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Test: Pressure Distribution in a Fluid- 2 - Mechanical Engineering MCQ


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15 Questions MCQ Test - Test: Pressure Distribution in a Fluid- 2

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Test: Pressure Distribution in a Fluid- 2 - Question 1

Three beakers 1, 2 and 3 of different shapes are kept on a horizontal table and filled with water up to a height h. If the pressure at the base of the beakers are P1, P2 and P3respectively, which one of the following will be the relation connecting the three?
 

Detailed Solution for Test: Pressure Distribution in a Fluid- 2 - Question 1

Explanation: The pressure on the surface of the liquid in the beakers is the same. Pressure varies in the downward direction according to the formula P = ρgh, where ρ is the density of the liquid and h is the height of the liquid column from the top.
P1 = ρgh
P2 = ρgh
P3 = ρgh
Since all the beakers contain water up to to the same height, P1 = P2 = P3.

Test: Pressure Distribution in a Fluid- 2 - Question 2

 A beaker is filled with a liquid of specific gravity S = 1.2 as shown. What will be the pressure difference (in kN/m2) between the two points A and B, 30 cm below and 10 cm to the right of point A?

Detailed Solution for Test: Pressure Distribution in a Fluid- 2 - Question 2

Explanation: Pressure increases in the vertically downward direction but remains constant in the horizontal direction. Thus,
PB = PA + ρgh
where PB = Pressure at B, PA = Pressure at A, ρ = density of the liquid, g = acceleration due to gravity and h = vertical distance separating the two points.
PB – PA = 1:2 * 103 * 9.81 * 0.3 N/m2 = 3.53 kN/m2

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Test: Pressure Distribution in a Fluid- 2 - Question 3

The arm of a teapot is 10 cm long and inclined at an angle of 60o to the vertical. The center of the arm base is 2 cm above the base of the beaker. Water is poured into the beaker such that half the arm is filled with it. What will be the pressure at the base of the beaker if the atmospheric pressure is 101.3 kPa?

Detailed Solution for Test: Pressure Distribution in a Fluid- 2 - Question 3

Explanation: Total height of the water in the beaker = 2 + 12 * 10 cos 60o cm = 4:5 cm. Pressure at the base of the beaker = 101.3 + 103 * 9.81 * 0.045 Pa = 101.3 + 0.44 kPa = 101.74 kPa.

Test: Pressure Distribution in a Fluid- 2 - Question 4

A beaker of height 10 cm is half-filled with water (Sw = 1) and half-filled with oil (So = 1). At what distance (in cm) from the base will the pressure be half the pressure at the base of the beaker?

Detailed Solution for Test: Pressure Distribution in a Fluid- 2 - Question 4

Explanation: Gauge pressure at the base of the beaker = So * 103 * 0.05 * g + Sw * 103 * 0.05 * g = 882.9Pa. Let the required height be h m from the base.
If 0.05 ≤ h < 0.1,
800(0.1 – h)g = 12 * 882.9
Thus, h = 0.04375 (out of the range considered).
If 0 < h ≤ 0:05,
800 * 0.05 * g + 103 * (0.05 – h) * g = 12 * 882.9
Thus, h = 0.045 (in the range considered). Hence, the correct answer will be 4.5 cm.

Test: Pressure Distribution in a Fluid- 2 - Question 5

A beaker of height 30 cm is filled with water (Sw = 1) up to a height of 10 cm. Now oil (So = 0.8) is poured into the beaker till it is completely filled. At what distance (in cm) from the base will the pressure be one-third the pressure at the base of the beaker?

Detailed Solution for Test: Pressure Distribution in a Fluid- 2 - Question 5

Explanation: Gauge pressure at the base of the beaker = So * 103 * 0.2 * g + Sw * 103 * 0.1 * g = 2550.6Pa. Let the required height be h m from the base.
If 0.1 ≤ h < 0.3,
800(0.3 – h)g = 13 * 2550.6
Thus, h = 0.192 (in the range considered).
Even if there’s no need to check for the other range, it’s shown here for demonstration purpose.If
0 < h ≤ 0.1,
800 * 0.2 * g + 103 * (0.2 – h) * g = 13 * 2550.6
Thus, h = 0.2733 (out of the range considered). Hence, the correct answer will be 19.2 cm.

