Question 1 |
The determinant of matrix A is 5 and the determinant of matrix B is 40. The
determinant of matrix AB is ______.
200 | |
100 | |
50 | |
45 |
Question 1 Explanation:
Determinant of A=5
Determinant of B=40
Determinant of AB=|A||B|
\begin{array}{l} \quad=5 \times 40 \\ \quad=200 \end{array}
Determinant of B=40
Determinant of AB=|A||B|
\begin{array}{l} \quad=5 \times 40 \\ \quad=200 \end{array}
Question 2 |
Let X be a random variable which is uniformly chosen from the set of positive
odd numbers less than 100. The expectation, E [X], is _____.
100 | |
50 | |
25 | |
10 |
Question 2 Explanation:
E[X]=\frac{1+2+3+\cdots 99}{50}=\frac{2500}{50}=50
Question 3 |
For 0\leq t \leq \infty
, the maximum value of the function f(t) =e^{-t}-2e^{-2t} occurs at
t=log_{e}4 | |
t=log_{e}2 | |
t=0 | |
t=log_{e}8 |
Question 3 Explanation:
\begin{aligned} f(t)&=e^{-t}-2 e^{-2 t} \\ f^{\prime}(t)&=-e^{-t}+4 e^{-2 t} \\ \text{For maximum value} &P(t)=0 \\ f^{\prime}(t)&=0=-e^{-t}+4 e^{-2 t} \\ \Rightarrow \quad 4 e^{-2 t}&=e^{t} \\ 4 e^{t}&=1 \\ \therefore \quad t&=\log _{e} 4 \end{aligned}
Question 4 |
The value of
\lim_{x\rightarrow \infty }(1+\frac{1}{x})^{x}
is
\lim_{x\rightarrow \infty }(1+\frac{1}{x})^{x}
is
ln 2 | |
1 | |
e | |
\infty |
Question 4 Explanation:
\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e^{\lim _{x \rightarrow \infty} \frac{1}{x} \cdot x}=e^{1}=e
Question 5 |
If the characteristic equation of the differential equation
\frac{d^{2}y}{dx^{2}}+2\alpha \frac{dy}{dx}+y=0
has two equal roots, then the values of a are
\frac{d^{2}y}{dx^{2}}+2\alpha \frac{dy}{dx}+y=0
has two equal roots, then the values of a are
\pm 1 | |
0,0 | |
\pm j | |
\pm 1/2 |
Question 5 Explanation:
\frac{d^{2} y}{d x^{2}}+2 \alpha\frac{d y}{d x}+y=0
The characteristic equation is given as
\begin{aligned} \left(m^{2}+2(x)+1\right) &=0 \\ m_{1}, m_{2} &=\frac{-2 x_{1} \pm \sqrt{4 x^{2}-4}}{2} \end{aligned}\\ \text{since both roots are equal i.e.} \\ \begin{aligned} m_{1}=& m_{2} \\ \frac{-2 \alpha+\sqrt{4 \alpha^{2}-4}}{2} &=\frac{-2\alpha \cdot-\sqrt{4 a^{2}-4}}{2} \\ \sqrt{4\left(1^{2}-4\right.} &=-\sqrt{4 c^{2}-4} \\ 2 \sqrt{4 c^{2}-4} &=0 \\ 4 \alpha^{2}-4 &=0 \\ \alpha^{2} &=1 \\ \alpha &=\pm 1 \end{aligned}
The characteristic equation is given as
\begin{aligned} \left(m^{2}+2(x)+1\right) &=0 \\ m_{1}, m_{2} &=\frac{-2 x_{1} \pm \sqrt{4 x^{2}-4}}{2} \end{aligned}\\ \text{since both roots are equal i.e.} \\ \begin{aligned} m_{1}=& m_{2} \\ \frac{-2 \alpha+\sqrt{4 \alpha^{2}-4}}{2} &=\frac{-2\alpha \cdot-\sqrt{4 a^{2}-4}}{2} \\ \sqrt{4\left(1^{2}-4\right.} &=-\sqrt{4 c^{2}-4} \\ 2 \sqrt{4 c^{2}-4} &=0 \\ 4 \alpha^{2}-4 &=0 \\ \alpha^{2} &=1 \\ \alpha &=\pm 1 \end{aligned}
Question 6 |
Norton's theorem states that a complex network connected to a load can be
replaced with an equivalent impedance
in series with a current source | |
in parallel with a voltage source | |
in series with a voltage source | |
in parallel with a voltage source |
Question 6 Explanation:
Norton's theorem states that a complex network
connected to a load can be replaced with an equivalent
impedance in parallel with a current source
Question 7 |
In the figure shown, the ideal switch has been open for a long time. If it is closed
at t = 0, then the magnitude of the current (in mA) through the 4 k\Omega resistance
at t = 0^{+} is _____.
