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# GATE’22 CE:- Daily Practice Quiz 25-Aug-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. A discharge of 1 cumec is flowing in a rectangular channel one meter wide at a depth of 20 cm. the bed slope of the channel is
(b) steep
(c) mild
(d) critical

Q2. A bar of 40 mm diameter and 400 mm length is subjected to an axial load of 100 kN. It elongates by 0.150 mm and that the diameter decreases by 0.005 mm. what is the Poission’s ratio of the material of the bar?
(a) 0.37
(b) 0.35
(c) 0.33
(d) 0.30

Q3. If a pipe of diameter 30 cm. running full of water with velocity 100 m/sec is changed by a pipe of diameter 15 cm, then the velocity of water flowing through the pipe will be
(a) 50 m/sec
(b) 100 m/sec
(c) 200 m/sec
(d) 400 m/sec

Q4. A ship has a metacentric height of 0.90 m and its period of rolling is 20 seconds. The relevant radius of gyration is nearly
(a) 5.5 m.
(b) 7.5 m
(c) 9.5 m
(d) 11.5m

Q5. For a pipe of radius r, flowing half full under the action of gravity, the hydraulic depth is
(a) πr/4
(b) r
(c) r/2
(d) 0.379 r

Solutions

S1. Ans.(b)
Sol.
q=Q/B=1/1=1 m^3 \/sec\/m
y_n=0.2 m
For rectangular channel, critical depth (y_c ) is given by-
y_c=(q^2/g)^(1\/3)
=(1^2/9.8)^(1\/3)
▭(y_c=0.467 m.)
Since y_c>y_n, hence steep slope.

S2. Ans.(c)
Sol. Given,
d = 40mm
∆d= -0.005 mm.
L=400 mm
∆L=0.150mm
μ= ?
Poission’s ratio (μ)=-(Latrel strain)/(Longitudinal strain)
=-((∆d/d))/((∆L/L) )
= -(((-0.005)/40))/((0.150/400) )
=1/3
▭(μ=0.33)

S3. Ans.(d)
Sol. given,
d_1=30 cm
V_1=100m\/sec
d_2=15 cm
V_2= ?
By continuity equation
A_1 V_1=A_2 V_2
π/4 (d_1 )^2.V_1=π/4 (d_2 )^2.V_2
(30)^2×100=(15)^2×V_2
▭(V_2=400 m\/sec )

S4. Ans.(c)
Sol. we know, ▭(T=2π √(k^2/(g.(GM) ̅ )))
Given,
T=20 sec.
(GM) ̅=0.90 m.
20= 2π √(k^2/(9.81×0.9))
K^2=((20)^2×9.81×0.9)/(2π)^2
▭(K=9.459m.)

S5. Ans.(a)
Sol. Area (A) = (πr^2)/2
Top width (T) = 2r
We know,
Hydraulic depth (D) = A/T
=(πr^2/2)/2r
▭(D=πr/4)

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