GATE CE 2016 SET-2

Question 1
The spot speeds (expressed in km/hr) observed at a road section are 66, 62, 45, 79, 32, 51, 56, 60, 53, and 49. The median speed (expressed in km/hr) is ________.
(Note: answer with one decimal accuracy)
A
54.5
B
51.5
C
53.5
D
56
Engineering Mathematics   Probability and Statistics
Question 1 Explanation: 
Median speed is the speed at the middle value in series of spot speeds that are arranged in ascending order. 50% of speed values will be greater than the median 50% will be less than the median.
Ascending order order of spot speed studies are 32, 39, 45, 51, 53, 56, 60, 62, 66, 79
Median speed=\frac{53+56}{2}=54.5 km/hr
Question 2
The optimum value of the function f(x)=x^{2}-4x+2 is
A
2 (maximum)
B
2 (minimum)
C
-2 (maximum)
D
-2 (minimum)
Engineering Mathematics   Calculus
Question 2 Explanation: 
\begin{aligned} {f}'&=0 \\ \Rightarrow \;\; 2x-4&=0 \\ \Rightarrow \;\; x&=2 \text{ (stationary point)}\\ {f}''\left ( x \right )&=2 \gt 0 \\ \Rightarrow\;\; f(x)& \text{ is minimum at } x=2\end{aligned}
i.e., \left ( 2 \right )^{2}-4\left ( 2 \right )+2=-2
\therefore The optimum value of f(x) is -2 (minimum).
Question 3
The Fourier series of the function,
\begin{matrix} f(x) & =0 & -\pi \lt x \leq 0 \\ f(x) &=\pi-x & 0 \lt x \lt \pi \end{matrix}

in the interval [-\pi ,\pi ] is

f(x)=\frac{\pi }{4}+\frac{2}{\pi }\left [ \frac{\cos x}{1^{2}}+\frac{\cos 3x}{3^{2}}+\cdots\: \cdots \ \cdots \right ] + \left [ \frac{\sin x}{1}+\frac{\sin 2x}{2}+\frac{\sin 3x}{3}+\cdots \: \cdots\: \cdot \right ]

The convergence of the above Fourier series at x = 0 gives
A
\sum_{n=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi ^{2}}{6}
B
\sum_{n=1}^{\infty }\frac{(-1)^{n+1}}{n^{2}}=\frac{\pi ^{2}}{12}
C
\sum_{n=1}^{\infty }\frac{1}{(2n-1)^{2}}=\frac{\pi ^{2}}{8}
D
\sum_{n=1}^{\infty }\frac{(-1)^{n+1}}{(2n-1)}=\frac{\pi}{4}
Engineering Mathematics   Partial Differential Equation
Question 3 Explanation: 
The function is f(x)=0
-p\lt x\leq 0
=p-x,\, 0 \lt x \lt \pi
And Fourier series is,
f\left ( x \right )=\frac{\pi }{4}+\frac{2}{\pi }\left [ \frac{\cos x}{1^{2}}+\frac{\cos 3x}{3^{2}}+\frac{\cos 5x}{5^{2}}+... \right ]+\left [ \frac{\sin x}{1}+\frac{\sin 2x}{2}+\frac{\sin 3x}{3}+... \right ] ...\left ( i \right )
At x=0, (a point of discontinuity), the fourier series converges to \frac{1}{2}\left [ f\left ( 0^{-1} \right )+f\left ( 0^{+} \right ) \right ]
where f\left ( 0^{-} \right )=\lim_{x\rightarrow 0}\left ( \pi -x \right )=\pi
Hence, eq. (i), we get,
\frac{\pi }{2}=\frac{\pi }{4}+\frac{2}{\pi }\left [ \frac{1}{1^{2}}+\frac{1}{3^{2}}+... \right ]
\Rightarrow \;\; \frac{1}{1}+\frac{1}{3^{2}}+\frac{1}{5^{2}}+...\frac{\pi ^{2}}{8}
Question 4
X and Y are two random independent events. It is known that P(X)=0.40 and P(X\cup Y^{C})=0.7. Which one of the following is the value of P(X\cup Y) ?
A
0.7
B
0.5
C
0.4
D
0.3
Engineering Mathematics   Probability and Statistics
Question 4 Explanation: 
\; \; \; \; P\left ( X\: \cup \: Y^{c} \right )=0.7
\Rightarrow \; \; P\left ( X \right )+P\left ( Y^{c} \right )-P\left ( X \right )P\left ( Y^{c} \right )=0.7
(Since X, Y are independent events)
\Rightarrow \; \; P\left ( X \right )+1-P\left ( Y \right )-P\left ( X \right )\left \{ 1-P\left ( Y \right ) \right \}=0
\Rightarrow \; \; P\left ( X \right )-P\left ( X\: \cap \: Y \right )=0.3\; \; \; \; \; \; ...\left ( i \right )
\; \; \; \; P\left ( X\: \cup \: Y \right )=P\left ( X \right )+P\left ( Y \right )-P\left ( X\: \cap \: Y \right )
\; \; \; \; =0.4+0.3=0.7
Question 5
What is the value of \lim_{\begin{matrix} x\rightarrow 0\\ y\rightarrow 0 \end{matrix}} \frac{xy}{x^{2}+y^{2}} ?
A
1
B
-1
C
0
D
Limit does not exist
Engineering Mathematics   Calculus
Question 5 Explanation: 
(i) \lim_{x\rightarrow \infty }\frac{xy}{x^{2}+y^{2}}\lim_{y\rightarrow \infty }\left ( \frac{0}{0^{2}+y^{2}} \right )=0
(i.e., put x=0 and then y=0)
(ii) \lim_{x\rightarrow 0 y\rightarrow 0}\frac{xy}{x^{2}+y^{2}}\lim_{x\rightarrow 0}\left ( \frac{0}{x^{2}+0} \right )=0
( i.e., put y=0 and then x=0)
(iii)\lim_{x\rightarrow 0 y\rightarrow 0}\frac{xy}{x^{2}+y^{2}}\lim_{x\rightarrow 0}\frac{x\left ( mx \right )}{x^{2}+m^{2}x^{2}}
(i.e., put y=mx)
\lim_{x\rightarrow \infty }\left ( \frac{m}{1+m^{2}} \right )=\frac{m}{1+m^{2}}
which depends on m.
Question 6
The kinematic indeterminacy of the plane truss shown in the figure is
A
11
B
8
C
3
D
0
Structural Analysis   Determinacy and Indeterminacy
Question 6 Explanation: 
Kinematic indeterminacy,
\begin{aligned} D_{k}&=2 j-r_{e} \\ &=2 \times 7-3=11 \end{aligned}
Question 7
As per IS 456-2000 for the design of reinforced concrete beam, the maximum allowable shear stress \tau _{cmax} depends on the
A
grade of concrete and grade of steel
B
grade of concrete only
C
grade of steel only
D
grade of concrete and percentage of reinforcement
RCC Structures   Shear, Torsion, Bond, Anchorage and Development Length
Question 8
An assembly made of a rigid arm A-B-C hinged at end A and supported by an elastic rope C-D at end C is shown in the figure. The members may be assumed to be weightless and the lengths of the respective members are as shown in the figure.

