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NHPC-JE’21 CE: Daily Practices Quiz. 17-Nov-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. Removal of excess chlorine resulting from super chlorination in part or completely, is called:
(a) Re-chlorination
(b) Pre-chlorination
(c) Post-chlorination
(d) De-chlorination

Q2. What will be the wetted perimeter of a regime channel as per Lacey’s theory:
Assume : total discharge = 100 cumecs
(a) 47.2 m
(b) 47.3 m
(c) 47.4 m
(d) 47.5 m

Q3. The expression used to calculate the safe allowable speed at a curve of radius ‘R’ is:
(a) √(27.94*R)
(b) √(29.74*R)
(c) √(42.97*R)
(d) √(24.97*R)

Q4. The formula for Fat lime is ………
(a) CaCO_2
(b) CaCO_3
(c) Ca(OH)_3
(d) Ca(OH)_2

Q5. Determine the present value of the building if the plinth area is 150 sq. m. and rate of plinth area = Rs 2500 sq. m. Assume: Straight line method of depreciation.
(a) 375000
(b) 365000
(c) 355000
(d) 356000

Q6. What should be the height (m) of a lighthouse, so that it can be visible from a distance of 3 km?
(a) 0.101
(b) 0.605
(c) 0.673
(d) 0.707

Solutions

S1. Ans.(d)
Sol. De- chlorination is the process of removing excess chlorine from water.
De – chlorinating agents
(i) SO_2 gas
(ii) Sodium thiosulphate (Na_2 S_2 O_3)
(iii) Activate carbon
(iv) Ammonia as (NH_4 OH)
(v) Sodium sulphate (Na_2 SO_3 )

S2. Ans.(d)
Sol. Given Q= 100 cumecs
As per Lacey’s theory, wetted perimeter (P) = 4.75 √Q
= 4.75 √100
= 47.5 m

S3. Ans.(a)
Sol. For Super-elevation
e + f = V^2/127R
For plain & rolling terrain
e = 7% = 0.07 & f = 0.15
Now,
0.07 + 0.15 = V^2/127R
V2 = 27.94R
V = √27.94R

S4. Ans.(d)
Sol. Quick lime ⇒ CaO
Limestone ⇒ CaCO_3
Gypsum ⇒ CaSO_4.2H_2 O
Fat lime/Hydrated lime ⇒ Ca(OH)_2

S5. Ans.(a)
Sol. Plinth area = 150 m²
Rate of plinth area = 2500 Rs / m²
Present value of building = 2500×150 = 375000 Rs.

S6. Ans (b)
Sol. Distance (d) = 3 km
” Height of lighthouse “(h)=0.0673d^2
=0.0673×3^2
=0.6057m

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