Question 1 |
Given that the determinant of the matrix
\begin{bmatrix} 1 & 3 & 0\\ 2 & 6 & 4\\ -1 & 0 & 2 \end{bmatrix}
is -12, the determinant of the matrix
\begin{bmatrix} 2 & 6 & 0\\ 4 & 12 & 8\\ -2 & 0 & 4 \end{bmatrix}
\begin{bmatrix} 1 & 3 & 0\\ 2 & 6 & 4\\ -1 & 0 & 2 \end{bmatrix}
is -12, the determinant of the matrix
\begin{bmatrix} 2 & 6 & 0\\ 4 & 12 & 8\\ -2 & 0 & 4 \end{bmatrix}
-96 | |
-24 | |
24 | |
96 |
Question 1 Explanation:
Let D=-12 for the given matrix
|A|=\left|\begin{array}{ccc}2 & 6 & 0 \\4 & 12 & 8 \\-2 & 0 & 4\end{array}\right|=(2)^{3}\left|\begin{array}{ccc}1 & 3 & 0 \\2 & 6 & 4 \\-1 & 0 & 2\end{array}\right|
(Taking 2 common from each row)
\begin{aligned}\therefore \text{Det}(A) &=(2)^{3} \times D \\=& 8 x-12=-96\end{aligned}
|A|=\left|\begin{array}{ccc}2 & 6 & 0 \\4 & 12 & 8 \\-2 & 0 & 4\end{array}\right|=(2)^{3}\left|\begin{array}{ccc}1 & 3 & 0 \\2 & 6 & 4 \\-1 & 0 & 2\end{array}\right|
(Taking 2 common from each row)
\begin{aligned}\therefore \text{Det}(A) &=(2)^{3} \times D \\=& 8 x-12=-96\end{aligned}
Question 2 |
\lim_{x\rightarrow 0} \frac{x-sin x}{1-cos x} is
0 | |
1 | |
3 | |
not define |
Question 2 Explanation:
\lim_{x \rightarrow 0} \frac{x-\sin x}{1-\cos x}
Applying L ' Hospital's rule
\lim_{x \rightarrow 0} \frac{\frac{d}{dx}(x-sinx)}{\frac{d}{dx}(1-cosx)}=\lim_{x \rightarrow 0} \frac{(1-\cos x)}{\sin x}
(It is still of \frac{0}{0} form)
Again applying L' Hospital's rule
\lim_{x \rightarrow 0} \frac{\frac{d}{d x}(1-\cos x)}{\frac{d}{d x}(\sin x)}=\lim_{x \rightarrow 0} \frac{\sin x}{\cos x}=\frac{0}{1}=0
Applying L ' Hospital's rule
\lim_{x \rightarrow 0} \frac{\frac{d}{dx}(x-sinx)}{\frac{d}{dx}(1-cosx)}=\lim_{x \rightarrow 0} \frac{(1-\cos x)}{\sin x}
(It is still of \frac{0}{0} form)
Again applying L' Hospital's rule
\lim_{x \rightarrow 0} \frac{\frac{d}{d x}(1-\cos x)}{\frac{d}{d x}(\sin x)}=\lim_{x \rightarrow 0} \frac{\sin x}{\cos x}=\frac{0}{1}=0
Question 3 |
The argument of the complex number \frac{1+i}{1-i},where i=\sqrt{-1}, is
-\pi | |
-\pi/2 | |
\pi/2 | |
\pi |
Question 3 Explanation:
\begin{aligned} Let z&=\frac{1+i}{1-i} \\ or z&=\frac{(1+i)(1+i)}{(1-i)(1+i)}=\frac{1+i^{2}+2 i}{1-i^{2}}=\frac{2 i}{2}=i \\ z&=x+i y=i \\ SO, x&=0 \\ y &=1 \end{aligned}
\begin{aligned} Arg(z) &=\tan ^{-1}\left(\frac{y}{x}\right) \\ &=\tan ^{-1}\left(\frac{1}{0}\right)=\tan ^{-1} \infty=\frac{\pi}{2} \end{aligned}
\begin{aligned} Arg(z) &=\tan ^{-1}\left(\frac{y}{x}\right) \\ &=\tan ^{-1}\left(\frac{1}{0}\right)=\tan ^{-1} \infty=\frac{\pi}{2} \end{aligned}
Question 4 |
The matrix form of the linear system \frac{dx}{dt}=3x-5y and \frac{dy}{dt}=4x+8y is
\frac{d}{dt} \begin{Bmatrix} x\\ y \end{Bmatrix}=\begin{bmatrix} 3 & -5\\ 4& 8 \end{bmatrix}\begin{Bmatrix} x\\ y \end{Bmatrix} | |
\frac{d}{dt} \begin{Bmatrix} x\\ y \end{Bmatrix}=\begin{bmatrix} 3 & 8\\ 4& -5 \end{bmatrix}\begin{Bmatrix} x\\ y \end{Bmatrix} | |
\frac{d}{dt} \begin{Bmatrix} x\\ y \end{Bmatrix}=\begin{bmatrix} 4 & -5\\ 3& 8 \end{bmatrix}\begin{Bmatrix} x\\ y \end{Bmatrix} | |
\frac{d}{dt} \begin{Bmatrix} x\\ y \end{Bmatrix}=\begin{bmatrix} 4 & 8\\ 3& -5 \end{bmatrix}\begin{Bmatrix} x\\ y \end{Bmatrix} |
Question 4 Explanation:
\begin{aligned} \frac{d x}{d t} &=3 x-5 y \\ \frac{d y}{d t} &=4 x+8 y \\ \frac{d}{d t}\left\{\begin{array}{l}x \\y\end{array}\right\}&=\left[\begin{array}{cc}3 & -5 \\4 & 8\end{array}\right]\left\{\begin{array}{l}x \\y\end{array}\right\}=\left[\begin{array}{cc} 3 x & -5 y \\4 x & +8 y\end{array}\right]\end{aligned}
Question 5 |
Which one of the following describes the relationship among the three vectors \hat{i}+\hat{j}+\hat{k}, 2\hat{i}+3\hat{j}+\hat{k} and 5\hat{i}+6\hat{j}+4\hat{k}
The vectors are mutually perpendicular | |
The vectors are linearly dependent | |
The vectors are linearly independent | |
The vectors are unit vectors |
Question 5 Explanation:
For linear dependency, det \left|\begin{array}{ccc}1 & 1 & 1 \\ 2 & 3 & 1 \\ 5 & 6 & 4\end{array}\right| must be zero.
