GATE ME 2003

Question 1
\lim_{x\rightarrow 0}\frac{\sin^{2}x}{x} is equal to
A
0
B
\infty
C
1
D
-1
Engineering Mathematics   Calculus
Question 1 Explanation: 
\lim_{x \to 0}frac{\sin^2 x}{x}
=\lim_{x \to 0}\left (frac{\sin x}{x} \right )^2 x
=1 \times 0 =0
Question 2
The accuracy of Simpson's rule quadrature for a step size h is
A
O(h^{2})
B
O(h^{3})
C
O(h^{4})
D
O(h^{5})
Engineering Mathematics   Numerical Methods
Question 3
For the matrix \begin{bmatrix} 4 & 1\\ 1 & 4 \end{bmatrix} the Eigen values are
A
3 and -3
B
-3 and -5
C
3 and 5
D
5 and 0
Engineering Mathematics   Linear Algebra
Question 3 Explanation: 
\small A=\left[\begin{array}{ll}4 & 1 \\ 1 & 4\end{array}\right]
\text{Now }\quad|A-\lambda I|=0
\text{where} \quad \lambda= \text{ eigen value}
\therefore \quad\left|\begin{array}{cc}4-\lambda & 1 \\ 1 & 4-\lambda \\ (4-\lambda)^{2}-1 & =0\end{array}\right|=0
\text{or,} \quad(4-\lambda)^{2}-(1)^{2}=0
\text{or,} (4-\lambda+1)(4-\lambda-1)=0
\text{or,} \quad(5-\lambda)(3-\lambda)=0
\therefore \quad \lambda=3,5
Question 4
The second moment of a circular area about the diameter is given by (D is the diameter).
A
\frac{\pi D^{4}}{4}
B
\frac{\pi D^{4}}{16}
C
\frac{\pi D^{4}}{32}
D
\frac{\pi D^{4}}{64}
Strength of Materials   Bending of Beams
Question 4 Explanation: 
Polar moment of inertia perpendicular to the plane of paper
I_{p}=\frac{\pi D^{4}}{32}


By 'Perpendicular Axis" theorem
\begin{aligned} I_{x x}+I_{y y} &=I_{p} \quad\left[\because I_{x x}=I_{y y}\right] \\ 2 I_{x x} &=\frac{\pi D^{4}}{32} \\ \therefore \quad I_{x x} &=I_{y y}=\frac{\pi D^{4}}{64} \end{aligned}
Question 5
A concentrated load of P acts on a simply supported beam of span L at a distance \frac{L}{3} form the left support. The bending moment at the point of application of the load is given by
A
\frac{PL}{3}
B
\frac{2PL}{3}
C
\frac{PL}{9}
D
\frac{2PL}{9}
Strength of Materials   Bending of Beams
Question 5 Explanation: 