0.75 | |
1.25 | |
0.5 | |
1.5 |
Question 7 Explanation:
At steady state t = 0_{-}
\begin{aligned} \therefore V_{c}(0) &=V_{c}\left(0^{+}\right)=5 \mathrm{V} \\ I_{L}\left(0^{-}\right) &=I_{L}\left(0^{+}\right)=1 \mathrm{mA} \end{aligned}
At, t=0, switch get closed
Thus, the current through 4\Omega resistance is
I=\frac{5}{4 \times 10^{3}}=1.25 \mathrm{mA}
\begin{aligned} \therefore V_{c}(0) &=V_{c}\left(0^{+}\right)=5 \mathrm{V} \\ I_{L}\left(0^{-}\right) &=I_{L}\left(0^{+}\right)=1 \mathrm{mA} \end{aligned}
At, t=0, switch get closed
Thus, the current through 4\Omega resistance is
I=\frac{5}{4 \times 10^{3}}=1.25 \mathrm{mA}
Question 8 |
A silicon bar is doped with donor impurities N_{D}=2.25\times10^{15} atom/cm^{3}. Given the intrinsic carrier concentration of silicon at T = 300 K is n_{i}=1.5\times10^{10}cm^{-3}. Assuming complete impurity ionization, the equilibrium election and hole concentrations are
n_{0}=1.5 \times 10^{16}cm^{-3}, p_{0}=1.5 \times 10^{5}cm^{-3} | |
n_{0}=1.5 \times 10^{10}cm^{-3}, p_{0}=1.5 \times 10^{15}cm^{-3} | |
n_{0}=2.25 \times 10^{15}cm^{-3}, p_{0}=1.5 \times 10^{10}cm^{-3} | |
n_{0}=2.25 \times 10^{15}cm^{-3}, p_{0}=1 \times 10^{5}cm^{-3} |
Question 8 Explanation:
since N_{D} \gt \gt n_{i}
therefore equilibrium electron concentration is
n \simeq N_{D}=2.25 \times 10^{15} \mathrm{cm}^{-3}
And equilibrium hole concentration is given by mass action law
p=\frac{n_{i}^{2}}{N_{D}}=\frac{\left(1.5 \times 10^{10}\right)^{2}}{2.25 \times 10^{15}}=1 \times 10^{5} \mathrm{cm}^{-3}
therefore equilibrium electron concentration is
n \simeq N_{D}=2.25 \times 10^{15} \mathrm{cm}^{-3}
And equilibrium hole concentration is given by mass action law
p=\frac{n_{i}^{2}}{N_{D}}=\frac{\left(1.5 \times 10^{10}\right)^{2}}{2.25 \times 10^{15}}=1 \times 10^{5} \mathrm{cm}^{-3}
Question 9 |
An increase in the base recombination of a BJT will increase
the common emitter dc current gain \beta | |
the breakdown voltage BV_{CEO} | |
the unity-gain cut-off frequency f_{T} | |
the transconductance g_{m} |
Question 10 |
In CMOS technology, shallow P-well or N -well regions can be formed using
low pressure chemical vapour deposition | |
low energy sputtering | |
low temperature dry oxidation | |
low energy ion-implantation |
Question 10 Explanation:
Ion implanation/diffusion is used for well
implantation.
There are 10 questions to complete.