# GATE ME 2016 SET-2

 Question 1
The condition for which the eigenvalues of the matrix
$A=\begin{bmatrix} 2&1 \\ 1& k \end{bmatrix}$
are positive, is
 A k$\gt$1/2 B k$\gt$-2 C k$\gt$0 D k$\lt$-1/2
Engineering Mathematics   Linear Algebra
Question 1 Explanation:
All Eigen values of $A=\left[\begin{array}{ll}2 & 1 \\ 1 & k\end{array}\right]$ are positive
$2>0$
$\therefore 2 \times 2$leading minor must be greater than zero
$\left|\begin{array}{ll}2 & 1 \\ 1 & k\end{array}\right|>0$
$2 k-1>0$
$2 k>1$
$k>\frac{1}{2}$
 Question 2
The values of x for which the function
$f( x)=\frac{x^{2}-3x-4}{x^{2}+3x-4}$
is NOT continuous are
 A 4 and -1 B 4 and 1 C -4 and 1 D -4 and -1
Engineering Mathematics   Calculus
Question 2 Explanation:
$f(x)=\frac{x^{2}-3 x-4}{x^{2}+3 x-4}$ is not continous
when
\begin{aligned} x^{2}+3 x-4 &=0 \\ (x+4)(x-1) &=0 \\ x &=-4,1 \end{aligned}
 Question 3
Laplace transform of $cos(\omega t)$ is
 A $\frac{s}{s^{2}+\omega ^{2}}$ B $\frac{\omega }{s^{2}+\omega ^{2}}$ C $\frac{s}{s^{2}\omega ^{2}}$ D $\frac{\omega }{s^{2}-\omega ^{2}}$
Engineering Mathematics   Calculus
Question 3 Explanation:
$L(\cos \omega t)=\frac{s}{s^{2}+\omega^{2}}$
 Question 4
A function of the Complex Variables z=x+iy, is given as f(x,y)=u(x,y)+i v(x,y) , where u(x,y)=2kxy and v(x,y)=$x^{2}-y^{2}$. The value of k, for which the function is analytic, is _____
 A -5 B -1 C -8 D 1
Engineering Mathematics   Complex Variables
Question 4 Explanation:
Given that $( f(z)=u+i v)$ is analytic
\begin{aligned} u(x, y) &=2 k x y & & v=x^{2}-y^{2} \\ u_{x} &=2 k y & & v_{y}=-2 y \\ u_{x} &=v_{y} & & \\ k &=-1 & & v_{x}=2 x \\ u_{y} &=2 k x & & & \\ u_{y} &=-v_{x} & & & \\ 2 k x &=-2 x & & & \\ k &=-1 & & \end{aligned}
 Question 5
Numerical integration using trapezoidal rule gives the best result for a single variable function
 A linear B parabolic C logarithmic D hyperbolic
Engineering Mathematics   Numerical Methods
Question 5 Explanation:
Trapezoidal rule gives the best result in single variable function when the function is linear (degree 1)
 Question 6
A point mass having mass M is moving with a velocity V at an angle $\theta$ to the wall as shown in the figure. The mass undergoes a perfectly elastic collision with the smooth wall and rebounds. The total change (final minus initial) in the momentum of the mass is A -2MV $\cos \theta \hat{J}$ B 2MV $\sin \theta \hat{J}$ C 2MV $\cos \theta \hat{J}$ D -2MV $\sin \theta \hat{J}$
Engineering Mechanics   Impulse and Momentum, Energy Formulations
Question 6 Explanation: \begin{aligned} P_{1}&=M V \cos \theta \hat{i}+M V \sin \theta \hat{j} \\ P_{f}&=M V \cos \theta \hat{i}-M V \sin \theta \hat{j} \\ P_{f}-P_{i}&=-2 M V \sin \theta \hat{j} \end{aligned}
 Question 7
A shaft with a circular cross-section is subjected to pure twisting moment. The ratio of the maximum shear stress to the largest principal stress is
 A 2 B 1 C 0.5 D 0
Strength of Materials   Stress-strain Relationship and Elastic Constants
Question 7 Explanation:
For the case of pure shear
\begin{aligned} \sigma_{1} &=\tau_{\max } \\ \sigma_{2} &=-\tau_{\max } \\ \text { Required ratio } &=\tau_{\max } / \sigma_{1}=\sigma_{1} / \sigma_{1}=1 \end{aligned}
 Question 8
A thin cylindrical pressure vessel with closed-ends is subjected to internal pressure. The ratio of circumferential (hoop) stress to the longitudinal stress is
 A 0.25 B 0.5 C 1 D 2
Strength of Materials   Thin Cylinder
Question 8 Explanation:
For thin cylinder
Circumferential stress
$\sigma_{h}=\frac{p d}{2 t}$
Longitudinal stress,
\begin{aligned} \sigma_{L} &=\frac{p d}{4 t} \\ \therefore \quad \frac{\sigma_{h} }{\sigma_{L}} &=2 \end{aligned}
 Question 9
The forces F1 and F2 in a brake band and the direction of rotation of the drum are as shown in the figure. The coefficient of friction is 0.25. The angle of wrap is $3\pi /2$ radians. It is given that R = 1 m and F2 = 1 N. The torque (in N-m) exerted on the drum is _________ A 2.24Nm B 6.25Nm C 8.9Nm D 4.5Nm
Machine Design   Brakes and Clutches
Question 9 Explanation: \begin{aligned} F_{1} &>F_{2} \\ \frac{F_{1}}{F_{2}} &=e^{\mu \theta} \\ \frac{F_{1}}{F_{2}} &=e^{0.25 \times \frac{3 \pi}{2}} \\ F_{1} &=F_{2} \cdot e^{0.25 \times \frac{3 \pi}{2}} \\ &=1 \times e^{0.25 \times \frac{3 \times \pi}{2}} \\ &=3.2482 \mathrm{N} \\ \text { Torque } &=\left(F_{1}-F_{2}\right) \times r \\ &=(3.2482-1) \times 11 \\ &=2.2482 \mathrm{Nm} \end{aligned}
 Question 10
A single degree of freedom mass-spring-viscous damper system with mass m, spring constant k and viscous damping coefficient q is critically damped. The correct relation among m, k, and q is
 A q=$\sqrt{2km}$ B q=2$\sqrt{km}$ C q=$\sqrt{\frac{2k}{m}}$ D q=2$\sqrt{\frac{k}{m}}$
Theory of Machine   Vibration
Question 10 Explanation:
Given :
1. single degree of freedom
2. Mass =m
3. Spring constant =k
4. Damping coefficient =q
5. Critically damped
Damping factor =1 Correct relation:
Damping factor =
$\begin{array}{c}\ \frac{\text { Damping coefficient }}{2 \times \text { mass } \times \text { Natural frequency of vibration }} \\ 1=\frac{q}{2 \times m \times \sqrt{\frac{k}{m}}} \\ q=2 \sqrt{km} \end{array}$
There are 10 questions to complete.