Skip to main content

Mechanics of Fluids - MCQ from AMIE Exam (Summer 2021)


Select the correct answer
 
1. Shear stress in fluid flow is proportional to
(a) ∂u/∂x
(b) ∂u/∂y
(c) ∂v/∂x
(d) ∂v/∂y
 
2. If the contact angle at the surface of a glass tube immersed in a liquid is 100⁰, Then the liquid in the glass tube will
(a) Rise
(b) Fall
(c) Neither rise nor fall
(d) None of these
 
Hint: when contact angle θ is obtuse, i.e. θ > 90°
In case of non – wetting fluid, the level in the capillary tube will fall, and the phenomenon is capillary fall. For mercury glass tube, θ = 128°.
 
3. The atmospheric pressure at mean sea level is equivalent in terms of Hg is
(a) 70 cm
(b) 86 cm
(c) 76 cm
(d) 75 cm
 
4. Piezometric head consists of
(a) Pressure head
(b) Potential head
(c) Pressure head + Potential head
(d) None of these
 
Hint:  The sum of pressure head and datum is known as piezometric head. It is given by
p/ρg + z
 
5. For uniform flow through pipe
(a) ∂u/∂x ≠ 0
(b) ∂u/∂x = 0
(c) ∂u/∂r = 0
(d) None of these
 
6. For non-Newtonian fluids, shear stress (τ) is equal to
(a) ∂u/∂x
(b) (∂u/∂x)n 
(c) (∂u/∂y)n 
(d) ∂u/∂y
 
7. For continuity equation for incompressible and steady flow, the discharge (Q) through pipe is
(a) 0
(b) Not constant
(c) Constant
(d) None of these
 
8. ln momentum equation, summation of forces on the control volume is equal to
(a) ρQV2 
(b) ρQV
(c) ρV
(d) ρQ
 
9. Energy correction factor for laminar flow through pipe is equal to
(a) 1.0
(b) 1.5
(c) 2.0
(d) 1.7
 
Hint: Its value for a fully developed laminar pipe flow is around 2. It is 1 for a turbulent flow.
 
10. Head loss for laminar flow between parallel plates is equal to
(a) 12μVL/D2
(b) 12μVL/γD2
(e) 32μVL/D2
(d) 10μVL/γD2

11. Nominal boundary layer thickness (ઠ) is that distance (y) from the plate, at which
(a) u =1.0U₀
(b) u = 0.95
(c) u = 0.99U₀
(d) u = 1.01U₀
Where U₀ is the free stream velocity.
 
12. ln laminar boundary layer with appropriate velocity distribution, ratio of displacement thickness and displacement thickness (ઠ*/ઠ) is
(a) 1/2
(b) 3/10
(c) 2/15
(d) 3/15
 
13. Laminar sub-layer occurred at plate surface in
(a) Laminar boundary layer
(b) Turbulent boundary layer
(c) Transition boundary layer
(d) None of these
 
Hint:
 

14. Separation of the boundary layer developed over the curved surface takes place after that point, where the pressure gradient (∂p/∂x) is
(a) Greater than zero
(b) Equal to zero
(c) Less than zero
(d) None of these
 
Hint:  When the pressure goes on increasing in the direction of flow, the pressure force acts against the direction of flow in the boundary layer and hence thickening the boundary layer more rapidly. This and the boundary shear bring the fluid in the boundary layer to rest and causes backflow. Due to this the boundary layer no more sticks to the boundary but is shifted away from the boundary. This phenomenon is called as “Boundary layer separation”.
 
15. The dimensionless laminar boundary layer (δi/x) developed over the flat plate just from the leading edge is equal to
(a) 3.0/(Rex)0.5
(b) 5.0/(Rex)0.5
(c) 7.0/(Rex)0.5
(d) 1.0/(Rex)0.5
 
Hint: δ=5x/√Rex
 
16. The local skin friction coefficient (cf) for a turbulent boundary layer developed over a flat plate just from the leading edge, is proportional to
(a) Re1/5
(b) Re-1/5
(c) Re-2/5
(d) Re2/5
Where x is the distance from the leading edge.
 
17. The ratio of friction factor (f) and skin friction coefficient (f') is equal to
(a) 4.0
(b) 2.0
(c) 1.0
(d) 0.25
 
Hint: Friction factor = 4 x friction coefficient

18. In Blasius equation, the friction factor (f) is given by
(a) 0.416Re1/4
(b) 0.316Re1/4
(c)  0.216Re1/4
(d)  0.516Re1/4
Where R is the Reynolds number.
 
