Question 1 |

Which one of the following matrices is singular?

\begin{bmatrix} 2 &5 \\ 1 & 3 \end{bmatrix} | |

\begin{bmatrix} 3 &2 \\ 2 & 3 \end{bmatrix} | |

\begin{bmatrix} 2 &4 \\ 3 &6 \end{bmatrix} | |

\begin{bmatrix} 4 &3 \\ 6 &2 \end{bmatrix} |

Question 1 Explanation:

Option (A): \left | A \right |=6-5=1

Option (B): \left | A \right |=9-4=5

Option (C): \left | A \right |=12-12=0

Option (D): \left | A \right |=8-18=-10

Hence matrix (C) is singular.

Option (B): \left | A \right |=9-4=5

Option (C): \left | A \right |=12-12=0

Option (D): \left | A \right |=8-18=-10

Hence matrix (C) is singular.

Question 2 |

For the given orthogonal matrix Q,

Q=\begin{bmatrix} 3/7 &2/7 &6/7\\ -6/7 &3/7 &2/7\\ 2/7 &6/7 &-3/7 \end{bmatrix}

The inverse is

Q=\begin{bmatrix} 3/7 &2/7 &6/7\\ -6/7 &3/7 &2/7\\ 2/7 &6/7 &-3/7 \end{bmatrix}

The inverse is

\begin{bmatrix} 3/7 &2/7 &6/7\\ -6/7 &3/7 &2/7\\ 2/7 &6/7 &-3/7 \end{bmatrix} | |

\begin{bmatrix} -3/7 &-2/7 &-6/7\\ 6/7 &-3/7 &-2/7\\ -2/7 &-6/7 &3/7 \end{bmatrix} | |

\begin{bmatrix} 3/7 &-6/7 &2/7\\ 2/7 &3/7 &6/7\\ 6/7 &2/7 &-3/7 \end{bmatrix} | |

\begin{bmatrix} -3/7 &6/7 &-2/7\\ -6/7 &-3/7 &-6/7\\ -2/7 &-2/7 &3/7 \end{bmatrix} |

Question 2 Explanation:

\begin{aligned} \left | Q \right |&=\frac{3}{7}\left ( -\frac{9}{49}-\frac{12}{49} \right )-\frac{2}{7}\left ( \frac{18}{49}-\frac{4}{49} \right )+\frac{6}{7}\left ( \frac{-36}{49}-\frac{6}{49} \right ) \\ &=-1 \\ Adj. \; Q&=\begin{bmatrix} -\frac{21}{49} & \frac{42}{49} &-\frac{14}{49} \\ -\frac{14}{49}& -\frac{21}{49} & -\frac{42}{49}\\ -\frac{42}{49} & -\frac{14}{49} & \frac{21}{42} \end{bmatrix} \\ \therefore\;\; Q^{-1}&=\frac{Adj\: Q}{\left | Q \right |}=\begin{bmatrix} \frac{3}{7} & -\frac{6}{7} & \frac{2}{7}\\ \frac{2}{7} & \frac{3}{7} & \frac{6}{7}\\ \frac{6}{7} & \frac{2}{7} & -\frac{3}{7} \end{bmatrix} \end{aligned}

Or \because Q is orthogonal

\therefore \;\; Q^{-1}=Q^{T}

Or \because Q is orthogonal

\therefore \;\; Q^{-1}=Q^{T}

Question 3 |

At the point x= 0, the function f(x)=x^{3} has

local maximum | |

local minimum | |

both local maximum and minimum | |

neither local maximum nor local minimum |

Question 3 Explanation:

f\left ( x \right )=x^{3} at x=0

At x=0, the function y=x^{3} has neither minima nor maxima.

At x=0, the function y=x^{3} has neither minima nor maxima.

Question 4 |

A column of height h with a rectangular cross-section of size ax2a has a buckling load of P. If the cross-section is changed to 0.5a x 3a and its height changed to 1.5h, the buckling load of the redesigned column will be

P/12 | |

P/4 | |

P/2 | |

3P/4 |

Question 4 Explanation:

\begin{aligned} \text { For column, } & P=\frac{\pi^{2} E I_{\min }}{L^{2}} \\ &=\frac{\pi^{2} E\left(\frac{2 a \times a^{3}}{12}\right)}{h^{2}}=\frac{\pi^{2} E a^{4}}{6 h^{2}}\\ \text{For new column, }P&=\frac{\pi^{2} E\left[\frac{3 a \times(0.5 a)^{3}}{12}\right]}{(1.5 h)^{2}} \\ &=\frac{1}{12} \times \frac{\pi^{2} E a^{4}}{6 h^{2}}=\frac{P}{12} \end{aligned}

