Quiz: Civil Engineering
Exam: GATE
Topic: Miscellaneous
Each question carries 1 mark
Negative marking: 1/3 mark
Time: 20 Minutes
Q1. The thickness of the laminar boundary layer on a flat plate at a point X is 20 mm and at a point Y downstream by 1000 mm of X is 30 mm. what is the distance of X from the leading edge of the plate?
(a) 500 mm
(b) 800 mm
(c) 1000 mm
(d) 1250 mm
Q2. The average surface area of a reservoir in the month of June is 20 km^2. In the same month, the average rate of inflow is 10m^3 \/s, outflow rate is 15 m^3 \/s, monthly rainfall is 10 cm, monthly seepage loss is 1.8 cm and the storage change is 16 million m^3. The evaporation (in cm) in that month is
(a) 46.8
(b) 136.0
(c) 13.6
(d) 23.4
Q3. Consider different types of soils i.e. fine sand (F), Homogeneous clay (C), Coarse gravel (G), Silty clays (S). Arrange the soils in the increasing order of their permeability (low to high values).
(a) C, F, S, G
(b) S, C, F, G
(c) C, S, F, G
(d) S, C, G, F
Q4. If design bond stress = 1.5 N/mm² is assumed, then the development length of Fe500 HYSD bar of nominal diameter 12 mm – which is fully stressed in tension – will be
(a) 544 mm
(b) 246 mm
(c) 634 mm
(d) 798 mm
Q5. The tie member in a truss girder is 250 mm × 14mm in size. It is welded to a gusset plate 10 mm thick by a fillet weld. The overlap of the member is 300 mm. the weld is 6 mm in size. Find the strength of the joint if the welding is done as shown in figure. Allowable shear stress in weld = 108 N/mm².
(a) 308.4 kN
(b) 385.6 kN
(c) 498.9 kN
(d) 514.3 kN
Q6. The following observations were made on 2% dilution of wastewater:
Dissolved oxygen of aerated water used for dilution = 0 mg/l
Dissolved oxygen of diluted sample after 5 days = 0.6 mg/l
Dissolved oxygen of original sample = 2.0 mg/l
The BOD of 5 days of the sample is:
(a) 80 mg/l
(b) 60 mg/l
(c) 100 mg/l
(d) 70 mg/l
Q7. The compensated gradient provided at the curve of radius of 60m with a ruling gradient of 6% is:
(a) 5.25%
(b) 4%
(c) 4.75%
(d) 4.5%
Q8. The bending moments at point A, B and C of the beam shown in the given figure will be
(a) 10 kNm, 10 kNm and 10 kNm
(b) 10 kNm, 10 kNm, and -10 kNm
(c) 20 kNm, 10 kNm and -10 kNm
(d) 10 kNm, -10 kNm and 20 kNm
Q9. 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
Q10. 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
Solutions
S1. Ans.(b)
Sol.
for laminar flow ▭(δ α √x)
δ_1/δ_2 =√(x_1/x_2 )
20/30=√(x/(1000+x))
(2/3)^2=x/(1000+x)
4x + 4000 = 9x
5x = 4000
▭(x=800 mm)
S2. Ans.(d)
Sol. we know,
▭(Total inflow-total outflow=change in storage )
→ Let ‘a’ cm. evaporation takes place in month of June
Total in flow = (10×30×24×60×60)/(20×10^6 )×100+10=139.6cm
Total out flow = (15×30×24×60×60)/(20×16^6 )×100+1.8+a=196.2+a cm.
Total in flow < Total out flow, since depression takes place.
Now,
Depression in storage = (-16×10^6)/(20×10^6 )×100= -80cm
139.6 – (196.2 + a) = -80
-a = -80 + 56.6
▭(a=23.4 cm)
S3. Ans.(c)
Sol.
| Type of soil | Coefficient of permeability |
| Gravel | > 1cm/s |
| Sand | 1 – 10^(-3) cm/s |
| Silt | 10^(-3)-10^(-6) cm/s |
| Clay | < 10^(-6) cm/s |
S4. Ans.(a)
Sol. Given, for HYSD bars, τ_bd=1.6×1.5=2.4 N\/mm^2
ϕ = 12 mm.
Development length (L_d ) =?
We know,
L_d=(0.87f_y ϕ)/(4τ_bd )
=(0.87×500×12)/(4×2.4)
▭(L_d=543.75 mm)
S5. Ans.(b)
Sol. Size of weld (S) = 6mm.
Throat thickness (t_t) = 0.7×S = 0.7 × 6 = 4.2 mm
Total length of weld (L_ω ) = 250 + 250 + 300 = 850 mm.
Allowable stress (σa ) = 108 MPa Strength of joint = t_t×Lω×σ_a
= 4.2 × 850 × 108/10^3 kN.
= 385. 56 kN.
S6. Ans.(d)
Sol. Given,
DO_i=2 mg\/l
DO_f=0.6 mg\/l
Dilution factor (DF) = 100/2 = 50
BOD_5 = (DO_i-DO_f ) × DF
= (2 – 0.6) × 50
= 70 mg/l
S7. Ans.(c)
Sol. Given, R = 60m.
Ruling gradient (R.G) = 6%
Grade compensation (G.C.) = min^m of [((30+R)/R)%,(75/R)%]
= min^m of [((30+60)/60)%,(75/60)%]
=min^m of [1.5%,1.25%]
Hence, ▭(G.C.=1.25%)
Now,
Compensated gradient (C.G) = R.G. – G.C
= 6% – 1.25%
▭(C.G=4.75%)
S8. Ans.(b)
Sol. Taking moment about Hinge→
R_D×7.5 -4×2.5-4×5=0
▭(R_D=4 kN)
Moment at point A
M_A=R_D×2.5 =4×2.5=10 kN-m
Moment at point B
M_B=R_D×5-4×2.5
=4×5-4×2.5=10 kN-m
Moment at point C
M_C=R_D×10-4×7.5-4×5
=4×10-4×7.5-4×5
= -10 kN-m
S9. 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.)
S10. 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)
