Question 1 |
\lim_{x\rightarrow 0}\frac{\sin^{2}x}{x}
is equal to
0 | |
\infty | |
1 | |
-1 |
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
=\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
O(h^{2}) | |
O(h^{3}) | |
O(h^{4}) | |
O(h^{5}) |
Question 3 |
For the matrix \begin{bmatrix} 4 & 1\\ 1 & 4 \end{bmatrix}
the Eigen values are
3 and -3 | |
-3 and -5 | |
3 and 5 | |
5 and 0 |
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
\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).
\frac{\pi D^{4}}{4} | |
\frac{\pi D^{4}}{16} | |
\frac{\pi D^{4}}{32} | |
\frac{\pi D^{4}}{64} |
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}
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
\frac{PL}{3} | |
\frac{2PL}{3} | |
\frac{PL}{9} | |
\frac{2PL}{9} |
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.