GATE ME 2008

Question 1
In the Taylor series expansion of e^{x} about x=2, the coefficient of (x-2)^4 is
A
\frac{1}{4!}
B
\frac{2^4}{4!}
C
\frac{e^2}{4!}
D
\frac{e^4}{4!}
Engineering Mathematics   Calculus
Question 1 Explanation: 
f(x)\text{in the neighbourhood of a}
f(x)=\Sigma b_{n}(x-a)^{n} \text{where n from 0 to 0}
\begin{aligned}\quad\quad\quad\quad b_{n}&=\frac{f^{n}(a)}{n !} \\ f^{4}(x)&=e^{x} ; f^{4}(2)=e^{2}\\ \end{aligned}
\therefore \text{Coefficient of} (x-2)^{4}=b_{4}=\frac{f^{4}(2)}{4 !}=\frac{e^{2}}{4 !}
Question 2
Given that \ddot{x}+3x=0, and x(0) = 1, \dot{x}(0)=0, what is x(1)?
A
-0.99
B
-0.16
C
0.16
D
0.99
Engineering Mathematics   Differential Equations
Question 2 Explanation: 
\ddot{x}+3 x=0
Auxiliary equation is
m^{2}+3=0
i.e. \qquad m=\pm \sqrt{3}
\therefore \qquad x =A \cos \sqrt{3} t+B \sin \sqrt{3} t
\dot{x} =\sqrt{3}(B \cos \sqrt{3} t-A \sin \sqrt{3} t)
\text{At} \quad t=0
\qquad 1=A \qquad ...(i)
\qquad 0=B \qquad ...(ii)
\therefore \qquad x =\cos \sqrt{3} t
x(1) =\cos (\sqrt{3} \times 57.3)=-0.16
Question 3
The value of \lim_{x\rightarrow 8}\frac{x^{1/3}-2}{(x-8)}
A
1/16
B
1/12
C
1/8
D
1/4
Engineering Mathematics   Calculus
Question 3 Explanation: 
\begin{aligned} \quad \quad(x-8) &=h(\text { say }) \\ \Rightarrow \quad x &=8+h \end{aligned}
\; \therefore \lim_{h \rightarrow 0} \frac{(8+h)^{1 / 3}-2}{h}
\text{Above form in the} \left(\frac{0}{0}\right) \text{by putting the value h=0 }
\text{Applying L hospital rule}
\lim _{h \rightarrow 0} \frac{\frac{1}{3}(8+h)\left(\frac{1}{3}-1\right)}{1}=\frac{1}{3}(8)^{-\frac{2}{3}}=\frac{1}{12}
Question 4
A coin is tossed 4 times. What is the probability of getting heads exactly 3 times?
A
1/4
B
1/2
C
3/8
D
3/4
Engineering Mathematics   Probability and Statistics
Question 4 Explanation: 
P(H)=0.5
Probability of getting head exactly 3 times is
^{4} C_{3}(0.5)^{3}(0.5)=\frac{1}{4}
Question 5
The matrix \begin{bmatrix} 1 &2 & 4\\ 3 & 0 &6 \\ 1& 1 & p \end{bmatrix} has one Eigenvalue equal to 3. The sum of the other two Eigenvalues is
A
p
B
p-1
C
p-2
D
p-3
Engineering Mathematics   Linear Algebra
Question 5 Explanation: 
Sum of the eigen values of matrix is
= sum of diagonal values present in the matrix
\therefore 1+0+p=3+\lambda_{2}+\lambda_{3} \\ \Rightarrow p+1=3+\lambda_{2}+\lambda_{3} \\ \Rightarrow \lambda_{2}+\lambda_{3}=p+1-3=p-2
Question 6
The divergence of the vector field (x-y)\hat{i}+(y-x)\hat{j}+(x+y+z)\hat{k} is
A
0
B
1
C
2
D
3
Engineering Mathematics   Calculus
Question 6 Explanation: 
\begin{aligned} &\operatorname{div}\{(x-y) \hat{i}+(y-x) \hat{j}+(x+y+z) \hat{k}\}\\ &=\frac{\partial}{\partial x}(x-y)+\frac{\partial}{\partial y}(y-x)+\frac{\partial}{\partial z}(x+y+z)=3 \end{aligned}
Question 7
The transverse shear stress acting in a beam of rectangular cross-section, subjected to a transverse shear load, is
A
Variable with maximum at the bottom of the beam
B
Variable with maximum at the top of the beam
C
Uniform
D
Variable with maximum at the neutral axis
Strength of Materials   Bending of Beams
Question 7 Explanation: 


For a rectangle cross-section:
\tau _v=\frac{FA\bar{Y}}{Ib}=\frac{6F}{bd^3}\left ( \frac{d^2}{4}-y^2 \right )F=\text{Transverse shear load}
Maximum values of \tau _v occurs at the neutral axis where, y=0
\text{Maximum }\tau _v=\frac{6F}{bd^3}\times \frac{d^2}{4}=\frac{3F}{2bd}=\frac{3}{2}T_{mean}
T_{mean}=\frac{F}{bd}

So, transverse shear stress is variable with maximum on the neutral axis.
Question 8
A rod of Length L and diameter D is subjected to a tensile load P. Which of the following is sufficient to calculate the resulting change in diameter?
A
Young's modulus
B
Shear modulus
C
Poisson's ratio
D
Both Young's modulus and shear modulus
Strength of Materials   Stress and Strain
Question 8 Explanation: 


From the application of load P, the length of the rod increases by an amount of \Delta L

\Delta L=\frac{PL}{AE}=\frac{PL}{\frac{\pi}{4}D^2E}=\frac{4PL}{\pi D^2E}

And increase in length due to applied load P in axial or longitudinal direction, the shear modulus is comes in action.

G=\frac{\text{Shearing stress}}{\text{Shearing strain}}=\frac{\tau _s}{\Delta L/L}=\frac{\tau _sL}{\Delta L}

So, for calculating the resulting change in diameter both young's modulus and shear modulus are used.
Question 9
A straight rod of length L(t), hinged at one end and freely extensible at the other end, rotates through an angle \theta (t) about the hinge. At time t, L(t)=1m, L(t)=1m/s, \theta (t)=\frac{\pi}{4}rad and \theta (t)=1 rad/s. The magnitude of the velocity at the other end of the rod is
A
1m/s
B
\sqrt{2}m/s
C
\sqrt{3}m/s
D
2m/s
Engineering Mechanics   Plane Motion
Question 9 Explanation: 


Velocity has two components


\begin{aligned} \bar{v} &=\sqrt{(1)^{2}+(1)^{2}} \\ &=\sqrt{2} \mathrm{m} / \mathrm{s} \end{aligned}
Question 10
A cantilever type gate hinged at Q is shown in the figure. P and R are the centers of gravity of the cantilever part and the counterweight respectively. The mass of the cantilever part is 75 kg. The mass of the counterweight, for static balance, is
A
75kg
B
150kg
C
225kg
D
300kg
Theory of Machine   Balancing
Question 10 Explanation: 
Let counterweight at R is m kg

For static condition
\begin{aligned} \Sigma M_{Q} &=0 \\ m \times 0.5 &=75 \times 2 \\ m &=300 \mathrm{kg} \end{aligned}
There are 10 questions to complete.

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