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}
Question 6
A link OB is rotating with a constant angular velocity of 2 rad/s in counter clockwise direction and a block is sliding radially outward on it with an uniform velocity of 0.75 m/s with respect to the rod, as shown in the figure below. If OA = 1 m, the magnitude of the absolute acceleration of the block at location A in m/s^{2} is

A
3
B
4
C
5
D
6
Theory of Machine   Displacement, Velocity and Acceleration
Question 6 Explanation: 
Acceleration of the block at A=\vec{a}_{c r}+\vec{a}_{r}
\begin{aligned} &=\overrightarrow{2 \omega V}+\overrightarrow{\omega^{2} r} \\ \Rightarrow \quad r&=O A\\ \Rightarrow \quad 2 \omega V&=2 \times 2 \times 0.75=3 \mathrm{m} / \mathrm{s}^{2}\\ \Rightarrow \quad \omega^{2} r&=2^{2} \times 1=4 \mathrm{m} / \mathrm{s}^{2}\\ \end{aligned}
As coriolis component and radial component are perpendicular to each other.
So, \left|\vec{a}_{a b s}\right|=\sqrt{3^{2}+4^{2}}=5 \mathrm{m} / \mathrm{s}
Question 7
For steady, fully developed flow inside a straight pipe of diameter D, neglecting gravity effects, the pressure drop \Delta p over a length L and the wall shear stress \tau _{w} are reated by
A
\tau _{w}=\frac{\Delta p D}{4 L}
B
\tau _{w}=\frac{\Delta p D^{2}}{4 L^{2}}
C
\tau _{w}=\frac{\Delta p D}{2 L}
D
\tau _{w}=\frac{4\Delta p L}{ D}
Fluid Mechanics   Flow Through Pipes
Question 7 Explanation: 
Shear stress on wall,
\begin{aligned} \tau_{w} &=-\frac{\partial p}{\partial x} \frac{R}{2} \\ \text { where } \quad-\frac{\partial p}{\partial x} &=\frac{\Delta p}{L} \\ \Delta p &=\text { Pressure drop }\\ \text{and }\quad R&=\frac{D}{2}\\ \therefore \tau_{w}&=\frac{\Delta p}{L} \times \frac{D}{2 \times 2}=\frac{\Delta p D}{4 L} \end{aligned}
Question 8
The pressure, dry bulb temperature and relative humidity of air in a room are 1 bar, 30^{\circ}C and 70%, respectively. If the saturated steam pressure at 30^{\circ}C is 4.25 kPa, the specific humidity of the room air in kg water vapour/kg dry air is
A
0.0083
B
0.0101
C
0.0191
D
0.0232
Refrigeration and Air-conditioning   Properties of Moist Air
Question 8 Explanation: 
\begin{aligned} p_{\mathrm{atn}} &=1 \mathrm{bar}=100 \mathrm{kPa} \\ \mathrm{DBT} &=30^{\circ} \mathrm{C} \\ \phi &=70 \%=0.7 \\ p_{\mathrm{vs}} &=4.25 \mathrm{kPa} \end{aligned}
Specific humidity: \omega=?
\begin{aligned} \phi &=\frac{p_{v}}{p_{v s}} \\ 0.7 &=\frac{p_{v}}{4.25} \\ p_{v} &=2.975 \mathrm{kPa} \end{aligned}
Specific humidity,
\begin{aligned} \omega&=0.622 \times \frac{p_{v}}{p-p_{v}}\\ &=0.622 \times \frac{2.975}{100-2.975} \\ &=00.0191 \end{aligned}
Question 9
Consider one-dimensional steady state heat conduction, without heat generation, in a plane wall; with boundary conditions as shown in the figure below. The conductivity of the wall is given by k=k_{0}+bT; where k_{0} and b are positive constants, and is T temperature.

As x increases, the temperature gradient ( dT/dx ) will
A
remain constant
B
be zero
C
increase
D
decrease
Heat Transfer   Conduction
Question 9 Explanation: 


Q=-k A \frac{d T}{d x} \ldots(i)
where \quad k=k_{0}+b T
As x increases, T increases, then k also increases.
i.e., k is +ve
Equation (i), indicates \frac{d T}{d x} decreases.
Question 10
In a rolling process, the state of stress of the material undergoing deformation is
A
pure compression
B
pure shear
C
compression and shear
D
tension and shear
Manufacturing Engineering   Forming Process
Question 10 Explanation: 
Most metal rolling operations are similar in that the work material is plastically deformed by compressive forces between two constantly spinning rolls. Thus in a Rolling process, the material undergoing deformation is in the state of pure biaxial compression

There are 10 questions to complete.

Leave a Comment

Like this FREE website? Please share it among all your friends and join the campaign of FREE Education to ALL.