Question 1 |

[A] is a square matrix which is neither symmetric nor skew-symmetric and [A]^{T} is its transpose. The sum and difference of these matrices are defined as [S] = [A]+[A]^{T} and [D] = [A]-[A]^{T}, respectively. Which of the following statements is TRUE?

Both [S] and [D] are symmetric | |

Both [S] and [D] are skew-symmetric | |

[S] is skew-symmetric and [D] is symmetric | |

[S] is symmetric and [D] is skew-symmetric |

Question 1 Explanation:

Since

\begin{aligned} \left ( A+{A}' \right )&=A^{t}+\left ( A^{t} \right )^{t} \\ &=A^{t}+A\\ \text{i.e. } S^{t}&=S \\ \therefore\;\; & \text{ S is symmetric}\\ \text{Since }\left ( A-A^{t} \right )^{t}&=A^{t}-\left ( A^{t} \right )^{t} \\ &=A^{t}-A \\ &=-\left ( A-A^{t} \right ) \\ \text{i.e. } D^{t}&=-D \end{aligned}

So D is Skew-Symmetric.

\begin{aligned} \left ( A+{A}' \right )&=A^{t}+\left ( A^{t} \right )^{t} \\ &=A^{t}+A\\ \text{i.e. } S^{t}&=S \\ \therefore\;\; & \text{ S is symmetric}\\ \text{Since }\left ( A-A^{t} \right )^{t}&=A^{t}-\left ( A^{t} \right )^{t} \\ &=A^{t}-A \\ &=-\left ( A-A^{t} \right ) \\ \text{i.e. } D^{t}&=-D \end{aligned}

So D is Skew-Symmetric.

Question 2 |

The square root of a number N is to be obtained by applying the Newton
Raphson iterations to the equation x^2-N=0. If i denotes the iteration index,
the correct iterative scheme will be

x_{i+1}=\frac{1}{2}(x_{i}+\frac{N}{x_{i}}) | |

x_{i+1}=\frac{1}{2}(x_{i}^{2}+\frac{N}{x_{i}^{2}}) | |

x_{i+1}=\frac{1}{2}(x_{i}+\frac{N^{2}}{x_{i}}) | |

x_{i+1}=\frac{1}{2}(x_{i}-\frac{N}{x_{i}}) |

Question 2 Explanation:

\begin{aligned} x_{i+1}&=x_{i}-\frac{f\left ( x_{i} \right )}{{f}'\left ( x_{i} \right )} \\ &=x_{i}-\left ( \frac{x_{i}^{2}-N}{2x_{i}} \right ) \\ &=\frac{x_{i}^{2}+N}{2x_{i}} \\ &=\frac{1}{2}\left [ x_{i}+\frac{N}{x_{i}} \right ] \end{aligned}

Question 3 |

There are two containers, with one containing 4 Red and 3 Green balls and the
other containing 3 Blue and 4 Green balls. One ball is drawn at random from
each container. The probability that one of the balls is Red and the other is Blue
will be

1/7 | |

9/49 | |

12/49 | |

3/7 |

Question 3 Explanation:

P(one ball is Red & another is blue)

=p(first is Red and second is Blue)

=\frac{4}{7}\times \frac{3}{7} =\frac{12}{49}

=p(first is Red and second is Blue)

=\frac{4}{7}\times \frac{3}{7} =\frac{12}{49}

Question 4 |

For the fillet weld of size 's' shown in the figure the effective throat
thickness is

0.61s | |

0.65s | |

0.7s | |

0.75s |

Question 4 Explanation:

Effective throat thickness is the shortest distance from the root of fillet weld to the face of the line joining the toes. The effective throat thickness should not be less than 3 mm.

Effective throat thickness =Ks

where s is the size of weld in mm and K is a constant. The value of K depends upon the angle between the fusion faces.

Effective throat thickness =Ks

where s is the size of weld in mm and K is a constant. The value of K depends upon the angle between the fusion faces.

Question 5 |

A 16 mm thick plate measuring 650 mm x 420 mm is used as a base plate for an
ISHB 300 column subjected to a factored axial compressive load of 2000 kN. As
per IS 456-2000, the minimum grade of concrete that should be used below the
base plate for safely carrying the load is

M15 | |

M20 | |

M30 | |

M40 |

Question 5 Explanation:

Working axial load

\begin{aligned} &=\frac{\text { Factored axial load }}{1.5} \\ &=\frac{2000}{1.5}=1333.33 \mathrm{kN} \end{aligned}

The allowable bearing pressure on concrete may be given as,

\begin{aligned} \sigma_{\text {all }} &=\frac{\text { Direct load }}{\text { Area of base plate }} \\ &=\frac{1333.33 \times 10^{3}}{650 \times 420}=4.88 \mathrm{N} / \mathrm{mm}^{2} \end{aligned}

