# 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}
 Question 6
Two mating spur gears have 40 and 120 teeth respectively. The pinion rotates at 1200 rpm and transmits a torque of 20 N.m. The torque transmitted by the gear is
 A 6.6 Nm B 20 Nm C 40 Nm D 60 Nm
Machine Design   Gears
Question 6 Explanation:
Power transmitted,
$P=\frac{2 \pi N_{G} T_{G}}{60}=\frac{2 \pi N_{P} T_{P}}{60}$

\begin{aligned} \Rightarrow \quad N_{G} T_{G}&=N_{p} T_{P} \;\; \text{(where T is torque)} \\ and, \quad \frac{N_{p}}{N_{G}}&=\frac{Z_{G}}{Z_{p}} \\ &\text{(where Z is number of teeths)}\\ \Rightarrow \quad \frac{1200}{N_{G}}&=\frac{120}{40} \\ N_{G}&=400 \mathrm{r} . \mathrm{p.m} \\ \text{Hence,}\\ 400 \times T_{G}&=1200 \times 20 \\ T_{G}&=60 \mathrm{N} . \mathrm{m} \end{aligned}
 Question 7
The figure shows the state of stress at a certain point in a stressed body. The magnitudes of normal stresses in the x and y direction are 100 MPa and 20 MPA respectively. The radius of Mohr's stress circle representing this state of stress is
 A 120 B 80 C 60 D 40
Strength of Materials   Mohr's Circle
Question 7 Explanation:
\begin{aligned} \text { Radius } &=\sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right)^{2}+\tau_{x y}} \\ \text{Given,} \quad \sigma_{x} &=100 \mathrm{MPa} \\ \sigma_{y}&=-20 \mathrm{MPa} \\ \tau_{x y} &=0 \\ \therefore \text { Radius } &=\sqrt{\left(\frac{100+20}{2}\right)^{2}+0} \\ &=60 \mathrm{MPa} \end{aligned}
 Question 8
For a mechanism shown below, the mechanical advantage for the given configuration is
 A 0 B 0.5 C 1 D $\infty$
Theory of Machine   Dynamic Analysis of Slider-crank
Question 8 Explanation:
$=\frac{T_{\text {output }}}{T_{\text {input }}}=\frac{\omega_{\text {input }}}{\omega_{\text {output }}}$
since output link is a slider
Hence,$\omega_{\text {output }}=0$
$\therefore$ Mechanical advantage $=\infty$
 Question 9
A vibrating machine is isolated from the floor using springs. If the ratio of excitation frequency of vibration of machine to the natural frequency of the isolation system is equal to 0.5, the transmissibility of ratio of isolation is
 A $\frac{1}{2}$ B $\frac{3}{4}$ C $\frac{4}{3}$ D 2
Theory of Machine   Vibration
Question 9 Explanation:
Given $\quad r=0.5$
Due to isolation
$\xi_{S}=0$
Transmissibility ratio
\begin{aligned} T R &=\frac{\sqrt{1+(2 \xi r)^{2}}}{\sqrt{\left(1-r^{2}\right)^{2}+(2 \xi r)^{2}}} \\ &=\frac{1}{\sqrt{\left(1-(0.5)^{2}\right)^{2}}} \\ &=\frac{1}{1-0.25}=\frac{4}{3} \end{aligned}
 Question 10
A torque of 10 Nm is transmitted through a stepped shaft as shown in figure. The torsional stiffnesses of individual sections of lengths MN, NO and OP are 20 Nm/rad, 30 Nm/rad and 60 Nm/rad respectively. The angular deflection between the ends M and P of the shaft is
\begin{aligned} \theta_{M N}&=\frac{10}{20}=\frac{10}{20} \text { radian } \\ \theta_{N O}&=\frac{10}{30}=\frac{10}{30} \text { radian } \\ \theta_{O P}&=\frac{10}{60}= \frac{10}{60} \text { radian } \end{aligned}
$\theta_{M P}=\theta_{M N}+\theta_{N O}+\theta_{O P}=1 \text { radian }$