\therefore \Delta=1(12-6)-1(8-5)+1(12-15)
=6-3-3=0
\therefore There three vectors are linearly dependent.
\therefore \Delta=1(12-6)-1(8-5)+1(12-15)
=6-3-3=0
\therefore There three vectors are linearly dependent.
Question 6 |
A circular rod of length 'L' and area of cross-section 'A' has a modulus of elasticity 'E' and coefficient of thermal expansion '\alpha'. One end of the rod is fixed and other end is free. If the temperature of the rod is increased by \Delta T, then
stress developed in the rod is E\alpha \Delta T and strain developed in the rod is \alpha \Delta T | |
both stress and strain developed in the rod are zero | |
stress developed in the rod is zero and strain developed in the rod is \alpha \Delta T | |
stress developed in the rod is E\alpha \Delta T and strain developed in the rod is zero |
Question 6 Explanation:
For thermal stress to be developed, there must be constraint in the system. So, although strain develops but there is no thermal stress.
Question 7 |
A mentalic rod of 500 mm length and 50mm diameter, when subjected to a tensile force of 100KN at the ends, experiences an increase in its length by 0.5 mm and a reduction in its diameter by 0.015 m. the poission's ratio of the rod material is_________
0.3 | |
0.7 | |
0.1 | |
0.9 |
Question 7 Explanation:
\begin{array}{l} v=-\frac{\text { Lateral strain }}{\text { Longitudinal strain }}=-\frac{\left(\frac{\Delta D}{D}\right)}{\left(\frac{\Delta L}{L}\right)} \\ v=\left(\frac{L}{D}\right)\left(-\frac{\Delta D}{\Delta L}\right) \\ \text{given,}L=500\mathrm{mm},D=50\mathrm{mm}\\ \Delta D=-0.015, \Delta L=+0.5 \mathrm{mm} \\ \qquad v=\frac{500}{50} \times \frac{0.015}{0.500}=0.3 \end{array}
Question 8 |
Critical damping is the
largest amount of damping for which no oscillation occurs in free vibration | |
smallest amount of damping for which no oscillation occurs in free vibration | |
largest amount of damping for which the motion is simple harmonic in free vibration | |
smallest amount of damping for which the motion is simple harmonic in free vibration |
Question 8 Explanation:
The system with critical damping returns to equilibrium in fastest time without any oscillation. In critically damped free vibrations, the damping force is just sufficient to dissipate the energy within One cycle of motion. The system never executes a cycle, it approaches equilibrium with exponentially decaying displacement.
Question 9 |
A circular object of radius r rolls without slipping on a horizontal level floor with the center having velocity V. The velocity at the point of contact between the object and the floor is
zero | |
V in the direction of motion | |
V opposite to the direction of motion | |
V vertically upward from the floor |
Question 9 Explanation:

\begin{aligned} V_{P}= & \omega \times \overline{O P}=2 \omega r \\ V_{C}= & \omega \times \overline{O C}=\omega r \\ V_{O}= & \omega \times 0=0 \end{aligned}
Question 10 |
For the given statements:
I. Mating spur gear teeth is an example of higher pair
II. A revolute joint is an example of lower pair
Indicate the correct answer
I. Mating spur gear teeth is an example of higher pair
II. A revolute joint is an example of lower pair
Indicate the correct answer
Both I and II are false | |
I is true and II is false | |
I is false and II is true | |
Both I an d II are true |
Question 10 Explanation:
Example of Lower Pair:
Revolute pair, Prismatic joint, Screw pair, cylindrical joint, spherical joint.
Example of higher pair:
Wheel rolling an surface, cam and follower contact, meshing teeth of two gears.
Revolute pair, Prismatic joint, Screw pair, cylindrical joint, spherical joint.
Example of higher pair:
Wheel rolling an surface, cam and follower contact, meshing teeth of two gears.
There are 10 questions to complete.