Quiz: Civil Engineering
Exam: RSMSSB-JEn
Topic: Miscellaneous
Each question carries 1 mark
Negative marking: 1/3 mark
Time: 8 Minutes
Q1. Void ratio of an undisturbed sample of soil is 0.6. the values of maximum and minimum possible void ratio are found as 0.8 and 0.4, respectively. The relative density in percentage, for this soil sample will be
(a) 25
(b) 50
(c) 75
(d) 90
Q2. Coulomb’s equation for shear strength can be represented by (symbols have their usual meanings)
(a) c=s+σ tanϕ
(b) s=c-σ tanϕ
(c) c=σ+c tanϕ
(d) s=c+σ tanϕ
Q3. Which of the following conditions is correct for the shafts connected in series to each other?
(a) θ= θ_1- θ_2
(b) T=T_1-T_2
(c) θ_1= θ_2
(d) T_1=T_2
Q4. The nominal shear stress (τ_v) in a reinforced concrete beam is given by
(a) bd/V_u
(b) V_u/(bd )
(c) V_u bd
(d) (V_u.b)/d
Q5. The major disadvantage of lime soda process of water softening is that
(a) It is unsuitable for turbid and acidic water
(b) Huge amount of precipitate is formed which creates a disposal problem
(c) The effluent cannot be reduced to zero hardness
(d) It is unsuitable for softening the water of excessive hardness
Q6. The length and breadth of a field of area 33600 m², on map is 12 cm and 7 cm respectively. The R.F. of the scale will be
(a) 1 : 400
(b) 1 : 20
(c) 1 : 800
(d) 1 : 2000
SOLUTIONS
S1. Ans.(b)
Sol. e_max=0.8
e_min=0.4
e=0.6
Relative density = (e_(max )-e)/(e_max-e_min )×100
(0.8-0.6)/(0.8-0.4)×100
0.2/0.4×100
=50%
S2. Ans.(d)
Sol. Coulomb’s equation for shear strength.
▭(τ=C+σ_n tanϕ)
τ → shear strength
c → cohesion
σ_n → Normal stress
ϕ → friction Angle.
S3. Ans.(d)
Sol. Shafts connected in series to each other having the same torque at each section and sum of twist at the end of section.
θ= θ_1+ θ_2
T=T_1=T_2
S4. Ans.(b)
Sol. Nominal Shear Stress (τ_v )=V/bd
V→Shear force
bd→Cross sectional area.
S5. Ans.(b)
Sol. The major disadvantage of lime-soda process is that huge amount of precipitate is formed, hence re-carbonation is done before filtration and after sedimentation which convert the precipitate into soluble form.
S6. Ans.(d)
Sol. Given, Area of field = 33600 m²
Length on map = 0.12 m.
Breadth on map = 0.07 m.
Area on map = 0.07 × 0.12
= 0.0084 m²
∵ 0.0084 m² = 33600 m²
∴ 1m² = 4000000 m²
∴ 1m = 2000 m.
Hence, ▭(R.F.=1∶2000)
