown. What will be the water force (in kN) on the lamina?

Detailed Solution for Test: Hydrostatic Force on Plane Area - 2 - Question 3

Explanation: Total liquid pressure on the lamina = F = γA, where γ = specific weight of the liquid,  = depth of centroid of the lamina from the free surface, A= area of the centroid. Now, γ = 9.81 * 103 N / m3;  = 1 + 3 – 1 / 2 = 2m, each side of the lamina =
Hence, F = 156:96 kN.

Test: Hydrostatic Force on Plane Area - 2 - Question 4
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The highest and lowest vertices of a diagonal of a square lamina (each side equal to 4m) are 1 m and 3 m respectively as shown. What will be the depth (in m) of it’s centre of pressure?
Detailed Solution for Test: Hydrostatic Force on Plane Area - 2 - Question 4

Explanation:
where I = the moment of inertia and A= area and θ = the angle of inclination of the lamina to the horizontal. Now,  = 1 + 3 – 1 / 2 = 2m, each side of the lamina = =8; sin θ = 3-1/4 = 1 ⁄ 2. Thus, yCP = 2.08m.

Test: Hydrostatic Force on Plane Area - 2 - Question 5
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A square lamina (each side equal to 2m) is submerged vertically in water such that the upper edge of the lamina is at a depth of 0.5 m from the water surface. If the pressure on the surface is 12 bar, what will be the total water pressure (in kN) on the lamina?
Detailed Solution for Test: Hydrostatic Force on Plane Area - 2 - Question 5

Explanation: Total liquid pressure on the lamina = F = γA, where γ = specific weight of the liquid,  = depth of centroid of the lamina from the free surface, A= area of the centroid. Now, γ = 9.81 * 103 N / m3;A = 2 * 2 = 4 m2. Hence, F = 63.65 kN.

Test: Hydrostatic Force on Plane Area - 2 - Question 6
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A container is lled with two liquids of densities ρ1 and ρ2 up to heights h1 and h2respectively. What will be the hydrostatic force (in kN) per unit width of the lower face AB?
Detailed Solution for Test: Hydrostatic Force on Plane Area - 2 - Question 6

Explanation: Total liquid pressure on the lamina = F = γA, where γ = specific weight of the liquid,  = depth of centroid of the lamina from the free surface, A= area of the centroid. Now,

Test: Hydrostatic Force on Plane Area - 2 - Question 7
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 A container is lled with two liquids of densities ρ and 2ρ up to heights h and eh respectively. What will be the ratio of the total pressure on the lower face AB and on the upper face BC?
Detailed Solution for Test: Hydrostatic Force on Plane Area - 2 - Question 7

Explanation: Total liquid pressure on the lamina = F = γA, where γ = specific weight of the liquid,  = depth of centroid of the lamina from the free surface, A= area of the centroid.

Test: Hydrostatic Force on Plane Area - 2 - Question 8
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A container is lled with two liquids of densities ρ and 2ρ up to heights h and eh respectively. What will be the ratio of the depths of the centres of pressure of the upper face BC and the lower face AB?
Detailed Solution for Test: Hydrostatic Force on Plane Area - 2 - Question 8

Explanation:

Test: Hydrostatic Force on Plane Area - 2 - Question 9
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A gate of length 5 m is hinged at A as shown to support a water column of height 2.5 m. What should be the minimum mass per unit width of the gate to keep it closed?
Detailed Solution for Test: Hydrostatic Force on Plane Area - 2 - Question 9

Explanation: To keep the gate closed, moment due to weight of the gate should be balanced by the moment due to the hydrostatic force.

where m = mass of the plate, θ = angle of inclination to the horizontal, Fhyd = hydrostatic force on
the plate,   = distance of the point of action of Fhyd from the hinge point = 2⁄3 * 5 = 10⁄3

Fhyd = γA, where γ = specific weight of the liquid = 9.81 * 103  = depth of the centre of pressure from the free surface = 2.5/2 = 1.25 and A = 5 * 1. Substituting all the values in the equation, we get m = 9622.5g.

Test: Hydrostatic Force on Plane Area - 2 - Question 10
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A large tank is lled with three liquids of densities ρ1, ρ2 and ρ3 up to heights of h1, h2 and h3 respectively. What will be the expression for the instantaneous velocity of discharge through a small opening at the base of the tank? (assume that the diameter of the opening is negligible compared to the height of the liquid column)
Detailed Solution for Test: Hydrostatic Force on Plane Area - 2 - Question 10

 

Explanation: Instantaneous velocity of discharge  where h= height of the liquid column.

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