The permissible stress in direct compression in various grades of concrete as per IS 456: 2000 are tabulated below:

The permissible stress in concrete should be more than the allowable bearing pressure. Thus the minimum grade of concrete which should be used is M20

\begin{aligned} &=\frac{\text { Factored axial load }}{1.5} \\ &=\frac{2000}{1.5}=1333.33 \mathrm{kN} \end{aligned}

The allowable bearing pressure on concrete may be given as,

\begin{aligned} \sigma_{\text {all }} &=\frac{\text { Direct load }}{\text { Area of base plate }} \\ &=\frac{1333.33 \times 10^{3}}{650 \times 420}=4.88 \mathrm{N} / \mathrm{mm}^{2} \end{aligned}

The permissible stress in direct compression in various grades of concrete as per IS 456: 2000 are tabulated below:

The permissible stress in concrete should be more than the allowable bearing pressure. Thus the minimum grade of concrete which should be used is M20

Question 6 |

Consider a reinforcing bar embedded in concrete. In a marine environment this bar
undergoes uniform corrosion, which leads to the deposition of corrosion products
on its surface and an increase in the apparent volume of the bar. This subjects the
surrounding concrete to expansive pressure. As a result, corrosion induced cracks
appear at the surface of concrete. Which of the following statements is TRUE?

Corrosion causes circumferential tensile stresses in concrete and the cracks
will be parallel to the corroded reinforcing bar. | |

Corrosion causes radial tensile stresses in concrete and the cracks will be
parallel to the corroded reinforcng bar. | |

Corrosion causes circumferential tensile stresses in concrete and the cracks
will be perpendicular to the direction of the corroded reinforcing bar. | |

Corrosion causes radial tensile stresses in concrete and the cracks will be
perpendicular to the direction of the corroded reinforcing bar. |

Question 6 Explanation:

Corrosion in steel occupies a volume several times
greater than volume of reinforcing steel hence
exerts radial pressure on adjoining concrete layer.
As a consequence, cracking results along
reinforcing bars and concrete starts spalling.

Question 7 |

The results for sieve analysis carried out for three types of sand, P, Q and R, are
given in the figure. If the fineness modulus values of the three sands are
given as FMP, FMQ and FMR, it can be stated that

FM_{Q}=\sqrt{FM_{P} \times FM_{R}} | |

FM_{Q}=0.5(FM_{P}+FM_{R}) | |

FM_{P}\gt FM_{Q}\gt FM_{R} | |

FM_{P}\lt FM_{Q}\lt FM_{R} |

Question 7 Explanation:

The term fineness modulus is used to indicate an
index number which roughly proportional to
average size of particle in entire quantity of
aggregate

F_{M Q}=\sqrt{F_{M P} \times F_{M R}} (Geometric mean)

F_{M Q}=\sqrt{F_{M P} \times F_{M R}} (Geometric mean)

Question 8 |

The cross-section of a thermo-mechanically treated (TMT) reinforcing bar has

soft ferrite-pearlite throughout | |

hard martensite throughout | |

a soft ferrite-pearlite core with a hard martensitic rim | |

a hard martensitic core with a soft pearlite-bainitic rim |

Question 9 |

Consider a simply supported beam with a uniformly distributed load having
a neutral axis (NA) as shown.

For points P (on the neutral axis) and Q (at the bottom of the beam) the state of stress is best represented by which of the following pairs?

For points P (on the neutral axis) and Q (at the bottom of the beam) the state of stress is best represented by which of the following pairs?

A | |

B | |

C | |

D |

Question 9 Explanation:

Point P:

Point P lies on NA, hence bending stress is zero at point P.

Point P also lies at mid span, so shear force, V=0

\Rightarrow Shear stress, \tau=0

\therefore State of stress of point P will be

Point Q :

At point Q flexural stress is maximum and nature of which is tensile due to downward loading. Point Q lies at the extreme of beam, therefore, shear stress at point Q is zero.

\therefore State of stress of point Q will be

Question 10 |

For a saturated sand deposit, the void ratio and the specific gravity of solids
are 0.70 and 2.67, respectively. The critical (upward) hydraulic gradient for the
deposit would be

0.54 | |

0.98 | |

1.02 | |

1.87 |

Question 10 Explanation:

Critical hydraulic gradient is given by

i_{c r}=\frac{G-1}{1+e}=\frac{2.67-1}{1+0.70}=0.98

i_{c r}=\frac{G-1}{1+e}=\frac{2.67-1}{1+0.70}=0.98

There are 10 questions to complete.