Question 1 |
The estimate of \int_{0.5}^{0.5}\frac{dx}{x} obtained using Simpson's rule with three-point function evaluation exceeds the exact value by
0.235 | |
0.068 | |
0.024 | |
0.012 |
Question 1 Explanation:
Exact value of \int_{0.5}^{1.5}\frac{dx}{x}=\left [ \log x \right ]_{0.5}^{1.5} = \log (1.5)- \log(0.5) =1.0986
Approximate value by Simpson's rule with 3 point is,
\begin{aligned} I&=\frac{h}{3}\left ( f\left ( 0 \right )+4f\left ( 1 \right )+f\left ( 2 \right ) \right ) \\ n_{i}&=n_{pt}-1=3-1=2 \end{aligned}
(n_{pt} is the number of pts and n_{i} is the number of intervals)
Here h=\frac{b-a}{n_{i}}=\frac{1.5-0.5}{2}=0.5
The table is
I=\frac{0.5}{3}\left ( \frac{1}{0.5}+4 \times 1+\frac{1}{1.5}\right ) = 1.1111
So the estimate exceeds the exact value by,
Approximate value-Exact value
=0.012499 \approx 0.012499
Approximate value by Simpson's rule with 3 point is,
\begin{aligned} I&=\frac{h}{3}\left ( f\left ( 0 \right )+4f\left ( 1 \right )+f\left ( 2 \right ) \right ) \\ n_{i}&=n_{pt}-1=3-1=2 \end{aligned}
(n_{pt} is the number of pts and n_{i} is the number of intervals)
Here h=\frac{b-a}{n_{i}}=\frac{1.5-0.5}{2}=0.5
The table is
I=\frac{0.5}{3}\left ( \frac{1}{0.5}+4 \times 1+\frac{1}{1.5}\right ) = 1.1111
So the estimate exceeds the exact value by,
Approximate value-Exact value
=0.012499 \approx 0.012499
Question 2 |
The annual precipitation data of a city is normally distributed with mean and standard deviation as
1000 mm and 200 mm, respectively. The probability that the annual precipitation will be more than
1200 mm is
\lt50% | |
50% | |
75% | |
100% |
Question 2 Explanation:
The annual precipitation is normally distributed with \mu =1000\: mm and \sigma =200\: mm
\begin{aligned} p\left ( x\gt 1200 \right ) &=p\left ( z\gt \frac{1200-1000}{200} \right ) \\ &=p\left ( z\gt 1 \right ) \end{aligned}
Where z is the standard normal variate.
In normal distribution
Now, since
p\left ( -1\lt z\lt 1 \right )\approx 0.68
(\simeq 68% of data is within one standard deviation of mean)
\begin{aligned} p\left ( 0\lt z\lt 1 \right ) &=\frac{0.68}{2}&=0.34 \\ \text{So} p\left ( z\gt 1 \right ) &=0.5-0.34 \\ &=0.16\simeq 16\% \end{aligned}
which is \lt 50\%
\begin{aligned} p\left ( x\gt 1200 \right ) &=p\left ( z\gt \frac{1200-1000}{200} \right ) \\ &=p\left ( z\gt 1 \right ) \end{aligned}
Where z is the standard normal variate.
In normal distribution
Now, since
p\left ( -1\lt z\lt 1 \right )\approx 0.68
(\simeq 68% of data is within one standard deviation of mean)
\begin{aligned} p\left ( 0\lt z\lt 1 \right ) &=\frac{0.68}{2}&=0.34 \\ \text{So} p\left ( z\gt 1 \right ) &=0.5-0.34 \\ &=0.16\simeq 16\% \end{aligned}
which is \lt 50\%
Question 3 |
The infinite series 1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+... corresponds to
sec x | |
e^{x} | |
cos x | |
1+sin^{2}x |
Question 3 Explanation:
e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}...
(By McLaurin's series expansion)
(By McLaurin's series expansion)
Question 4 |
The Poisson's ratio is defined as
|\frac{\text{Axial tress}}{\text{Lateral stress}}| | |
|\frac{\text{Lateral strain}}{\text{Axial strain}}| | |
|\frac{\text{Lateral stress}}{\text{Axial stress}}| | |
|\frac{\text{Axial strain}}{\text{Lateral strain}}| |
Question 4 Explanation:
\mu =\left | \frac{\text{Lateral strain}}{\text{Axial strain}} \right |
Question 5 |
The following statements are related to bending of beams:
I The slope of the bending moment diagram is equal to the shear force.
