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
Each question carries 1 mark
Negative marking: 1/4 mark
Time: 8 Minutes
Q1. Frog is provided into the bricks to:
i. Indicate the manufacture’s name
ii. Provide a key for mortar
Correct among these is/are correct?
(a) Only i
(b) Only ii
(c) Both I and ii
(d) Neither I nor ii
Q2. What is the least window opening for 30 m³ inside volume of the room?
(a) 30 m²
(b) 3 m²
(c) 1 m²
(d) 10 m²
Q3. The value of limiting moment of resistance of a RCC beam for M25 grade of concrete and Fe 500 grade of steel is given by (Notational have their usual meaning) –
(a) 3.33 bd²
(b) 3.38 bd²
(c) 3.35 bd²
(d) 3.44 bd²
Q4. Alkaline soils are reclaimed by
(a) Leaching only
(b) Addition of gypsum and leaching
(c) Addition of gypsum only
(d) Provision of drainage
Q5. The maximum shear stress of steel beam should not exceed
(a) 0.45 f_y
(b) 0.55 f_y
(c) 0.66 f_y
(d) 0.50 f_y
Q6. If fore bearing of a line is N 30° E, the back bearing of the line is:
(a) N 30° W
(b) N 30° E
(c) S 30° W
(d) S 30° E
Sol. Frog is a depression made on the surface of brick. The main reason to provide frog is to form keyed joint between brick and mortar and also used to indicate the manufacture’s name. It is generally 1-2 cm deep.
Sol. For residential building→
1. Area of window = 1/8×floor area of the room
2. Total area of window opening is taken as 10 to 20% floor area of the room.
3. Total area of window opening is taken as 1m² for every 30 to 40 m³ of floor area of room.
Sol. Limiting value of moment.
[■(Fe 250&→& Mu=0.148 fck bd^2@Fe 415&→& Mu=0.138 fck bd^2@Fe 500 &→& Mu=0.133 fck bd^2 )]
For M25 & Fe500
Mu_lim= 3.325 bd^2
Sol. Alkaline soils are reclaimed by addition of gypsum and leaching. Leaching is the process of application of excess irrigation water in order to avoid building up salinity in soil.
Sol. The average shear stress in steel beams = 0.4〖 f〗_y.
The maximum shear stress in steel beam = 0.45 f_y.
Sol. (FB) = N 30°E
(BB) = ?
Note → If the fore bearing of a line is given as quadrantal bearing then back bearing is numerically same as fore bearing except direction change means just change N for S and vice versa and E for W and vice versa.
Hence, for fore bearing N 30° E, the back bearing will be S 30° W.