Question 1 |
A random variable X has probability density function f(x) as given below:
f(x)=\left\{\begin{matrix} a+bx & for \; 0 \lt x \lt 1\\ 0& otherwise \end{matrix}\right.
If the expected value E[X]=2/3, then Pr[X \lt 0.5] is _____________.
f(x)=\left\{\begin{matrix} a+bx & for \; 0 \lt x \lt 1\\ 0& otherwise \end{matrix}\right.
If the expected value E[X]=2/3, then Pr[X \lt 0.5] is _____________.
0.25 | |
0.5 | |
0.75 | |
1 |
Question 1 Explanation:
\begin{aligned} f(x)&=\left\{\begin{matrix} a+bx & \text{for }0 \lt x \lt 1\\ 0 & \text{otherwise} \end{matrix}\right.&\\ \text{Now given }E(X)&=2/3\\ \int_{0}^{1}xf(x)dx&=\frac{2}{3}\\ \int_{0}^{1}x(a+bx)dx&=\frac{2}{3}\\ a\left ( \frac{x^2}{2} \right )_0^1+b\left ( \frac{x^3}{3} \right )_0^1&=\frac{2}{3}\\ a\left ( \frac{1}{2} \right )+b\left ( \frac{1}{3} \right )&=\frac{2}{3}\\ 3a+2b&=4\;\;...(i)\\ \text{Now, }\int_{0}^{1}f(x)dx &=1\\ (\text{total probability} & \text{ is always equal to 1})\\ \int_{0}^{1}(a+bx)dx&=\left ( ax+\frac{bx^2}{2} \right )_0^1=1\\ a+\frac{b}{2}&=1\\ 2a+b&=2\;\;...(ii)\\ \text{Now solving } & \text{ (i) and (ii), we get}\\ a&=0,b=2\\ So,\; \;f(x)&=\left\{\begin{matrix} 2x & \text{for }0 \lt x lt 1\\ 0 & \text{otherwise} \end{matrix}\right.&\\ \text{Now we need}&\\ P(x \lt 0.5)&=\int_{0}^{0.5}2xdx=2\left ( \frac{x^2}{2} \right )_0^{0.5}\\ &=0.5^2-0^2=0.25 \end{aligned}
Question 2 |
If a continuous function f(x) does not have a root in the interval [a,b], then which one of the following statements is TRUE?
f(a)\cdot f(b)=0 | |
f(a)\cdot f(b) \lt 0 | |
f(a)\cdot f(b) \gt 0 | |
f(a)/ f(b)\leq 0 |
Question 2 Explanation:
Intermediate value theorem states that if a function is continious and f(a) \cdot f(b) \lt 0, then surely there is a root in (a,b). The contrapositive of this theorem is that if a function is continious and has no root in (a,b) then surely f(a) \cdot f(b) \geq 0. But since it is given that there is no root in the closed interval [a,b] it means f(a) \cdot f(b) \neq 0.
So surely f(a) \cdot f(b) \gt 0 which is choise(C).
So surely f(a) \cdot f(b) \gt 0 which is choise(C).
Question 3 |
If the sum of the diagonal elements of a 2x2 matrix is -6, then the maximum possible value of determinant of the matrix is ________.
6 | |
8 | |
9 | |
12 |
Question 3 Explanation:
Consider a symmetric matrix A =\begin{bmatrix} a & b\\ b & d \end{bmatrix}
Given a+d=-6
|A|=ad-b^2
Now since b^2 is always non-negative, maximum determinant will come when b^2=0.
So we need to maximize
\begin{aligned} |A|&=ad-0=ad=a \times -(6+a) \\ &= -a^2-6a\\ \frac{d|A|}{da} &=-2a-6=0 \\ \Rightarrow \;\; a &=-3 \text{ is the only stationary point} \end{aligned}
Since, \left [\frac{d^2|A|}{da} \right ]_{a=-3}=-2 \lt 0, we have a maximum at a=-3.
Since a+d=-6, corresponding value of d=-3.
|A|=ad=-3 \times -3=9
Given a+d=-6
|A|=ad-b^2
Now since b^2 is always non-negative, maximum determinant will come when b^2=0.
So we need to maximize
\begin{aligned} |A|&=ad-0=ad=a \times -(6+a) \\ &= -a^2-6a\\ \frac{d|A|}{da} &=-2a-6=0 \\ \Rightarrow \;\; a &=-3 \text{ is the only stationary point} \end{aligned}
Since, \left [\frac{d^2|A|}{da} \right ]_{a=-3}=-2 \lt 0, we have a maximum at a=-3.
Since a+d=-6, corresponding value of d=-3.
|A|=ad=-3 \times -3=9
Question 4 |
Consider a function \vec{f}=\frac{1}{r^{2}}\hat{r}, where r is the distance from the origin and \hat{r} is the unit vector in the radial direction. The divergence of this function over a sphere of radius R, which includes the origin, is
0 | |
2\pi | |
4\pi | |
R\pi |
Question 4 Explanation:
\bar{f}=\frac{1}{r^2}\hat{r}
From divergence theorem as we know,
\begin{aligned} \int _{vol.} (\triangledown \cdot \bar{f})dV &=\oint \oint _S \bar{f}\cdot d\bar{S}\\ \oint \oint _S \bar{f}\cdot d\bar{S} &= \oint \oint _S \left ( \frac{1}{r^2} \cdot \hat{r}\right ) \times r^2 \sin \theta \cdot d\theta \cdot d\phi \cdot \hat{r} \\ &= \oint \oint _S \sin \theta \cdot d\theta \cdot d\phi=4\pi \end{aligned}
From divergence theorem as we know,
\begin{aligned} \int _{vol.} (\triangledown \cdot \bar{f})dV &=\oint \oint _S \bar{f}\cdot d\bar{S}\\ \oint \oint _S \bar{f}\cdot d\bar{S} &= \oint \oint _S \left ( \frac{1}{r^2} \cdot \hat{r}\right ) \times r^2 \sin \theta \cdot d\theta \cdot d\phi \cdot \hat{r} \\ &= \oint \oint _S \sin \theta \cdot d\theta \cdot d\phi=4\pi \end{aligned}
Question 5 |
When the Wheatstone bridge shown in the figure is used to find the value of resistor R_X, the galvanometer G indicates zero current when R_1=50\Omega ,R_2=65\Omega and R_3=100\Omega. If R_3 is known with \pm 5% tolerance on its nominal value of 100 \Omega , what is the range of R_X in Ohms?


