Question 1 |
\lim_{x\rightarrow \infty }\left ( \frac{x+\sin x}{x} \right ) equals to
-\infty | |
0 | |
1 | |
\infty |
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________
12 | |
23 | |
45 | |
54 |
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) _____________
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) _____________
0.2 | |
0.4 | |
0.8 | |
0.6 |
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}
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}
where \left [ M \right ]=\begin{bmatrix} 215 &650 &795\\ 655 &150 &835 \\ 485 &355 &550 \end{bmatrix}
915 | |
1355 | |
1640 | |
2180 |
Question 4 Explanation:
Sum of eigen values = trace of matrix
=215+150+550 =915
=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
\frac{3}{2} | |
\frac{3}{4} | |
\frac{4}{5} | |
\frac{5}{2} |
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.
P-1; Q-2; R-3; S-4 | |
P-2; Q- ; R-4; S-3 | |
P-3; Q-4; R-2; S-1 | |
P-4; Q-3; R-2; S -1 |
Question 7 |
The possible location of shear centre of the channel section, shown below, is
P | |
Q | |
R | |
S |
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
P=\frac{2M_{p}}{L} | |
P=\frac{4M_{p}}{L} | |
P=\frac{6M_{p}}{L} | |
P=\frac{8M_{p}}{L} |
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 ________________
4 | |
7 | |
9 | |
11 |
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
V \gt 0.6V_{s} | |
V \gt 0.7V_{s} | |
V \gt 0.8V_{s} | |
V \gt 0.9V_{s} |
Question 10 Explanation:
Clause 9.2.1 IS 800:2007
There are 10 questions to complete.