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RSMSSB-JE’21 CE: Daily Practices Quiz. 10-Aug-2021

Quiz: Civil Engineering
Exam: RSMSSB-JEn
Topic: Miscellaneous

Each question carries 1 mark
Negative marking: 1/3 mark
Time: 8 Minutes

Q1. 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

Q2. 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

Q3. The ‘Center Line Method’ is specially adopted for estimating-
(a) Polygonal Buildings only
(b) Circular, hexagonal and other geometric
(c) Circular buildings only
(d) Hexagonal buildings only

Q4. 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

Q5. The development length in compression for a 20 mm diameter deformed bar of grade Fe 415 embedded in concrete of grade M 25 whose design bond stress is 1.40 N/mm², is
(a) 1489 mm
(b) 1289 mm
(c) 806 mm
(d) 645 mm

Q6. Calculate the value of shear stress (MPa) in the solid circular shaft of diameter 0.1 m which is subjected to the torque of 10 kN-m.
(a) 40.5
(b) 50.93
(c) 60.5
(d) 70.5

Solutions

S1. Ans.(b)
Sol. 1 cm = 4 m
1 cm = 400 cm
▭(Fractional scale=1/400=1:400)

S2. 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.)

S3. Ans.(b)
Sol. Center line method→ This method is suitable for symmetrical cross-sections like circular, hexagonal and other geometric shapes. This method is suitable for walls have the same thickness. In this method the center line for each type worked out and then multiplied by the breadth and depth of respective item to get the total quantity.

S4. 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

S5. Ans.(d)
Sol. According to IS 456 : 2000
The development length (L_d ) = (ϕσ_st)/(4τ_bd )
→ For deformed bars τ_bd shall be increases by 60%.
→ for bars in compression τ_bd increased by 25%
L_d=(20×0.87×415)/(4×1.4×1.6×1.25)
▭(L_d= 645 mm)

S6. Ans.(b)
Sol. Shear stress in solid circular shaft
τ=16T/(πd^3 )=(16×10×10^3)/(π(0.1)^3 )
=50.93 MPa

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