GATE Civil Engineering 2021 SET-1

Question 1
The rank of matrix \left[\begin{array}{llll} 1 & 2 & 2 & 3 \\ 3 & 4 & 2 & 5 \\ 5 & 6 & 2 & 7 \\ 7 & 8 & 2 & 9 \end{array}\right] is
A
1
B
2
C
3
D
4
Engineering Mathematics   Linear Algebra
Question 1 Explanation: 
Using R_{2} \rightarrow R_{2} \rightarrow 3 R_{1}, R_{3} \rightarrow R_{3}-5 R_{1}, R_{4} \rightarrow R_{4}-7 R_{1}
A=\left[\begin{array}{cccc} 1 & 2 & 2 & 3 \\ 0 & -2 & -4 & -4 \\ 0 & -4 & -8 & -8 \\ 0 & -6 & -12 & -12 \end{array}\right]
Using R_{3} \rightarrow R_{3}-2 R_{2}, R_{4} \rightarrow R_{4}-3 R_{2}
A=\left[\begin{array}{cccc} 1 & 2 & 2 & 3 \\ 0 & -2 & -4 & -4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]
So, \rho(A)= No. of non-zero rows = 2.
Question 2
If P=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] and Q=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] then Q^{T} P^{T} is
A
\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]
B
\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right]
C
\left[\begin{array}{ll} 2 & 1 \\ 4 & 3 \end{array}\right]
D
\left[\begin{array}{ll} 2 & 4 \\ 1 & 3 \end{array}\right]
Engineering Mathematics   Linear Algebra
Question 2 Explanation: 
\begin{array}{l} \quad P Q=\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]=\left[\begin{array}{ll} 2 & 4 \\ 1 & 3 \end{array}\right] \\ (P Q)^{\top}=\left[\begin{array}{ll} 2 & 4 \\ 1 & 3 \end{array}\right] \end{array}
Now using Reversal law
Q^{\top} P^{\top}=(P Q) T=\left[\begin{array}{ll} 2 & 4 \\ 1 & 3 \end{array}\right]
Question 3
The shape of the cumulative distribution function of Gaussian distribution is
A
Horizontal line
B
Straight line at 45 degree angle
C
Bell-shaped
D
S-shaped
Engineering Mathematics   Probability and Statistics
Question 3 Explanation: 


PDF:f(x)=\frac{1}{\sigma \sqrt{2 \pi}}e^{-(x-\mu )^2/(2\sigma ^2)}
CDF:F(x)=\frac{1}{2}\left [ 1+eff\left ( \frac{x-\mu }{\sigma \sqrt{2}} \right ) \right ]
Question 4
A propped cantilever beam EF is subjected to a unit moving load as shown in the figure (not to scale). The sign convention for positive shear force at the left and right sides of any section is also shown.

The CORRECT qualitative nature of the influence line diagram for shear force at G is
A
B
C
D
Structural Analysis   Influence Line Diagram and Rolling Loads
Question 4 Explanation: 


