Question 1 |
There is no value of x that can simultaneously satisfy both the given equations. Therefore, find the
'least squares error' solution to the two equations, i.e., find the value of x that minimizes the sum of
squares of the errors in the two equations. __________
2x= 3
4x= 1
2x= 3
4x= 1
0.5 | |
1.5 | |
2 | |
4 |
Question 2 |
What is the minimum number of multiplications involved in computing the matrix product PQR?
MatrixP has 4 rows and 2 columns, matrixQ has 2 rows and 4 columns, and matrixR has 4 rows and 1 column. __________
MatrixP has 4 rows and 2 columns, matrixQ has 2 rows and 4 columns, and matrixR has 4 rows and 1 column. __________
256 | |
32 | |
16 | |
128 |
Question 2 Explanation:
If we multiply QR first then,
Q_{2\times 4}\times R_{\left ( 2\times 1 \right )} having multiplication number 8.
Therefore P_{\left ( 4\times 2 \right )}QR_{\left ( 2\times 1 \right )} will have minimum number of multiplication =(8+8)=16.
Q_{2\times 4}\times R_{\left ( 2\times 1 \right )} having multiplication number 8.
Therefore P_{\left ( 4\times 2 \right )}QR_{\left ( 2\times 1 \right )} will have minimum number of multiplication =(8+8)=16.
Question 3 |
A 1-h rainfall of 10 cm magnitude at a station has a return period of 50 years. The probability that
a 1-h rainfall of magnitude 10 cm or more will occur in each of two successive years is:
0.04 | |
0.2 | |
0.02 | |
0.0004 |
Question 3 Explanation:
Return period of rainfall,
T&=50 years
\therefore probability of occurrence once in 50 years,
p&=\frac{1}{50}&=0.02
Probability of occurrence in each of 2 successive year =p^{2}=\left ( 0.02 \right )^{2}=0.0004
T&=50 years
\therefore probability of occurrence once in 50 years,
p&=\frac{1}{50}&=0.02
Probability of occurrence in each of 2 successive year =p^{2}=\left ( 0.02 \right )^{2}=0.0004
Question 4 |
Maximum possible value of Compacting Factor for fresh (green) concrete is:
0.5 | |
1 | |
1.5 | |
2 |
Question 5 |
As per IS 800:2007, the cross-section in which the extreme fiber can reach the yield stress, but
cannot develop the plastic moment of resistance due to failure by local buckling is classified as
plastic section | |
compact section | |
semi-compact section | |
slender section |
Question 5 Explanation:
As per clause 3.7.2 of IS 800:2007
Question 6 |
The creep strains are
caused due to dead loads only | |
caused due to live loads only | |
caused due to cyclic loads only | |
independent of loads |
Question 6 Explanation:
Creep strains occur with time at a constant level
of stress. So they occur due to permanent load
i.e., Dead Load.
Question 7 |
As per IS 456:2000 for M20 grade concrete and plain barsin tension the design bond stress \tau _{bd}=1.2 MPa.Further, IS 456:2000 permits this design bond stress value to be increased by 60% for HSD bars. The stress in the HSD reinforcing steel barsin tension, \sigma _{s}=360MPa. Find the required development length, L_{d}, for HSD barsin terms of the bar diameter, \phi. __________
34.54 | |
46.87 | |
89.25 | |
93.47 |
Question 7 Explanation:
\begin{aligned} L_{d} &=\frac{\phi \sigma_{s t}}{4 \tau_{b d}}=\frac{\phi \times 360}{4 \times 1.2 \times 1.60} \\ &=46.875 \phi \end{aligned}
Question 8 |
The 'plane section remains plane' assumption in bending theory implies:
strain profile is linear | |
stress profile is linear | |
both strain and stress profiles are linear | |
shear deformations are neglected |
Question 9 |
Two steel columns P (length L and yield strength f_{y}= 250 MPa) and Q (length 2L and yield strength f_{y}= 500 MPa) have the same cross-sections and end-conditions. The ratio of buckling load of column P to that of column Q is:
0.5 | |
1 | |
2 | |
4 |
Question 9 Explanation:
Buckling load,
\begin{aligned} P &=\frac{\pi^{2} E \mid}{\left(l_{e f f}\right)^{2}} \\ \therefore \quad P_{p} &=\frac{\pi^{2} E \mid}{(L)^{2}} ; P_{Q}=\frac{\pi^{2} E \mid}{(2 L)^{2}} \\ \therefore \quad \frac{P_{P}}{P_{a}} &=4 \end{aligned}
\begin{aligned} P &=\frac{\pi^{2} E \mid}{\left(l_{e f f}\right)^{2}} \\ \therefore \quad P_{p} &=\frac{\pi^{2} E \mid}{(L)^{2}} ; P_{Q}=\frac{\pi^{2} E \mid}{(2 L)^{2}} \\ \therefore \quad \frac{P_{P}}{P_{a}} &=4 \end{aligned}
Question 10 |
The pin-jointed 2-D truss is loaded with a horizontal force of 15 kN at joint S and another 15 kN
vertical force at joint U, as shown. Find the force in member RS (in kN) and report your answer
taking tension as positive and compression as negative. __________
1 | |
2 | |
4 | |
0 |
Question 10 Explanation:
\begin{aligned} \Sigma M_{T}&=0 \\ \Rightarrow R_{v} \times 8-15 \times 4+15 \times 4 & =0 \\ \therefore \quad R_{V} & =0\\ \text{Take moment about V, } M_{V}&=0 \\ \Rightarrow \quad R_{H} \times 4&=0 \\ \therefore \quad R_{H}&=0 \end{aligned}
We know that if two members meet at a joint which are not collinear and also there is no external forces acting on that joint then both members will carry zero forces.
\therefore F_{O V}=F_{Q R}=0
Now, consider joint R ,
\Sigma F_{y}=0
\begin{aligned} \Rightarrow F_{R U} \sin 45^{\circ}&=0 \\ \Rightarrow F_{R U}&=0\\ \text{Now},\quad \Sigma F_{x}&=0 \\ \Rightarrow \quad F_{R S}+F_{R U} \cos 45^{\circ}&=0 \\ \therefore F_{R S}&=0 \end{aligned}
So, force is member is "RS" will be zero.
We know that if two members meet at a joint which are not collinear and also there is no external forces acting on that joint then both members will carry zero forces.
\therefore F_{O V}=F_{Q R}=0
Now, consider joint R ,
\Sigma F_{y}=0
\begin{aligned} \Rightarrow F_{R U} \sin 45^{\circ}&=0 \\ \Rightarrow F_{R U}&=0\\ \text{Now},\quad \Sigma F_{x}&=0 \\ \Rightarrow \quad F_{R S}+F_{R U} \cos 45^{\circ}&=0 \\ \therefore F_{R S}&=0 \end{aligned}
So, force is member is "RS" will be zero.
There are 10 questions to complete.