GATE ME 2016 SET-3

A is skew-symmetric
\therefore \quad A^{T}=-A

\lim_{x \rightarrow 0} \frac{\ln (1+4 x)}{e^{3 x}-1}
\lim_{x \rightarrow 0} \frac{\frac{1}{1+4 x} \cdot 4}{3 e^{3 x}}=\frac{4}{3}

Solution of laplace equation having continuous
Second order partial derivating
\therefore \nabla^{2} \phi =0
\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{\partial^{2} \phi}{\partial y^{2}} =0
\therefore \; \phi is harmonic function.

\begin{aligned} f(x)&=x^3+x-1\\ f(1)&=1+1-1=1 \\ f'(x)&=3x^2+1\\ f'(1)&=3+1=4\\ x_1&=x_0-\frac{f(x_0)}{f'(x_0)}\\ &=1-\frac{1}{4}=1-0.25=0.75 \end{aligned}

When we draw a free body diagram we remove all the supports and force applied due to that support are drawn and force or moment will apply in that manner so that it resists forces in any direction as well as any tendency of rotation.
Option (B) is wrong because the ground should not be shown in FBD as the forces due to ground are already depicted.
Option (C) is wrong because x-component is not shown.
Option (D) is wrong because the rotation due to the force acting eccentrically causes moment which is not shown.

\begin{aligned} \therefore \quad b^{2}&=\frac{\pi}{4} d^{2} \\ \Rightarrow \quad \frac{b^{2}}{d^{2}}&=\frac{\pi}{4} \\ I_{1} &=\frac{b^{4}}{12} \\ I_{2} &=\frac{\pi d^{4}}{64} \\ \therefore \qquad \frac{I_{1}}{I_{2}} &=\frac{b^{4}}{12} \times \frac{64}{\pi d^{4}}=\frac{16}{3 \pi} \frac{b^{4}}{d^{4}} \\ &=\frac{16}{3 \pi} \cdot \frac{\pi^{2}}{16}=\frac{\pi}{3} \end{aligned}

The given plane is principal plane
\left(\sigma_{1}=P, \sigma_{2}=-P\right) . At 45^{\circ} from principal plane, plane of max shear occurs. On the plane of max shear.
\begin{aligned} \operatorname{both}, \quad \tau_{x y} &=\tau_{y y}=\frac{\sigma_{1}+\sigma_{2}}{2}=0 \\ \tau_{x y} &=\frac{\sigma_{1}+\sigma_{2}}{2}=P \end{aligned}

To find relative velocity direction of V_{p} is reversed
and placed at fail of \vec{V}_{Q}

Ans. (D) Resultant V_{PQ} is perpendicular to link P Q.

Degrees of freedom is given by 3(n-1) - 2j