Question 1 |

Solution for the system defined by the set of equations

4y + 3z = 8;

2x - z = 2 and

3x + 2y = 5 is

4y + 3z = 8;

2x - z = 2 and

3x + 2y = 5 is

x=0; y=1; z=4/3 | |

x=0; y=1/2; z=2 | |

x=1; y=1/2; z=2 | |

non-existent |

Question 1 Explanation:

The augmented matrix for given system is,

\left [ \left.\begin{matrix} 0 & 4 & 3\\ 2 & 0 & -1\\ 3 & 2 & 0 \end{matrix}\right|\begin{matrix} 8\\ 2\\ 5 \end{matrix} \right ]\xrightarrow[]{exchange 1st and 2nd row}\left [ \left.\begin{matrix} 2 &0 & -1\\ 0 & 4 & 3\\ 3 & 2 & 0 \end{matrix}\right|\begin{matrix} 2\\ 8\\ 5 \end{matrix} \right ]

then by Gauss elimination procedure

\left [ \left.\begin{matrix} 2 & 0 & -1\\ 0 & 4 & 3\\ 3 & 2 & 0 \end{matrix}\right|\begin{matrix} 2\\ 8\\ 5 \end{matrix} \right ]\xrightarrow[]{R_{3}-\frac{3}{2}R_{1}}\left [ \left.\begin{matrix} 2 &0 & -1\\ 0 & 4 & 3\\ 0 & 2 & 3/2 \end{matrix}\right|\begin{matrix} 2\\ 8\\ 2 \end{matrix} \right ]\xrightarrow[]{R_{3}-\frac{2}{4}R_{2}}\left [ \left.\begin{matrix} 2 & 0 & -1\\ 0 & 4 & 3\\ 0 & 0 & 0 \end{matrix}\right|\begin{matrix} 8\\ 8\\ -2 \end{matrix} \right ]

For last row we see 0=-2 which is inconsistent

\therefore Solution is non-existent for above system.

\left [ \left.\begin{matrix} 0 & 4 & 3\\ 2 & 0 & -1\\ 3 & 2 & 0 \end{matrix}\right|\begin{matrix} 8\\ 2\\ 5 \end{matrix} \right ]\xrightarrow[]{exchange 1st and 2nd row}\left [ \left.\begin{matrix} 2 &0 & -1\\ 0 & 4 & 3\\ 3 & 2 & 0 \end{matrix}\right|\begin{matrix} 2\\ 8\\ 5 \end{matrix} \right ]

then by Gauss elimination procedure

\left [ \left.\begin{matrix} 2 & 0 & -1\\ 0 & 4 & 3\\ 3 & 2 & 0 \end{matrix}\right|\begin{matrix} 2\\ 8\\ 5 \end{matrix} \right ]\xrightarrow[]{R_{3}-\frac{3}{2}R_{1}}\left [ \left.\begin{matrix} 2 &0 & -1\\ 0 & 4 & 3\\ 0 & 2 & 3/2 \end{matrix}\right|\begin{matrix} 2\\ 8\\ 2 \end{matrix} \right ]\xrightarrow[]{R_{3}-\frac{2}{4}R_{2}}\left [ \left.\begin{matrix} 2 & 0 & -1\\ 0 & 4 & 3\\ 0 & 0 & 0 \end{matrix}\right|\begin{matrix} 8\\ 8\\ -2 \end{matrix} \right ]

For last row we see 0=-2 which is inconsistent

\therefore Solution is non-existent for above system.

Question 2 |

The differential equation \frac{dy}{dx}=0.25y^2 is to be solved using the backward
(implicit) Euler's method with the boundary condition y=1 at x=0 and with
a step size of 1. What would be the value of y at x=1 ?

1.33 | |

1.67 | |

2 | |

2.33 |

Question 2 Explanation:

\frac{\mathrm{d} y}{\mathrm{d} x}=0.25y^{2} \;\;\left ( y=1\: at\: x=0 \right )

h=1

Iterative equation for backward (implicit) Euler methods for above equation would be,

\frac{y_{k+1}-y_{k}}{h}=0.25y^{2}_{k+1}

\Rightarrow \;\; y_{k+1}-y_{k}=0.25h y^{2}_{k+1}

\Rightarrow \;\; 0.25h y^{2}_{k+1}-y_{k+1}+y_{k}=0

putting k=0 in above equation

0.25h y^{2}_{1}-y_{1}+y_{0}=0

since y_{0}=1 and h=1

0.25 y_{1}^{2}-y_{1}+1=0

y_{1}=2

h=1

Iterative equation for backward (implicit) Euler methods for above equation would be,

