Question 1 |
The value of \lim _{x \rightarrow \infty} \frac{x \ln (x)}{1+x^{2}} is
0 | |
1 | |
0.5 | |
\infty |
Question 1 Explanation:
\begin{aligned} &\lim _{x \rightarrow \infty}\left(\frac{x \ln x}{x^{2}+1}\right) \qquad \qquad \qquad \qquad \qquad \left(\frac{\infty}{\infty} \text { form }\right)\\ &=\lim _{x \rightarrow \infty}\left(\frac{x\left(\frac{1}{x}\right)+\ln x}{2 x}\right) \qquad \qquad \qquad \left(\frac{\infty}{\infty} \text { form }\right)\\ \lim _{x \rightarrow \infty}\left(\frac{0+\frac{1}{x}}{2}\right)&=\lim _{x \rightarrow \infty}\left(\frac{1}{2 x}\right)=\frac{1}{2 \times \infty}=0 \end{aligned}
Question 2 |
The rank of the matrix \left[\begin{array}{cccc} 5 & 0 & -5 & 0 \\ 0 & 2 & 0 & 1 \\ -5 & 0 & 5 & 0 \\ 0 & 1 & 0 & 2 \end{array}\right] is
1 | |
2 | |
3 | |
4 |
Question 2 Explanation:
\begin{aligned} \left[\begin{array}{cccc} 5 & 0 & 1 & 0 \\ 0 & 2 & 0 & 1 \\ -5 & 0 & -1 & 0 \\ 0 & 1 & 0 & 2 \end{array}\right] & \stackrel{R_{1} \longleftrightarrow R_{1}+R_{3}}{\longrightarrow}\left[\begin{array}{llll} 5 & 0 & 1 & 0 \\ 0 & 2 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 2 \end{array}\right] \\ & \stackrel{R_{4} \longleftrightarrow R_{4}-\frac{1}{2} R_{2}}{\longrightarrow}\left[\begin{array}{llll} 5 & 0 & 1 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{3}{2} \end{array}\right]\\ &R_{3} \longleftrightarrow R_{4}\left[\begin{array}{llll}5 & 0 & 1 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & \frac{3}{2} \\ 0 & 0 & 0 & 0\end{array}\right] \end{aligned}
Rank(A) = 3
Rank(A) = 3
Question 3 |
The unit normal vector to the surface X^{2}+Y^{2}+Z^{2}-48=0 at the point (4,4,4) is
\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} | |
\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} | |
\frac{2}{\sqrt{2}}, \frac{2}{\sqrt{2}}, \frac{2}{\sqrt{2}} | |
\frac{1}{\sqrt{5}}, \frac{1}{\sqrt{5}}, \frac{1}{\sqrt{5}} |
Question 3 Explanation:
\begin{aligned} \phi &=x^{2}+y^{2}+z^{2}-48, P(4,4,4) \\ \operatorname{grad} \phi &=\vec{\nabla} \phi=\hat{i} \frac{\partial \phi}{\partial x}+\hat{j} \frac{\partial \phi}{\partial y}+\hat{k} \frac{\partial \phi}{\partial z} \\ &=(2 x) \hat{i}+(2 y) \hat{j}+(2 z) \hat{k} \\ \vec{n} &=(\operatorname{grad} \phi)_{P}=8 \hat{i}+8 \hat{j}+8 \hat{k} \\ \hat{n} &=\frac{\vec{n}}{|\vec{n}|}=\frac{8 \hat{i}+8 \hat{j}+8 \hat{k}}{\sqrt{64+64+64}}=\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}} \\ & \simeq\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}},\right) \end{aligned}
Question 4 |
If A is a square matrix then orthogonality property mandates
A A^{T}=I | |
A A^{T}=0 | |
A A^{T}=A^{-1} | |
A A^{T}=A^{2} |
Question 4 Explanation:
\text { If, } \qquad \qquad A A^{\top}=I \quad \text { or } A^{-1}=A^{T}
The matrix is orthogonal.
The matrix is orthogonal.
