Question 1 |
The rank of matrix \left[\begin{array}{llll} 1 & 2 & 2 & 3 \\ 3 & 4 & 2 & 5 \\ 5 & 6 & 2 & 7 \\ 7 & 8 & 2 & 9 \end{array}\right] is
1 | |
2 | |
3 | |
4 |
Question 1 Explanation:
Using R_{2} \rightarrow R_{2} \rightarrow 3 R_{1}, R_{3} \rightarrow R_{3}-5 R_{1}, R_{4} \rightarrow R_{4}-7 R_{1}
A=\left[\begin{array}{cccc} 1 & 2 & 2 & 3 \\ 0 & -2 & -4 & -4 \\ 0 & -4 & -8 & -8 \\ 0 & -6 & -12 & -12 \end{array}\right]
Using R_{3} \rightarrow R_{3}-2 R_{2}, R_{4} \rightarrow R_{4}-3 R_{2}
A=\left[\begin{array}{cccc} 1 & 2 & 2 & 3 \\ 0 & -2 & -4 & -4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]
So, \rho(A)= No. of non-zero rows = 2.
A=\left[\begin{array}{cccc} 1 & 2 & 2 & 3 \\ 0 & -2 & -4 & -4 \\ 0 & -4 & -8 & -8 \\ 0 & -6 & -12 & -12 \end{array}\right]
Using R_{3} \rightarrow R_{3}-2 R_{2}, R_{4} \rightarrow R_{4}-3 R_{2}
A=\left[\begin{array}{cccc} 1 & 2 & 2 & 3 \\ 0 & -2 & -4 & -4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]
So, \rho(A)= No. of non-zero rows = 2.
Question 2 |
If P=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] and Q=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] then Q^{T} P^{T} is
\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] | |
\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right] | |
\left[\begin{array}{ll} 2 & 1 \\ 4 & 3 \end{array}\right] | |
\left[\begin{array}{ll} 2 & 4 \\ 1 & 3 \end{array}\right] |
Question 2 Explanation:
\begin{array}{l} \quad P Q=\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]=\left[\begin{array}{ll} 2 & 4 \\ 1 & 3 \end{array}\right] \\ (P Q)^{\top}=\left[\begin{array}{ll} 2 & 4 \\ 1 & 3 \end{array}\right] \end{array}
Now using Reversal law
Q^{\top} P^{\top}=(P Q) T=\left[\begin{array}{ll} 2 & 4 \\ 1 & 3 \end{array}\right]
Now using Reversal law
Q^{\top} P^{\top}=(P Q) T=\left[\begin{array}{ll} 2 & 4 \\ 1 & 3 \end{array}\right]
Question 3 |
The shape of the cumulative distribution function of Gaussian distribution is
Horizontal line | |
Straight line at 45 degree angle | |
Bell-shaped | |
S-shaped |
Question 3 Explanation:
PDF:f(x)=\frac{1}{\sigma \sqrt{2 \pi}}e^{-(x-\mu )^2/(2\sigma ^2)}
CDF:F(x)=\frac{1}{2}\left [ 1+eff\left ( \frac{x-\mu }{\sigma \sqrt{2}} \right ) \right ]
Question 4 |
A propped cantilever beam EF is subjected to a unit moving load as shown in the figure (not to scale). The sign convention for positive shear force at the left and right sides of any section is also shown.
The CORRECT qualitative nature of the influence line diagram for shear force at G is
The CORRECT qualitative nature of the influence line diagram for shear force at G is
 | |
 | |
 | |
 |
Question 4 Explanation:
As per Muller Breslau principle ILD for stress function (shear -V_{G}) will be a combination of curves (3^{\circ} curves).
Question 5 |
Gypsum is typically added in cement to
prevent quick setting | |
enhance hardening | |
increase workability | |
decrease heat of hydration |
Question 5 Explanation:
The Gypsum is added to cement at the end of grinding clinker it is added to prevent quick setting.