Test: Pressure Distribution in a Fluid- 2 - Question 6

An oil tank of height 6 m is half-filled with oil and the air above it exerts a pressure of 200 kPa on the upper surface. The density of oil varies according to the given relation:

What will be the percentage error in the calculation of the pressure at the base of the tank if the density is taken to be a constant equal to 800?

Detailed Solution for Test: Pressure Distribution in a Fluid- 2 - Question 6

 

Explanation: The change of pressure with the vertical direction y is given by
dP/dy = – ρg
dP = -ρg dy
If Pa and Pb be the pressures at the top and bottom surfaces of the tank,


Thus, Pb = 223.5746kPa. If the density is assumed to be constant,
Pb = 200 + 800 * 9.81 * 3 * 103 = 223.544 kPa. Hence, precentage error

Test: Pressure Distribution in a Fluid- 2 - Question 7

If a gas X be confined inside a bulb as shown, by what percent will the pressure of the gas be higher or lower than the atmospheric pressure? (Take the atmospheric pressure equal to 101.3 kPa)

Detailed Solution for Test: Pressure Distribution in a Fluid- 2 - Question 7

Explanation: Pa = Patm = 101.3
Pb = Pa + 0.9 * 9.81 * 0.03 = 101.56
Pc = Pb + 13.6 * 9.81 * 0.04 = 106.9
Pd = Pc – 1 * 9.81 * 0.05 = 106.41
Pe = Pd – 0.9 * 9.81 * 0.04 = 106.1
PX = Pe = 106.1
Since, PX > Patm, the percentage by which the pressure of the gas is higher than the atmospheric pressure will be

Test: Pressure Distribution in a Fluid- 2 - Question 8

A tank of height 3 cm is completely filled with water. Now two-third of the liquid is taken out and an equal amount of two other immiscible liquids of specific gravities 0.8 and 1.2 are poured into the tank. By what percent will the pressure at the base of the tank change?

Detailed Solution for Test: Pressure Distribution in a Fluid- 2 - Question 8

Explanation: Pressure at the base initially = 1 * 9.81 * 3 = 29.43 kPa; Pressure at the base after adding the other two liquids= 0.8 * 9.81 * 1 + 1 * 9.81 * 1 + 1.2 * 9.81 * 1 kPa; Thus the pressure at the base remains the same.

Test: Pressure Distribution in a Fluid- 2 - Question 9

In Fig. P2.12 the tank contains water and immiscible oil at 20°C. What is h in centimeters if the density of the oil is 898 kg/m3?

Detailed Solution for Test: Pressure Distribution in a Fluid- 2 - Question 9

Explanation: For water take the density =998 kg/m3. Apply the hydrostatic relation from the oil surface to the water surface, skipping the 8-cm part:

Patm + (898)(g)(h + 0.12) - (998)(g)(0.06+0.12) = Patm,

Solve for h ≈ 0.08m ≈ 8.0 cm

Test: Pressure Distribution in a Fluid- 2 - Question 10

All fluids in Figure below are at 20°C. If atmospheric pressure = 101.33 kPa and the bottom pressure is 242 kPa absolute, what is the specific gravity of fluid X?

Detailed Solution for Test: Pressure Distribution in a Fluid- 2 - Question 10

 

Explanation:Simply apply the hydrostatic equation from top to bottom:

Pbottom = Ptop + ∑ γ h

or: 242000 = 101330+ (8720)(1.0)+(9790)(2.0)+ γx(3.0)+ (133100)(0.5)

Solve for  γx = 15273N/m2

or: SGx = 15273/9790 = 1.56

Test: Pressure Distribution in a Fluid- 2 - Question 11

A beaker, partially filled with a liquid is rotated by an angle 30o as shown. If the pressure at point B becomes 12 bar, what will be the height (in cm) of the beaker?

