Quiz: Civil Engineering
Exam: SSC-JE
Topic: Miscellaneous
Each question carries 1 mark
Negative marking: 1/4 mark
Time: 8 Minutes
Q1. What is the tensile stress (in MPa) in a rod of cross section 20 mm × 30 mm, carrying an axial tensile load of 20 kN?
(a) 0.03
(b) 0.33
(c) 33.33
(d) 333.33
Q2. The plasticity to mould bricks in suitable shape is contributed by
(a) Alumina
(b) Lime
(c) Magnesia
(d) Silica
Q3. As per the general rules for the measurement of a building, all works shall be measured subject to certain tolerance. In this regard the cubic contents shall be worked out to the nearest.
(a) 0.1 cu m
(b) 0.01 cu m
(c) 0.001 cu m
(d) 0.0001 cu m
Q4. The R.F. for 2 cm to a meter is –
(a) 1/100
(b) 1/20
(c) 1/50
(d) 1/10
Q5. Specific gravity of sand deposit is 2.50 having the void ratio 0.65 the critical hydraulic gradient for the sand deposit:
(a) 0.85
(b) 0.90
(c) 0.75
(d) 1.0
Q6. The velocity distribution for flow over a flat plate is given by u = (y-y²) in which u is velocity in meters per second at a distance ‘y’ meters above the plate. What is the shear stress value at y = 0.15m? the dynamic viscosity of fluid is 8.0 poise.
(a) 12.4N/m²
(b) 1.24 N/m²
(c) 0.56 N/m²
(d) 5.6 N/m²
Solutions
S1. Ans.(c)
Sol. Given that
Tensile load (σ)=20 kN
Cross-section (A) = 20 mm × 30 mm
Tensile stress =(tensile load)/Area
=(20×10^3)/(10×30×10^(-6) )
=33.33×10^6
=33.33 MPa
S2. Ans.(a)
Sol. Alumina imparts plasticity to bricks. If it is in excess then shrinkage and warping of bricks takes place. The percentage of alumina in good brick earth is 20-30 %.
S3. Ans.(b)
Sol.
1.Linear measurement shall be measured nearest up to 0.01m.
2.Area shall be measured to the nearest 0.01 m²
3.Cubic contents shall be worked out to be nearest 0.01 m³
4.Wood work shall be measured. Nearest to 2 mm or 0.002m.
5.RCC work in slab/beams/columns measured nearest to 0.5 cm or 0.005m.
S4. Ans.(c)
Sol. Representative fraction (R.F) = 2cm/1m
=2cm/100cm
▭(R.F.=1/50)
S5. Ans.(b)
Sol. G=2.50
e=0.65
Critical hydraulic gradient (i)=(G-1)/(1+e)=(2.50-1)/(1+0.65)
=0.909
S6. Ans.(c)
Sol. We know that the shear stress is
τ= μ dv/dy
∵[dv/dy]_(y=0.15)=d/dy [ y-y^2 ]
= [1-2y]^0.15
=1-2[0.15-0]
=1-030
=0.70
τ=μ dv/dy
τ=(8×10^(-1) )×0.70
=0.56 N/m²
