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# GATE QUIZ : CIVIL ENGINEERING (31 MAY 2021)

Quiz: Civil Engineering
Exam: GATE
Topic: Miscellaneous

Each question carries 1 mark
Negative marking: 1/3 mark
Time: 15 Minutes

Q1. A simple supported beam of length 9 m. carries a concreted load W at its centre such that BM at centre of the beam is 9 KNm. If EI is the flexural rigidity of the beam then deflection at the centre is
(a) 54/EI
(b) 24/EI
(c) 61/EI
(d) 70/EI

Q2. The equilibrium discharge of an S-curve obtained by 4 – hr. unit hydrograph summation is _. When the catchment area is 40 km²
(a) 25 m³/sec
(b) 10 m³/sec
(c) 28 m³/sec
(d) 50 m³/sec

Q3. The downhill end of a 50 m tape is held 100 cm too low. What is the horizonal distance measure
(a) 49.999 m
(b) 48.999 m
(c) 48 m
(d) 49 m

Q4. If one litre of sample settle in 30 minutes and the measuring cylinder showed a sludge vol. of 300 ml, then the sludge volume index would be nearly __? If the MLSS concentration in the aeration tank of extended aeration activated sludge process is 5000 mg/lit.
(a) 200
(b) 150
(c) 100
(d) 60

Q5. If the stability number is 0.07, unit weight of soil is 30 kN/m³ concession is 20 kN/m², factor of safety is 2.5 then safe max. height of the slope is
(a) 5.0 m
(b) 2.0 m
(c) 3.81 m
(d) 4.50 m

Q6. The following data are given for a soil sample
σ_0’=275 kPa,∈_0’=1.2
σ_0’+∆σ_0’=400 kPa,∈=0.8
If thickness of the clay specimen in 30 mm, the value of coefficient of volume compressibility is?
(a) 7.54×10^(-4) m²\/kN
(b) 1.45 × 10^(-3) m²\/kN
(c) 1.45 × 10^(-4) m²\/kN
(d) 7.54×10^(-3) m²\/kN

Q7. The failure strain was 40% in an unconfined compression test An initial cross section area of a clay sample was 18 sq.cm. the corrected area of the sample of failure would be
(a) 20 sq.cm
(b) 25 sq.cm
(c) 35 sq.cm
(d) 30 sq.cm

Q8. A portal frame is shown in the given fig. if Q_B=Q_e=500/EI radian, then the value of moment at B will be

(a) 150 kN-m
(b) 240 kN-m
(c) 200 kN-m
(d) 280 kN-m

Q9. The maximum depth of neutral axis for a beam with ‘d’. as the effective depth in limit state method of design for Fe415 Steel is
(a) 0.46 d
(b) 0.48 d
(c) 0.50 d
(d) 0.53 d

Q10. In a rectangle strain gauge rosette, the strain recorded are ∈0 = 500μ strain, ∈(45°) = 400μ strain and ∈_(90°) = 300μ strain, what is the maximum principal strain at the point.
(a) 500μ strain
(b) 400μ strain
(c) 300μ strain
(d) 200μ strain

Solutions

S1. Ans.(c)
Sol.

BM at centre = 9 kN-m
∵WL/4=9
⇒(W×9)/4=9
⇒W=4 kN
∵ Deflection at centre = (WL^3)/48EI=(4×9^3)/48EI=61/EI

S2. Ans.(c)
Sol. Equilibrium discharge of S-curve = (Total discharge)/(Unit hr.duration)
=(40×10^6×0.01)/(4×3600)
= 28 m³/sec

S3. Ans.(a)
Sol. Correction of slope = h^2/2L=〖0.10〗^2/(2×50)=0.0001 m
∵ Horizontal distance
= 50 – 0.0001 = 49.999 m.

S4. Ans.(d)
Sol. Sludge vol. index (S V I) = (Vob (ml\/l))/(X_ob (mg\/l) )×1000
=300/5000×1000
= 60 ml/gm

S5. Ans.(c)
Sol. Stability number S_N=Cm/γ_H
Where, C_m=C/FOS=20/2.5=8 kN\/m^2
∵ Maximum safe height
H = 8/(30×0.07)=3.81 m.

S6. Ans.(b)
Sol. coeff. Of volume compressibility
m_v=a_v/(1+e_o )=∆e/((∆σ ̅)/(1+e_0 ))
=(1.2-0.8)/((400-275)/(1+1.2))
=1.45×10^(-3) m^2 \/kN

S7. Ans.(d)
Sol. Corrected area at failure is given by
A_f=A_0/(1-ϵ_l )=18/(1-0.40)=30 sq.m

S8. Ans.(a)
Sol. The deformed shaped of structure will have translation at B and C
MBC=0+2EI/20 (2θ_B+θ_C )
=2EI/20×3×500/EI
= 150 kN-m

S9. Ans.(d)
Sol. X_u max=0.0035/(0.0055+0.87fy/ES )×d
For fy = 415 N/mm²
And Es = 2 × 10^5 N/mm²
Xu, max = 0.48d and
Fy = 250 N/mm²
Xu, max = 0.53 d

S10. Ans.(a)
Sol. ϕxy = 2ϵ(45°)-(∈0+∈(90°) )
2×400-(500+300)
∵ϕ_xy = 0
∵ϵ_1,2=(ϵ_0+ϵ_90)/2± √(((∈_0-∈_90)/2 )^2+(ϕxy/2)^2 )
=400±√((100)^2 )
=400±100 μ strain
∵ Max. Principal strain = 500μ strain.

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