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NHPC-JE’21 CE: Daily Practices Quiz. 07-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
Topic: Miscellaneous

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

Q1. The true length of line is known to be 300m. when this line is again measured with a new 30 m tape, the length is found to be 300.90 m. the correct length of the tape is
(a) 28.90 m
(b) 29.91 m
(c) 30.19 m
(d) 28.30 m

Q2. Find the plinth area of building if a building built up area 350m² is constructed with a courtyard of 5.5 × 6.5m inside the building
(a) 314.24 m²
(b) 314.25 m²
(c) 314.26 m²
(d) 314.27 m²

Q3. The defects of a tree caused by injuries sustained during its young age are called _________.
(a) burls
(b) knots
(c) shakes
(d) upsets

Q4. A material has identical properties in all directions is said to be
(a) Homogeneous
(b) Isotropic
(c) Anisotropic
(d) Orthotropic

Q5. Partial safety factors for concrete and steel respectively may be taken as:
(a) 1.5 and 1.15
(b) 1.5 and 1.78
(c) 3 and 1.78
(d) 3 and 1.2

Q6. The R.F. for 2 cm to a meter is –
(a) 1/100
(b) 1/20
(c) 1/50
(d) 1/10


S1. Ans.(b)
Sol. True length of line (L) = 300 m
Measured length of line (L’) = 300.90 m
Length of tape (l) = 30 m.
True length of tape (l’) =?
L × l = L’ × l’
300 × 30 = 300.9 × l’

S2. Ans.(b)
Sol. Built up area = 350 m²
Courtyard area = 5.5 × 6.5 m² = 35.75 m²
Plinth area = 350 – 35.75 = 314.25 m²

S3. Ans.(a)
Sol. Burls → these are the uneven projections on the body of the tree during its growth. These are mainly due to effects of injuries and shocks during its young age.

S4. Ans.(b)
Sol. Isotropic → A material is said to be isotropic when it has same properties in all direction.
Anisotropic → A material is said to be anisotropic when it has different properties in all direction.
Homogenous → Same property at each cross-section.
Orthotropic → Different property in three mutually perpendicular direction.

S5. Ans.(a)
Sol. In limit state method.
Partial safety factor for RCC = 1.5
Partial safety factor for steel = 1.15
In working state method.
Partial safety factor for R.C.C. = 3
Partial safety factor for steel = 1.78

S6. Ans.(c)
Sol. Representative fraction (R.F) = 2cm/1m

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