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

4 | |

3 | |

2 | |

1 |

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}

\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

100% | |

50% | |

49% | |

None of these |

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}

\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

3 | |

5 | |

7 | |

9 |

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

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

16 | |

20 | |

25 | |

30 |

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}.

=\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

drawing influence line diagram for any force function | |

writing virtual work equation | |

super position of load effects | |

none of these |

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."

"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.