II The slope of the shear force diagram is equal to the load intensity.
III The slope of the curvature is equal to the flexural rotation.
IV The second derivative of the deflection is equal to the curvature.
The only FALSE statement is
I The slope of the bending moment diagram is equal to the shear force.
II The slope of the shear force diagram is equal to the load intensity.
III The slope of the curvature is equal to the flexural rotation.
IV The second derivative of the deflection is equal to the curvature.
The only FALSE statement is
I | |
II | |
III | |
IV |
Question 5 Explanation:
We know that
\begin{aligned} \frac{d S}{d x} &=W ; \frac{d M}{d X}=S_{x} \\ E 1 \cdot \frac{d^{2} y}{d x^{2}} &=M \\ \therefore\quad \frac{d^{2} y}{d x^{2}} &=\frac{M}{E I} \\ \text{Also}\quad \frac{M}{I} &=\frac{f}{y}=\frac{E}{R} \quad \therefore \frac{M}{E I}=\frac{1}{R} \\ \therefore\quad \frac{d^{2} y}{d x^{2}} &=\frac{1}{R} \end{aligned}
\begin{aligned} \frac{d S}{d x} &=W ; \frac{d M}{d X}=S_{x} \\ E 1 \cdot \frac{d^{2} y}{d x^{2}} &=M \\ \therefore\quad \frac{d^{2} y}{d x^{2}} &=\frac{M}{E I} \\ \text{Also}\quad \frac{M}{I} &=\frac{f}{y}=\frac{E}{R} \quad \therefore \frac{M}{E I}=\frac{1}{R} \\ \therefore\quad \frac{d^{2} y}{d x^{2}} &=\frac{1}{R} \end{aligned}
Question 6 |
If a small concrete cube is submerged deep in still water in such a way that the pressure exerted on
all faces of the cube is p, then the maximum shear stress developed inside the cube is
0 | |
p/2 | |
p | |
2p |
Question 6 Explanation:
Maximum shear stress,
\tau=\frac{\sigma_{1}-\sigma_{2}}{2}=\frac{P-P}{2}=0
Question 7 |
As per IS 456:2000, in the Limit State Design of a flexural member, the strain in reinforcing bars
under tension at ultimate state should not be less than
\frac{f_{y}}{E_{s}} | |
\frac{f_{y}}{E_{s}}+0.002 | |
\frac{f_{y}}{1.15E_{s}} | |
\frac{f_{y}}{1.15E_{s}}+0.002 |
Question 7 Explanation:
Question 8 |
Which one of the following is categorised as a long-term loss of prestress in a prestressed concrete
member?
Loss due to elastic shortening | |
Loss due to friction | |
Loss due to relaxation of strands | |
Loss due to anchorage slip |
Question 9 |
In a steel plate with bolted connections, the rupture of the net section is a mode of failure under
tension | |
compression | |
flexure | |
shear |
Question 10 |
The ratio of the theoretical critical buckling load for a column with fixed ends to that of another
column with the same dimensions and material, but with pinned ends, is equal to
0.5 | |
1 | |
2 | |
4 |
Question 10 Explanation:
Eulars critical load,
\begin{aligned} P_{e}&=\frac{\pi^{2}EI}{I^{2}_{\text{eft}}}\\ \frac{P_{e1}}{P_{e2}}=\left[\frac{(I_{\text{elf}})_{2}}{(I_{\text{elf}})_{1}} \right ]^{2}&=\left[\frac{I}{1/2}\right ]^{2}=4 \end{aligned}
\begin{aligned} P_{e}&=\frac{\pi^{2}EI}{I^{2}_{\text{eft}}}\\ \frac{P_{e1}}{P_{e2}}=\left[\frac{(I_{\text{elf}})_{2}}{(I_{\text{elf}})_{1}} \right ]^{2}&=\left[\frac{I}{1/2}\right ]^{2}=4 \end{aligned}
There are 10 questions to complete.