NHPC-JE'21 CE: Daily Practices Quiz. 12-Oct-2021 |_00.1
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NHPC-JE’21 CE: Daily Practices Quiz. 12-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. Kharif crop sown in
(a) In June
(b) In September
(c) In October-November
(d) None of the above

Q2. The correction for curvature and combined correction for curvature and refraction for a distance 1,500 m are:
(a) 176 m, 151 m
(b) 176 cm, 151 cm.
(c) 176 mm, 151 mm.
(d) 176 km, 151 km.

Q3. Calculate detention time of the tank whose breadth is 2 m, length is 4 m and depth is 2.5m. the rate of flow is 4 x 10³ liters per hour
(a) 5 hours
(b) 4 hours
(c) 5.5 hours
(d) 2.5 hours

Q4. Estimate the quantity of plastering (two faces) for 4 m long, 3m high and 30 cm thick wall.
(a) 72 cum
(b) 2.4 sqm
(c) 24 sqm
(d) 36 cum

Q5. Void ratio of an undisturbed sample of soil is 0.6. the values of maximum and minimum possible void ratio are found as 0.8 and 0.4, respectively. The relative density in percentage, for this soil sample will be
(a) 25
(b) 50
(c) 75
(d) 90

Q6. Calculate the value of shear stress (MPa) in the solid circular shaft of diameter 0.1 m which is subjected to the torque of 10 kN-m.
(a) 40.5
(b) 50.93
(c) 60.5
(d) 70.5

Solutions

S1. Ans.(a)
Sol. Kharif crops sown in June and harvested in September and October. The major kharif crops are bajra, maize, cotton, sugarcane, paddy, jowar, rice, pulses, and ground-nut etc.

S2. Ans.(c)
Sol. Given, d = 1500m = 1.5 km
Correction for curvature (C_C ) = -0.0785 d²
= -0.0785 × (1.5) ²
= -0.176625 m.
= 176 mm. (Negative)
Combined correction (C) = -0.0673 d²
= -0.0673 × (1.5) ²
= 0.151425 m.
= 151 mm. (Negative)

S3. Ans.(a)
Sol. Given,
Length (L) = 4m.
Breadth (B) = 2 m.
Depth (D) = 2.5 m.
Discharge (Q) = 4 × 10³ l/hour
Detention time of tank (t_d) =?
t_d = (Volume of tank (V))/(Discharge (Q) )
= (4×2×2.5 (m^3 ))/(4×10^3×10^(-3) (m^3 \/hour) )
= 5 hours.

S4. Ans.(c)
Sol. Given, Length of wall (L) = 4m.
Height of wall (H) = 3 m.
Area of wall = L × H
= 4 × 3 = 12 m^2
Quantity of plaster (two faces) = 2 × 12
= 24 m^2

S5. Ans.(b)
Sol. e_max=0.8
e_min=0.4
e=0.6
Relative density = (e_(max )-e)/(e_max-e_min )×100
(0.8-0.6)/(0.8-0.4)×100
0.2/0.4×100
=50%

S6. Ans.(b)
Sol. Shear stress in solid circular shaft
τ=16T/(πd^3 )=(16×10×10^3)/(π(0.1)^3 )
=50.93 MPa

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