# 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,
$M=\frac{2 P}{3} \times \frac{L}{3}=\frac{2 P L}{9}$

There are 5 questions to complete.