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. Which type of bond can be made for 10 cm thick wall-
(a) Heading bond
(b) Flemish bond
(c) English bond
(d) Stretching bond
Q2. According to Terzaghi’s equation, the net ultimate bearing capacity of strip footing resting on c1ohesive soil (C=10 kN/m²) per unit depth and unit width (Nc = 5.7) will be –
(a) 47 kN/m²
(b) 64 kN/m²
(c) 57 kN/m²
(d) 77 kN/m²
Q3. Plain cement concrete is strong in taking:
(a) Tensile stresses
(b) Compressive stresses
(c) Shear stresses
(d) Tear stresses
Q4. If a surveying drawing is prepared at a scale of 1 cm = 4m, then what will be the fractional scale?
(a) 1 : 0.025
(b) 1 : 400
(c) 1 : 40
(d) 1 : 0.25
Q5. The distance between two points measured using 20m chain was recorded as 327 m. it was found that the chain is 3 cm too long. The true length of the line is
(a) 326.55 m
(b) 327.49 m
(c) 327.55 m
(d) 326.49 m
Q6. If the porosity of a soil sample is 40%, void ratio for this sample is
(a) 0.50
(b) 0.70
(c) 0.60
(d) None of these
Solutions
S1. Ans.(d)
Sol. Stretcher bond consist stretcher in each course. This bond is used in the construction of 10 cm. thick wall or partition wall or cavity wall etc.
S2. Ans.(c)
Sol. Ultimate bearing capacity (q_u ) = CN_c+ qN_q + 0.5B_r N_r
For cohesive soil
C=10 kN\/m²
N_c=5.7
N_q=1
Net ultimate bearing capacity (q_nu ) =?
q_nu=CN_c
= 10 ×5.7
=57 kN/m²
S3. Ans.(b)
Sol. The concrete in which no reinforcement is provided called plain concrete. Plain concrete is strong in taking Compressive stresses. The tensile strength of concrete is 10 to 15% of compressive strength.
S4. Ans.(b)
Sol. 1 cm = 4 m
1 cm = 400 cm
▭(Fractional scale=1/400=1:400)
S5. Ans.(b)
Sol. Measured length of line (L’) = 327 m.
Length of tape (l) = 20 m.
True length of tape (l’) = 20 + 0.03 = 20.03 m.
True length of line (L) =?
L × l = L’ × l’
L × 20 = 327 × 20.03
L = (327×20.03)/20
▭(L=327.49m.)
S6. Ans.(d)
Sol. Given, Porosity (n) = 40% = 0.4
Void ratio (e) = ?
e=n/(1-n)=0.4/(1-0.4)=0.4/0.6=0.67
