Quiz: Civil Engineering 6 May 2021

Quiz: Civil Engineering
Exam: DFCCIL
Topic: Hydraulics

Each question carries 1 mark
Negative marking: 1/4 mark
Time: 10 Minutes

Q1. If C_v, C_c, C_d and C_r are the hydraulic coefficients of an orifice, then
(a) C_d = Cc. C_v
(a) C_r = 1 + C_v/C_d
(c) C_v = Cc + C_d
(d) Cc = C_v/C_d

Q2. The critical depth of flow in a most economical triangular channel section for a discharge of 1 m³/sec is given by
(a) (1/9.8)^(1/5)metre
(a) (1/9.8)^(1/3)metre
(c) (1/4.9)^(1/5)metre
(d) (1/4.9)^(1/3)metre

Q3. The increase in meta centric height
A. Increases stability
B. Decreases stability
C. Increases comfort for passengers
D. Decreases comfort for passengers
The correct answer is
(a) A and C
(a) A and D
(c) B and C
(d) C and D

Q4. When two identical centrifugal pumps are operating in series on a common rising main, then?
(a) The pressure is the rising main will be nearly doubled, while discharge will remain same.
(a) The discharge will be nearly doubled, while the pressure remains the same
(c) Discharge as well as the pressure in the rising main will be doubled
(d) Discharge as well as the pressure in the rising main will increase but not become double.

Q5. At a hydraulic jump, the depths at the two sides are 0.4 m and 1.4 m. The head loss in the jump is nearly?
(a) 1.0 m
(a) 0.9 m
(c) 0.7 m
(d) 0.45 m

Q6. If a pipe of diameter 30 cm running full of water with velocity 100 m/sec is changed by a pipe of diameter 15 cm, then the velocity of water flowing through the pipe will be
(a) 50 m/sec
(a) 25 m/sec
(c) 200 m/sec
(d) 400 m/sec

Q7. The density of water is 1000 kg/m3 at

(a) 0oC

(a) 0oK

(c) 4oC

(d) 20oC

Q8. Unit of kinematic viscosity is
(a) m2/sec
(a) Newton-sec/m2
(c) Newton-sec/m3
(d) Kg-sec/m2.

Q9. Inside pressure in a hollow soap bubble in the air is: (where d is the diameter of the bubble)
(a) 4σ/d
(a) 2σ/d
(c) 6σ/d
(d) 8σ/d

Q10. A two-dimensional flow field is given by stream function Ψ = x² – y². The magnitude of absolute velocity at a point (1,1)
(a) 2
(a) 4
(c) 8
(d) 2√2

Solutions

S1. Ans.(a)
Sol. Coefficient of discharge is defined as the ratio of the actual discharge from an orifice to the theoretical discharge from the orifice. It is denoted by C_d.

C_d=(Actual discharge)/(Theoritical discharge)
=(Actual velocity)/(Theoritical velocity)×(Actual area)/(Theoritical Area)
▭(C_d=C_v×C_C )
C_V= coefficient of velocity
C_C= coefficient of contraction.

S2. Ans.(c)
Sol. For most economical triangular channel section, the side slope m = tanθ = tan 45
▭(m=1)
Given, Q = 1 m³/sec.
Now, critical depth for triangular channel
Section –
y_c=((2Q^2)/(gm^2 ))^(1\/5)
=[(2×(1)^2)/(9.8×(1)^2 )]^(1\/5)
▭(y_c=(1/4.9)^(1\/5) metre)

S3. Ans.(b)
Sol. Meta centric height is the vertical distance between centroid of the submerged body and the metacenter of the submerged body.
Now,
For ship (time period of oscillation under stability)
▭(T=2π√(k^2/g(GM) ))
GM = metacentric height
k = radius of gyration
∵ T α 1/√GM α 1/(Stability )
If GM is large, then time period of oscillation is small hence oscillation is more comfort of passenger is decreased.

S4. Ans.(a)
Sol. When two pumps are operated in series then discharge remains same and head produced is doubled while if they operated in parallel then discharge is doubled and head remains the same.

S5. Ans.(d)
Sol. Given, conjugate depths y_1=0.4 m.
y_2=1.4 m.
E_L= ?
E_L=(y_2-y_1 )^3/(4y_1 y_2 )
=(1.4-0.4)^3/(4×1.4×0.4)
▭(E_L=0.446≈0.45m)

S6. Ans.(d)
Sol. Given,
d_1=30 cm
V_1=100 m\/sec⁡
d_2=15 cm
V_2= ?
By continuity equation
A_1 V_1=A_2 V_2
π/4 (d_1 )^2×100=π/4 (d_2 )^2×V_2
(30)^2×100=(15)^2×V_2
▭(V_2=400 m\/sec )

S7. Ans.(c)
Sol. The density of water is 1000 kg/m³ at 4°C or 39°F.

S8. Ans.(a)
Sol. Unit of kinematic viscosity is m^2/sec or (cm^2)/sec
▭(1 stokes =1 (cm^2)/sec.=10^(-4) m^2/sec.)

S9. Ans.(d)
Sol. → The pressure inside in a hollow soap bubble in the air is 8σ/d.
→ The pressure inside in water droplet is 4σ/d.
→ The pressure inside in liquid jet is 2σ/d.
Where, d = diameter
σ= Surface tension

S10. Ans.(d)
Sol. Given, stream function Ψ = (x²-y²)
Magnitude of absolute velocity at point (1, 1) = ?
Now,
V_x=∂Ψ/∂y
V_x= -2y
▭(〖V_x〗_((1,1) )= -2)
V_y=(-∂Ψ)/∂x
V_y= -2x
▭(〖V_y〗_1,1= -2)
V=√((V_x )^2+(V_y )^2 )
=√((-2)^2+(-2)^2 )
=√8
▭(V=2√2)

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