GATE ME 2006

Question 1
Match the items in columns I and II.
A
P-1 Q-4 R-3 S-2
B
P-1 Q-4 R-2 S-3
C
P-1 Q-3 R-2 S-4
D
P-4 Q-1 R-2 S-3
   
Question 2
The solution of the differential equation \frac{\mathrm{d} y}{\mathrm{d} x}+ 2xy = e^{-x^{2}} with y(0)=1 is:
A
(1+x) e^{+x^{2}}
B
(1+x) e^{-x^{2}}
C
(1-x) e^{+x^{2}}
D
(1-x) e^{-x^{2}}
Engineering Mathematics   Differential Equations
Question 2 Explanation: 
Given equation
\frac{d y}{d x}+2 x y=\sigma^{x^{2}}
Integrating factor
\begin{aligned} \text { I.F. } &=\mathrm{e}^{\int 2 x \;dx}=e^{x^{2}} \\ \text { Solution is } y x^{e^{x^{2}}}&=\int e^{-x^{2}} \cdot e^{x^{2}} d x \\ \therefore \quad y x^{ e^{x^{2}}}&=x+c\\ d t x &=0 \\ y &=1\\ \therefore \; \; c&=1 \\ y&=(1+x) e^{-x^{2}} \end{aligned}
Question 3
Let x denote a real number. Find out the INCORRECT statement.
A
S = \left \{ x : x \gt 3 \right \} represents the set of all real numbers greater than 3
B
S=\left \{ x : x^{2} \lt 0 \right \} represents the empty set
C
S=\left \{ x : x \in A \; and \; x \in B \right \} represents the union of set A and set B
D
S = \left \{ x : a \lt x \lt b \right \} represents the set of all real numbers between a and b, where a and b are real numbers.
Engineering Mathematics   Linear Algebra
Question 3 Explanation: 
The incorrect statement is, S=\left \{ x : x \in A \; and \; x \in B \right \} represents the union of set A and set B
As and represents the intersection not union.
Question 4
A box contains 20 defective items and 80 non-defective items. If two items are selected at random without replacement, what will be the probability that both items are defective.?
A
\frac{1}{5}
B
\frac{1}{25}
C
\frac{20}{99}
D
\frac{19}{495}
Engineering Mathematics   Probability and Statistics
Question 4 Explanation: 
Probability of first item being defective is
P_{1}=\frac{20}{100}
Probability that second item being defective is
P_{2}=\frac{19}{99}
Probability that both are defective
P=P_{1} P_{2}=\frac{20}{100} \times \frac{19}{99}=\frac{19}{495}
Question 5
For a circular shaft of diameter d subjected to torque T, the maximum value of the shear stress is:
A
\frac{64T}{\pi d^{3}}
B
\frac{32T}{\pi d^{3}}
C
\frac{16T}{\pi d^{3}}
D
\frac{8T}{\pi d^{3}}
Strength of Materials   Torsion of Shafts
Question 5 Explanation: 
\begin{aligned} \frac{T}{J} &=\frac{\tau}{d / 2} \\ \Rightarrow \qquad \frac{T}{\frac{\pi}{32} d^{4}} &=\frac{\tau}{d / 2} \\ \therefore \qquad & \tau=\frac{16 T}{\pi d^{3}} \end{aligned}
Question 6
For a four-bar linkage in toggle position, the value of mechanical advantage is:
A
0
B
0.5
C
1
D
\infty
Theory of Machine   Dynamic Analysis of Slider-crank
Question 6 Explanation: 


\omega_{4} of the output link DC becomes zero at the extreme positions. The extreme positions of the linkage are known as "Toggle position".
\therefore Mechanical advantage =\frac{\omega_{\text {input }}}{\omega_{\text {output }}}
\because \quad \omega_{\text {output }}=0
\therefore Mechanical advantage =\infty
Question 7
The differential equation governing the vibrating system is:
A
m\ddot{x} + c\dot{x} + k(x-y) = 0
B
m(\ddot{x}-\ddot{y}) + c(\dot{x}-\dot{y}) + kx = 0
C
m \ddot{x} + c(\dot{x}-\dot{y}) + kx = 0
D
m(\ddot{x}-\ddot{y}) + c(\dot{x}-\dot{y}) + k(x-y) = 0
Theory of Machine   Vibration
Question 7 Explanation: 


Differential equation governing the above vibration
system is given by
\begin{array}{l} \Rightarrow \frac{m d^{2} x}{d t^{2}}+C\left(\frac{d x}{d t}-\frac{d y}{d t}\right)+k x=0 \\ \Rightarrow m \ddot{x}+c(\dot{x}-\dot{y})+k x=0 \end{array}
Question 8
A pin-ended column of length L, modulus of elasticity E and second moment of the cross-sectional area I is loaded centrically by a compressive load P. the critical buckling load (P_{cr}) is given by
A
P_{cr}= \frac{EI}{\pi ^{2}L^{2}}
B
P_{cr}= \frac{\pi ^{2} EI}{3L^{2}}
C
P_{cr}= \frac{\pi EI}{L^{2}}
D
P_{cr}= \frac{\pi^{2} EI}{L^{2}}
Strength of Materials   Euler's Theory of Column
Question 8 Explanation: 
P_{E}=\frac{n \pi^{2} E I}{L^{2}}
For pin-ended column
n=1
Here n= End fixity coefficient
E = Modulus of elasticity
I = Second moment of area
L = Actual length of column
Question 9
The number of inversions for a slider crank mechanism is:
A
6
B
5
C
4
D
3
Theory of Machine   Planar Mechanisms
Question 9 Explanation: 
No. of links of a slider crank mechanism =4
So there are four inversion of slider crank mechanism.
Question 10
For a Newtonian fluid
A
shear stress is proportional to shear strain
B
rate of shear stress is proportional to shear strain
C
shear stress is proportional to rate of shear strain
D
rate of shear stress is proportional to rate of shear strain
Fluid Mechanics   Fluid Properties
Question 10 Explanation: 
For a Newtonian fluid
\tau \propto \frac{d u}{d y}
\tau=\mu \frac{d u}{d y}
where \quad \tau= shear stress
\frac{d u}{d y}= rate of shear strain
There are 10 questions to complete.

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