Question 1 |
The order of the differential equation \frac{d^{2}y}{dt^{2}}+(\frac{dy}{dt})^{3}+y^{4}=e^{-t} is
1 | |
2 | |
3 | |
4 |
Question 1 Explanation:
Highest derivative of differential equation is 2.
Question 2 |
The Fourier series of a real periodic function has only
(P) cosine terms if it is even
(Q) sine terms if it is even
(R) cosine terms if it is odd
(S) sine terms if it is odd
Which of the above statements are correct ?
(P) cosine terms if it is even
(Q) sine terms if it is even
(R) cosine terms if it is odd
(S) sine terms if it is odd
Which of the above statements are correct ?
P and S | |
P and R | |
Q and S | |
Q and R |
Question 2 Explanation:
The Fourier series of a real periodic function has
only cosine terms if it is even and only sine terms
if it is odd.
Question 3 |
A function is given by f(t)=sin^{2}t+cos2t. Which of the following is true ?
f has frequency components at 0 and \frac{1}{2\pi } Hz | |
f has frequency components at 0 and \frac{1}{\pi } Hz | |
f has frequency components at \frac{1}{2\pi } \; and \; \frac{1}{\pi}Hz | |
f has frequency components at 0, \frac{1}{2\pi } \; and \; \frac{1}{\pi}Hz |
Question 3 Explanation:
f(t)=\frac{1}{2}(1-\cos 2 t)+\cos 2 t
frequency components are
\begin{aligned} f_{1}&=0 \\ f_{2}&=\frac{\omega_{2}}{2 \pi}=\frac{2}{2 \pi}=\frac{1}{\pi} \mathrm{Hz} \end{aligned}
frequency components are
\begin{aligned} f_{1}&=0 \\ f_{2}&=\frac{\omega_{2}}{2 \pi}=\frac{2}{2 \pi}=\frac{1}{\pi} \mathrm{Hz} \end{aligned}
Question 4 |
A fully charged mobile phone with a 12 V battery is good for a 10 minute talktime. Assume that, during the talk-time the battery delivers a constant current of 2 A and its voltage drops linearly from 12 V to 10 V as shown in the figure. How much energy does the battery deliver during this talk-time?
220J | |
12kJ | |
13.2kJ | |
14.4J |
Question 4 Explanation:
\begin{aligned} P &=V I \\ \text { Energy } &=P \cdot t=V \cdot|t=(V . t)| \\ I &=2 \mathrm{A} \quad(\text { given }) \\ V \cdot t &=\text { Area under } V-t \text { curve } \\ V \cdot t &=\left(\frac{1}{2} \times 2 \times 600\right)+(10 \times 600) \\ &=600+6000 \\ V \cdot t &=6600 \\ E &=(6600) \times 2=13200=13.2 \mathrm{kJ} \end{aligned}
Question 5 |
In an n-type silicon crystal at room temperature, which of the following can have a concentration of 4 \times 10^{19}cm^{-3}?
Silicon atoms | |
Holes | |
Dopant atoms | |
Valence electrons |
There are 5 questions to complete.