Question 1 |
\oint _c \vec{F}(x,y)\cdot (dx\vec{i}+dy\vec{j})

0 | |
1 | |
8+2 \pi | |
-1 |
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 |
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 |
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 |

1.25 \times 10^{-3}A | |
0.75 \times 10^{-3}A | |
-0.5 \times 10^{-3}A | |
1.16 \times 10^{-3}A |


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 |



1A | |
0A | |
0.1A | |
-0.1A |
Current always flow in loop.
Question 5 |
\frac{\pi}{2j}\omega e^{-|\omega|} | |
\frac{\pi}{2}\omega e^{-|\omega|} | |
\frac{\pi}{2j} e^{-|\omega|} | |
\frac{\pi}{2} 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 |}
Question 6 |
0.37 \times 10^{14} cm^{-3} | |
0.63 \times 10^{13} cm^{-3} | |
3.7 \times 10^{14} cm^{-3} | |
0^{3} cm^{-3} |


From continuity equation of electrons
\frac{dn}{dt}=n\mu _n\frac{dE}{dx}+\mu _nE\frac{dn}{dx}+G_n-R_n+x_n\frac{d^2x}{dx^2} \;\;\;...(i)
[Because \vec{E} is not mentioned hence
\frac{dE}{dx}=0
For x \gt 0, G_n is also zero
n=\frac{n_i^2}{N_A}=\frac{10^{20}}{10^{17}}=10^3
n=n_0+\delta n=10^3+10^{14}=10^{14}
at steady state, \frac{db}{dt}=0
Hence equation (i) becomes:
O=D_n\frac{d^2\delta n}{dx^2}-\frac{\delta n}{\tau _n}
\frac{d^2\delta n}{dx^2}=\frac{\delta n}{L_n^2} \;\;\;...(ii)
From solving equation (ii)
\delta _n(x)=\delta _n(0)e^{-x/L_n}
at x=2\mu m
\delta _n(2\mu m)=10^{14}e^{-2/2}=10^{14}e^{-1}=0.37 \times 10^{14}
Question 7 |
226 meV | |
174 meV | |
218 meV | |
182 meV |
As we know,
E_C-E_F=kT \ln\left ( \frac{N_c}{n} \right ) \;\;\;...(i)
E_C-E_{F1}=kT \ln\left ( \frac{N_c}{n_1} \right ) \;\;\;...(ii)
E_C-E_{F2}=kT \ln\left ( \frac{N_c}{n_2} \right ) \;\;\;...(iii)
Equation (ii) - Equation (iii)
(E_C-E_{F1})-(E_C-E_{F2})=kT \ln \left ( \frac{\frac{N_c}{n_1}}{\frac{N_c}{n_2}} \right )=kT \ln \frac{n_2}{n_1}
\Rightarrow 200meV-(E_C-E_{F2})=26meV \times \ln \left ( \frac{0.5 \times 10^{16}}{1 \times 10^{16}} \right )
200meV-(E_C-E_{F2})=+26meV \ln (0.5)=-18
(E_C-E_{F2})=200+8=218meV
Question 8 |


equal to 0 V | |
more than 2 V | |
less than 2 V | |
equal to 2 V |


\begin{aligned} \mu _PCO_x\left ( \frac{\omega }{L} \right )_1[4-V_0-1]^2&=\mu _nCO_x\left ( \frac{\omega }{L} \right )_2[V_0-0-1]^2\\ \Rightarrow \frac{300}{40}\times \frac{1}{5}(V_0-1)^2&=(3-V_0)^2\\ \Rightarrow \sqrt{1.5} (V_0-1)&=3-V_0\\ \Rightarrow V_0&=\frac{3+\sqrt{1.5}}{\sqrt{1.5}+1} \lt 2V \end{aligned}
Question 9 |


A_0=0,A_1=0,A_2=1,A_3=1 | |
A_0=1,A_1=0,A_2=1,A_3=0 | |
A_0=0,A_1=1,A_2=1,A_3=0 | |
A_0=1,A_1=1,A_2=0,A_3=0 |


f=\bar{C}\bar{D}I_0+\bar{C}DI_1+C\bar{D}I_2+CDI_3
For this
A_0=A_3=0
A_1=A_2=1
Question 10 |


V_1=5 V, V_2=5 V | |
V_1=5 V, V_2=4 V | |
V_1=4 V, V_2=5 V | |
V_1=4V, V_2=-5 V |