19. The exit loss coefficient for a pipe discharging freely into the atmosphere is
(a) 1.0
(b) 0.75
(c) 0.5
(d) 0.25
 
Hint: 
At entry,  hL, entry = 0.5v2/2g
At exit =  hL, exit = (1.0)v2/2g
 
20. Adiabatic constant for diatomic gases is equal to
(a) 1.0
(b) 1.4
(c) 1.2
(d) 0.8
 
Hint: 
γ for monatomic =  1.67
γ for diatomic =  1.40
 
---

The study material for AMIE/Junior Engineer exams is available at https://amiestudycircle.com

Comments

Popular posts from this blog

Mechanics of Fluids (Solved Numerical Problems)

Numerical The surface Tension of water in contact with air at 20°C is 0.0725 N/m. The pressure inside a droplet of water is to be 0.02 N/cm² greater than the outside pressure. Calculate the diameter of the droplet of water. (7 marks) (AMIE Summer 2023) Solution Surface tension, σ = 0.0725 N/m Pressure intensity, P = 0.02 N/m 2 P = 4σ/d Hence, the Diameter of the dropd = 4 x 0.0725/200 = 1.45 mm Numerical Find the surface tension in a soap bubble of 40 mm diameter when the inside pressure is 2.5 N/m² above atmospheric pressure. (7 marks) (AMIE Summer 2023) Answer: 0.0125 N/m Numerical The pressure outside the droplet of water of diameter 0.04 mm is 10.32 N/cm² (atmospheric pressure). Calculate the pressure within the droplet if surface tension is given as 0.0725 N/m of water. (AMIE Summer 2023, 7 marks) Answer: 0.725 N/cm 2   Numerical An open lank contains water up to a depth of 2 m and above it an oil of specific gravity 0.9 for a depth of 1 m. Find the pressure intensity (i) at t...

Design of Electrical Systems (Solved Numerical Problems)

Important note There is something wrong with this question paper. It seems that instead of "Design of Electrical Systems" the IEI has given problems from "Electrical Machines". You should raise a complaint to director_eea@ieindia.org in this regard. Numerical A 120 V DC shunt motor draws a current of 200A. The armature resistance is 0.02 ohms and the shunt field resistance is 30 ohms. Find back emf. If the lap wound armature has 90 slots with 4 conductors per slots, at what speed will the motor run when flux per pole is 0.04 Wb?​ (AMIE Summer 2023, 8 marks) Solution The back EMF (E b ) of a DC motor can be calculated using the formula: E b = V - I a R a   Given: V = 120 V I a = 200 A R a = 0.02 ohms Substituting the values into the formula: E b = 120 − 200 × 0.02 = 120 − 4​ = 116 V Now, let's calculate the speed (N) at which the motor will run using the given flux per pole (φ p ). The formula to calculate the speed of a DC motor is: N = 60×E b /(P×φ p ) Wh...

Energy Systems (Solved Numerical Problems)

Wind at 1 standard atmospheric pressure and \({15^0}C\) has velocity of 15 m/s, calculate (i) the total power density in the wind stream (ii) the maximum obtainable power density (iii) a reasonably obtainable power density (iv) total power (v) torque and axial thrust Given: turbine diameter = 120 m, and turbine operating speed = 40 rpm at maximum efficiency. Propeller type wind turbine is considered. (AMIE Winter 2023) Solution For air, the value of gas constant is R = 0.287 kJ/kg.K 1 atm = 1.01325 x 105 Pa Air density \(\rho  = \frac{P}{{RT}} = \frac{{1.01325x{{10}^5}}}{{287}}(288) = 1.226\,kg/{m^3}\) Total Power \({P_{total}} = \rho A{V_1}^3/2\) Power density \(\begin{array}{l}\frac{{{P_{total}}}}{A} = \frac{1}{2}\rho {V_1}^3\\ = \frac{1}{2}(1.226){(15)^3}\\ = 2068.87{\mkern 1mu} W/{m^2}\end{array}\) Maximum power density \(\begin{array}{l}\frac{{{P_{\max }}}}{A} = \frac{8}{{27}}\rho A{V^3}_1\\ = \frac{8}{{27}}(1.226){(15)^3}\\ = 1226{\mkern 1mu} W/{m^2}\end{array}\) Assuming eff...