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 |}
Question 6 |
Consider a long rectangular bar of direct bandgap p-type semiconductor. The
equilibrium hole density is 10^{17}cm^{-3} and the intrinsic carrier concentration is 10^{10}cm^{-3}. Electron and hole diffusion lengths are 2\mu mand 1\mu m, respectively.
The left side of the bar (x=0) is uniformly illuminated with a laser having photon
energy greater than the bandgap of the semiconductor. Excess electron-hole pairs
are generated ONLY at x=0 because of the laser. The steady state electron density
at x=0 is 10^{14}cm^{-3}
due to laser illumination. Under these conditions and ignoring
electric field, the closest approximation (among the given options) of the steady state
electron density at x=2 \mu m, is _____
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} |
Question 6 Explanation:
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 |
In a non-degenerate bulk semiconductor with electron density n=10^{16}cm^{-3}, the
value of E_C-E_{Fn}=200meV, where E_C and E_{Fn} denote the bottom of the
conduction band energy and electron Fermi level energy, respectively. Assume
thermal voltage as 26 meV and the intrinsic carrier concentration is 10^{10}cm^{-3}. For n=0.5 \times 10^{16}cm^{-3}, the closest approximation of the value of (E_C-E_{Fn}), among
the given options, is ______.
226 meV | |
174 meV | |
218 meV | |
182 meV |
Question 7 Explanation:
Here we have to find the value of
E_c-E_{fn}
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
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 |
Consider the CMOS circuit shown in the figure (substrates are connected to their
respective sources). The gate width (W) to gate length (L) ratios \frac{W}{L} of the
transistors are as shown. Both the transistors have the same gate oxide capacitance
per unit area. For the pMOSFET, the threshold voltage is -1 V and the mobility of
holes is 40\frac{cm^2}{V.s}. For the nMOSFET, the threshold voltage is 1 V and the mobility
of electrons is 300\frac{cm^2}{V.s}. The steady state output voltage V_o is ________.
equal to 0 V | |
more than 2 V | |
less than 2 V | |
equal to 2 V |
Question 8 Explanation:
\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 |
Consider the 2-bit multiplexer (MUX) shown in the figure. For OUTPUT to be the
XOR of C and D, the values for A_0,A_1,A_2 \text{ and }A_3 are _______
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 |
Question 9 Explanation:
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 |
The ideal long channel nMOSFET and pMOSFET devices shown in the circuits
have threshold voltages of 1 V and -1 V, respectively. The MOSFET substrates are
connected to their respective sources. Ignore leakage currents and assume that the
capacitors are initially discharged. For the applied voltages as shown, the steady
state voltages are ______
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 |
Question 10 Explanation:
There are 10 questions to complete.