Question 1 |
If v_1, v_2,..., v_6 are six vectors in \mathbb{R}^4, which one of the following statements is False?
It is not necessary that these vectors span \mathbb{R}^4. | |
These vectors are not linearly independent. | |
Any four of these vectors form a basis for \mathbb{R}^4. | |
If {v_1, v_3,v_5, v_6} spans \mathbb{R}^4, then it forms a basis for \mathbb{R}^4. |
Question 1 Explanation:
v_1, v_2,..., v_6 are six vectors in \mathbb{R}^4.
For a 4-dimensional vector space,
(i) any four linearly independent vectors form a basis (or)
(ii) Any set of four vectors in \mathbb{R}^4 spans \mathbb{R}^4, then it forms a basis.
Therefore, clearly options (A), (B), (D) are true.
Option (C) is FALSE
For a 4-dimensional vector space,
(i) any four linearly independent vectors form a basis (or)
(ii) Any set of four vectors in \mathbb{R}^4 spans \mathbb{R}^4,
Therefore, clearly options (A), (B), (D) are true.
Option (C) is FALSE
Question 2 |
For a vector field \vec{A}, which one of the following is False?
\vec{A} is solenoidal if \bigtriangledown \cdot \vec{A}=0 | |
\bigtriangledown \times \vec{A} is another vector field. | |
\vec{A} is irrotational if \bigtriangledown ^2 \vec{A}=0. | |
\bigtriangledown \times(\bigtriangledown \times \vec{A})=\bigtriangledown (\bigtriangledown \cdot \vec{A})-\bigtriangledown ^2 \vec{A} |
Question 2 Explanation:
Divergence and curl operator is performed on a vector field \vec{A}
Curl operation provides a vector orthogonal to the given vector field \vec{A}
\bigtriangledown \times(\bigtriangledown \times \vec{A})=\bigtriangledown (\bigtriangledown \cdot \vec{A})-\bigtriangledown ^2 \vec{A}
If a vector field is irrortational then \bigtriangledown \times \vec{A}=0
If a vector field is solenoidal then \bigtriangledown \cdot \vec{A}=0
If a field is scalar A, then \bigtriangledown ^2 \vec{A}=0, is a laplacian equation.
Hence option (C) is incorrect
Curl operation provides a vector orthogonal to the given vector field \vec{A}
\bigtriangledown \times(\bigtriangledown \times \vec{A})=\bigtriangledown (\bigtriangledown \cdot \vec{A})-\bigtriangledown ^2 \vec{A}
If a vector field is irrortational then \bigtriangledown \times \vec{A}=0
If a vector field is solenoidal then \bigtriangledown \cdot \vec{A}=0
If a field is scalar A, then \bigtriangledown ^2 \vec{A}=0, is a laplacian equation.
Hence option (C) is incorrect
Question 3 |
The partial derivative of the function
f(x,y,z)=e^{1-x \cos y}+xze^{-1/(1+y^2)}
with respect to x at the point (1,0,e) is
f(x,y,z)=e^{1-x \cos y}+xze^{-1/(1+y^2)}
with respect to x at the point (1,0,e) is
-1 | |
0 | |
1 | |
\frac{1}{e} |
Question 3 Explanation:
\begin{aligned} \text{Given, } f(x,y,z)&=e^{1-x\cos y}+xze^{-1/(1+y^{2})} \\ \frac{\partial f}{\partial x}&=e^{1-x\cos y}(0-\cos y)+ze^{-1/1+y^{2}} \\ \left ( \frac{\partial f }{\partial x} \right )_{(1,0,e)}&=e^{0}(0-1)+e\cdot e^{-1/(1+0)} \\ &=-1+1=0
\end{aligned}
Question 4 |
The general solution of \frac{d^2y}{dx^2}-6\frac{dy}{dx}+9y=0 is
y=C_1e^{3x}+C_2e^{-3x} | |
y=(C_1+C_2x)e^{-3x} | |
y=(C_1+C_2x)e^{3x} | |
y=C_1e^{3x} |
Question 4 Explanation:
Taking \frac{\mathrm{d} }{\mathrm{d} x}=D
Given, D^{2}-6D+9=0
(D-3)^2=0
D=3,3
So, Solution of the given Differential equation
y=(C_{1}+C_{2}x)e^{3x}
Given, D^{2}-6D+9=0
(D-3)^2=0
D=3,3
So, Solution of the given Differential equation
y=(C_{1}+C_{2}x)e^{3x}
Question 5 |
The output y[n] of a discrete-time system for an input x[n] is
y[n]=\begin{matrix} max\\ -\infty \leq k\leq n \end{matrix}\; |x[k]|.
The unit impulse response of the system is
y[n]=\begin{matrix} max\\ -\infty \leq k\leq n \end{matrix}\; |x[k]|.
The unit impulse response of the system is
0 for all n | |
1 for all n | |
unit step signal u[n]. | |
unit impulse signal \delta[n]. |
Question 6 |
A single crystal intrinsic semiconductor is at a temperature of 300 K with effective density
of states for holes twice that of electrons. The thermal voltage is 26 mV. The intrinsic
Fermi level is shifted from mid-bandgap energy level by
18.02 meV | |
9.01 meV | |
13.45 meV | |
26.90 meV |
Question 6 Explanation:
\frac{E_{c}+E_{v}}{2}-E_{F_{i}}=\frac{KT}{2}\ln \left ( \frac{N_{C}}{N_{V}}\right )\, \, \, \, \, \, \left ( \because N_{C}=\frac{N_{V}}{2} \right )
=\frac{0.026}{2}\ln 0.5=-9.01\, meV
=\frac{0.026}{2}\ln 0.5=-9.01\, meV
Question 7 |
Consider the recombination process via bulk traps in a forward biased pn homojunction
diode. The maximum recombination rate is U_{max}. If the electron and the hole capture
cross-section are equal, which one of the following is False?
With all other parameters unchanged, U_{max} decreases if the intrinsic carrier density
is reduced. | |
U_{max} occurs at the edges of the depletion region in the device. | |
U_{max} depends exponentially on the applied bias. | |
With all other parameters unchanged,U_{max} increases if the thermal velocity of the
carriers increases. |
Question 8 |
The components in the circuit shown below are ideal. If the op-amp is in positive feedback
and the input voltage V_i is a sine wave of amplitude 1 V, the output voltage V_o is


