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: 12 Minutes
Q1. A trapezoidal channel with base of 6 m and side slope of two horizontal to one vertical conveys water at 17 m³/sec with a depth of 1.5 m. The flow situation in the channel is:
(a) Critical
(b) Supercritical
(c) Subcritical
(d) None of the above
Q2. A bolt designated as Hex bolt M16 × 70NL will have
(a) Cross – section area of 16 × 70 〖cm〗^2
(b) Length of 16 mm
(c) Diameter of 70 mm
(d) Diameter of 16 mm
Q3. A test plate 30 cm × 30 cm resting on a sand deposit settles by 10 mm. under a certain loading intensity. A footing 150 × 200cm. resting on the on the same sand deposit and loaded to the same load intensity settles by
(a) 50 mm
(b) 30.2 mm
(c) 27.8 mm
(d) 2.0 mm
Q4. Calculate the safe stopping sight distance for a design speed of 70 kmph for two-way traffic on a single lane road. The reaction time of deriver is 2.5 sec.
(a) 136.23 m.
(b) 207.37 m
(c) 674. 24m
(d) 164.42 m
Q5. If tomato juice is having a pH of 4.2, the hydrogen ion concentration will be
(a) 6.31 × 10^(-5) mol/l
(b) 8.94 × 10^(-5) mol/l
(c) 10.31 × 10^(-5) mol/l
(d) 9.94 × 10^(-5) mol/l
Solutions
S1. Ans.(c)
Sol. Given,
B = 6m.
Q = 17 m³/s
y = 1.5 m.
2H:1V = mH:1V ⇒ m = 2
T = B + 2my
T = 6 + (2 × 2 × 1.5)
▭(T=12m)
A = (B + my) y
= [6 + (2 × 1.5)] × 1.5
= 13.5 m²
We know,
Froude no.(F)=(Q^2 T)/(gA^3 )
=((17)^2×12)/(9.81×(13.5)^3 )
▭(F=0.14368<1) Hence,flow is Subcritical
S2. Ans.(d)
Sol. Hex bolt designation M16 ×70 NL means bolt have nominal diameter of 16 mm and length of 70 mm.
S3. Ans.(c)
Sol.
S_P=10 mm.
B_P=0.3 m
B_F=1.5 m
S_F= ?
For sandy soils
S_P/S_F =[(B_P (B_F+0.3))/(B_F (B_P+0.3) )]^2
10/S_F =[0.3(1.5+0.3)/1.5(0.3+0.3) ]^2
▭(S_F=27.8 mm)
S4. Ans.(b)
Sol. Given, velocity (V) = 70 kmph
=70/3.6=19.44 m\/sec
Reaction time (t) = 2.5 sec.
Assuming longitudinal friction (f) = 0.35
Note →
For single lane, two-way, SSD will be = 2SSD
Now,
SSD=2 [v.t+v^2/2gf]
=2[19.44×2.5+(19.44)^2/(2×9.8×0.35)]
▭(SSD=207.37 m.)
S5. Ans.(a)
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)
