NHPC-JE'21 CE: Daily Practices Quiz. 21-Oct-2021 |_00.1
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NHPC-JE’21 CE: Daily Practices Quiz. 21-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. The true length of a line is known to be 200 m. When this is measured with a 20 m tape, the length is 200.80 m.
What is the correct length of the 20 m tape?
(a) 19.98 m
(b) 20.08 m
(c) 20.04 m
(d) 19.92 m

Q2. Calculate the capitalized value of a building having annual rent of Rs. 20,000 and highest rate of interest is 5%
(a) 1000
(b) 21000
(c) 220000
(d) 400000

Q3. According to IS 399-1963, the weight of the timber is to be recommended at a moisture content of ________.
(a) 4%
(b) 12%
(c) 8%
(d) 5%

Q4. A rectangular plot of 16 km² in area is shown on a map by a similar rectangular area of 1 cm², R.F. of the scale to measure a distance of 40 km will be :
(a) 1/1600
(b) 1/400000
(c) 1/400
(d) 1/16000

Q5. Shear reinforcement in a RCC beam is provided to resist
(a) Diagonal tension
(b) Bending stress
(c) Compressive stress
(d) Punching shear

Q6. The concrete cubes are prepared, cured and tested according to Indian standards code number
(a) IS: 515
(b) IS: 516
(c) IS: 517
(d) IS: 518

Solutions

S1. Ans.(d)
Sol. Given, True length of line (L) = 200 m.
True length of Tape (l) = 20 m.
Wrong length of line (L’) = 200.80m.
Actual length of Tape (l’) =?
L × l = L’ × l’
200 × 20 = 200.8 × l’
▭(l^’=19.92 m.)

S2. Ans.(d)
Sol. Annual rent (R) = 20,000
Year purchase (Y.P) = 100/(i%)
= 100/5
▭(Y.P=20)
Capitalized value = Annual rent × Year’s Purchase
= 20,000 × 20
= 400,000 Rs.

S3. Ans.(b)
Sol. Seasoning is the process of adjusting the moisture content of wood or remove the sap from the wood. The moisture content in a well-seasoned timber is about 10-12%
Objective of seasoning →
– Reduce the weight
– Reduction in Shrinkage and warping.
– Increase in strength and durability.
– Remove sap.
– Reduce the tendency of split and decay.

S4. Ans.(b)
Sol. Area of plot =16 km^2
Area of plot in map = 1 cm²
1 cm² = 16 km²
1 cm = 4 km
Representative fraction (R.F.) = 1cm/(4×100000 cm)
▭(R.F.1/400000)

S5. Ans.(a)
Sol. Shear reinforcement in RCC beam is provided to resist diagonal tension. Stirrups is provided in the beam for shear reinforcement. At corner spacing between stirrups decreases and increase towards center.

S6. Ans.(b)
Sol. According to IS :516, concrete cubes are prepared, cured and tested.

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