a non-inverted sine wave of 2 V amplitude | |
an inverted sine wave of 1 V amplitude | |
a square wave of 5 V amplitude | |
a constant of either +5 or -5V |
Question 8 Explanation:

Given circuit is a Schmitt trigger of non-inverting type.
V_{o}=\pm 5\, V
V^{+}=\frac{V_{o}\times 1+V_{i}\times 1}{1+1}=\frac{V_{o}+V}{2}
let, V_{o}=-5\, V,\, \, \, \,V^{+}=\frac{-5+V_{i}}{2}
V_{o} can change from -5 V to +5 V if V^{+} \gt 0
i.e. \frac{-5+V_{i}}{2} \gt 0\Rightarrow V_{i} \gt 5\, V
similarly, V_{o} can change from -5 V to +5 V if V_{i} \lt -5\, V
But given input has peak value 1 V. Hence output cannot change from +5 V to -5 V or -5 V to +5 V.
Output remain constant at +5 V or -5 V.
Question 9 |
In the circuit shown below, the Thevenin voltage V_{TH} is


2.4 V | |
2.8 V | |
3.6 V | |
4.5 V |
Question 9 Explanation:
By applying the Source Transformation

Question 10 |
The figure below shows a multiplexer where S_1 \; and \; S_0 are the select lines, I_0 \; to \; I_3 are
the input data lines, EN is the enable line, and F(P, Q, R) is the output, F is


PQ+\bar{Q}R | |
P+Q\bar{R} | |
P\bar{Q}R+\bar{P}Q | |
\bar{Q}+PR |
Question 10 Explanation:
Output,F=\bar{P}\bar{Q}R+P\bar{Q}R+PQ\, \, \, \,
F=\bar{Q}R+PQ
F=\bar{Q}R+PQ

There are 10 questions to complete.