Question 1 |
Consider the two-dimensional vector field \vec{F}(x,y)=x\vec{i}+y\vec{j}, where \vec{i} and \vec{j} denote
the unit vectors along the x-axis and the y-axis, respectively. A contour C in the x-y plane, as shown in the figure, is composed of two horizontal lines connected at the
two ends by two semicircular arcs of unit radius. The contour is traversed in the
counter-clockwise sense. The value of the closed path integral
\oint _c \vec{F}(x,y)\cdot (dx\vec{i}+dy\vec{j})
\oint _c \vec{F}(x,y)\cdot (dx\vec{i}+dy\vec{j})
0 | |
1 | |
8+2 \pi | |
-1 |
Question 1 Explanation:
\oint \vec{F} (x,y)\cdot [dx\vec{i}+dy\vec{j}]
Given \vec{F} (x,y)=x\vec{i}+y\vec{j}
\therefore \int_{c}xdx+ydy=0
Because here vector is conservative.
If the integral function is the total derivative over the closed contoure then it will be zero
Given \vec{F} (x,y)=x\vec{i}+y\vec{j}
\therefore \int_{c}xdx+ydy=0
Because here vector is conservative.
If the integral function is the total derivative over the closed contoure then it will be zero
Question 2 |
Consider a system of linear equations Ax=b, where
A=\begin{bmatrix} 1 & -\sqrt{2} & 3\\ -1& \sqrt{2}& -3 \end{bmatrix},b=\begin{bmatrix} 1\\ 3 \end{bmatrix}
This system of equations admits ______.
A=\begin{bmatrix} 1 & -\sqrt{2} & 3\\ -1& \sqrt{2}& -3 \end{bmatrix},b=\begin{bmatrix} 1\\ 3 \end{bmatrix}
This system of equations admits ______.
a unique solution for x | |
infinitely many solutions for x | |
no solutions for x | |
exactly two solutions for x |
Question 2 Explanation:
Here equation will be
x-\sqrt{2}y+3z=1
-x+\sqrt{2}y-3z=3
therefore inconsistant solution i.e. there will not be any solution.
x-\sqrt{2}y+3z=1
-x+\sqrt{2}y-3z=3
therefore inconsistant solution i.e. there will not be any solution.
Question 3 |
The current I in the circuit shown is ________
1.25 \times 10^{-3}A | |
0.75 \times 10^{-3}A | |
-0.5 \times 10^{-3}A | |
1.16 \times 10^{-3}A |
Question 3 Explanation:
Applying Nodal equation at Node-A
\begin{aligned} \frac{V_A}{2k}+\frac{V_A-5}{2k}&=10^{-3}\\ \Rightarrow 2V_A-5&=2k \times 10^{-3}\\ V_A&=3.5V\\ Again,&\\ I&=\frac{5-V_A}{2k}\\ &=\frac{5-3.5}{2k}\\ &=0.75 \times 10^{-3}A \end{aligned}
Question 4 |
Consider the circuit shown in the figure. The current I flowing through the 10\Omega resistor is _________.
1A | |
0A | |
0.1A | |
-0.1A |
Question 4 Explanation:
Here, there is no any return closed path for Current (I) . Hence I=0
Current always flow in loop.
Current always flow in loop.
Question 5 |
The Fourier transform X(j\omega ) of the signal x(t)=\frac{t}{(1+t^2)^2} is _________.
\frac{\pi}{2j}\omega e^{-|\omega|} | |
\frac{\pi}{2}\omega e^{-|\omega|} | |
\frac{\pi}{2j} e^{-|\omega|} | |
\frac{\pi}{2} e^{-|\omega|} |
Question 5 Explanation:
x(t)=\frac{t}{(1+t^2)^2}
As we know that FT of te^{-|t|} \; \underleftrightarrow{FT} \;\frac{-j4\omega }{(1+\omega ^2)^2}
Duality \frac{-j4\omega }{(1+t ^2)^2} \leftrightarrow 2 \pi(-\omega )e^{-|-\omega |}
\Rightarrow \frac{t}{(1+t^2)^2} \underrightarrow{FT} \frac{-2\pi}{-j4}\omega e^{-|\omega |}
\Rightarrow \;\;\;\rightarrow\frac{\pi}{j2} \omega e^{-|\omega |}
As we know that FT of te^{-|t|} \; \underleftrightarrow{FT} \;\frac{-j4\omega }{(1+\omega ^2)^2}
Duality \frac{-j4\omega }{(1+t ^2)^2} \leftrightarrow 2 \pi(-\omega )e^{-|-\omega |}
\Rightarrow \frac{t}{(1+t^2)^2} \underrightarrow{FT} \frac{-2\pi}{-j4}\omega e^{-|\omega |}
\Rightarrow \;\;\;\rightarrow\frac{\pi}{j2} \omega e^{-|\omega |}
There are 5 questions to complete.