Question 1 |
The condition for which the eigenvalues of the matrix
A=\begin{bmatrix} 2&1 \\ 1& k \end{bmatrix}
are positive, is
A=\begin{bmatrix} 2&1 \\ 1& k \end{bmatrix}
are positive, is
k\gt1/2 | |
k\gt-2 | |
k\gt0 | |
k\lt-1/2 |
Question 1 Explanation:
All Eigen values of A=\left[\begin{array}{ll}2 & 1 \\ 1 & k\end{array}\right] are positive
2>0
\therefore 2 \times 2 leading minor must be greater than zero
\left|\begin{array}{ll}2 & 1 \\ 1 & k\end{array}\right|>0
2 k-1>0
2 k>1
k>\frac{1}{2}
2>0
\therefore 2 \times 2 leading minor must be greater than zero
\left|\begin{array}{ll}2 & 1 \\ 1 & k\end{array}\right|>0
2 k-1>0
2 k>1
k>\frac{1}{2}
Question 2 |
The values of x for which the function
f( x)=\frac{x^{2}-3x-4}{x^{2}+3x-4}
is NOT continuous are
f( x)=\frac{x^{2}-3x-4}{x^{2}+3x-4}
is NOT continuous are
4 and -1 | |
4 and 1 | |
-4 and 1 | |
-4 and -1 |
Question 2 Explanation:
f(x)=\frac{x^{2}-3 x-4}{x^{2}+3 x-4} is not continous
when
\begin{aligned} x^{2}+3 x-4 &=0 \\ (x+4)(x-1) &=0 \\ x &=-4,1 \end{aligned}
when
\begin{aligned} x^{2}+3 x-4 &=0 \\ (x+4)(x-1) &=0 \\ x &=-4,1 \end{aligned}
Question 3 |
Laplace transform of cos(\omega t) is
\frac{s}{s^{2}+\omega ^{2}} | |
\frac{\omega }{s^{2}+\omega ^{2}} | |
\frac{s}{s^{2}\omega ^{2}} | |
\frac{\omega }{s^{2}-\omega ^{2}} |
Question 3 Explanation:
L(\cos \omega t)=\frac{s}{s^{2}+\omega^{2}}
Question 4 |
A function of the Complex Variables z=x+iy, is given as f(x,y)=u(x,y)+i v(x,y) , where u(x,y)=2kxy and v(x,y)=x^{2}-y^{2}. The value of k, for which the function is analytic, is _____
-5 | |
-1 | |
-8 | |
1 |
Question 4 Explanation:
Given that ( f(z)=u+i v) is analytic
\begin{aligned} u(x, y) &=2 k x y & & v=x^{2}-y^{2} \\ u_{x} &=2 k y & & v_{y}=-2 y \\ u_{x} &=v_{y} & & \\ k &=-1 & & v_{x}=2 x \\ u_{y} &=2 k x & & & \\ u_{y} &=-v_{x} & & & \\ 2 k x &=-2 x & & & \\ k &=-1 & & \end{aligned}
\begin{aligned} u(x, y) &=2 k x y & & v=x^{2}-y^{2} \\ u_{x} &=2 k y & & v_{y}=-2 y \\ u_{x} &=v_{y} & & \\ k &=-1 & & v_{x}=2 x \\ u_{y} &=2 k x & & & \\ u_{y} &=-v_{x} & & & \\ 2 k x &=-2 x & & & \\ k &=-1 & & \end{aligned}
Question 5 |
Numerical integration using trapezoidal rule gives the best result for a single variable function
linear | |
parabolic | |
logarithmic | |
hyperbolic |
Question 5 Explanation:
Trapezoidal rule gives the best result in single variable function when the function is linear (degree 1)
Question 6 |
A point mass having mass M is moving with a velocity V at an angle \theta to the wall as shown in the figure. The mass undergoes a perfectly elastic collision with the smooth wall and rebounds. The total change (final minus initial) in the momentum of the mass is
-2MV \cos \theta \hat{J} | |
2MV \sin \theta \hat{J} | |
2MV \cos \theta \hat{J} | |
-2MV \sin \theta \hat{J} |
Question 6 Explanation:
\begin{aligned} P_{1}&=M V \cos \theta \hat{i}+M V \sin \theta \hat{j} \\ P_{f}&=M V \cos \theta \hat{i}-M V \sin \theta \hat{j} \\ P_{f}-P_{i}&=-2 M V \sin \theta \hat{j} \end{aligned}
Question 7 |