[123.50, 136.50] | |
[125.89, 134.12] | |
[117.00, 143.00] | |
[120.25, 139.75] |
Question 5 Explanation:
R_1=50\Omega
R_2=60\Omega
R_3=100\pm 5%\Omega
The value of R_3 \text{with} \pm 5% of tollerance,
\begin{aligned} R_3&= 100\pm 5\% \\ &= 100+100 \times \frac{5}{100}=105 \Omega \\ &= 100-100 \times \frac{5}{100}=95 \Omega \end{aligned}
In both condition, the bridge is balance, so under balance condition,
\begin{aligned} \frac{R_1}{R_3}&=\frac{R_2}{R_x} \\ R_x&=\frac{R_2R_3}{R_1} \\ (i) \; \text{when}, \; R_3&=105\Omega \\ \therefore \;\;R_x &= \frac{105 \times 65}{50}=136.50\Omega \\ (ii)\; \text{when}, \; R_3&=95 \Omega \\ \therefore \;\;R_x &= \frac{95 \times 65}{50}=123.50\Omega \end{aligned}
R_2=60\Omega
R_3=100\pm 5%\Omega
The value of R_3 \text{with} \pm 5% of tollerance,
\begin{aligned} R_3&= 100\pm 5\% \\ &= 100+100 \times \frac{5}{100}=105 \Omega \\ &= 100-100 \times \frac{5}{100}=95 \Omega \end{aligned}
In both condition, the bridge is balance, so under balance condition,
\begin{aligned} \frac{R_1}{R_3}&=\frac{R_2}{R_x} \\ R_x&=\frac{R_2R_3}{R_1} \\ (i) \; \text{when}, \; R_3&=105\Omega \\ \therefore \;\;R_x &= \frac{105 \times 65}{50}=136.50\Omega \\ (ii)\; \text{when}, \; R_3&=95 \Omega \\ \therefore \;\;R_x &= \frac{95 \times 65}{50}=123.50\Omega \end{aligned}
Question 6 |
A (0-50 A) moving coil ammeter has a voltage drop of 0.1 V across its terminals at full scale deflection. The external shunt resistance (in milliohms) needed to extend its range to (0-500 A) is _______.
0.11 | |
0.22 | |
0.45 | |
0.68 |
Question 6 Explanation:

