Engineering Jobs   »   Civil Engineering quizs   »   Gate Quiz

# GATE’22 CE:- Daily Practice Quiz 16-Oct-2021

IIT Kharagpur has released the GATE 2022 exam dates. As per the GATE 2022 exam date, the exam will be conducted on February 5, 6, 12 & 13. Candidates preparing for GATE 2022 must attempt this Civil Engineering Quiz to boost your GATE exam preparation.

Quiz: Civil Engineering
Exam: GATE
Topic: Miscellaneous

Each question carries 2 marks
Negative marking: 1/3 mark
Time: 20 Minutes

Q1. A nozzle is so shaped that the average flow velocity changes linearly from 1.5 m/s at the beginning to 15 m/s at its end in a distance of 0.375 m. the magnitude of the convective acceleration (in m/s²) at the end of the nozzle is ____________________.
(a) 3240
(b) 340
(c) 440
(d) 540

Q2. 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) 246 mm
(b) 544 mm
(c) 634 mm
(d) 798 mm

Q3. A soil has bulk unit weight of 20 kN/m³ and water content of 17%. Calculate the water content if the soil particle dries to a unit weight of 19 kN/m³ and the void ratio remains constant. (Given the answer rounded to the nearest integer value)
(a) 8%
(b) 10%
(c) 11%
(d) 20%

Q4. The ratio of flexural rigidity of a beam (b × d) to another one (b × 2d) of similar material will be
(a) 1/16
(b) 1/8
(c) 1/4
(d) 1/2

Q5. A plan of an area drawn with the original scale of 1 cm = 10 m, the shrunk such that a line, originally 15 cm long on the plan, measures now 14.5 cm. the shrunk scale is given by 1 cm equal to:
(a) 10.97m
(b) 10.34 m
(c) 0.97 m
(d) 9.70 m

Q6. While designing a hydraulic structure, the piezometric head at bottom of the floor is computed as 10m. the datum is 3m below floor bottom. The assured standing water depth above the floor is 2m. the specific gravity of the floor material is 2.5 the floor thickness should be
(a) 4.40 m.
(b) 3.33 m.
(c) 2.33 m.
(d) 2.00 m.

Q7. The extra widening required for a two-lane national highway at a horizontal curve of 300 m radius considering a wheel base of 8 m and a design speed of 100 kmph is
(a) 0.82 m
(b) 0.72 m
(c) 0.62 m
(d) 0.52 m

Q8. The Ca^(2+) concentration and Mg^(2+) concentration of a water sample are 180 mg/lit and 40 mg/lit as their ions respectively. the total hardens of the water sample in terms of CaCO_(3 )in mg/lit is approximately equal to
(a) 316.67 mg/lit
(b) 416.67 mg/lit
(c) 516.67 mg/lit
(d) 616.67 mg/lit

Q9. Considering beam as axially rigid, the degree of freedom of a plane frame shown below is

(a) 9
(b) 8
(c) 7
(d) 6

Q10. The diameter of a rivet connecting plate of thickness 16 mm given by Unwin’s formula is_________:
(a) 28 mm
(b) 23 mm
(c) 24 mm
(d) None of these

Solutions

S1. Ans.(d)
Sol. Distance (∂x) = 0.375m
(∂u) = 15-1.5
= 13.5 m/s
u = 15 m/s
Convective acceleration = u∂u/∂x
=15×13.5/0.375
=540m\/s^2

S2. Ans.(b)
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)

S3. Ans.(c)
Sol. Case – I
y_b1=20 kN\/m^3
w_1=17%
Case – II
y_b2=19 kN\/m^3
w_2= ?
y_d, e is constant
We know,
▭(y_d=y_b/(1+w))
y_d1=y_d2
y_b1/(1+w_1 )=y_b2/(1+w_2 )
20/(1+0.17 )=19/(1+w_2 )
1+w_2=1.1115
▭(w_2=0.1115=11.15%)

S4. Ans.(b)
Sol. We know,
▭(flexural Rigidity (FR)=EI )
FR_1=EI_1 (for section=b×d)
FR_1=E×(bd^3)/12—(1)
FR_2=EI_2 (for section=b×2d)
=E×(b(2d)^3)/12—–(2)
Ratio→(FR_1)/(FR_2 )=(E×(bd^3)/12)/(E×(8bd^3)/12)
▭((FR_1)/(FR_2 )=1/8)

S5. Ans.(b)
Sol. Original scale (S) ⇒ 1 cm = 10 m ⇒ 1/1000
Shrinkage factor (S.F) = (Shrunk length)/(Original length )
=14.5/15
▭(S.F.=0.96666)
Revised or shrunk scale (S.S) = Original sale (S) × Shrinkage factor (S.F)
=1/1000×0.96666
=1/1034.5
1 cm=1034.5 cm
▭(1cm=10.345m)

S6. Ans.(b)
Sol. floor thickness (t) = h/(G-1)
h=10-(3+2)
▭(h=5m.)
t=5/(2.5-1)
▭(t=3.33m)

S7. Ans.(a)
Sol. Given,
n=2
R=300 m
l=8 m
V=100 kmph
E_W= ?
E_W=(nl^2)/2R+V/(9.5 √R)
=(2×(8)^2)/(2×300)+100/(9.5 √300)
▭(E_W=0.82 m)

S8. Ans.(d)
Sol. Given,
Concentration of [Ca^(2+) ] = 180 mg/lit
Concentration of [Mg^(2+) ] = 40 mg/lit
Total hardness in terms of CaCO_3 is given by –
= [Ca^(2+) ]×(eq.wt of CaCO_3)/(eq.wt of Ca^(2+) )+ [Mg^(2+) ]×(eq.wt of CaCO_3)/(eq.wt of Mg^(2+) )
=180×50/20+40×50/12
=450+166.67
=616.67 mg/lit

S9. Ans.(b)
Sol. No. of joints (j) = 4
No of external reaction (re) = 3
Axially rigid member (m) = 1
Now, Degree of kinematic indeterminacy or degree of freedom –
D_k=3j-re-m
=(3×4)-3-1
▭(D_k=8 )

S10. Ans.(c)
Sol. Given, t=16mm.
Unwin’s formula
ϕ=6.05√tmin
=6.05√16
=6.05×4
=24.2≃24mm.

Sharing is caring!