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QUESTION: 1

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?

Solution:

Explanation: Hydrostatic force per unit width on a vertical side of a beaker = ^{1} ⁄ _{2} * ρgh^{2}, 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} = 2ρgh^{2}.

Thus, percentage increase in hydrostatic force = = 300%.

QUESTION: 2

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?

Solution:

Hydrostatic force per unit width on a vertical side of a beaker = ^{1} ⁄ _{2} * ρgh^{2}, 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.

QUESTION: 3

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 F_{1} and F_{2}respectively. If ρ1 > ρ2, which one will be the correct relation between F_{1} and F_{2}?

Solution:

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

QUESTION: 4

Which of the following is the correct relation between centroid (G) and the centre of pressure (P) of a plane submerged in a liquid?

Solution:

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} =

QUESTION: 5

A beaker contains water up to a height of h. What will be the location of the centre of pressure?

Solution:

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 = bh^{3}/12 ;A = bh, then

QUESTION: 6

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?

Solution:

Explanation: Hydrostatic force per unit width on a vertical side of a beaker F_{v} = ^{1} ⁄ _{2} * ρgh^{2}, 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 = ρgh^{2}. Thus, F_{b} : F_{v} = 2 : 1.

QUESTION: 7

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)?

Solution:

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}.

QUESTION: 8

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?

Solution:

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 = bh^{3}/12 ;A = bd. thus,

QUESTION: 9

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?

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}; = 0.5 + ^{1} ⁄ _{2} * 2m = 1.5 m, A = 2 * 2 m^{2} = 4 m^{2}. Hence, F = 58.86 kN.

QUESTION: 10

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?

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}; = 0.5 + ^{1} ⁄ _{2} * 2m = 1.5 m, A = 2 * 2 – ^{π} ⁄ _{4} * 1^{2} m^{2} = 3.215 m^{2}Hence, F = 47.31 kN.

QUESTION: 11

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?

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. Now,

QUESTION: 12

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?

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 * 0.9 * 10^{3} N / m^{3}; = 2.5m, A = ^{1} ⁄ _{2} * 2^{2} = 2 m^{2}. Hence, F = 44.1 kN.

QUESTION: 13

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?

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,

= 2:17m.

QUESTION: 14

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?

Solution:

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 * 10^{3} N / m^{3}; = 1 + 3-1 / 2 = 2m, A = 4 * 4 = 16 m^{2}. Hence, F = 313.92 kN.

QUESTION: 15

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?

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

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