\frac{y_{k+1}-y_{k}}{h}=0.25y^{2}_{k+1}

\Rightarrow \;\; y_{k+1}-y_{k}=0.25h y^{2}_{k+1}

\Rightarrow \;\; 0.25h y^{2}_{k+1}-y_{k+1}+y_{k}=0

putting k=0 in above equation

0.25h y^{2}_{1}-y_{1}+y_{0}=0

since y_{0}=1 and h=1

0.25 y_{1}^{2}-y_{1}+1=0

y_{1}=2

Question 3 |

The necessary and sufficient condition for a surface to be called as a free surface is

no stress should be acting on it | |

tensile stress acting on it must be zero | |

shear stress acting on it must be zero | |

no point on it should be under any stress |

Question 4 |

Mohr's circle for the state of stress defined by \begin{bmatrix} 30 &0 \\ 0&30 \end{bmatrix} MPa is a circle with

centre at (0, 0) and radius 30 MPa | |

centre at (0, 0) and radius 60 MPa | |

centre at (30, 0) and radius 30 MPa | |

centre at (30, 0) and zero radius |

Question 4 Explanation:

The maximum and minimum principal stresses are
same. So, radius of circle becomes zero and centre
is at (30, 0). The circle is represented by a point.

Question 5 |

The buckling load P=P_{cr} for the column AB shown in figure, as K_T approaches infinity,
becomes \alpha \frac{\pi ^{2}EI}{L^{2}}, where \alpha is equal to

0.25 | |

1 | |

2.05 | |

4 |

Question 6 |

A long shaft of diameter d is subjected to twisting moment T at its ends. The
maximum normal stress acting at its cross-section is equal to

zero | |

\frac{16T}{\pi d^{3}} | |

\frac{32T}{\pi d^{3}} | |

\frac{64T}{\pi d^{3}} |

Question 6 Explanation:

Cross-section of the shaft is the section
perpendicular to the longitudinal axis of shaft. In
case of twisting moment acting at its ends, normal
stress is zero on the cross-section only shear
stress acts.

Question 7 |

If the characteristic strength of concrete f_{ck} is defined as the strength below which not more than 50% of the test results are expected to fall, the expression for f_{ck} in terms of mean strength f_{m} and standard deviation S would be

f_{m}- 0.1645 S | |

f_{m} - 1.645 S | |

f_{m} | |

f_{m} + 1.645S |

Question 8 |

The range of void ratio between which quick sand condition occurs in cohesionless
granular soil deposits is

0.4 - 0.5 | |

0.6 - 0.7 | |

0.8 - 0.9 | |

1.0 - 1.1 |

Question 8 Explanation:

The specific gravity of cohesionless granular soils
(sands) does not vary much and for all practical
purposes it is taken to be 2.65. Critical hydraulic
gradient should be nearly 1 for quick sand condition.

i.e.\quad i_{cr}=\frac{G-1}{1+e}

From the above equation, the void ratio range is found to be between 0.6 and 0.7.

i.e.\quad i_{cr}=\frac{G-1}{1+e}

From the above equation, the void ratio range is found to be between 0.6 and 0.7.

Question 9 |

Figure given below shows a smooth vertical gravity retaining wall with cohesionless soil backfill having an angle of internal friction \phi. In the graphical representation
of Rankine's active earth pressure for the retaining wall shown in figure, length OP represents

vertical stress at the base | |

vertical stress at a height H/3 from the base | |

lateral earth pressure at the base | |

lateral earth pressure at a height H/3 from the base |

Question 9 Explanation:

\begin{array}{l} O P=\text {Major stress}=\sigma_{1} \\ O Q=\text {Mirror stress}=\sigma_{3} \end{array}

Active condition:

OP = Vertical stress at base

=\sigma_{v} \cos \beta=\gamma+\cos \beta

OQ = Lateral earth pressure at base

=\mathrm{K}_{\mathrm{a}} \cdot \sigma_{\mathrm{v}} \cos \beta

Note: QR and QR' are failure plane for active case.

Question 10 |

Which of the following statement is NOT TRUE in the context of capillary pressure in solids ?

Water is under tension in capillary zone | |

Pore water pressure is negative in capillary zone | |

Effective stress increases due to the capillary zone | |

Capillary pressure is more in coarse grained soils |

Question 10 Explanation:

Capillary rise is more in fine grained soils and as
a result of this, the capillary pressure is also more
than the coarse grained soils.

There are 10 questions to complete.