Quiz: Civil Engineering
Exam: Punjab PSC
Topic: MISCELLANEOUS
Each question carries 1 mark
Negative marking: No negative marking
Time: 10 Minutes
Q1. For a sandy soil, the angle of internal friction is 30. If the major principal stress is 50 KN/m² at failure the value of minor principal stress will be-
(a) 20.00 KN/m²
(b) 16.66 KN/m²
(c) 20.8 KN/m²
(d) 27.2 KN/m²
Q2. The back site reading taken from a level at a bench mark is 2.54m and a fore site at a point A is taken on an inverted staff is 1.45m calculate the reduced level of the point A.
If RL of Bench mark is 100m.
(a) 146.79
(b) 105.79
(c) 103.99
(d) 102.30
Q3. Calculate the additive & multiplying constant. If the focal length of the objective glass is 300mm, stadia intercept is 3mm and distance of the instrument axis from the centre of the objective glass is 200 mm.
(a) 125, 440
(b) 100, 500
(c) 95, 440
(d) 100, 550
Q4. Which of the following set of internal Angles (degree) of a triangle does not show well-conditioned triangle?
(a) 20, 90, 70
(b) 25, 45, 110
(c) 40, 125, 15
(d) 40, 75, 65
Q5. The water is flowing through 1000 m long circular pipe of diameter 50 cm. With the velocity of 0.26 m/s. friction factor for the pipe is given as 0.016 what is the head loss (cm) in the pipe due to friction?
(a) 20 cm
(b) 14 cm
(c) 18 cm
(d) 24 cm
Q6. The relation between shear stress Z and velocity gradient of a fluid is given by where A and n are constants. If n = 1, what type of fluid will it be?
a) Newtonian fluid
b) Non-Newtonian fluid
c) Pseudoplastic
d) Bingham plastic
Q7. The head of water over a rectangular notch is 900 mm. The discharge is 300 liters/s. Find the length of the notch, when CD =.62
a) 0.192 m
b) 0.250 m
c) 0.205 m
d) 0.175 m
Q8. The main property that affects a boundary layer is__________
a) Temperature
b) Pressure
c) Viscosity
d) Surface tension
Q9. Which among the following force is developed due to resistance of a fluid flow?
a) Viscous force
b) Inertial force
c) Gravity force
d) Pressure force
Q10. Unit weight of brick work is about
(a) 17-18 kN/m³
(b) 18-19 kN/m³
(c) 19-20 kN/m³
(d) 20-21 kN/m³
SOLUTION
S1. Ans.(a)
Sol. for sandy soil C=0
Internal friction ∅=38°
σ_1= 60 KN/m²
σ_3 = ?
σ_1= σ_3 tan²(45+∅/2)
σ_1= σ_3tan²(45+30/2)
σ_3= 20 KN/m²
S2. Ans.(c)
Sol. Back site = 2.54 m
RL of Bench Mark= 100m
Hight of instrument = 100+2.54
= 102.54 m.
Fore site at A (invested staff) ⇒ 1.45 m
RL of point ‘A’ = 102.54+1.45= 103.99 m
S3. Ans.(b)
Sol. focal length (f) = 300 mm
Intercept (i)= 3 mm
d= 200 mm
Multiplying constant = f/i=300/3=100 mm
Additive constant= f+ d= 300+200= 500 mm
S4. Ans.(d)
Sol. Well-conditioned triangle are those triangles in which at least one angle is less than 30 degree or greater than 120 degree.
S5. Ans.(c)
Sol. length of pipe= 100 m.
Dia of pipe= 30 cm= 0.30 m.
Velocity = 0.26 m/sec
Friction factor (f) = 0.016
Head loss (hf) due to friction= flv²/2gd
= (0.016×1000(0.26)²)/(2×9.81×0.30)=0.18 m
= 18 cm
S6. Ans (a)
Sol: When n = 1, the relation reduces to Newton’s law of viscosity: z = A * , where A will represent the viscosity of the fluid. The fluid following this relation will be a Newtonian fluid
S7. Ans (a)
Sol: L = 1.5 * Q / (Cd * √2g * H 1.5)
= 1.5 * .3 / (.62 * √19.62 * .91.5)
= 0.192 m.
S8. Ans (c)
Explanation: A boundary layer is an important concept that refers to the layer of fluid. The fluid that is in the immediate vicinity of a bounding surface. The main property that affects a boundary layer is viscosity.
S9. Ans (a)
Sol: Viscous force is the force that is developed due to resistance of a fluid flow. Viscous force is equal to the product of shear stress due to viscosity and surface area of the fluid. It acts in the opposite direction to that of the acceleration.
S10. Ans.(c)
Sol.
Material Unit Weight (kN/m³)
Brick masonry 19-20
Lime mortar 17-18
Cement mortar 20-21
Reinforced Cement concrete 24-25
