# GATE ME 2013

 Question 1
The partial differential equation $\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}=\frac{\partial^2 u}{\partial x^2}$ is a
 A linear equation of order 2 B non-linear equation of order 1 C linear equation of order 1 D non-linear equation of order 2
Engineering Mathematics   Differential Equations
Question 1 Explanation:
A differential equation in the form $\frac{d y}{d x}+P y=Q$
where, P and Q are functions of x i.e., f(x) is said
to be linear equation.
$\frac{\partial u}{\partial t}+u \frac{\partial u}{\partial x}=\frac{\partial^{2} u}{\partial x^{2}}$
The given equation is not complying with the definition of linear equation, therefore it is a nonlinear equation of order 2.
 Question 2
The eigenvalues of a symmetric matrix are all
 A complex with non-zero positive imaginary part. B complex with non-zero negative imaginary part. C real. D pure imaginary.
Engineering Mathematics   Linear Algebra
Question 2 Explanation:
(i) The Eigen values of symmetric matrix
$\left[A^{T}=A\right]$ are purely real
(ii) The Eigen value of skew-symmetric matrix
$\left[A^{T}=-A\right]$ are either purely imaginary or zeros

 Question 3
Match the CORRECT pairs. A P-2, Q-1, R-3 B P-3, Q-2, R-1 C P-1, Q-2, R-3 D P-3, Q-1, R-2
Engineering Mathematics   Numerical Methods
 Question 4
A rod of length L having uniform cross-sectional area A is subjected to a tensile force P as shown in the figure below. If the Young's modulus of the material varies linearly from $E_{1}$ to $E_{2}$ along the length of the rod, the normal stress developed at the section-SS is A $\frac{P}{A}$ B $\frac{P(E_{1}-E_{2})}{A(E_{1}+E_{2})}$ C $\frac{PE_{2}}{AE_{1}}$ D $\frac{PE_{1}}{AE_{2}}$
Strength of Materials   Stress and Strain
Question 4 Explanation:
Normal stress at any section is independent of modulus of elasticity.
 Question 5
Two threaded bolts A and B of same material and length are subjected to identical tensile load. If the elastic strain energy stored in bolt A is 4 times that of bolt B and the mean diameter of bolt A is 12 mm, the mean diameter of bolt B in mm is
 A 16 B 24 C 36 D 48
Strength of Materials   Strain Energy and Thermal Stresses
Question 5 Explanation:
Given:
\begin{aligned} P_{1}&=P_{2}=P\\ &\text{(identical tensile load on bolt A \& B )} \\ &\qquad \text{(same length)}\\ L_{1} &=L_{2}=L \\ d_{A} &=12 \mathrm{mm} \\ U_{A} &=\text { strain energy in bolt } A \\ U_{B} &=\text { Strain energy in bolt } B \\ U_{A} &=4 U_{B}(\text { Given }) \qquad \cdots(i)\\ \therefore \quad d_{B} &=? \end{aligned}
Strain energy is given by
$U_{A}=\frac{1}{2} P \times \delta=\frac{1}{2} \frac{P^{2} L}{A E}$
$\therefore$ Eq. (i)
\begin{aligned} \frac{1 P_{1}^{2} L_{1}}{2} &=4 \times \frac{1}{2} \frac{P_{2}^{2} L_{2}}{A_{2} E} \\ \frac{1}{A_{1}} &=4 \times \frac{1}{A_{2}} \\ \frac{4}{\pi(12)^{2}} &=4 \times \frac{4}{\pi\left(d_{B}\right)^{2}} \\ d_{B}^{2} &=576 \\ d_{B} &=24 \mathrm{mm} \end{aligned}

There are 5 questions to complete.