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Test: Pressure & Fluid Statics Level - 3 - Mechanical Engineering MCQ


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20 Questions MCQ Test - Test: Pressure & Fluid Statics Level - 3

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Test: Pressure & Fluid Statics Level - 3 - Question 1

What is the intensity of pressure (in MPa) in the ocean at a depth of 1500 m. The salt water is compressible and weighs 10050 N/m3 at the free surface? E (bulk modulus of elasticity of salt water) = 2070 MN/m2 (constant).


Detailed Solution for Test: Pressure & Fluid Statics Level - 3 - Question 1

By hydro-static law

dp / dh = ρg ⋯ ①

By the definition of Bulk Modulus

K = dp / dρ/ρ

⇒ dp = Kdρ / ρ ⋯②

Substituting ② in ①

Kdρ / ρdh = ρg

⇒Kdρ / ρ 2 = g dh

⇒−K / ρ = g h + C

Given at h = 0, ρ = ρ0

Given at h = 0, ρ = ρ0 ⇒

−K/ ρ0 = C

⇒−K / ρ = gh − K / ρ0

K / ρ =K / ρ0 − gh

ρ0K / K − ρ0gh = ρ ⋯ ③

Substituting the same in ①

dp / dh =ρ0Kg / K − ρ0gh

Given, K = 2070 MN/m2

ρ0g = 10050 N/m3

∴ at h = 1500

p = 2070 ln ( 2070 × 106 / 2070 × 106 − 10050 × 1500)

p = 15.13 MPa

Test: Pressure & Fluid Statics Level - 3 - Question 2

Treating the atmospheric air to be an ideal gas, what shall be the atmospheric pressure (in kPa) at the top of the skyscraper Bhurj Khalifa in Dubai standing tallest since 2008 with a height of 828 m. Consider the given variation of atmospheric condition with altitude. The characteristic gas constant for atmospheric air can be considered as 287 J / kg-­K


Detailed Solution for Test: Pressure & Fluid Statics Level - 3 - Question 2
Based on the graph, the temperature varies linearly (decreases) ∴ T = T0 − βy ⋯ ① where T is the temperature at an altitude y from zero sea level. β is the lapse rate, T0 is temperature at zero sea level Using hydro-static law

dp / dy = −ρg ⋯ ②

But as the atmospheric air may be treated as an ideal gas ρ = p / RT

Substituting the same in ② dp / dy = −pg / RT

Using ①

dp / dy =−pg / R(T0 − βy )

∴dp / p = −g dy / R(T0 − βy )

Given, p0 = 101.3 kPa

T0 = 15°C i. e. 288 K

R = 287 J / kg‒ K

Also β can be predicted from the graph.

β = −56.5 − 15 / 11000

= 0.0065°C/m or 0.0065 K/m

∴ at y = 828 m

p = 101.3 ( 1 − 0.0065 × 828 / 288) 9.81/9.81/0.0065x287

p = 101.3 × 0.98135.256

p = 91.73 kPa

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Test: Pressure & Fluid Statics Level - 3 - Question 3

An aluminium cube 150 mm on a side is suspended by a string in oil and water as shown in figure. The cube is submerged with half of it being in oil and the other half in water. Find the tension in the string if the specific gravity of oil is 0.8 and the specific weight of aluminium is 25.93 kN/m3 .


Detailed Solution for Test: Pressure & Fluid Statics Level - 3 - Question 3
At equilibrium

T + FB = mg

FB = FB,Oil + FB,water

FB = (0.8 × 103) × g × Vsolid / 2 + 103 × g × Vsolid / 2

FB = (400 + 500)g VSolid

FB = 900 g VSolid

T = mg − FB

T = (25.93 × 103 − 900 g)VSolid

T = 57.71 N

Test: Pressure & Fluid Statics Level - 3 - Question 4

Gate AB in figure is 1.2 m wide (in a direction perpendicular to the plane of the figure) and is hinged at A. Gauge G reads −0.147 bar and oil in the right hand tank is having a relative density 0.75. What horizontal force must be applied at B for equilibrium of gate AB?