As per Muller Breslau principle ILD for stress function (shear -V_{G}) will be a combination of curves (3^{\circ} curves).
Question 5
Gypsum is typically added in cement to
A
prevent quick setting
B
enhance hardening
C
increase workability
D
decrease heat of hydration
Construction Materials and Management   
Question 5 Explanation: 
The Gypsum is added to cement at the end of grinding clinker it is added to prevent quick setting.
Question 6
The direct and indirect costs estimated by a contractor for bidding a project is Rs.160000 and Rs.20000 respectively. If the mark up applied is 10% of the bid price, the quoted price (in Rs.) of the contractor is
A
200000
B
198000
C
196000
D
182000
Construction Materials and Management   
Question 6 Explanation: 
Direct Costs = 160000
Indirect costs = 20000
Mark up applied is 10% of the Bid price
Bid price = Direct cost + Indirect cost + Markup
Bid price = Direct cost + Indirect cost + \frac{10}{100} \times Bid price
\text{Bid price}\left (1-\frac{10}{100} \right )=DC+IC
\text{Bid price}=\frac{160000+20000}{0.9}=2,00,000
Question 7
In an Oedometer apparatus, a specimen of fully saturated clay has been consolidated under a vertical pressure of 50 \mathrm{kN} / \mathrm{m}^{2} and is presently at equilibrium. The effective stress and pore water pressure immediately on increasing the vertical stress to 150 \mathrm{kN} / \mathrm{m}^{2}, respectively are
A
150 \mathrm{kN} / \mathrm{m}^{2} and 0
B
100 \mathrm{kN} / \mathrm{m}^{2} and 50 \mathrm{kN} / \mathrm{m}^{2}
C
50 \mathrm{kN} / \mathrm{m}^{2} and 100 \mathrm{kN} / \mathrm{m}^{2}
D
0 and 150 \mathrm{kN} / \mathrm{m}^{2}
Geotechnical Engineering   Effective Stress and Permeability
Question 7 Explanation: 
Stress is increased suddenly, hence entire change will be taken by water \Delta \bar{\sigma}=\Delta U=100 \mathrm{kPa}.
There will be no change in effective stress
\therefore \qquad \qquad\bar{\sigma}=50 \mathrm{kPa}
Question 8
A partially-saturated soil sample has natural moisture content of 25% and bulk unit weight of 18.5 \mathrm{kN} / \mathrm{m}^{3}. The specific gravity of soil solids is 2.65 and unit weight of water is 9.81 \mathrm{kN} / \mathrm{m}^{3}. The unit weight of the soil sample on full saturation is
A
21.12 \mathrm{kN} / \mathrm{m}^{3}
B
19.03 \mathrm{kN} / \mathrm{m}^{3}
C
20.12 \mathrm{kN} / \mathrm{m}^{3}
D
18.50 \mathrm{kN} / \mathrm{m}^{3}
Geotechnical Engineering   Properties of Soils
Question 8 Explanation: 
\begin{aligned} \mathrm{w} &=0.25, \gamma_{\mathrm{t}}=18.5 \mathrm{kN} / \mathrm{m}^{3} \\ \mathrm{G}_{\mathrm{s}} &=2.65, \gamma_{\mathrm{w}}=9.81 \\ \gamma_{\mathrm{t}} &=\frac{G_{\mathrm{S}} \gamma_{W}(1+w)}{1+e} \\ \Rightarrow \qquad \qquad \qquad \qquad \mathrm{e} &=\frac{2.65 \times 9.81 \times 1.25}{18.5}-1\\ \Rightarrow \qquad \qquad \qquad \qquad e&=0.756\\ \text{At full saturation}, \quad \mathrm{S}&=1\\ \Rightarrow \qquad \qquad \qquad \quad \gamma_{\mathrm{sat}}&=\frac{\left(G_{\mathrm{S}}+e\right) \gamma_{\mathrm{W}}}{1+e}\\ \gamma_{\text {sat }} &=\frac{(2.65+0.756) \times 9.81}{1.756} \\ &=19.03 \mathrm{kN} / \mathrm{m}^{3} \end{aligned}
Question 9
If water is flowing at the same depth in most hydraulically efficient triangular and rectangular channel sections then the ratio of hydraulic radius of triangular section to that of rectangular section is
A
\frac{1}{\sqrt{2}}
B
\sqrt{2}
C
1
D
2
Fluid Mechanics and Hydraulics   Open Channel Flow
Question 9 Explanation: 
Efficient channel section


\begin{aligned} A & =y^{2} & & A=2 y^{2} \\ P & =2 \sqrt{2} y & P & =4 y \\ R_{I} & =\frac{y}{2 \sqrt{2}} & R_{I I}&=\frac{y}{2} \\ \therefore \qquad \qquad \qquad \frac{R_{I}}{R_{I I}} & =\frac{1}{\sqrt{2}} & \end{aligned}
Question 10
Kinematic viscosity' is dimensionally represented as
A
\frac{M}{LT}
B
\frac{M}{L^{2} T}
C
\frac{T^{2}}{L}
D
\frac{L^{2}}{T}
Fluid Mechanics and Hydraulics   Dimensional Analysis
Question 10 Explanation: 
Kinematic viscosity
v=\frac{\mu }{\rho }=\frac{kg/m\cdot s}{kg/m^3}=m^2/s
[v]=\frac{m^2}{s }=\frac{L^2}{T}
There are 10 questions to complete.

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