A shaft with a circular cross-section is subjected to pure twisting moment. The ratio of the maximum shear stress to the largest principal stress is
2 | |
1 | |
0.5 | |
0 |
Question 7 Explanation:
For the case of pure shear
\begin{aligned} \sigma_{1} &=\tau_{\max } \\ \sigma_{2} &=-\tau_{\max } \\ \text { Required ratio } &=\tau_{\max } / \sigma_{1}=\sigma_{1} / \sigma_{1}=1 \end{aligned}
\begin{aligned} \sigma_{1} &=\tau_{\max } \\ \sigma_{2} &=-\tau_{\max } \\ \text { Required ratio } &=\tau_{\max } / \sigma_{1}=\sigma_{1} / \sigma_{1}=1 \end{aligned}
Question 8 |
A thin cylindrical pressure vessel with closed-ends is subjected to internal pressure. The ratio of circumferential (hoop) stress to the longitudinal stress is
0.25 | |
0.5 | |
1 | |
2 |
Question 8 Explanation:
For thin cylinder
Circumferential stress
\sigma_{h}=\frac{p d}{2 t}
Longitudinal stress,
\begin{aligned} \sigma_{L} &=\frac{p d}{4 t} \\ \therefore \quad \frac{\sigma_{h} }{\sigma_{L}} &=2 \end{aligned}
Circumferential stress
\sigma_{h}=\frac{p d}{2 t}
Longitudinal stress,
\begin{aligned} \sigma_{L} &=\frac{p d}{4 t} \\ \therefore \quad \frac{\sigma_{h} }{\sigma_{L}} &=2 \end{aligned}
Question 9 |
The forces F1 and F2 in a brake band and the direction of rotation of the drum are as shown in the figure. The coefficient of friction is 0.25. The angle of wrap is 3\pi /2 radians. It is given that R = 1 m and F2 = 1 N. The torque (in N-m) exerted on the drum is _________
2.24Nm | |
6.25Nm | |
8.9Nm | |
4.5Nm |
Question 9 Explanation:
\begin{aligned} F_{1} &>F_{2} \\ \frac{F_{1}}{F_{2}} &=e^{\mu \theta} \\ \frac{F_{1}}{F_{2}} &=e^{0.25 \times \frac{3 \pi}{2}} \\ F_{1} &=F_{2} \cdot e^{0.25 \times \frac{3 \pi}{2}} \\ &=1 \times e^{0.25 \times \frac{3 \times \pi}{2}} \\ &=3.2482 \mathrm{N} \\ \text { Torque } &=\left(F_{1}-F_{2}\right) \times r \\ &=(3.2482-1) \times 11 \\ &=2.2482 \mathrm{Nm} \end{aligned}
Question 10 |
A single degree of freedom mass-spring-viscous damper system with mass m, spring constant k and
viscous damping coefficient q is critically damped. The correct relation among m, k, and q is
q=\sqrt{2km} | |
q=2\sqrt{km} | |
q=\sqrt{\frac{2k}{m}} | |
q=2\sqrt{\frac{k}{m}} |
Question 10 Explanation:
Given :
1. single degree of freedom
2. Mass =m
3. Spring constant =k
4. Damping coefficient =q
5. Critically damped
Damping factor =1
Correct relation:
Damping factor =
\begin{array}{c}\ \frac{\text { Damping coefficient }}{2 \times \text { mass } \times \text { Natural frequency of vibration }} \\ 1=\frac{q}{2 \times m \times \sqrt{\frac{k}{m}}} \\ q=2 \sqrt{km} \end{array}
1. single degree of freedom
2. Mass =m
3. Spring constant =k
4. Damping coefficient =q
5. Critically damped
Damping factor =1
Correct relation:
Damping factor =
\begin{array}{c}\ \frac{\text { Damping coefficient }}{2 \times \text { mass } \times \text { Natural frequency of vibration }} \\ 1=\frac{q}{2 \times m \times \sqrt{\frac{k}{m}}} \\ q=2 \sqrt{km} \end{array}
There are 10 questions to complete.