Detailed Solution for Test: Pressure & Fluid Statics Level - 3 - Question 4
Force on gate due to water

= ρghC(Area) + (pair Gauge)(Area)

= Area (103 × g × (3.7 + 0.9) − 0.147 × 105)

= 65.72 kN

Force on gate due to oil

= ρoil × g × hC × Area

= 0.75 × 103 × 9.81 × 1.8 / 2 × 1.2 × 1.8

= 14.30 kN

The center of pressure for water force can be calculated using pressure triangle

yP = h / 3×(2 × 21.6 + 39.25) / (21.6 + 39.25)

= 0.81 m from bottom

Similarly centre of pressure on oil side

= 0.6 m from the bottom i. e

For equilibrium about A Fw × (1.8 − 0.81)

= Foil (1.8 − 0.6) + FH (1.8)

⇒ 65.72 × 0.99

= 14.3(1.2) + FH × 1.8

FH = 26.61 kN

Test: Pressure & Fluid Statics Level - 3 - Question 5

Consider the gasoline tank of a Maruti Suzuki Swift. Somehow water enters the fuel tank as shown. If for the given configuration, the reading on the fuel meter is full, Find out height of air column (h in cm) in the tank.


Detailed Solution for Test: Pressure & Fluid Statics Level - 3 - Question 5
The gauge reads full when there is 25 cm of gasoline in the tank. Therefore in the given situation, the gauge pressure is equivalent to 25 cm of gasoline.

h cm of air +(23.5 − h) cm of gasoline +1.5 cm of water= 25 cm of gasoline

∴ 0.0188 × h + 6.65 × (23.5 − h) + 9.81 × 1.5 = 6.65 × 225

∴ h = 0.713 cm

Test: Pressure & Fluid Statics Level - 3 - Question 6

Find out the value of x?

Manometric fluid is mercury


Detailed Solution for Test: Pressure & Fluid Statics Level - 3 - Question 6
Equation for manometer E

pC + ρg × 0.2 = 0

pC = −0.2 × ρg

Equation for manometer D

pB = pC + ρgx

2.1 × 105 = −0.2 × ρg + ρgx

ρ = 13600 kg/m3

2.1 × 105 + 2.66 × 105 = 13.6 × 103 × g × x

x = 1.77 m

Test: Pressure & Fluid Statics Level - 3 - Question 7

Consider the given configuration, if the pressure in pipe A reduces by 5 kN/m2, find out the new manometric height.


Detailed Solution for Test: Pressure & Fluid Statics Level - 3 - Question 7

Using hydro-static law pA = pB

⇒ p + 10 cm of water = 10 cm of Hg ⋯ ① If p decreases by 5 kPa, level of Hg in left limb i.e. A will rise up to A′ and C will fall to C′ . Thus new configuration will be as follows.

Equating pressure at !′ and B′ pA′ = pB′ p′ + (10 − x)cm of water = (10 − 2x)cm of Hg p′ + 10 cm of water − x cm of water = 10 cm of Hg − 2x cm of Hg Also p′ = p − 5 kPa ∴ p − 5 kPa + 10 cm of water − x cm of water = 10 cm of Hg − 2x cm of Hg ⋯ ② Substituting ① in ②

−5 kPa − x cm of water = −2x cm of Hg

−5 kPa = x cm of water − 2x cm of Hg

∴ −5000 =x / 100 × 1000 × 9.81 − 2x / 100 × 13600 × 9.81

∴ 5000 = 2x / 100 × 13600 × 9.81 − x / 100 × 1000 × 9.81

5000 = x ( 2/100 × 13600 × 9.81 −1 / 100× 1000 × 9.81)

x = 5000 / 2570.22

x = 1.9453 cm

∴ New Manometric height

10 − 2x = 6.1 cm

Test: Pressure & Fluid Statics Level - 3 - Question 8

The 200-kg, 5-m-wide rectangular gate shown in figure is hinged at B and leans against the floor at A making an angle of 45° with the horizontal. The gate is to be opened from its lower edge by applying a normal force at its center. Determine the minimum force F required to open the water gate.

