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

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.

Explanation: Pressure increases in the vertically downward direction but remains constant in the horizontal direction. Thus,
P_B = P_A + ρgh
where P_B = Pressure at B, P_A = 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

Explanation: Total height of the water in the beaker = 2 + ¹⁄₂ * 10 cos 60^o 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.

Explanation: Gauge pressure at the base of the beaker = S_o * 10³ * 0.05 * g + S_w * 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 = ¹⁄₂ * 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 = ¹⁄₂ * 882.9
Thus, h = 0.045 (in the range considered). Hence, the correct answer will be 4.5 cm.

Explanation: Gauge pressure at the base of the beaker = S_o * 10³ * 0.2 * g + S_w * 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 = ¹⁄₃ * 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 = ¹⁄₃ * 2550.6
Thus, h = 0.2733 (out of the range considered). Hence, the correct answer will be 19.2 cm.

Explanation: The change of pressure with the vertical direction y is given by
dP/dy = – ρg
dP = -ρg dy
If P_a and P_b 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

Explanation: P_a = P_atm = 101.3
P_b = P_a + 0.9 * 9.81 * 0.03 = 101.56
P_c = P_b + 13.6 * 9.81 * 0.04 = 106.9
P_d = P_c – 1 * 9.81 * 0.05 = 106.41
P_e = P_d – 0.9 * 9.81 * 0.04 = 106.1
P_X = P_e = 106.1
Since, P_X > P_atm, the percentage by which the pressure of the gas is higher than the atmospheric pressure will be

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.

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:

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

Solve for h ≈ 0.08m ≈ 8.0 cm

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

P_{bottom =} P_top + ∑ γ h

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

Solve for  γ_{x =} 15273N/m²

or: SG_x = 15273/9790 = 1.56

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

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 = 70.12^o.

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).

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.

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