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NHPC-JE’21 CE: Daily Practices Quiz. 13-Dec-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. Specific gravity of sand deposit is 2.50 having the void ratio 0.65 the critical hydraulic gradient for the sand deposit:
(a) 0.85
(b) 0.90
(c) 0.75
(d) 1.0

Q2. The estimate earth work by the center line method for total center length 40.10 m, breadth = 1.10 m and depth = 1m is
(a) 22.22 cum
(b) 38.89 cum
(c) 42.90 cum
(d) 44.11 cum

Q3. The natural bedding plane of stones and the direction of pressure in stone masonry is –
(a) At 45°
(b) Parallel
(c) At 30°
(d) Normal

Q4. 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

Q5. A single load of 100 kN rolls along a girder of 20 m simply supported span, the maximum bending moment is –
(a) 100 kNm
(b) 500 kNm
(c) 150 kNm
(d) 600 kNm

Q6. The pressure in Pascals at a depth of 1 m below the free surface of a body of water will be equal to
(a) 1 Pa
(b) 98.1 Pa
(c) 981 Pa
(d) 9810 Pa

Solutions

S1. Ans.(b)
Sol. G=2.50
e=0.65
Critical hydraulic gradient (i)=(G-1)/(1+e)=(2.50-1)/(1+0.65)
=0.909

S2. Ans.(d)
Sol. Given, Total center Length (L) = 40.10m.
Breadth (B) = 1.10 m
Depth (d) = 1.0 m.
Estimate earthwork by center line method
= L × B × d
= 40.10 × 1.10 × 1.0
= 44.11 m³

S3. Ans.(d)
Sol. In stone masonry, the stones are placed in position such that the natural bedding plane is normal to the direction of pressure they carry. The stone used in stone masonry should be hard, strong and durable.

S4. 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.

S5. Ans.(b)
Sol. The maximum bending moment due to moving load on a simply supported beam occurs under the load and it will be maximum at mid span, hence
Maximum bending moment 〖(M〗_(max⁡)) = wl/4.
Given W=100 kN ,l=20 m
M_max=(100×20)/4
▭(M_max=500 kN-m)

S6. Ans.(d)
Sol. Pressure (P) = ρ.gh
= 1000×9.81×1
= 9810 Pa.

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