Detailed Solution for Test: Pressure Distribution in a Fluid- 2 - Question 11

Explanation: If the height of the beaker is h, the pressure at point B = 103 * g * h * cos 30o = 12 * 103kPa; h = 24.5 cm.

Test: Pressure Distribution in a Fluid- 2 - Question 12

A beaker of height 15 cm is partially filled with a liquid and is rotated by an angle θ as shown.
If the pressure at point B becomes 5 bar, what will be the value of θ?

Detailed Solution for Test: Pressure Distribution in a Fluid- 2 - Question 12

Explanation: If the angle of inclination is taken to be θ, the pressure at point B = 103 * g * 0.15 * cos θ = 5 * 103 kPa; θ = 70.12o.

Test: Pressure Distribution in a Fluid- 2 - Question 13

A liquid compressed in a cylinder has a volume of 0.04 m2 at 50 N/cm2 and a volume of 0.039 m3 at 150 N/cm2. The bulk modulus of elasticity of liquid i

Detailed Solution for Test: Pressure Distribution in a Fluid- 2 - Question 13

The bulk modulus of elasticity, denoted by K, is a measure of the resistance of a material to compressibility. It is defined as the ratio of the change in pressure to the fractional change in volume. The formula for the bulk modulus is:

K = -V(dP/dV)

where V is the initial volume of the liquid, dP/dV is the pressure-volume relationship, and the negative sign indicates that pressure and volume change in opposite directions.

In this problem, we are given the initial and final volumes of the liquid at two different pressures. Using these values, we can calculate the change in pressure and the fractional change in volume, and then use the formula to find the bulk modulus.

Initial volume, V1 = 0.04 m^3 Initial pressure, P1 = 50 N/cm^2

Final volume, V2 = 0.039 m^3 Final pressure, P2 = 150 N/cm^2

Change in pressure, dP = P2 - P1 = 150 - 50 = 100 N/cm^2 Fractional change in volume, dV/V = (V2 - V1)/V1 = (0.039 - 0.04)/0.04 = -0.025

Substituting the values into the formula for bulk modulus:

K = -V1(dP/dV) = -0.04 m^3(100 N/cm^2/-0.025) = 4000 N/cm^2

Therefore, the bulk modulus of elasticity of the liquid is 4000 N/cm^2, which corresponds to option (b).

Test: Pressure Distribution in a Fluid- 2 - Question 14

The velocity distribution for flow over a flat plate is given by u = (y-y2) in which u is velocity in metres per second at a distance y metres above the plate. What is the shear stress value at y = 0.15 m? The dynamic viscosity of fluid is 8.0 poise.

Detailed Solution for Test: Pressure Distribution in a Fluid- 2 - Question 14

The shear stress value can be calculated using the formula:

τ = μ(dU/dy)

where τ is the shear stress, μ is the dynamic viscosity, and dU/dy is the velocity gradient.

We are given the velocity distribution as:

u = (y - y^2)

Taking the derivative with respect to y:

dU/dy = 1 - 2y

At y = 0.15 m:

dU/dy = 1 - 2(0.15) = 0.7

Substituting the values into the formula:

τ = (8.0 poise) (0.7) = 0.56 N/m^2

Therefore, the correct option is C) 0.56 N/m^2.

Test: Pressure Distribution in a Fluid- 2 - Question 15

 

For what height of the mercury column will the pressure inside the gas be 40 cm Hg vacuum?
 

Detailed Solution for Test: Pressure Distribution in a Fluid- 2 - Question 15

Answer: b
Explanation:
Pgas = Patm – ρgH
Taking gauge pressure in terms of cm of Hg,
-40 = 0 – H; H = 40.

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