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# GATE’22 CE:- Daily Practice Quiz 17-Sep-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: 12 Minutes

Q1. The depth of flow in an alluvial channel is 1.6 m. if critical velocity ratio is 1.1 and manning’s (n) is 0.018, the critical velocity of channel as per kennedy’s method is
(a) 0.713 m/sec
(b) 0.784 m/sec
(c) 1.108 m/sec
(d) 0.817 m/sec.

Q2. Group symbols assigned to clayey sand and silty sand are respectively
(a) CS and SS
(b) SC and SM
(c) CS and SM
(d) CS and MS

Q3. 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) 24 MPa
(d) Zero

Q4. Staff reading on the floor of a verandah of a school building is 1.816 m. and staff reading when held with bottom of staff touching the ceiling over the verandah is 2.871 m. R.L of the floor is 74.500m. Height of the ceiling above floor is
(a) 4.920m.
(b) 4.687m.
(c) 4.270m.
(d) 4.585m.

Q5. If the working stress in bearing in power driven rivets is 300 N\/mm^2 and a double riveted double cover butt joint is used to connect plates of 12 mm thick, then the strength of rivet in bearing is (Given; The nominal diameter of rivet= 22 mm)
(a) 94.6 kN
(b) 84.6 kN
(c) 72.6 kN
(d) 68.4 Kn

Solutions

S1. Ans.(d)
Sol. Given,
Depth (y) = 1.6 m.
Manning’s coefficient (n) = 0.018
Critical velocity ratio (m) = 1.1
Critical velocity (V_O )= ?
Now,
The critical velocity as per Kennedy’s method is given by
V_O=0.55 m y^0.64
=0.55×1.1×(1.6)^0.64
=0.817 m\/sec

S2. Ans.(b)
Sol. Group symbols assigned to clayey sand and silty sand are respectively SC and SM.

S3. Ans.(c)
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

S4. Ans.(b)
Sol. Height of ceiling above floor = (1.816 + 2.871) m
= 4.687m.

S5. Ans.(b)
Sol. Given, bearing stress (f_b )=300 N\/mm^2
Thickness of plates (t)=12mm
Nominal dia of rivet (ϕ)=22mm
Gross dia. of rivet (d)=ϕ+1.5
=22+1.5
=23.5 mm.
Bearing strength of rivet (P_b )=d×t×f_b
=23.5×12×300
=84600 N
▭(P_b=84.6 KN.)

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