Taking moment about A, we have
\begin{aligned} R_{B} \times L &=P \times \frac{L}{3} \\ \therefore \quad R_{B} &=\frac{P}{3}\\ \text{and }\quad R_{A}=P-R_{B}&=P-\frac{P}{3}\\ R_{A}&=\frac{2 P}{3} \end{aligned}
Bending moment at the point of application of load,
M=\frac{2 P}{3} \times \frac{L}{3}=\frac{2 P L}{9}
Question 6
Two identical circular rods of same diameter and same length are subjected to same magnitude of axial tensile force. One of the rods is made out of mild steel having the modulus of elasticity of 206 GPa. The other rod is made out of cast iron having the modulus of elasticity of 100 GPa. Assume both the materials to be homogeneous and isotropic and the axial force causes the same amount of uniform stress in both the roads. The stresses developed are within the proportional limit of the respective materials. Which of the following observation is correct?
A
Both rods elongate by the same amount
B
Mild steel rod elongates more than the cast iron rod
C
Cast iron rod elongates more than the mild steel rod
D
As the stresses are equal strains are also equal in both the rods
Strength of Materials   Stress and Strain
Question 6 Explanation: 
\begin{aligned} \text{Given,}\quad E_{S}&=206 \mathrm{GPa} \\ \text{and}\quad E_{I}&=100 \mathrm{GPa} \end{aligned}
Elongation in mild steel,
\Delta L_{s}=\frac{P L}{A E_{s}} \qquad\cdots(i)
Elongation in cast Iron,
\Delta L_{1}=\frac{P L}{A E_{I}} \qquad\cdots(ii)
Dividing Eq. (ii) by Eq. (i), we get
\frac{\Delta L_{s}}{\Delta L_{l}}=\frac{E_{I}}{E_{S}}
since diameter and length of both shafts are equal
\therefore \quad \frac{\Delta L_{S}}{\Delta L_{I}}=\frac{100}{206} \lt 1
Question 7
Two beams, one having square cross section and another circular cross-section, are subjected to the same amount of bending moment. If the cross sectional area as well as the material of both the beams are the same then
A
maximum bending stress developed in both the beams is the same
B
the circular beam experiences more bending stress than the square one
C
the square beam experiences more bending stress than the circular one
D
as the material is same both the beams will experience same deformation
Strength of Materials   Bending of Beams
Question 7 Explanation: 
Let, d= diameter of circular cross-section
a= side of square cross-section
since cross-sectional area of square and circular cross-section are equal
\therefore \quad \frac{\pi}{4} d^{2}=a^{2}\quad \cdots(i)
For circular cross-section,
\sigma_{c}=\frac{M y}{I}=\frac{M \cdot \frac{d}{2}}{\frac{\pi d^{4}}{64}}=\frac{32 M}{\pi d^{3}}
Square cross-section,
\begin{aligned} \sigma_{s} &=\frac{M \cdot y}{I} \\ &=\frac{M \frac{a}{2}}{\frac{a^{4}}{12}}=\frac{6 M}{a^{3}} \\ \text{Now}\quad \frac{\sigma_{c}}{\sigma_{s}} &=\frac{\frac{32 M}{\pi d^{3}}}{\frac{6 M}{a^{3}}}=\frac{32 a^{3}}{6 \pi d^{3}}\\ &=\frac{32}{6 \pi}\left(\frac{a}{d}\right)^{2}\left(\frac{a}{d}\right) \\ &=\frac{32}{6 \pi} \times\left(\frac{\pi}{4}\right) \times\left(\sqrt{\frac{\pi}{4}}\right) \\ &=\frac{32}{24} \sqrt{\frac{\pi}{4}}=\frac{32 \sqrt{\pi}}{48}>1 \\ \sigma_{c}&> \sigma_{s} \end{aligned}
Question 8
The mechanism used in a shaping machine is
A
a closed 4-bar chain having 4 revolute pairs
B
a closed 6-bar chain having 6 revolute pairs
C
a closed 4-bar chain having 2 revolute and 2 sliding pairs
D
an inversion of the single slider-crank chain
Theory of Machine   Planar Mechanisms
Question 8 Explanation: 
Crank and slotted lever mechanism is mostly used in shaping machines, which is the inversion of single slider-crank chain.
Question 9
The lengths of the links of a 4-bar linkage with revolute pairs only are p,q,r and s units. Given that p \lt q \lt r \lt s. Which of these links should be the fixed one, for obtaining a "double crank" mechanism?
A
link of length p
B
link of length q
C
link of length r
D
link of length s
Theory of Machine   Planar Mechanisms
Question 9 Explanation: 
Double crank mechanism is obtained when the shortest link is fixed. Here, the shortest link is p. Therefore link of length p should be fixed.
Question 10
Consider the arrangement shown in the figure below where J is the combined polar mass moment of inertia of the disc and the shafts. K_1,K_2,K_3 are the torsional stiffness of the respective shafts. The natural frequency of torsional oscillation of the disc is given by
A
\sqrt{\frac{{K_{1}+K_{2}+K_{3}}}{J}}
B
\sqrt{\frac{{K_{1}K_{2}+K_{2}K_{3}+K_{3}K_{1}}}{J(K_{1}+K_{2})}}
C
\sqrt{\frac{{K_{1}K_{2}K_{3}}}{J(K_{1}K_{2}+K_{2}K_{3}+K_{3}K_{1})}}
D
\sqrt{\frac{{K_{1}K_{2}+K_{2}K_{3}+K_{3}K_{1}}}{J(K_{2}+K_{3})}}
Theory of Machine   Vibration
Question 10 Explanation: 
Equivalent stiffness
k_{\mathrm{eq}}=\frac{k_{1} k_{2}}{k_{1}+k_{2}}+k_{3}=\frac{k_{1} k_{2}+k_{1} k_{3}+k_{2} k_{3}}{k_{1}+k_{2}}
Now natural frequency
=\sqrt{\frac{k_{e q}}{J}}=\sqrt{\frac{k_{1} k_{2}+k_{1} k_{3}+k_{2} k_{3}}{J\left(k_{1}+k_{2}\right)}}
There are 10 questions to complete.

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