I_{fs}=50A,
V_m=0.1V,
R_m=\frac{0.1}{50}=2 \times 10^{-3}\Omega
\because \; m=10
R_{sh}=\frac{R_m}{(m-1)}
\;\;\;=\frac{2 \times 10^{-3}}{9}=0.22\Omega
Question 7 |
Of the four characteristics given below, which are the major requirements for an instrumentation amplifier?
P. High common mode rejection ratio
Q. High input impedance
R. High linearity
S. High output impedance
P. High common mode rejection ratio
Q. High input impedance
R. High linearity
S. High output impedance
P, Q and R only | |
P and R only | |
P, Q and S only | |
Q, R and S only |
Question 8 |
In the following chopper, the duty ratio of switch S is 0.4. If the inductor and capacitor are sufficiently large to ensure continuous inductor current and ripple free capacitor voltage, the charging current (in Ampere) of the 5 V battery, under steady-state, is_____.


1 | |
1.5 | |
2.5 | |
3 |
Question 8 Explanation:

Step down chopper
\begin{aligned} \Rightarrow \; V_0&=\alpha V_s\\ V_0&=I_0R+E\\ I_0&=\frac{V_0-E}{R}\\ I_0&=\frac{\alpha V_s-E}{R}\\ &=\frac{(0.4)20-5}{3}=1A \end{aligned}
Question 9 |
A moving average function is given by y(t)=\frac{1}{T}\int_{t-T}^{t}u(\tau )d\tau. If the input u is a sinusoidal signal of frequency \frac{1}{2\tau } Hz, then in steady state, the output y will lag u (in degree) by ______ .
30 | |
60 | |
90 | |
120 |
Question 9 Explanation:
System input: \sin \omega _0 t
\begin{aligned} f_0&=\frac{1}{2T}Hz\\ \text{Therefore,}\\ \omega _0&=2 \pi f_0=\frac{2 \pi}{2T}=\frac{\pi}{T}\; rad/sec\\ y(t)&=\frac{1}{T}\int_{t-T}^{t}\sin \omega _0\tau \; d\tau\\ &=\frac{1}{T}[-\cos\omega _0 t ]_{t-T}^t\\ &=-\frac{1}{T}[\cos\omega _0 t -\cos\omega _0 (t-T) ]\\ &\left [ \because \; \omega _0=\frac{t}{T} \right ]\\ &=-\frac{1}{T}[\cos\omega _0 t -\cos\left ( \frac{\pi t}{T}- \pi \right ) ]\\ \\ &=-\frac{1}{T}[\cos\omega _0 t +\cos\omega _0 t ]\\ &=-\frac{2}{T} \cos\omega _0 t \\ &=\frac{2}{T}\sin (\omega _0 t -90^{\circ}) \\ &=\text{Output is lagging by }90^{\circ} \text{ w.r.t. input.} \end{aligned}
\begin{aligned} f_0&=\frac{1}{2T}Hz\\ \text{Therefore,}\\ \omega _0&=2 \pi f_0=\frac{2 \pi}{2T}=\frac{\pi}{T}\; rad/sec\\ y(t)&=\frac{1}{T}\int_{t-T}^{t}\sin \omega _0\tau \; d\tau\\ &=\frac{1}{T}[-\cos\omega _0 t ]_{t-T}^t\\ &=-\frac{1}{T}[\cos\omega _0 t -\cos\omega _0 (t-T) ]\\ &\left [ \because \; \omega _0=\frac{t}{T} \right ]\\ &=-\frac{1}{T}[\cos\omega _0 t -\cos\left ( \frac{\pi t}{T}- \pi \right ) ]\\ \\ &=-\frac{1}{T}[\cos\omega _0 t +\cos\omega _0 t ]\\ &=-\frac{2}{T} \cos\omega _0 t \\ &=\frac{2}{T}\sin (\omega _0 t -90^{\circ}) \\ &=\text{Output is lagging by }90^{\circ} \text{ w.r.t. input.} \end{aligned}
Question 10 |
The impulse response g(t) of a system, G , is as shown in Figure (a). What is the maximum value attained by the impulse response of two cascaded blocks of G as shown in Figure (b)?


\frac{2}{3} | |
\frac{3}{4} | |
\frac{4}{5} | |
1 |
Question 10 Explanation:
\begin{aligned} g(t)&=u(t)-u(t-1)\\ G(s)&=\frac{1}{s}-\frac{e^{-s}}{s}\\ G(s) \times G(s)&=g(t)* g(t) \end{aligned}

Maximum value =1

Maximum value =1
There are 10 questions to complete.