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NHPC-JE’21 CE: Daily Practices Quiz. 29-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
Topic: Miscellaneous

Each question carries 1 mark
Negative marking: 1/4 mark
Time: 8 Minutes

Q1. The locus of elevation that water will rise in a series of pitot tube is called ______.
(a) Hydraulic grade line
(b) Pressure head
(c) Energy grade line
(d) Head loss

Q2. EI(d³y/dx³) for a beam represent:
(a) Deflection
(b) Slope
(c) Moment
(d) Shear

Q3. In determining the bearing capacity of piles, empirical factor C is assumed………….. for drop hammers.
(a) 0.25
(b) 1.5
(c) 2.5
(d) 3.5

Q4. Lacey’s silt factor for medium silt whose average gain size is 0.25 mm, is likely to be:
(a) 0.66
(b) 0.77
(c) 0.88
(d) 0.99

Q5. Lacings in the built-up steel columns are designed to resist
(a) Bending Moment due to 2.5% of column load
(b) Shear Force due to 2.5% of column load
(c) Axial force due to 2.5% of column load
(d) Both (a) and (b)

Q6. The weight of 10 mm diameter mild steel rod per meter length is equal to________:
(a) 0.22 kg
(b) 0.32 kg
(c) 0.42 kg
(d) 0.62 kg


S1. Ans.(c)
Sol. Pitot tube gives head at stagnation point. Pitot tube gives pressure and velocity head both. At a particular datum locus of pitot tube gives energy grade line.

S2. Ans.(d)
y → deflection
EI d²y/dx²→moment
EI d³y/dx³→shear
EI (d^4 y)/(dx^4 )→Loading

S3. Ans.(c)
Sol. Engineers news record formula
▭(Q_up=WH/(S+C)) (Drop hammer & Single acting steam hammer)
▭(Q_up=(W+aP)H/(S+C)) (Drop acting steam hammer)
Elastic constant (C).
C = 2.5 cm (Drop hammer)
C = 0.25 cm (steam hammer)

S4. Ans.(c)
Sol. Given, dmm = 0.25 mm
Lacey’s silt factor (f) = 1.76 √dmm
= 1.76 √0.25
= 0.88

S5. Ans.(b)
Sol. Lacing member do not transfer any load from the column but in design of lacing member it is assumed that shear force due to 2.5% of column load will transfer through the lacing member.

S6. Ans.(d)
Sol. Quantity of steel (in kg.) per meter length is given by = M/L=d^2/162=0.00618d^2
Where d = diameter of steel rod (in mm.)
L = length (in m.)
M = Quantity of steel (in kg.)
Now = M/L=0.00618d^2
= 0.00618×(10)²
= 0.618 ≃ 0.62 kg

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