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# NHPC-JE’21 CE: Daily Practices Quiz. 05-Oct-2021

Know your strengths and practice your concepts with this quiz on NHPC JE Recruitment 2021. This quiz for NHPC JE Recruitment 2021 is designed specially according to NHPC Syllabus 2021.

Quiz: Civil Engineering
Exam: NHPC-JE
Topic: Miscellaneous

Each question carries 1 mark
Negative marking: 1/4 mark
Time: 8 Minutes

Q1. If the radial acceleration of transition curve is 30cm/sec², radius is 200m and the velocity is 14m/sec. the length of the transition curve is
(a) 46.0 m
(b) 46.5 m
(c) 45.0 m
(d) 45.5 m

Q2. As per the working stress method of reinforced concrete design, if a concrete has permissible compressive stress of p, then the modular ratio is approximately given as:
(a) 93/p
(b) 140/p
(c) 210/p
(d) 280/p

Q3. When waste water is disposed of into a running stream, four zones are formed. In which of the following zones will the minimum level of dissolved oxygen be found?
(b) Zone of active decomposition
(c) Zone of recovery
(d) Zone of clear water

Q4. Which one of the following defines Aridity index (AI)?
(a) AI=(PET-AET)/PET×100
(b) AI=PET/AET×100
(c) AI=AET/PET×100
(d) AI=(AET-PET)/AET×100
(where AET = Actual Evapotranspiration and PET = Potential Evapotranspiration)

Q5. A soil sample is having a specific gravity 2.60 and a void ratio of 0.78. The water content required to fully saturate, the soil at the void ratio would be
(a) 10%
(b) 30%
(c) 50%
(d) 70%

Q6. The maximum depth of neutral axis in a RC beam for Fe 415 grade of steel is:
(a) 0.53d
(b) 0.46d
(c) 0.67d
(d) 0.48d

Solutions

S1. Ans.(a)
Sol. C = 30 cm/sec²
= 0.30 m/sec²
R = 200 m.
V = 14 m/sec.
Length of transition curve (L) = V^3/(C.R)
= (14)^3/(0.3×200)
= 45.7 m.
= 46 m.

S2. Ans.(a)
Sol. modular ratio (m) =280/3σcbc
▭(σcbc=p)
M=280/3p
▭(m=93/p)

S3. Ans.(b)
Sol. There are four zones are formed when waste water is disposed off into a running stream
2. Zone of active decomposition
3. Zone of recovery
4. Zone of clear water
In zone of active decomposition water is more dark and more turbid and dissolved oxygen may go up to zero.

S4. Ans.(a)
Sol. ▭(Aridity Index (AI)=(PET-AET)/PET×100)
Where, AET = Actual Evapotranspiration
PET = Potential Evapotranspiration

S5. Ans.(b)
Sol.
Specific gravity (G)=2.60
Void ratio (e)=0.78
Water content (w)=?
Fully saturated i.e. S=1
se=wG
1×0.78=w×2.60
▭(w=30%)

S6. Ans.(d)
Sol.

 Grade of steel Depth of neutral axis () Fe 250 0.53d Fe 415 0.48d Fe 500 0.46d

Where, d= effective depth

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