# GATE CE 2003

 Question 1
Given Matrix $[A]=\begin{bmatrix} 4 & 2 & 1 &3 \\ 6&3 &4 & 7\\ 2 &1 & 0& 1 \end{bmatrix}$, the rank of the matrix is
 A 4 B 3 C 2 D 1
Engineering Mathematics   Linear Algebra
Question 1 Explanation:
Consider first $3\times 3$ minors, since maximum possible rank is 3
\begin{aligned} \begin{vmatrix} 4 & 2 &1 \\ 6 & 3 & 4\\ 2 & 1 & 0 \end{vmatrix}&=0 \\ \begin{vmatrix} 2 & 1 &3 \\ 3 & 4& 7\\ 1 & 0& 1 \end{vmatrix}&=0 \\ \begin{vmatrix} 4 & 1 & 3\\ 6 & 4 & 7\\ 2 & 0 & 1 \end{vmatrix}&=0\\ \begin{vmatrix} 4 & 2 &3 \\ 6 & 3 & 7\\ 2 & 1 & 1 \end{vmatrix}&=0 \end{aligned}
Since all $3\times 3$ minors are zero, now try $2\times 2$ minors.
\begin{aligned} \begin{vmatrix} 4 &2 \\ 6 & 3 \end{vmatrix}&=0 \\ \begin{vmatrix} 2 &1 \\ 3 & 4 \end{vmatrix}&=8-3=5\neq 0 \\ \therefore\;\; \text{rank }&=2 \end{aligned}
 Question 2
A box contains 10 screws, 3 of which are defective. Two screws are drawn at random with replacement. The probability that none of the two screws is defective will be
 A 100% B 50% C 49% D None of these
Engineering Mathematics   Probability and Statistics
Question 2 Explanation:
This problem is to be solved by hypergeometric distribution, since population is finite. $\therefore$ p(none of 2 screws is defective)
\begin{aligned} &=\frac{3C_{0}\times 7C_{2}}{10C_{2}}\times 100\% \\ &=\frac{7}{15}\times 100\%=46.6\simeq 47\ \% \end{aligned}

 Question 3
If P, Q and R are three points having coordinates (3,-2,-1), (1,3,4), (2,1,-2) in XYZ space, then the distance from point P to plane OQR (O being the origin of the coordinate system) is given by
 A 3 B 5 C 7 D 9
Engineering Mathematics   Calculus
Question 3 Explanation:
Given, \begin{aligned} P\left ( 3,-2,-1 \right ) &\\ Q\left ( 1,3,4 \right ) &\\ R\left ( 2,1,-2 \right ) &\\ O\left ( 0,0,0 \right )&\\ \text{Equation of plane OQR is,}&\\ \begin{vmatrix} x-x_{1} & y-y_{1} & z-z_{1}\\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1}\\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{vmatrix}&=0 \\ \begin{vmatrix} x-0 & y-0 & z-0\\ 1 & 3 & 4\\ 2 & 1 & -2 \end{vmatrix}&=0\\ \text{i.e., }2x-2y+z&=0 \end{aligned}
Now $\perp$ distance of $\left ( x_{1},\: y_{1},\: z_{1} \right )$ from $ax + by + cz + d=0$ is given by,
$\left | \frac{ax_{1}+by_{1}+cz_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}} \right |$
Therefore, $\perp$ distance of (3, -2, 1) from plane $2x-2y+z=0$ is given by,
$\left | \frac{2\times 3-2\times \left ( -2 \right )+\left ( -1 \right )}{\sqrt{2^{2}+\left ( -2 \right )^{2}+1^{2}}} \right |=3$
 Question 4
A bar of varying square cross-section is loaded symmetrically as shown in the figure. Loads shown are placed on one of the axes of symmetry of cross-section. Ignoring self weight, the maximum tensile stress in $N/mm^{2}$ anywhere is A 16 B 20 C 25 D 30
Solid Mechanics   Properties of Metals, Stress and Strain
Question 4 Explanation:
The stress in lower bar
$=\frac{50 \times 1000}{50 \times 50}=20 \mathrm{N} / \mathrm{mm}^{2}$
The stress in upper bar
$=\frac{250 \times 1000}{100 \times 100}=25 \mathrm{N} / \mathrm{mm}^{2}$
Thus the maximum tensile stress any where in the bar is $25 \mathrm{N} / \mathrm{mm}^{2}.$
 Question 5
Muller Breslau's principle in structural analysis is used for
 A drawing influence line diagram for any force function B writing virtual work equation C super position of load effects D none of these
Structural Analysis   Influence Line Diagram and Rolling Loads
Question 5 Explanation:
Muller Breslau's principle is a principle on the basis of which it is possible to draw influence lines for various quantities pertaining to a structure. It states that:
"The ordinates of the influence lines for any stress element (such as axial stress, moment or reaction) of any structure are proportional to those of deflection curve, which is obtained by removing the restraint corresponding to that element from the structure and introducing in its place a deformation into the primary structure which remains."

There are 5 questions to complete.