Detailed Solution for Test: Pressure & Fluid Statics Level - 3 - Question 8

For the surface the gate

hc = 2 m

∴ hp = hc + IG / Ahc sin2θ

hp = 2 + 5×(3√2) 3 / 12 / 5 × 3√2 × 2 × 1 / 2

hp = 2 + 9 × 2 / 12 × 2×1 / 2

hp = 2.375 m

Also Fp = ρgAhc

Fp = 1000 × 9.81 × (5 × 3√2) × 2

Fp = 416.2 kN

Fp × 1.875√2 + w × 1.5 = F × 1.5√2

F = 521 kN

Test: Pressure & Fluid Statics Level - 3 - Question 9

Consider a U-tube filled with mercury except the 18-cm-high portion at the top, as shown in figure. The diameter of the right arm of theU-tube is D = 2 cm, and the diameter of the left arm is 2D. Oil with a specific gravity of 2.72 is poured into the left arm, forcing some mercury from the left arm into the right one. Determine the maximum amount of oil in liters that can be added into the left arm.


Detailed Solution for Test: Pressure & Fluid Statics Level - 3 - Question 9
When oil is poured in left limb, the level of mercury goes down by h1 and in right limb goes up by h2

After adding the oil, ρoil × g(18 + h1) = ρHg × g(h1 + h2)

2.72 × (18 + h1) = 13.6 (h1 + h2)

Also h1 × (2D)2 = D2 × h2

h1 × 4 = h2

2.72 × (18 + h1) = 13.6(5h1)

48.96 + 2.72 h1 = 68 h1

h1 = 0.75 cm

Quantity of oil added

=π / 4 × (0.04)2 × (0.1875

= 2.35 × 102-4m3

= 0.2356 lts.

Test: Pressure & Fluid Statics Level - 3 - Question 10

The 4-m–diameter circular gate of the figure is located in the incline wall of a large reservoir containing water (γ = 9.80 kN/m3). The gate is mounted on a shaft along its horizontal diameter, and the water depth is 10 m above the shaft. Determine the moment in kN-m that would have to be applied to the shaft to open the gate.


Detailed Solution for Test: Pressure & Fluid Statics Level - 3 - Question 10

Hydrostatic forces on the circular gate

= ρ × g × hC × Area

= 103 × g × 10 × π / 4× 42

= 1232.76 kN

yP = IxC / A × yC + yC = πD4 /64 /π 4D2 × 11.54 + 11.54

= 11.62 m

(Note: this is ?? and not the depth from free surface)

Fp = 1232.76 kN acts 0.08 m Below the centroid, measured along the plane of the gate The moment required to turn the gate about the shaft = Fp × (yp − yc)

= 1232.76 × (11.62 − 11.54) × 103

= 98.62 kN-m

Test: Pressure & Fluid Statics Level - 3 - Question 11

Consider a cuboidal tank of dimensions 20 cm × 20 cm × 20 cm. Two immiscible liquids (S1 = 1.2, S3 = 1.4), equal amount of each are poured into the container. Find the hydrostatic force on any of the side surfaces of the container and the depth of corresponding center of pressure.

Detailed Solution for Test: Pressure & Fluid Statics Level - 3 - Question 11

The pressure diagram on any side of the tank will be

Using the concept of pressure prism, the total force is equal to total volume of the pressure prism.

i.e. volume (a + b + c)

Fp = ρ1g × h /4 × ( h/ 2× 0.20) + ρ2g × h /4 (h/ 2 × 0.20) +ρ1g h /2× (h/ 2× 0.20)

Fp = (1.2 + 1.4 + 2.4)g = 49.05 N

To calculate the center of pressure Fp × yP = Moment due to a + Moment of b +Moment of c

Fp × yP = (1.2 × g × 2 /3×h/ 2) + (1.4 × g × [ 2/3×h/ 2+h/ 2]) + (2.4g × [ h /2+0/h 4])

All distances measured from free surface 49.05 × yP = 0.7848 + 2.289 + 3.5316 yP = 0.134 m = 13.46 cm from top.

