GATE ME 2014 SET-1

Let D=-12 for the given matrix
|A|=\left|\begin{array}{ccc}2 & 6 & 0 \\4 & 12 & 8 \\-2 & 0 & 4\end{array}\right|=(2)^{3}\left|\begin{array}{ccc}1 & 3 & 0 \\2 & 6 & 4 \\-1 & 0 & 2\end{array}\right|
(Taking 2 common from each row)
\begin{aligned}\therefore \text{Det}(A) &=(2)^{3} \times D \\=& 8 x-12=-96\end{aligned}

\lim_{x \rightarrow 0} \frac{x-\sin x}{1-\cos x}
Applying L ' Hospital's rule
\lim_{x \rightarrow 0} \frac{\frac{d}{dx}(x-sinx)}{\frac{d}{dx}(1-cosx)}=\lim_{x \rightarrow 0} \frac{(1-\cos x)}{\sin x}
(It is still of \frac{0}{0} form)
Again applying L' Hospital's rule
\lim_{x \rightarrow 0} \frac{\frac{d}{d x}(1-\cos x)}{\frac{d}{d x}(\sin x)}=\lim_{x \rightarrow 0} \frac{\sin x}{\cos x}=\frac{0}{1}=0

\begin{aligned} Let z&=\frac{1+i}{1-i} \\ or z&=\frac{(1+i)(1+i)}{(1-i)(1+i)}=\frac{1+i^{2}+2 i}{1-i^{2}}=\frac{2 i}{2}=i \\ z&=x+i y=i \\ SO, x&=0 \\ y &=1 \end{aligned}
\begin{aligned} Arg(z) &=\tan ^{-1}\left(\frac{y}{x}\right) \\ &=\tan ^{-1}\left(\frac{1}{0}\right)=\tan ^{-1} \infty=\frac{\pi}{2} \end{aligned}

\begin{aligned} \frac{d x}{d t} &=3 x-5 y \\ \frac{d y}{d t} &=4 x+8 y \\ \frac{d}{d t}\left\{\begin{array}{l}x \\y\end{array}\right\}&=\left[\begin{array}{cc}3 & -5 \\4 & 8\end{array}\right]\left\{\begin{array}{l}x \\y\end{array}\right\}=\left[\begin{array}{cc} 3 x & -5 y \\4 x & +8 y\end{array}\right]\end{aligned}

For linear dependency, det \left|\begin{array}{ccc}1 & 1 & 1 \\ 2 & 3 & 1 \\ 5 & 6 & 4\end{array}\right| must be zero.
\therefore \Delta=1(12-6)-1(8-5)+1(12-15)
=6-3-3=0
\therefore There three vectors are linearly dependent.

For thermal stress to be developed, there must be constraint in the system. So, although strain develops but there is no thermal stress.

\begin{array}{l} v=-\frac{\text { Lateral strain }}{\text { Longitudinal strain }}=-\frac{\left(\frac{\Delta D}{D}\right)}{\left(\frac{\Delta L}{L}\right)} \\ v=\left(\frac{L}{D}\right)\left(-\frac{\Delta D}{\Delta L}\right) \\ \text{given,}L=500\mathrm{mm},D=50\mathrm{mm}\\ \Delta D=-0.015, \Delta L=+0.5 \mathrm{mm} \\ \qquad v=\frac{500}{50} \times \frac{0.015}{0.500}=0.3 \end{array}

The system with critical damping returns to equilibrium in fastest time without any oscillation. In critically damped free vibrations, the damping force is just sufficient to dissipate the energy within One cycle of motion. The system never executes a cycle, it approaches equilibrium with exponentially decaying displacement.

\begin{aligned} V_{P}= & \omega \times \overline{O P}=2 \omega r \\ V_{C}= & \omega \times \overline{O C}=\omega r \\ V_{O}= & \omega \times 0=0 \end{aligned}

Example of Lower Pair:
Revolute pair, Prismatic joint, Screw pair, cylindrical joint, spherical joint.
Example of higher pair:
Wheel rolling an surface, cam and follower contact, meshing teeth of two gears.