Question 1 |
For what value of p the following set of equations will have no solution?
2x+3y=5
3x+py=10
2x+3y=5
3x+py=10
2.5 | |
5.25 | |
4.5 | |
None |
Question 1 Explanation:
Given system of equations has no solution if the lines are parallel i.e., their slopes are equal
\begin{aligned} \frac{2}{3}&=\frac{3}{p} \\ \Rightarrow \;\; p&=4.5 \end{aligned}
\begin{aligned} \frac{2}{3}&=\frac{3}{p} \\ \Rightarrow \;\; p&=4.5 \end{aligned}
Question 2 |
The integral \int_{x_1}^{x_2}x^{2}dx with {x_1} \gt {x_2} \gt 0 is evaluated analytically as well as numerically using a single application of the trapezoidal rule. If I is the exact value of the integral obtained analytically and J is the approximate value obtained using the trapezoidal rule, which of the following statements is correct about their relationship?
J \gt I | |
J \lt I | |
J=I | |
Insufficient data to determine the relationship |
Question 2 Explanation:
Exact value is computed by integration which follows the exact shape of graph while computing the area.
Whereas, in Trapezoidal rule, the lines joining each points are considered straight lines which is not the exact variation of graph all the time like as shown figure.
\therefore J\gt I
OR
\text{Error } =-\frac{h^{3}}{12}{f}''\left ( \Im \right )\times n_{i}
Here, f\left ( x \right )=x^{2}
or, {f}'\left ( x \right )=2x
or, {f}''\left ( x \right )=2\gt 0
Since {f}''\left ( x \right ) is positive, the error is negative.
Since error = exact-approximate.
=I-J
and since error is negative in this case J\gt I is true.
Question 3 |
Consider the following probability mass function (p.m.f) of a random variable X :
p(x,q)=\begin{cases} q & \text{ if } X=0\\ 1-q& \text{ if } X=1 \\ 0 & \text otherwise \end{cases}
If q = 0.4, the variance of X is _______.
p(x,q)=\begin{cases} q & \text{ if } X=0\\ 1-q& \text{ if } X=1 \\ 0 & \text otherwise \end{cases}
If q = 0.4, the variance of X is _______.
0.15 | |
0.24 | |
1.25 | |
2.52 |
Question 3 Explanation:
Given, \; \; \; \; q=0.4
Required value =V\left ( X \right )=E\left ( X^{2} \right )-\left [ E\left ( X \right ) \right ]^{2}
E\left ( X \right )=\sum_{i}^{}\, X_{i}p_{i}=0\times 0.4+1\times 0.6=0.6
E\left ( X^{2} \right )=\sum_{i}^{}\, X_{i}^{2}p_{i}=0^{2}\times 0.4+1^{2}\times 0.6=0.6
\therefore \; \; \; V\left ( X \right )=E\left ( X^{2} \right )-\left [ E\left ( X \right ) \right ]^{2}=0.6-0.36=0.24
Required value =V\left ( X \right )=E\left ( X^{2} \right )-\left [ E\left ( X \right ) \right ]^{2}
E\left ( X \right )=\sum_{i}^{}\, X_{i}p_{i}=0\times 0.4+1\times 0.6=0.6
E\left ( X^{2} \right )=\sum_{i}^{}\, X_{i}^{2}p_{i}=0^{2}\times 0.4+1^{2}\times 0.6=0.6
\therefore \; \; \; V\left ( X \right )=E\left ( X^{2} \right )-\left [ E\left ( X \right ) \right ]^{2}=0.6-0.36=0.24
Question 4 |
Workability of concrete can be measured using slump, compaction factor and Vebe time. Consider the following statements for workability of concrete:
(i) As the slump increases, the Vebe time increases
(ii) As the slump increases, the compaction factor increases
Which of the following is TRUE?
(i) As the slump increases, the Vebe time increases
(ii) As the slump increases, the compaction factor increases
Which of the following is TRUE?
Both (i) and (ii) are True | |
Both (i) and (ii) are False | |
(i) is True and (ii) is False | |
(i) is False and (ii) is True |
Question 4 Explanation:
As slump increase, water content increases means
workability increases.
Compaction factor
=\frac{\text { Weight of partially compacted concrete }}{\text { Weight of fully compacted concrete }}
\therefore Compaction factor will also increase.
The time required for complete remoulding in seconds is considered as a measure of workability and is expressed as the number of Vee-Bee seconds.
\therefore As workability increases, Vee-Bee time decreases.
Compaction factor
=\frac{\text { Weight of partially compacted concrete }}{\text { Weight of fully compacted concrete }}
\therefore Compaction factor will also increase.
