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.
There are 5 questions to complete.