Question 6 |
The direct and indirect costs estimated by a contractor for bidding a project is Rs.160000 and Rs.20000 respectively. If the mark up applied is 10% of the bid price, the quoted price (in Rs.) of the contractor is
200000 | |
198000 | |
196000 | |
182000 |
Question 6 Explanation:
Direct Costs = 160000
Indirect costs = 20000
Mark up applied is 10% of the Bid price
Bid price = Direct cost + Indirect cost + Markup
Bid price = Direct cost + Indirect cost + \frac{10}{100} \times Bid price
\text{Bid price}\left (1-\frac{10}{100} \right )=DC+IC
\text{Bid price}=\frac{160000+20000}{0.9}=2,00,000
Indirect costs = 20000
Mark up applied is 10% of the Bid price
Bid price = Direct cost + Indirect cost + Markup
Bid price = Direct cost + Indirect cost + \frac{10}{100} \times Bid price
\text{Bid price}\left (1-\frac{10}{100} \right )=DC+IC
\text{Bid price}=\frac{160000+20000}{0.9}=2,00,000
Question 7 |
In an Oedometer apparatus, a specimen of fully saturated clay has been consolidated under a vertical pressure of 50 \mathrm{kN} / \mathrm{m}^{2} and is presently at equilibrium. The effective stress and pore water pressure immediately on increasing the vertical stress to 150 \mathrm{kN} / \mathrm{m}^{2}, respectively are
150 \mathrm{kN} / \mathrm{m}^{2} and 0 | |
100 \mathrm{kN} / \mathrm{m}^{2} and 50 \mathrm{kN} / \mathrm{m}^{2} | |
50 \mathrm{kN} / \mathrm{m}^{2} and 100 \mathrm{kN} / \mathrm{m}^{2} | |
0 and 150 \mathrm{kN} / \mathrm{m}^{2} |
Question 7 Explanation:
Stress is increased suddenly, hence entire change will be taken by water \Delta \bar{\sigma}=\Delta U=100 \mathrm{kPa}.
There will be no change in effective stress
\therefore \qquad \qquad\bar{\sigma}=50 \mathrm{kPa}
There will be no change in effective stress
\therefore \qquad \qquad\bar{\sigma}=50 \mathrm{kPa}
Question 8 |
A partially-saturated soil sample has natural moisture content of 25% and bulk unit weight of 18.5 \mathrm{kN} / \mathrm{m}^{3}. The specific gravity of soil solids is 2.65 and unit weight of water is 9.81 \mathrm{kN} / \mathrm{m}^{3}. The unit weight of the soil sample on full saturation is
21.12 \mathrm{kN} / \mathrm{m}^{3} | |
19.03 \mathrm{kN} / \mathrm{m}^{3} | |
20.12 \mathrm{kN} / \mathrm{m}^{3} | |
18.50 \mathrm{kN} / \mathrm{m}^{3} |
Question 8 Explanation:
\begin{aligned} \mathrm{w} &=0.25, \gamma_{\mathrm{t}}=18.5 \mathrm{kN} / \mathrm{m}^{3} \\ \mathrm{G}_{\mathrm{s}} &=2.65, \gamma_{\mathrm{w}}=9.81 \\ \gamma_{\mathrm{t}} &=\frac{G_{\mathrm{S}} \gamma_{W}(1+w)}{1+e} \\ \Rightarrow \qquad \qquad \qquad \qquad \mathrm{e} &=\frac{2.65 \times 9.81 \times 1.25}{18.5}-1\\ \Rightarrow \qquad \qquad \qquad \qquad e&=0.756\\ \text{At full saturation}, \quad \mathrm{S}&=1\\ \Rightarrow \qquad \qquad \qquad \quad \gamma_{\mathrm{sat}}&=\frac{\left(G_{\mathrm{S}}+e\right) \gamma_{\mathrm{W}}}{1+e}\\ \gamma_{\text {sat }} &=\frac{(2.65+0.756) \times 9.81}{1.756} \\ &=19.03 \mathrm{kN} / \mathrm{m}^{3} \end{aligned}
Question 9 |
If water is flowing at the same depth in most hydraulically efficient triangular and rectangular channel sections then the ratio of hydraulic radius of triangular section to that of rectangular section is
\frac{1}{\sqrt{2}} | |
\sqrt{2} | |
1 | |
2 |
Question 9 Explanation:
Efficient channel section
\begin{aligned} A & =y^{2} & & A=2 y^{2} \\ P & =2 \sqrt{2} y & P & =4 y \\ R_{I} & =\frac{y}{2 \sqrt{2}} & R_{I I}&=\frac{y}{2} \\ \therefore \qquad \qquad \qquad \frac{R_{I}}{R_{I I}} & =\frac{1}{\sqrt{2}} & \end{aligned}
\begin{aligned} A & =y^{2} & & A=2 y^{2} \\ P & =2 \sqrt{2} y & P & =4 y \\ R_{I} & =\frac{y}{2 \sqrt{2}} & R_{I I}&=\frac{y}{2} \\ \therefore \qquad \qquad \qquad \frac{R_{I}}{R_{I I}} & =\frac{1}{\sqrt{2}} & \end{aligned}
Question 10 |
Kinematic viscosity' is dimensionally represented as
\frac{M}{LT} | |
\frac{M}{L^{2} T} | |
\frac{T^{2}}{L} | |
\frac{L^{2}}{T} |
Question 10 Explanation:
Kinematic viscosity
v=\frac{\mu }{\rho }=\frac{kg/m\cdot s}{kg/m^3}=m^2/s
[v]=\frac{m^2}{s }=\frac{L^2}{T}
v=\frac{\mu }{\rho }=\frac{kg/m\cdot s}{kg/m^3}=m^2/s
[v]=\frac{m^2}{s }=\frac{L^2}{T}
There are 10 questions to complete.