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QUESTION: 1

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?

Solution:

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 * 10^{3} N / m^{3}; = 1 + 3 – 1 / 2 – 2m, A = ^{π} ⁄ _{4} * 4^{2} = 4π m^{2}. Hence, F = 246.55 kN.

QUESTION: 2

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?

Solution:

Explanation: The depth of the centroid y and the centre of pressure y_{CP} 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} * 4^{2} = 4π, A = ^{π} ⁄ _{4} * 4^{2} = 4π, sin θ = ^{1} ⁄ _{2} Thus, y_{CP} = 2.125m.

QUESTION: 3

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?

Solution:

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 * 10^{3} N / m^{3}; = 1 + 3 – 1 / 2 = 2m, each side of the lamina =

Hence, F = 156:96 kN.

QUESTION: 4

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?

Solution:

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, y_{CP} = 2.08m.

QUESTION: 5

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?

Solution:

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 * 10^{3} N / m^{3};A = 2 * 2 = 4 m^{2}. Hence, F = 63.65 kN.

QUESTION: 6

A container is lled with two liquids of densities ρ1 and ρ2 up to heights h_{1} and h_{2}respectively. What will be the hydrostatic force (in kN) per unit width of the lower face AB?

Solution:

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,

QUESTION: 7

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?

Solution:

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.

QUESTION: 8

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?

Solution:

Explanation:

QUESTION: 9

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?

Solution:

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, F_{hyd} = hydrostatic force on

the plate, = distance of the point of action of F_{hyd} from the hinge point = ^{2}⁄_{3} * 5 = ^{10}⁄_{3}

F_{hyd} = γA, where γ = specific weight of the liquid = 9.81 * 10^{3} = 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.

QUESTION: 10

A large tank is lled with three liquids of densities ρ1, ρ2 and ρ3 up to heights of h_{1}, h_{2} and h_{3} 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)

Solution:

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

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