Under the action of a concentrated load P at C as shown, the magnitude of tension developed in the rope is
A
\frac{3P}{\sqrt{2}}
B
\frac{P}{\sqrt{2}}
C
\frac{3P}{8}
D
\sqrt{2}P
Engineering Mechanics   
Question 8 Explanation: 



\begin{aligned} \sum M_{A}&=0\\ \Rightarrow \; R_{D}\times 2 L-P\times L&=0\\ \Rightarrow R_{D}&=\frac{P}{2} \end{aligned}
At joint D:


\begin{aligned} \sum F_{y}&=0\\ \Rightarrow \; T \cos 45^{\circ}&=\frac{P}{2}\\ \therefore \; T&=\frac{P}{\sqrt{2}} \end{aligned}
Question 9
As per Indian standards for bricks, minimum acceptable compressive strength of any class of burnt clay bricks in dry state is
A
10.0MPa
B
7.5MPa
C
5.0MPa
D
3.5MPa
Construction Materials and Management   
Question 9 Explanation: 
As per IS 1077 : 1992 clause 4.1, minimum strength of burnt clay bricks is 3.5 Mpa
Question 10
A construction Project consists of twelve activities.The estimated duration (in days) required to complete each of the activities along with the corresponding network diagram is shown below.

Total floats (in days) for the activities 5-7 and 11-12 for the project are, respectively,
A
25 and 1
B
1 and 1
C
0 and 0
D
81 and 0
Construction Materials and Management   
Question 10 Explanation: 
Total float can be determined once the activity times i.e. EST, EFT, LST and LFT are known.
total float,\begin{aligned} F_{T}&=LST-EST\\ &=LFT-EFT \end{aligned}

For activity 5-7,
\begin{aligned} EST&=38\\ EFT&=63\\ LFT&=63\\ LST&=38\\ F_{T}&=0 \end{aligned}
for activity 11-12,
\begin{aligned} EST&=80\\ EFT&=81\\ LFT&=81 \\ LST&=80 \\ F_T&=0 \end{aligned}
\textbf{Note}:It can be seen directly that since the slack of all events are zero, there is not margin left for the occurence of events and therefore.
Maximum available line=Time required for completion of activity
\therefore \; F_{t} for all activities is zero.
There are 10 questions to complete.

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