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Mechanical Engineering Exam  >  Mechanical Engineering Tests  >  Test: Hydrostatic Force on Plane Area - 1 - Mechanical Engineering MCQ

Test: Hydrostatic Force on Plane Area - 1 - Mechanical Engineering MCQ


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15 Questions MCQ Test - Test: Hydrostatic Force on Plane Area - 1

Test: Hydrostatic Force on Plane Area - 1 for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Test: Hydrostatic Force on Plane Area - 1 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Test: Hydrostatic Force on Plane Area - 1 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 - 1 below.
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Test: Hydrostatic Force on Plane Area - 1 - Question 1
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 A cuboidal beaker is half filled with water. By what percent will the hydrostatic force on one of the vertical sides of the beaker increase if it is completely filled?

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

Explanation: Hydrostatic force per unit width on a vertical side of a beaker = 1 ⁄ 2 * ρgh2, where ρ = density of the liquid and h= height of liquid column. The hydrostatic force when the beaker is completely filled = 1 ⁄ 2 ρg(2h)2 = 2ρgh2.
Thus, percentage increase in hydrostatic force =  = 300%.

Test: Hydrostatic Force on Plane Area - 1 - Question 2
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By what factor will the hydrostatic force on one of the vertical sides of a beaker decrease if the height of the liquid column is halved?

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

Hydrostatic force per unit width on a vertical side of a beaker = 1 ⁄ 2 * ρgh2, where ρ = density of the liquid and h= height of liquid column. Thus, if the liquid column is halved, the hydrostatic force on the vertical face will become one-fourth.

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Test: Hydrostatic Force on Plane Area - 1 - Question 3
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 Equal volume of two liquids of densities ρ1 and ρ2 are poured into two identical cuboidal beakers. The hydrostatic forces on the respective vertical face of the beakers are F1 and F2respectively. If ρ1 > ρ2, which one will be the correct relation between F1 and F2? 

Detailed Solution for Test: Hydrostatic Force on Plane Area - 1 - Question 3

Explanation: Hydrostatic force per unit width on a vertical side of a beaker = 1 ⁄ 2 * ρgh2, where ρ = density of the liquid and h= height of liquid column. Thus if ρ1 > ρ2, F1 > F2 and F1 ≠ F2, when the h is constant.

Test: Hydrostatic Force on Plane Area - 1 - Question 4
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Which of the following is the correct relation between centroid (G) and the centre of pressure (P) of a plane submerged in a liquid?
Detailed Solution for Test: Hydrostatic Force on Plane Area - 1 - Question 4

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

where I = the moment of inertia and A= area. None of the quantities I, A and  can be negative. Thus, YCP > . For horizontal planes, I = 0, hence YCP = 

Test: Hydrostatic Force on Plane Area - 1 - Question 5
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A beaker contains water up to a height of h. What will be the location of the centre of pressure?
Detailed Solution for Test: Hydrostatic Force on Plane Area - 1 - Question 5

 

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

where I = the moment of inertia and A = area. If  = h ⁄ 2; I = bh3/12 ;A = bh, then

Test: Hydrostatic Force on Plane Area - 1 - Question 6
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 A cubic tank is completely filled with water. What will be the ratio of the hydrostatic force exerted on the base and on any one of the vertical sides?
Detailed Solution for Test: Hydrostatic Force on Plane Area - 1 - Question 6

Explanation: Hydrostatic force per unit width on a vertical side of a beaker Fv = 1 ⁄ 2 * ρgh2, where ρ = density of the liquid and h= height of the liquid column. Hydrostatic force per unit width on the base of the beaker = Fb = ρgh * h = ρgh2. Thus, Fb : Fv = 2 : 1.

Test: Hydrostatic Force on Plane Area - 1 - Question 7
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A rectangular lamina of width b and depth d is submerged vertically in water, such that the upper edge of the lamina is at a depth h from the free surface. What will be the expression for the depth of the centroid (G)?
Detailed Solution for Test: Hydrostatic Force on Plane Area - 1 - Question 7

Explanation: The centroid of the lamina will be located at it’s centre. ( d ⁄ 2). Thus, the depth of the centre of pressure will be h + d ⁄ 2.

Test: Hydrostatic Force on Plane Area - 1 - Question 8
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 A rectangular lamina of width b and depth d is submerged vertically in water, such that the
upper edge of the lamina is at a depth h from the free surface. What will be the expression for the depth of the centre of pressure?
Detailed Solution for Test: Hydrostatic Force on Plane Area - 1 - Question 8

 

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

where I = the moment of inertia and A = area. If  = h + d ⁄ 2; I = bh3/12 ;A = bd. thus,

Test: Hydrostatic Force on Plane Area - 1 - Question 9
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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 free surface. What will be the total water pressure (in kN) on the lamina?
Detailed Solution for Test: Hydrostatic Force on Plane Area - 1 - Question 9

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;  = 0.5 + 1 ⁄ 2 * 2m = 1.5 m, A = 2 * 2 m2 = 4 m2. Hence, F = 58.86 kN.

Test: Hydrostatic Force on Plane Area - 1 - Question 10
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 A square lamina (each side equal to 2m) with a central hole of diameter 1m is submerged vertically in water such that the upper edge of the lamina is at a depth of 0.5 m from the free surface. What will be the total water pressure (in kN) on the lamina?
Detailed Solution for Test: Hydrostatic Force on Plane Area - 1 - Question 10

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; = 0.5 + 1 ⁄ 2 * 2m = 1.5 m, A = 2 * 2 – π ⁄ 4 * 12 m2 = 3.215 m2Hence, F = 47.31 kN.

Test: Hydrostatic Force on Plane Area - 1 - Question 11
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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 free surface. What will be the depth (in m) of the centre of pressure?
Detailed Solution for Test: Hydrostatic Force on Plane Area - 1 - Question 11

 

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. Now,

Test: Hydrostatic Force on Plane Area - 1 - Question 12
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What will be the total pressure (in kN) on a vertical square lamina submerged in a tank of oil (S=0.9) as shown in the figure?
 
Detailed Solution for Test: Hydrostatic Force on Plane Area - 1 - Question 12

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 * 0.9 * 103 N / m3;  = 2.5m, A = 1 ⁄ 2 * 22 = 2 m2. Hence, F = 44.1 kN.

Test: Hydrostatic Force on Plane Area - 1 - Question 13
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The upper and lower edges of a square lamina of length 4 m are at a depths of 1 m and 3 m respectively in water. What will be the depth (in m) of the centre of pressure?
Detailed Solution for Test: Hydrostatic Force on Plane Area - 1 - Question 13

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,

= 2:17m.

Test: Hydrostatic Force on Plane Area - 1 - Question 14
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 The upper and lower edges of a square lamina of length 4 m are at a depths of 1 m and 3 m respectively in water. What will be the total pressure (in kN) on the lamina?
Detailed Solution for Test: Hydrostatic Force on Plane Area - 1 - Question 14

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 * 0.9 * 103 N / m3;  = 1 + 3-1 / 2 = 2m, A = 4 * 4 = 16 m2. Hence, F = 313.92 kN.

Test: Hydrostatic Force on Plane Area - 1 - Question 15
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What will be the depth (in m) of the centre of pressure for a vertical square lamina submerged in a tank of oil (S=0.8) as shown in the figure?
Detailed Solution for Test: Hydrostatic Force on Plane Area - 1 - Question 15

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. Each side of the lamina = 3/&sqrt;2 Now, y = 1 + 3 ⁄ 2 = 1.5,

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