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
There are 5 questions to complete.