NHPC-JE'21 CE: Daily Practices Quiz. 13-Oct-2021 |_00.1
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NHPC-JE’21 CE: Daily Practices Quiz. 13-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. The ‘Center Line Method’ is specially adopted for estimating-
(a) Polygonal Buildings only
(b) Circular, hexagonal and other geometric
(c) Circular buildings only
(d) Hexagonal buildings only

Q2. The development length in compression for a 20 mm diameter deformed bar of grade Fe 415 embedded in concrete of grade M 25 whose design bond stress is 1.40 N/mm², is
(a) 1489 mm
(b) 1289 mm
(c) 806 mm
(d) 645 mm

Q3. Which of the following defect appears due to presence of alkalis in the bricks?
(a) Bloating
(b) Black core
(c) Crackers
(d) Efflorescence

Q4. The back sight taken on a B.M. (R.L. = 500.000m.) was 3.415 m. The fore sight taken on the next point (a) was 2.125m. The R.L. of point A will be:
(a) 503.415 m
(b) 502.125 m
(c) 505.540 m
(d) 501.290 m

Q5. What does Chemical Oxygen Demand (COD) indicate?
(a) Biodegradability of the waste water
(b) Potential for recycling the wastewater
(c) Age of the sewage
(d) Strength of a sewage

Q6. A compression member is termed as column or strut if the ratio of its effective length to the least lateral dimension is more than
(a) 1
(b) 2
(c) 3
(d) 5

Solutions

S1. Ans.(b)
Sol. Center line method→ This method is suitable for symmetrical cross-sections like circular, hexagonal and other geometric shapes. This method is suitable for walls have the same thickness. In this method the center line for each type worked out and then multiplied by the breadth and depth of respective item to get the total quantity.

S2. Ans.(d)
Sol. According to IS 456 : 2000
The development length (L_d ) = (ϕσ_st)/(4τ_bd )
→ For deformed bars τ_bd shall be increases by 60%.
→ for bars in compression τ_bd increased by 25%
L_d=(20×0.87×415)/(4×1.4×1.6×1.25)
▭(L_d= 645 mm)

S3. Ans.(d)
Sol. Defects in bricks-
Efflorescence→ This defect appears in bricks due to the presence of alkalis in the bricks.
Chuffs → This is the defect in bricks in which shape of the bricks gets deformed due to rain water falling on hot brick.
Bloating → Bloating of bricks is swollen spongy mass over the surface of burned bricks.
Black core→ The brick converts in black core because of improper burning.

S4. Ans.(d)
Sol. Given, BM = 500.00 m. (R.L)
BS = 3.415 m
FS at point A = 2.125 m.
RL of point A = ?
H.I = RL of BM + BS
= 500 + 3.415
= 503.415 m.
R.L at point ‘A’ = H.I – FS
= 503.415 – 2.125
= 501.290 m.

S5. Ans.(a)
Sol. Chemical oxygen demand indication the amount of oxygen required to carry out the decomposition of both biodegradable or non-biodegradable organic matter present in the system.

S6. Ans.(c)
Sol. The ratio of effective length & least lateral dimension for column is greater than ‘3’.
Slenderness ratio (λ)= leff/(L.L.D.)
λ < 12 → short column
λ ≥12 → Long column

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