# GATE ME 2004

 Question 1
If $x=a(\theta +\sin \theta )$ and $y=a(1-\cos \theta )$, then $\frac{\mathrm{d} y}{\mathrm{d} x}$ will be equal to
 A $\sin \left ( \frac{\theta }{2} \right )$ B $\cos \left ( \frac{\theta }{2} \right )$ C $\tan \left ( \frac{\theta }{2} \right )$ D $\cot \left ( \frac{\theta }{2} \right )$
Engineering Mathematics   Calculus
Question 1 Explanation:
\begin{array}{l} Given, \quad \begin{array}{l} x=a(\theta+\sin \theta) \\ y=a(1-\cos \theta) \end{array} \\ \begin{aligned} \therefore \quad \frac{d x}{d \theta} &=a(1+\cos \theta), \frac{d y}{d \theta}=a \sin \theta \\ \frac{d y}{d x} &=\frac{d y / d \theta}{d x / d \theta}=\frac{a \sin \theta}{a(1+\cos \theta)} \\ &=\frac{2 \operatorname{asin}\left(\frac{\theta}{2}\right) \cdot \cos \left(\frac{\theta}{2}\right)}{a \times 2 \cos ^{2}\left(\frac{\theta}{2}\right)} \\ &=\tan \theta / 2 \end{aligned} \end{array}
 Question 2
The angle between two unit-magnitude coplanar vectors P(0.86, 0.500,0) and Q(0.259, 0.956, 0) will be
 A $0^{\circ}$ B $30^{\circ}$ C $45^{\circ}$ D $60^{\circ}$
Engineering Mathematics   Calculus
Question 2 Explanation:
\begin{aligned} \vec{P}=0.866 \hat{i}+0.500 \hat{j}+0 \hat{k} &\\ \vec{Q}=0.259 \hat{i}+0.966 \hat{j}+0 \hat{k} \end{aligned}
$\therefore \vec{P} \cdot \vec{Q}=|\vec{P}||\vec{Q}| \cos \theta$
Or, $(0.866 \hat{i}+0.5 \hat{j}+0 \hat{k}) .(0.259 \hat{i}+0.966 \hat{j}+0 \hat{k})$
Or, \begin{aligned}=& \sqrt{(0.866)^{2}+(0.5)^{2}} \\ & \times \sqrt{(0.259)^{2}+(0.966)^{2}} \cdot \cos \theta \\ \therefore \cos \theta=& \frac{0.866 \times 0.259+0.5 \times 0.966}{\sqrt{0.99} \times \sqrt{1.001}} \\ \cos \theta &= 0.707 \\ \theta& = 45\end{aligned}

 Question 3
The sum of the eigen values of the matrix given below is $\begin{pmatrix} 1 & 1 & 3\\ 1 & 5 & 1\\ 3 & 1 & 1 \end{pmatrix}$
 A 5 B 7 C 9 D 18
Engineering Mathematics   Linear Algebra
Question 3 Explanation:
Sum of eigen values of given matrix
= sum of diagonal element of given matrix
=1+5+1=7
 Question 4
The figure shows a pin-jointed plane truss loaded at the point M by hanging a mass of 100 kg. The member LN of the truss is subjected to a load of
 A 0 Newton B 490 Newtons in compression C 981 Newtons in compression D 981 Newtons in tension
Engineering Mechanics   FBD, Equilirbium, Plane Trusses and Virtual work
Question 4 Explanation:

$\Sigma F_H=0$ and $\Sigma F_V=0$
At joint "L"
$\therefore \;\; F_{LK}-F_{LM}=0\;\;(\Sigma F_H=0)$
$F_{LN}=0\;\;\;(\Sigma F_V=0)$
Hence no force is acting on the truss number LN.
 Question 5
In terms of Poission's ratio (v) the ratio of Young's Modulus (E) to Shear Modulus (G) of elastic materials is
 A 2(1 + v) B 2(1 - v) C $\frac{1}{2}(1+v)$ D $\frac{1}{2}(1-v)$
Strength of Materials   Stress-strain Relationship and Elastic Constants
Question 5 Explanation:
We know that
\begin{aligned} E &=2 G(1+v) \\ \text{or}\quad \frac{E}{G} &=2(1+v) \end{aligned}

There are 5 questions to complete.