Quiz: Civil Engineering
Exam: GATE
Topic: Miscellaneous
Each question carries 1 mark
Negative marking: 1/3 mark
Time: 20 Minutes
Q1. If principal stresses in a two-dimensional case are -20 MPa and 30 MPa respectively, then maximum shear stress at the point is
(a) 10 MPa
(b) 15 MPa
(c) 25 MPa
(d) 30 MPa
Q2. The four cross – section shown below are required to be ordered in the increasing order of their respective shape factor
Which of the following order is correct?
(a) iv, ii, iii, i
(b) i, ii, iii, iv
(c) ii, iv, iii, i
(d) iii, I, iv, ii
Q3. The sequent depth ratio of a hydraulic jump in a rectangular horizontal channel is 10.32. the Froude number at the beginning of the jump is
(a) 7.64
(b) 5.64
(c) 13.61
(d) 8.05
Q4. If tomato juice is having a pH of 4.2, the hydrogen ion concentration will be
(a) 9.94 × 10^(-5) mol/l
(b) 8.94 × 10^(-5) mol/l
(c) 10.31 × 10^(-5) mol/l
(d) 6.31 × 10^(-5) mol/l
Q5. A 1:50 scale model of a spillway is to be tested in the laboratory. The discharge in the prototype is 1010 m³/sec. The discharge to be maintained in the model test is
(a) 5.7 m³/sec
(b) 0.057 m³/sec
(c) 0.57 m³/sec
(d) 57 m³/sec
Q6. The following data was obtained from a liquid test conducted on a soil sample
| Number of blows | 17 | 22 | 25 | 28 | 34 |
| Water content (%) | 53.8 | 53.1 | 51.9 | 50.6 | 50.5 |
The liquid limit of the soil is
(a) 53.1%
(b) 52.8%
(c) 51.9%
(d) 50.6%
Q7. A metal bar of length 150 mm is inserted between two rigid supports and its temperature is increased by 10°C. if the coefficient of thermal expansion is 12 × 10^(-6)/°C and young’s modulus is 2 × 10^5 MPa, the stress in the bar is
(a) 240 MPa
(b) 2.4 MPa
(c) Zero
(d) 24 MPa
Q8. A levelling is carried out to establish the reduces levels (RL) of point R with respect to bench mark (BM) at P. the staff reading taken are given below:
| Staff Station | BS | IS | FS | RL |
| P | 1.656 m | 100.000m | ||
| Q | -0.951m. | -1.510m | ||
| R | 0.751m |
If RL of P is +100.000m, then RL (in m) of R is
(a) 103.464
(b) 101.564
(c) 100.464
(d) 101.464
Q9. A fixed beam is loaded as in figure. The fixed end moment at support B.
(a) (WL^2)/20
(b) (WL^2)/30
(c) (WL^2)/8
(d) (WL^2)/10
Q10. A simply supported isotropically reinforced square slab of side 4m. is subjected to a service load of 7 kpa. The thickness of the slab is 120mm. The moment of resistance required as per yield line theory is
(a) 9 kN-m
(b) 9 kN-m/m
(c) 10 kN-m
(d) 10 kN-m/m
Solutions
S1. Ans.(c)
Sol. given σ_1=30 MPa ,σ_2= -20MPa
Maximum shear stress (τ) = (σ_1-σ_2)/2
=(30—(-20))/2
= 25 MPa
S2. Ans.(a)
Sol.
| S.N. | Shape | Shape factor |
| 1 | I-Section | 1.12-1.15 |
| 2. | Rectangle | 1.5 |
| 3. | Circle | 1.7 |
| 4. | Diamond | 2.0 |
S3. Ans.(a)
Sol. Sequent depth ratio
y_2/y_1 =1/2 [-1+√(1+8 f_(r_1)^2 )]
Given, y_2/y_1 =10.32,〖 f〗(r_1 )= ? ⇒10.32=1/2 [-1+√(1+8 f(r_1)^2 )]
⇒20.64= -1+ √(1+8 f_(r_1)^2 )
⇒(21.64)^2=(1+8 f_(r_1)^2 )
By solving the above equation, we get
▭(〖f_r〗_1=7.64)
S4. Ans.(d)
Sol. ▭(pH= -log_10 [H^+ ] )
Given, pH = 4.2
4.2= -log_10 [H^+ ]
[H^+ ]=10^(-4.2)
▭([H^+ ]=6.31×10^(-5) mol\/l)
S5. Ans.(b)
Sol. Froude model law will be applicable in this case.
Q_m/Q_p =(L_m/L_p )^(5\/2)
Given,
(L_m/L_P )=(1/50)
Q_P=1010 m^3 \/sec.
Now,
Q_m/1010=(1/50)^(5\/2)
Q_M=1010×(1/50)^2.5
▭(Q_m=0.057 m^3 \/sec)
S6. Ans.(c)
Sol. the liquid limit is the water content corresponding to 25 number of blows
W_L=51.9%
S7. Ans.(d)
Sol. Given, Length (L) = 150 mm.
Coefficient of thermal expansion (α) = 12×10^(-6)/°C
Young’s modulus of elasticity (E) = 2×10^5 MPa
Change in temp. (∆T)=10°C
Thermal stress (σ_th )= ?
Now,
Thermal stress(σ_th ) = E α ∆T
=2×10^5×12×10^(-6)×10
=24 MPa
S8. Ans.(d)
Sol
| Staff station | BS | IS | FS | HI | RL |
| P | 1.656 | – | – | 101.656 | 100.000 |
| Q | -0.951 | – | -1.51 | 102.215 | 103.166 |
| R | – | – | 0.751 | – | 101.464 |
Hence,RL of R=101.464
S9. Ans. (b)
Sol. fixed end moment for beam loaded with uniformly varying load.
S10. Ans. (d)
Sol. Given, service load = 7 kpa = 7 kN/m²
Dead load = 25 × 0.12 = 3 kN/m²
Total factored load on the slab (W_u) = 1.5 × (7+3)
= 15 kN/m²
Now,
Moment of resistance as per yield line theory is given by–
M = (W_u L^2)/24 kN-m/m.
= (15×(4)^2)/24
= (15×16)/24
= 240/24
▭(M=10 kN-m/m.)