Question 5 |
In general, the CORRECT sequence of surveying operations is
Field observations\rightarrow
Reconnaissance\rightarrow
Data analysis\rightarrow
Map making | |
Data analysis\rightarrow
Reconnaissance\rightarrow
Field observations \rightarrow
Map making | |
Reconnaissance\rightarrow
Field observations \rightarrow
Data analysis \rightarrow
Map making | |
Reconnaissance\rightarrow
Data analysis \rightarrow
Field observations \rightarrow
Map making |
Question 5 Explanation:
Reconnaissance\rightarrowField observations\rightarrowData analysis\rightarrowMap making
Question 6 |
Strain hardening of structural steel means
experiencing higher stress than yield stress with increased deformation | |
strengthening steel member externally for reducing strain experienced | |
strain occurring before plastic flow of steel material | |
decrease in the stress experienced with increasing strain |
Question 6 Explanation:
Strain hardening is experiencing higher stress than yield stress with increased deformation
In the figure AB = Strain hardening zone
OA = Linear elastic zone
Stress corresponding to point 'A' is yield stress.
In the figure AB = Strain hardening zone
OA = Linear elastic zone
Stress corresponding to point 'A' is yield stress.
Question 7 |
A single story building model is shown in the figure. The rigid bar of mass 'm' is supported by three massless elastic columns whose ends are fixed against rotation. For each of the columns, the applied lateral force (P) and corresponding moment (M) are also shown in the figure. The lateral deflection (\delta) of the bar is given by \delta=\frac{P L^{3}}{12 E I}, where L is the effective length of the column, E is the Young's modulus of elasticity and I is the area moment of inertia of the column cross-section with respect to its neutral axis.
For the lateral deflection profile of the columns as shown in the figure, the natural frequency of the system for horizontal oscillation is
For the lateral deflection profile of the columns as shown in the figure, the natural frequency of the system for horizontal oscillation is
6 \sqrt{\frac{E I}{m L^{3}}} \mathrm{rad} / \mathrm{s} | |
\frac{1}{L} \sqrt{\frac{2 E I}{m}} \mathrm{rad} / \mathrm{s} | |
6 \sqrt{\frac{6 E I}{m L^{3}}} \mathrm{rad} / \mathrm{s} | |
\frac{2}{L} \sqrt{\frac{E I}{m}} \mathrm{rad} / \mathrm{s}
|
Question 7 Explanation:
As the deflection will be same in all the 3 columns, so it represents a parallel connection.
\begin{aligned} k_{e q} &=3 k=\frac{36 E I}{L^{3}} \\ \text { Natural frequency }(\omega) &=\sqrt{\frac{k}{m}} \\ &=\sqrt{\frac{36 E I}{m L^{3}}}=6 \sqrt{\frac{E I}{m L^{3}}} \mathrm{rad} / \mathrm{s} \end{aligned}
Question 8 |
Seasoning of timber for use in construction is done essentially to
increase strength and durability | |
smoothen timber surfaces | |
remove knots from timber logs | |
cut timber in right season and geometry |
Question 8 Explanation:
Option 1 Increase strength and durability.
The process of drying of timber is known as seasoning.
Natural tree has more the 50% weight of water of its dry weight.
If we directly use this timber the because of irregular drying internal stresses will develop between fibres of timber and it will develop lots of defects (warps, shakes etc).
The process of drying of timber is known as seasoning.
Natural tree has more the 50% weight of water of its dry weight.
If we directly use this timber the because of irregular drying internal stresses will develop between fibres of timber and it will develop lots of defects (warps, shakes etc).
Question 9 |
In case of bids in Two-Envelop System, the correct option is
Technical bid is opened first | |
Financial bid is opened first | |
Both (Technical and Financial) bids are opened simultaneously | |
Either of the two (Technical and Financial) bids can be opened first |
Question 9 Explanation:
Option 1 technical bid is opened first
Opening of Tender
First technical bid is opened and after ensuring that all the technical aspects of a contractor are in order than only financial bid is opened
1. Envelope 1 ( Technical bid )
1. Cover letter
2. Registration Details
3. Pre-qualification documents
4. Earnest money deposit
5. Assumptions & Deviations in making of tender
6. Drawings
2. Envelope 2 (Financial Bid)
1. Forms of tender
Opening of Tender
First technical bid is opened and after ensuring that all the technical aspects of a contractor are in order than only financial bid is opened
1. Envelope 1 ( Technical bid )
1. Cover letter
2. Registration Details
3. Pre-qualification documents
4. Earnest money deposit
5. Assumptions & Deviations in making of tender
6. Drawings
2. Envelope 2 (Financial Bid)
1. Forms of tender
Question 10 |
The most appropriate triaxial test to assess the long-term stability of an excavated clay slope is
consolidated drained test | |
unconsolidated undrained test | |
consolidated undrained test | |
unconfined compression test |
Question 10 Explanation:
To assess the long term stability of clayey soil, the results of consolidated drained (CD) test are used.
There are 10 questions to complete.