# GATE CE 2014 SET-1

 Question 1
$\lim_{x\rightarrow \infty }\left ( \frac{x+\sin x}{x} \right )$ equals to
 A $-\infty$ B 0 C 1 D $\infty$
Engineering Mathematics   Calculus
Question 1 Explanation:
\begin{aligned}\text{Put } x= \frac{1}{h} & \text{ as } x\rightarrow \infty \: \Rightarrow \: h\rightarrow 0 \\ \lim_{x\rightarrow \infty }\left ( \frac{x+\sin x}{x} \right )&=\lim_{h\rightarrow 0}\left ( \frac{\frac{1}{h}+\sin \frac{1}{h}}{\frac{1}{h}} \right ) \\ &=\lim_{h\rightarrow 0}\: 1+\left ( \frac{\sin \frac{1}{h}}{\frac{1}{h}} \right )=1 \end{aligned}
 Question 2
Given the matrices $J=\begin{bmatrix} 3 &2 &1 \\ 2 &4 &2 \\ 1 &2 &6 \end{bmatrix}$ and $K=\begin{bmatrix} 1\\ 2\\ -1 \end{bmatrix}$, the product $K^{T}JK$ is________
 A 12 B 23 C 45 D 54
Engineering Mathematics   Linear Algebra
Question 2 Explanation:
\begin{aligned} J&=\begin{bmatrix} 3 & 2 & 1\\ 2 & 4 & 2\\ 1 & 2 & 6 \end{bmatrix} \\ K&=\begin{bmatrix} 1\\ 2\\ -1 \end{bmatrix} \\ K^{T}\: JK&=\begin{bmatrix} 1 & 2 & -1 \end{bmatrix}\begin{bmatrix} 3 & 2 & 1\\ 2 & 4 & 2\\ 1 & 2 & 6 \end{bmatrix}\begin{bmatrix} 1\\ 2\\ -1 \end{bmatrix} \\ &=\begin{bmatrix} 6 & 8 &-1 \end{bmatrix}\begin{bmatrix} 1\\ 2\\ -1 \end{bmatrix}\\ &=6+16+1=23\end{aligned}
 Question 3
The probability density function of evaporation E on any day during a year in a watershed is given by

$f(E)=\left\{\begin{matrix} \frac{1}{5}&0\leq E\leq 5\; mm/day\\ 0& \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! otherwise \end{matrix}\right.$

The probability that E lies in between 2 and 4 mm/day in a day in the watershed is (in decimal) _____________
 A 0.2 B 0.4 C 0.8 D 0.6
Engineering Mathematics   Probability and Statistics
Question 3 Explanation:
$f\left ( E \right )=\begin{cases} \frac{1}{5} & \text{ if } \; 0\leq E\leq mm/day \\ 0 & \; Otherwise \end{cases}$
$P\left ( 2\lt E\lt 4 \right )$
\begin{aligned} &=\int_{2}^{4}f\left ( E \right )dE=\int_{2}^{4}\frac{1}{5}dE=\frac{1}{5}\left [ E \right ]_{2}^{4} \\ &=\frac{1}{5}\left ( 4-2 \right )=\frac{2}{5}=0.4 \end{aligned}
 Question 4
The sum of Eigen values of the matrix, [M] is
where $\left [ M \right ]=\begin{bmatrix} 215 &650 &795\\ 655 &150 &835 \\ 485 &355 &550 \end{bmatrix}$
 A 915 B 1355 C 1640 D 2180
Engineering Mathematics   Linear Algebra
Question 4 Explanation:
Sum of eigen values = trace of matrix
=215+150+550 =915
 Question 5
With reference to the conventional Cartesian (x, y) coordinate system, the vertices of a triangle have the following coordinates: $\left ( x_{1},y_{1} \right )=\left ( 1,0 \right )$; $\left ( x_{2},y_{2} \right )=\left ( 2,2 \right )$; and $\left ( x_{3},y_{3} \right )=\left ( 4,3\right )$. The area of the triangle is equal to
 A $\frac{3}{2}$ B $\frac{3}{4}$ C $\frac{4}{5}$ D $\frac{5}{2}$
Engineering Mathematics   Linear Algebra
Question 5 Explanation:

Area of triangle is
\begin{aligned} &=\frac{1}{2}\left [ x_{1}\left ( y_{1}-y_{3} \right )+x_{2}\left ( y_{3}-y_{1} \right )+x_{2}\left ( y_{1}-y_{2} \right ) \right ] \\ &=\frac{1}{2}\left [ 1\left ( 2-3 \right )+2\left ( 3-0 \right )+4\left ( 0-2 \right ) \right ] \\ &=\frac{1}{2}\left [ -1+6-8 \right ]=\frac{3}{2} \end{aligned}
 Question 6
Match the information given in Group - I with those in Group - II.
 A P-1; Q-2; R-3; S-4 B P-2; Q- ; R-4; S-3 C P-3; Q-4; R-2; S-1 D P-4; Q-3; R-2; S -1
RCC Structures   Working Stress and Limit State Method
 Question 7
The possible location of shear centre of the channel section, shown below, is
 A P B Q C R D S
Solid Mechanics   Theory of Columns and Shear centre
Question 7 Explanation:

For no twisting,
\begin{aligned} V \times e &=H \times h\\ \Rightarrow \;\; e&=\frac{Hh}{V}\\ \end{aligned}
Hence, possible location of shear center is P.
 Question 8
The ultimate collapse load (P) in terms of plastic moment $M_{p}$ by kinematic approach for a propped cantilever of length L with P acting at its mid-span as shown in the figure, would be
 A $P=\frac{2M_{p}}{L}$ B $P=\frac{4M_{p}}{L}$ C $P=\frac{6M_{p}}{L}$ D $P=\frac{8M_{p}}{L}$
Design of Steel Structures   Plastic Analysis
Question 8 Explanation:

From principal of virtual work
$-M_{P}\theta -M_{P}\theta -M_{P}\theta +W_{u}\frac{L}{2}\theta =0$
$\Rightarrow \; \; \; W_{u}=\frac{6M_{P}}{L}$
 Question 9
While designing, for a steel column of Fe250 grade, a base plate resting on a concrete pedestal of M20 grade, the bearing strength of concrete (in $N/mm^{2}$) in limit state method of design as per IS:456-2000 is ________________
 A 4 B 7 C 9 D 11
RCC Structures   Working Stress and Limit State Method
Question 9 Explanation:
\begin{aligned} &\text { Permissible bearing stress } \\ &=0.45 \mathrm{f}_{\mathrm{ck}} \\ &=0.45 \times 20=9 \mathrm{N} / \mathrm{mm}^{2} \\ \end{aligned}
 Question 10
A steel section is subjected to a combination of shear and bending actions. The applied shear force is V and the shear capacity of the section is $V_{s}$. For such a section, high shear force (as per IS:800-2007) is defined as
 A $V \gt 0.6V_{s}$ B $V \gt 0.7V_{s}$ C $V \gt 0.8V_{s}$ D $V \gt 0.9V_{s}$
Design of Steel Structures   Structural Fasteners
Question 10 Explanation:
Clause 9.2.1 IS 800:2007
There are 10 questions to complete.