The time required for complete remoulding in seconds is considered as a measure of workability and is expressed as the number of Vee-Bee seconds.
\therefore As workability increases, Vee-Bee time decreases.
Question 5 |
Consider the following statements for air-entrained concrete.
(i) Air-entrainment reduces the water demand for a given level of workability
(ii) Use of air-entrained concrete is required in environments where cyclic freezing and thawing is expected
Which of the following is TRUE?
(i) Air-entrainment reduces the water demand for a given level of workability
(ii) Use of air-entrained concrete is required in environments where cyclic freezing and thawing is expected
Which of the following is TRUE?
Both (i) and (ii) are True | |
Both (i) and (ii) are False | |
(i) is True and (ii) is False | |
(i) is False and (ii) is True |
Question 5 Explanation:
An air-entraining agent introduces air in the form
of bubbles that occupy upto 5% of the volume of
concrete distributed uniformly throughout cement
paste. Thus, for the same amount of water/cement
ratio, we get higher workability.
Question 6 |
Consider the singly reinforced beam shown in the figure below:
At cross-section XX, which of the following statements is TRUE at the limit state?
At cross-section XX, which of the following statements is TRUE at the limit state?
The variation of stress is linear and that of strain is non-linear | |
The variation of strain is linear and that of stress is non-linear | |
The variation of both stress and strain is linear | |
The variation of both stress and strain is non- linear |
Question 6 Explanation:
Question 7 |
For the beam shown below, the stiffness coefficient K_{22} can be written as
\frac{6EI}{L^{2}} | |
\frac{12EI}{L^{3}} | |
\frac{3EI}{L} | |
\frac{EI}{6L^{2}} |
Question 7 Explanation:
By giving unit displacement in 2^{\text {nd }} direction without giving displacement in any other direction, the force developed R_{6}, is,
K_{22}=\frac{12 E I}{L^{3}}
Question 8 |
The development length of a deformed reinforncement bar can be expressed as (1/k) (\phi \sigma _{s}/\tau _{bd}). From the IS: 456-2000, the value of k can be calculated as________.
6.4 | |
9.6 | |
3.4 | |
8.3 |
Question 8 Explanation:
Bond strength of concrete
\begin{aligned} &=\text{ Tensile force in steel}\\ \tau_{b d} \times\left(L_{d} \pi \phi\right)&=0.87 f_{y} \times\left(\frac{\pi}{4} \phi^{2}\right)\\ \Rightarrow \quad L_{d}&=\frac{0.87 f_{y} \phi}{4 \tau_{b d}} \\ \end{aligned}
For deformed bars \tau_{\text {bd }} value is increased by 60 %
\begin{aligned} \therefore \quad L_{d}&=\frac{\phi \sigma_{s}}{4 \times\left(1.6 \tau_{b d}\right)}=\frac{\phi \sigma_{s}}{6.4 \tau_{b d}}\\ \therefore \quad k&=6.4 \end{aligned}
\begin{aligned} &=\text{ Tensile force in steel}\\ \tau_{b d} \times\left(L_{d} \pi \phi\right)&=0.87 f_{y} \times\left(\frac{\pi}{4} \phi^{2}\right)\\ \Rightarrow \quad L_{d}&=\frac{0.87 f_{y} \phi}{4 \tau_{b d}} \\ \end{aligned}
For deformed bars \tau_{\text {bd }} value is increased by 60 %
\begin{aligned} \therefore \quad L_{d}&=\frac{\phi \sigma_{s}}{4 \times\left(1.6 \tau_{b d}\right)}=\frac{\phi \sigma_{s}}{6.4 \tau_{b d}}\\ \therefore \quad k&=6.4 \end{aligned}
Question 9 |
For the beam shown below, the value of the support moment M is __________ kN-m.
3 | |
9 | |
5 | |
7 |
Question 9 Explanation:
By symmetry we can consider two parts of beam separately with half the load
Moment at joint B=10 \mathrm{kNm}
Carry over moment at joint A due to 10 kNm moment at propped end =\frac{10}{2}=5 \mathrm{kN}-\mathrm{m}
Moment at joint B=10 \mathrm{kNm}
Carry over moment at joint A due to 10 kNm moment at propped end =\frac{10}{2}=5 \mathrm{kN}-\mathrm{m}
Question 10 |
Two triangular wedges are glued together as shown in the following figure. The stress acting normal to the interface, \sigma _{n} is _____ MPa.
0 | |
1 | |
2 | |
3 |
Question 10 Explanation:
As plane AB and BC are principle planes, therefore
Mohr's circle for the given condition is
Here, normal stress is zero at 45^{\circ} to the principle plane.
Here, normal stress is zero at 45^{\circ} to the principle plane.
There are 10 questions to complete.