Quiz: Civil Engineering 3 May 2021

Quiz: Civil Engineering
Exam: GATE
Topic: Miscellaneous

Each question carries 1 mark
Negative marking: 1/3 mark
Time: 30 Minutes

Q1. A 4 hr storm had 4 cm of rainfall and the resulting direct runoff was 2.0 cm. if the ϕ-index remains at the same value, the runoff due to 10 cm of rainfall in 8 hr in the catchment is
(a) 6.0 cm
(b) 7.5 cm
(c) 2.3 cm
(d) 2.8 cm

Q2. The simply supported beam as shown in the figure.

Find reaction at support A?
(a) 18 kN
(b) 22 kN
(c) 28 kN
(d) 30 kN

Q3. Match list I (Admixture) with List – I (Action in concrete) and select the correct answer using the code given below the lists:

List – IList – II
Calcium lignosulphonateAccelerators
Aluminum powdersRetarder
Tartaric acidAir entrainer
Sodium silicateWater reducer

 Codes:

            A          B          C          D

(a)        1          3          2          4

(b)        4          3          2          1

(c)        4          2          3          1

(d)        1          2          3          4

Q4. The pH value of the incoming water is 7.2 and outgoing water is 8.4, assuming a linear variation of pH with time determine average pH value of water:
(a) 7.800
(b) 7.474
(c) 7.2
(d) 8.4

Q5. The degree of static indeterminacy of a rigid jointed frame PQR supported as shown in figure is

(a) zero
(b) one
(c) two
(d) unstable

Q6. Crude oil of kinematic viscosity 2.25 stokes flows through a 20 cm diameter pipe, the rate of flow being 1.5 litres/s. the flow will be-
(a) Uncertain
(b) Laminar
(c) Turbulent
(d) Transition

Q7. In the situation shown in the given figure, the coefficient of permeability of the soil is 10⁻³cm/s.
This discharge in one minute will be

(a) 2 ml
(b) 3 ml
(c) 6 ml
(d) 12 ml

Q8. A liquid compressed in cylinder has a volume of 0.04 m³ at 50 N/cm² and a volume of 0.039 m³ at 150 N/cm². The bulk modulus of elasticity of liquid is:
(a) 400 N/cm²
(b) 4000 N/cm²
(c) 40000 N/cm²
(d) 40 N/cm²

Q9. Maximum strains in an extreme fibre in concrete and in the tension reinforcement (Fe-415 grade and E_S = 200 kN/mm²) in a balanced section at limit state of flexure are respectively
(a) 0.0035 and 0.0038
(b) 0.002 and 0.0018
(c) 0.0035 and 0.0041
(d) 0.002 and 0.0031

Q10. The permissible stresses in rivets under wind load conditions as per IS : 800 can be exceeded by about
(a) 15%
(b) 33%
(c) 25%
(d) 50%

Solutions

S1. Ans.(a)
Sol. Case – I
Duration (t) = 4 hr.
Rainfall (P) = 4 cm.
Direct Runoff (R) = 2 cm.
ϕ-index=(P-R)/t=(4-2)/4=0.5 cm\/hr.
Case – II
ϕ – index is same as in Case – I = 0.5 cm/hr
Rainfall (P) = 10 cm
Runoff (R) = ?
Duration (t) = 8hr
ϕ – index = (P-R)/t
0.5=(10-R)/8
▭(R=6cm.)

S2. Ans.(b)
Sol. R_A+R_B=40 kN
Taking moment about point A →
ΣM_A=0
R_B×10-4×8-12×7-6×5-8×3-10×1=0
10R_B-32-84-30-24-10=0
10 R_B=180
▭(R_B=18 kN)
Reaction at A = (R_A ) = 40 – R_B
= 40 – 18
▭(R_A=22 kN)

S3. Ans.(b)

Sol.

AdmixtureChemical
1. AcceleratorCalcium Chloride, Silicate, Tri-ethanol amine etc.
2. RetarderGypsum, Sugar, Starch, Tartaric acid etc.
3. Air entrainerAluminium powder, hydrogen per oxide, neutralized vinsol resin etc.
4. Water reducerLignosulphonate, Polyglycol ester, Carboxylic acid 

S4. Ans.(b)
Sol. pH of incoming water = 7.2
pH of outgoing water = 8.4
Average pH value of water =?
pH = – log [H+]
7.2 = – log [H+]
Tanking anti log-
[H+] = 10^(-7.2)———— (1)
pH = – log [H+]
8.4 = – log [H+]
Tanking anti log-
[H+] = 10^(-8.4)———— (2)
Average concentration of [H+] = (10^(-7.2)+10^(-8.4))/2

Average pH = – log (10^(-7.2)+10^(-8.4))/2
= – log [3.35384×10^(-8) ]
= -(-7.47445762)
= 7.474

S5. Ans.(a)
Sol. No.of members (m) =3
No.of external reactions (r_e) =4
No.of joints (j) =4
Internal hinged reactions (r_r) =1
Degree of static indeterminacy (D_s)=3m+r_e-3j-r_r
=(3×3)+4-(3×4)-1
=13-13
▭(D_s=0)

S6. Ans.(b)
Sol. Reynold’s no. = ρVd/μ=Vd/υ
V=Q/A=(1.5×10^(-3))/(π/4 〖(0.20)〗^2 )
V=(4×1.5×10^(-3))/(0.04×π)
V=0.0477 m\/sec
Re =Vd/υ=(0.0477×0.20)/(2.25×10^(-4) )=42.48
It is below than 2000 so the flow is laminar.

S7. Ans.(b)
Sol. Q=K h/L A
The cross-sectional area A = 10² 〖cm〗^2
Q = 10⁻³ × 1/2 × 100 = 0.05 cm³/s.
= 3 cm³/minute = 3 ml/minute

S8. Ans.(b)
Sol. Bulk modulus
K = – ∆P/(∆V/V)
K= -[(0.04×[150-50])/(0.039-0.04)]=4000 N\/cm^2

S9. Ans.(a)
Sol. Given f_y=415,E_S=200×10^3 N\/mm^2
→ Maximum strain in extreme fibres for Concrete = 0.0035
→ Maximum strain in extreme fibres in the tension reinforcement = 0.002 + 0.87fy/E_S
=0.002+(0.87×415)/(200×10^3 )
=0.002+0.0018
=0.0038

S10. Ans. (c)
Sol. As per IS : 800, when the effect of wind or seismic load is taken in to account the permissible stresses for bolts, rivets and tension rods are increased by 25%.

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