GATE CE 2018 SET-1

Option (A): \left | A \right |=6-5=1
Option (B): \left | A \right |=9-4=5
Option (C): \left | A \right |=12-12=0
Option (D): \left | A \right |=8-18=-10
Hence matrix (C) is singular.

\begin{aligned} \left | Q \right |&=\frac{3}{7}\left ( -\frac{9}{49}-\frac{12}{49} \right )-\frac{2}{7}\left ( \frac{18}{49}-\frac{4}{49} \right )+\frac{6}{7}\left ( \frac{-36}{49}-\frac{6}{49} \right ) \\ &=-1 \\ Adj. \; Q&=\begin{bmatrix} -\frac{21}{49} & \frac{42}{49} &-\frac{14}{49} \\ -\frac{14}{49}& -\frac{21}{49} & -\frac{42}{49}\\ -\frac{42}{49} & -\frac{14}{49} & \frac{21}{42} \end{bmatrix} \\ \therefore\;\; Q^{-1}&=\frac{Adj\: Q}{\left | Q \right |}=\begin{bmatrix} \frac{3}{7} & -\frac{6}{7} & \frac{2}{7}\\ \frac{2}{7} & \frac{3}{7} & \frac{6}{7}\\ \frac{6}{7} & \frac{2}{7} & -\frac{3}{7} \end{bmatrix} \end{aligned}
Or \because Q is orthogonal
\therefore \;\; Q^{-1}=Q^{T}

f\left ( x \right )=x^{3} at x=0



At x=0, the function y=x^{3} has neither minima nor maxima.

\begin{aligned} \text { For column, } & P=\frac{\pi^{2} E I_{\min }}{L^{2}} \\ &=\frac{\pi^{2} E\left(\frac{2 a \times a^{3}}{12}\right)}{h^{2}}=\frac{\pi^{2} E a^{4}}{6 h^{2}}\\ \text{For new column, }P&=\frac{\pi^{2} E\left[\frac{3 a \times(0.5 a)^{3}}{12}\right]}{(1.5 h)^{2}} \\ &=\frac{1}{12} \times \frac{\pi^{2} E a^{4}}{6 h^{2}}=\frac{P}{12} \end{aligned}

\begin{aligned} &\text{Area required for base plate}\\ &=\frac{\text{Factored load}}{\text{Bearing capacity of concrete}}\\ &=\frac{2000\times 10^{3}}{0.45\times 20}=222222.222mm^{2} \end{aligned}
So, side of base plate =\sqrt{\text{Area}}=471.4 mm= 47.14 cm
Since, provided area must be more than required
So, answer should be 48 cm.

Le Chatelier Apparatus is used to detemine the soundness of cement as per IS code4031 (part 3); this cement testing procedure is called Le Chatelier test for determining the unsoundness properties of cement due to presence of "free lime".

Creep is inelastic deformation with time due to sustained loading

The only non-zero stresses are \tau_{\theta z}=\tau_{z \theta}=\tau
if \theta is 90^{\circ} then \theta=y
\begin{aligned} \text{Hence}\quad\tau_{z y} &=\tau_{y z}=\tau_{\max } \\ &=16 T / \pi \mathrm{d}^{3}=0.815 \mathrm{MPa} \end{aligned}
But in rest of the planes shear stresses are zero, hence, \tau_{x y}=\tau_{y x}=0

Due to presence of more compression steel in section Y, NA of section of Y is above than as of X. It means Y is more under-reinforced than X so ductility of Y is more.
Since compression steel of Y is more so flexure resistance of X is less than as of Y.

For strip footing on sand (c=0)
q_{u}=\gamma D_{f} N_{q}+0.5 \mathrm{B} \gamma N_{\gamma}
In flooding condition water level rises to base of footing hence IIIrd term unit weight of soil will change and IInd term unit weight will be unaffected.
\begin{array}{ll} \therefore & q_{u}=\gamma D_{f} N_{q}+0.5 B \gamma N_{\gamma} \\ \because & \gamma \simeq \frac{1}{2} \gamma_{s a t} \end{array}
Hence third term reduced and second term will be same thereby percentage reduction will not be 50%.
According to option approach answer should be 25%
Note: If water table rises to ground level then both \gamma will reduce to \gamma. Hence, percentage reduction would be approximately 50%.