Quiz: Civil Engineering
Exam: RSMSSB-JEn
Topic: Miscellaneous
Each question carries 1 mark
Negative marking: 1/3 mark
Time: 8 Minutes
Q1. The minimum freeboard recommended for lined canal carrying discharge greater than 10 m³/s
(a) 0.15 m
(b) 0.60 m
(c) 1.25 m
(d) 0.75 m
Q2. Estimate for electrical wiring is prepared on the basis of?
(a) Voltage
(b) Number of points
(c) Number of appliances
(d) Power
Q3. In the design of a sedimentation tank the essential factor to be considered is
(a) depth of tank
(b) turbidity of water
(c) concentration of suspended solids in the water
(d) surface loading of tank
Q4. Newton’s law of viscosity is a relationship between-
(a) Rate of shear strain & temperature
(b) Shear stress & rate of shear strain
(c) Pressure, velocity & temperature
(d) Shear stress & velocity
Q5. The angle of the failure plane with the major principal plane is given by
(a) 45° + ϕ’
(b) 45° +(ϕ’)/2
(c) 45° – (ϕ’)/2
(d) 45° – ϕ’
Q6. A 30 m metric chain is found to be 0.1 m too short throughout the measurement. If the distance measured is recorded as 300 m, then the actual distance measured will be
(a) 300.1 m
(b) 301.0 m
(c) 299.0 m
(d) 310.0 m
Solutions
S1. Ans.(d)
Sol.
| Discharge (m³/sec.) | Freeboard (m.) |
| 1 to 5 | 0.50 |
| 5 to 10 | 0.60 |
| 10 to 30 | 0.75 |
| 30 to 150 | 0.90 |
S2. Ans.(b)
Sol. Estimate for electrical wiring is prepared on the basis of number of points.
S3. Ans.(d)
Sol. Surface loading rate of tank or surface overflow rate is the important parameter for designing of tank as it should be less than settling velocity of discrete particles for effective sedimentation.
S4. Ans.(b)
Sol. Newton’s law states that the shear stress is directly proportional to the rate of shear strain for Newtonian fluid.
τ α dv/dy
▭(τ=μ.dv/dy)
S5. Ans.(b)
Sol. Angle of the failure plane with the major principal plane is
[45+ϕ/2]
S6. Ans.(c)
Sol. Correct length of chain (l) = 30 m.
Wrong length of chain (l’) = 30-0.1
= 29.9m.
Wrong length of measured line (L’) = 300 m.
Correct length of measured line (L) = ?
L×l×l^’×L^’
L×30=29.9×300
L=29.9/30×300
▭(L=299m.)
