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

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.

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 * 4² = 4π, A = ^π ⁄ ₄ * 4² = 4π, sin θ = ¹ ⁄ ₂ Thus, y_CP = 2.125m.

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.

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.

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.

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,

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.

Explanation:

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 * 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.

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