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

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%.

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.

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.

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

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

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

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

Explanation: Hydrostatic force per unit width on a vertical side of a beaker F_v = 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 = F_b = ρgh * h = ρgh². Thus, F_b : F_v = 2 : 1.

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.

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

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

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.

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.

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. 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. Now, γ = 9.81 * 0.9 * 10³ N / m³;  = 2.5m, A = ¹ ⁄ ₂ * 2² = 2 m². Hence, F = 44.1 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,

= 2:17m.

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.

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