Test: Pressure & Fluid Statics Level - 3 - Question 12

A dam is having a curved surface as shown in the figure. The height of the water retained by the dam is 20 m, density of water is 1000 kg/m3. Assuming g as 9.81 m/s2, the vertical force in kN acting on the dam per unit length is


Detailed Solution for Test: Pressure & Fluid Statics Level - 3 - Question 12

The vertical component of pressure force is equal to the weight of liquid above the curved surface. i.e. the volume between the curved surface and y-axis

⇒ Fv = ρg ∫y=020 x dy × 1

⇒ Fv = 4 × 9810 / 3× 203/2

⇒ Fv = 1169.9 × 103 N Or 1169.9 kN

Test: Pressure & Fluid Statics Level - 3 - Question 13

A 50-ton, 6-m-diameter hemispherical dome on a level surface is filled with water, as shown in figure. Someone claims that he can lift this dome by making use of Pascal’s law by attaching a long tube to the top and filling it with water. Determine the required height of water in the tube to lift the dome. Disregard the weight of the tube and the water in it.

Common Data for Question No. 14 & 15 Consider two tumblers, A & B, initially empty. An ice block is placed in one of the tumblers, say A. Now slowly pour water in both the tumblers, such that the final water level is the same in the two.


Detailed Solution for Test: Pressure & Fluid Statics Level - 3 - Question 13
The hydrostatic force on the dome can be calculated by weight of the virtual fluid considered upto free surface.

FHS = [π /4× d2 × (h + 3) − 2 /3 πr3] × ρW × g

To lift the dome FHS = mg

So,

[28.27(h + 3) − 56.54]ρW × g = 50 × 103 × g

28.27 (h + 3) = 106.54

h + 3 = 3.768

h = 0.77 m

Test: Pressure & Fluid Statics Level - 3 - Question 14

Which tumbler weighs more?

Detailed Solution for Test: Pressure & Fluid Statics Level - 3 - Question 14

WA = W1 + W2

WA = ρwV1g + ρiV2g WB = ρwVbg

Consider A, as ice is floating, weight of ice is equal to weight of water displaced. ρiV2g = ρwV4g ∴ WA = ρwV1g + ρwV4g

WA = ρw(V1 + V4)g

WA = ρwV3g

WA = WB

Hence the two tumblers weigh the same.

Test: Pressure & Fluid Statics Level - 3 - Question 15

As the ice in tumbler A melts, in which tumbler level will be more?

Detailed Solution for Test: Pressure & Fluid Statics Level - 3 - Question 15
Once ice melts, it gets converted into water, ρiV2g = ρwV5g Where V5 is the additional volume of water formed in tumbler A due to melting of ice.

Also by buoyancy

ρiV2g = ρwV4g

∴ ρwV5g = ρwV4g

i.e. V5 = V4

After ice melts, total volume of water in A,

VT = V1 + V5

VT = V1 + V4

VT = V3

Hence volume of water is the same in both tumblers even if ice in A melts.

Test: Pressure & Fluid Statics Level - 3 - Question 16

A 1-m-diameter, 2-m-high vertical cylinder is completely filled with gasoline whose density is 740 kg/m3. The tank is now rotated about its vertical axis at a rate of 90 rpm, while being accelerated upward at 5 m/s2. Determine

(i) the difference between the pressure at the centers of the bottom and top surfaces

(ii) the difference between the pressures at the center and the edge of the bottom surface.

Detailed Solution for Test: Pressure & Fluid Statics Level - 3 - Question 16

Considering a small fluid element

Resolving forces in z-direction, pdA − (p + ∂p / ∂z dz) dA − ρdAdzg = ρdAdz. a

⇒ − ∂p / ∂z dz = ρ(g + a)dz

Again resolving forces along r − (p + ∂p ∂r dr) dA + pdA + ρdAdrω2r = 0

[ρdAdrω2 r is the centrifugal force]

⇒∂p / ∂r dr = ρω2 rdr

Also we can write dp = ∂p / ∂r dr + ∂p / ∂z dz

∴ dp = ρω2rdr − ρ(a + g)dz

Which gives p = ρω2r2 / 2− ρ(a + g)z + C

(a) The difference of pressure at the center (r = 0) of top & bottom surface

Δp = Pbottom − Ptop

⇒ Δp = [−ρ(a + g)0 + C] − [−ρ(a + g)(2) + C]