Question 5 |

A steel column of ISHB 350 @72.4 kg/m is subjected to a factored axial compressive load of 2000 kN. The load is transferred to a concrete pedestal of grade M20 through a square base plate. Consider bearing strength of concrete as 0.45f_{ck}, where f_{ck} is the characteristic strength of concrete. Using limit state method and neglecting the self weight of base plate and steel column, the length of a side of the base plate to be provided is

39 cm | |

42 cm | |

45 cm | |

48 cm |

Question 5 Explanation:

\begin{aligned}
&\text{Area required for base plate}\\ &=\frac{\text{Factored load}}{\text{Bearing capacity of concrete}}\\ &=\frac{2000\times 10^{3}}{0.45\times 20}=222222.222mm^{2}
\end{aligned}

So, side of base plate =\sqrt{\text{Area}}=471.4 mm= 47.14 cm

Since, provided area must be more than required

So, answer should be 48 cm.

So, side of base plate =\sqrt{\text{Area}}=471.4 mm= 47.14 cm

Since, provided area must be more than required

So, answer should be 48 cm.

Question 6 |

The Le Chatelier apparatus is used to determine

compressive strength of cement | |

fineness of cement | |

setting time of cement | |

soundness of cement |

Question 6 Explanation:

Le Chatelier Apparatus is used to detemine the soundness of cement as per IS code4031 (part 3); this cement testing procedure is called Le Chatelier test for determining the unsoundness properties of cement due to presence of "free lime".

Question 7 |

The deformation in concrete due to sustained loading is

creep | |

hydration | |

segregation | |

shrinkage |

Question 7 Explanation:

Creep is inelastic deformation with time due to
sustained loading

Question 8 |

A solid circular beam with radius of 0.25 m and length of 2 m is subjected to a twisting moment of 20 kNm about the z-axis at the free end, which is the only load acting as shown in the figure. The shear stress component \tau_{xy} at Point 'M' in the cross-section of the beam at a distance of 1 m from the fixed end is

0.0 MPa | |

0.51 MPa | |

0.815 MPa | |

2.0 MPa |

Question 8 Explanation:

The only non-zero stresses are \tau_{\theta z}=\tau_{z \theta}=\tau

if \theta is 90^{\circ} then \theta=y

\begin{aligned} \text{Hence}\quad\tau_{z y} &=\tau_{y z}=\tau_{\max } \\ &=16 T / \pi \mathrm{d}^{3}=0.815 \mathrm{MPa} \end{aligned}

But in rest of the planes shear stresses are zero, hence, \tau_{x y}=\tau_{y x}=0

Question 9 |

Two rectangular under-reinforced concrete beam sections X and Y are similar in all aspects except that the longitudinal compression reinforcement in section Y is 10% more. Which one of the following is the correct statement?

Section X has less flexural strength and is less ductile than section Y | |

Section X has less flexural strength but is more ductile than section Y | |

Sections X and Y have equal flexural strength but different ductility | |

Sections X and Y have equal flexural strength and ductility |

Question 9 Explanation:

Due to presence of more compression steel in section Y, NA of section of Y is above than as of X. It means Y is more under-reinforced than X so ductility of Y is more.

Since compression steel of Y is more so flexure resistance of X is less than as of Y.

Question 10 |

The percent reduction in the bearing capacity of a strip footing resting on sand under flooding condition (water level at the base of the footing) when compared to the situation where the water level is at a depth much greater than the width of footing, is approximately

0 | |

25 | |

50 | |

100 |

Question 10 Explanation:

For strip footing on sand (c=0)

q_{u}=\gamma D_{f} N_{q}+0.5 \mathrm{B} \gamma N_{\gamma}

In flooding condition water level rises to base of footing hence IIIrd term unit weight of soil will change and IInd term unit weight will be unaffected.

\begin{array}{ll} \therefore & q_{u}=\gamma D_{f} N_{q}+0.5 B \gamma N_{\gamma} \\ \because & \gamma \simeq \frac{1}{2} \gamma_{s a t} \end{array}

Hence third term reduced and second term will be same thereby percentage reduction will not be 50%.

According to option approach answer should be 25%

Note: If water table rises to ground level then both \gamma will reduce to \gamma. Hence, percentage reduction would be approximately 50%.

q_{u}=\gamma D_{f} N_{q}+0.5 \mathrm{B} \gamma N_{\gamma}

In flooding condition water level rises to base of footing hence IIIrd term unit weight of soil will change and IInd term unit weight will be unaffected.

\begin{array}{ll} \therefore & q_{u}=\gamma D_{f} N_{q}+0.5 B \gamma N_{\gamma} \\ \because & \gamma \simeq \frac{1}{2} \gamma_{s a t} \end{array}

Hence third term reduced and second term will be same thereby percentage reduction will not be 50%.

According to option approach answer should be 25%

Note: If water table rises to ground level then both \gamma will reduce to \gamma. Hence, percentage reduction would be approximately 50%.

There are 10 questions to complete.