NHPC-JE'21 CE: Daily Practices Quiz. 18-Nov-2021 |_00.1
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NHPC-JE’21 CE: Daily Practices Quiz. 18-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. A sewer is commonly designed to attain self-cleansing velocity at
(a) Sewer running half full
(b) Peak hourly rate of flow
(c) Minimum hourly rate of flow
(d) Average hourly rate of flow

Q2. 10 divisions of verniers scale are equal to 11 division of a main scale of each 0.1 mm. what is the least count of the verniers scale?
(a) 0.009
(b) 0.01
(c) 0.1
(d) 1.1

Q3. After storage, the strength of cement-
(a) Decrease
(b) Increase
(c) Remains same
(d) May increase or decrease

Q4. Minimum pitch of the rivets shall not be less than:
(a) 2.0d
(b) 1.5d
(c) 2.5d
(d) 3.0d

Q5. The ratio between adopted centrifugal ratios for roads and railways is
(a) 3 : 1
(b) 2 : 1
(c) 5 : 1
(d) 4 : 1

Q6. A summit curve is formed at the intersection of a 3% upgrade and a 5% downgrade what is the length of the summit curve in order to provide a stopping distance of 130 m.
(a) 271 m
(b) 298 m
(c) 307 m
(d) 340 m

Solutions

S1. Ans.(c)
Sol. Self-cleansing velocity is such flow velocity as would be sufficient to flush out any deposited solids in the sewer. A sewer is commonly designed to attain the self-cleansing velocity at minimum hourly rate of flow.

S2. Ans. (b)
Sol. Least count = (Smallest division of one main scale)/(no.of division on vernier scale ) = 0.1/10 = 0.01

S3. Ans.(a)
Sol. The strength of cement decreases after storage.

Time Reduction in strength
3 months

6 months

1 year

2 years

10%

20%

40%

50%

S4. Ans.(c)
Sol. Minimum pitch of rivets should be 2.5 times the nominal diameter or rivet i.e. 2.5d.
d = nominal diameter of rivet.

S5. Ans.(b)
Sol.
→ Maximum value of centrifugal ratio for roads = 1/4
→ Maximum value of centrifugal ratio for railways = 1/8
So, ratio between these two = (1/4)/(1/8)
= (2 : 1)

S6. Ans.(c)
Sol. Deviation angle = N = 3 + 5= 8% = 0.08
Assuming L > S
L = (NS^2)/4.4=(0.08×(130)^2)/4.4
= 307.2 m ≃ 307 m

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