⇒ Δp = ρ(a + g)(2)

⇒ Δp = 21.91 kPa

(b) The difference of pressure at the center and edge of bottom surface (Z = 0)

Δp = ρω2(0.5) 2/ 2

⇒ Δp = ρ × (N/60× 2π)2× (0.5)2 / 2

⇒ Δp = 8.21 kPa

Test: Pressure & Fluid Statics Level - 3 - Question 17

The distance between the centers of the two arms of a U-tube open to the atmosphere is 25 cm, and the U-tube contains 20-cm-high alcohol in both arms. Now the U-tube is rotated about the left arm at 4.2 rad/s. Determine the elevation difference between the fluid surfaces in the two arms.


Detailed Solution for Test: Pressure & Fluid Statics Level - 3 - Question 17
When the U-Tube will rotate

p1 = p2 = patm For forced vortex

p = ρω2r2 / 2 − ρgZ + C

Equating pressures at 1 & 2

ρω2(0)2 /2− ρg(0) = ρω2r2/2 /2 − ρgh

i. e ω2r2/2 / 2 = gh

(4.2)2(0.25)2 /2 = gh

h = 5.61 cm

Test: Pressure & Fluid Statics Level - 3 - Question 18

A U-tube contains water in the right arm, and another liquid in the left arm. It is observed that when the U–tube rotates at 30 rpm about an axis that is 15 cm from the right arm and 5 cm from the left arm, the liquid levels in both arms become the same. Determine the density of the fluid in the left arm.


Detailed Solution for Test: Pressure & Fluid Statics Level - 3 - Question 18
p = ρω2r2 / 2 − ρgZ + C

For right arm, ρ = 103, r = 0.15, Z = 0.1 and pguage = 0

0 = 103 ( N/602π)2× (0.15)2 /2− 103 × g(0.1) + C

0 = 111.03 − 981 + C ⟹ C = 870

For left arm, p = 0, Z = 0.1 cm, r = 0.05m

0 = ρ × 0.012 − ρ(0.981) + 870

ρ(0.969) = 870 ρ

= 897.83 kg/m3

Test: Pressure & Fluid Statics Level - 3 - Question 19

A circular cylinder partly filled with a liquid is rotated about its axis at ω rad/s without spilling. At the walls, the rise of liquid surface above the original level will be

Detailed Solution for Test: Pressure & Fluid Statics Level - 3 - Question 19

Volume of parallelepiped

=1 / 2× πr2Z2 = πr2Z1

Z2 = 2Z1

Taking 0 as origin at gauge pressure

p = ρω2r2 / 2 − ρg Z + C

atO,p = 0,r = 0,Z = 0so,C = 0

at A p = 0 = ρω2r2 / 2 − lgZ1 ⟹ Z2 = ω2r2 / 2g

he rise of liquid above original level (Z1)

= Z2 − Z1 ⟹ Z2 − Z2 / 2 = Z2 / 2 = ω2r2 / 4g

Test: Pressure & Fluid Statics Level - 3 - Question 20

Figure shows a conical vessel having its outlet at A to which a U-tube manometer is connected. The reading of the manometer indicated in the figure pertains to the situation when the vessel is empty, i.e. the water surface is at A. Find the reading of the manometer when the vessel is completely filled with water.


Detailed Solution for Test: Pressure & Fluid Statics Level - 3 - Question 20
Case (a) when the vessel is empty Equating pressure values at the points 1 and 2 which lie at the same horizontal plane; patm + (13.6 × 9810 × 0.2) = 9810 × h + Patm

∴ h = 13.6 × 9810 × 0.2 / 9810

= 2.72 m

For case (b) when the vessel is full of water: Let the mercury level fall by a distance δ in the right limb with a corresponding rise δ in the left limb. Again writing the governing manometric equation for the points 1′ and 2′:

Patm + 13.6 × 9810 × (0.2 + 2δ)

= 9810 (3 + 2.72 + δ) + Patm

Solution gives: δ = 0.1145 m Hence the manometer reading would be

0.20 + 2 × 0.1145 = 0.429 m

= 42.90 cm

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