BEL Trainee Engineer Electronics Mock Test - 3 - Electronics and Communication Engineering (ECE) MCQ

Concept:

For a series RLC circuit, the net/effective impedance is given by:

X_L = Inductive reactance 

X_C = Capacitive reactance

The magnitude of the effective/net impedance is given by:

Also, the phase angle of the effective impedance is given by:

Calculation:

For a series R_L circuit, X_C = 0 Ω

Given, R = 5 kΩ and X_L = 5 kΩ

The magnitude of the net/effective impedance will be:

Putting on the respective values, we get:

Z = 5√2 = 7.07 kΩ

The phase angle of the effective impedance will be:

ϕ = 45° 

• The radar altimeter measures absolute altitude, i.e. the altitude that the aircraft is above the ground (Above ground level)
• The radar altimeter system sends out a radio wave, and that radio wave bounces off the ground and is received.
• AGL is computed by taking the speed of the radio wave and the amount of time for it to return.
• Unlike radar, the signal is not pulsed, but is Frequency modulated and is transmitted at 4300 MHz at 350 milliwatts.
• Radar Altimeters operate between - 20 and 2500 feet.

FOM of a logic Family is defined as the product of the propagation delay (t_pd) and the power dissipation (P_Davg), i.e.

FOM = t_pd × P_D(avg)

For the best IC operation, FOM should be as small as possible.

Units: ns × mW

= pJ (pico Joule)

Propagation delay (t_pd­):

t_PHL = delay time in going form High to low logic

t_PLH = Delay time in going from low to High logic

Power dissipation (P_D):

P_D(avg) = I_cc × V_cc

V_CC = power supply

I_CC = avg collector current calculated as the average of the High and low current, i.e.

Calculation:

The given situation can be visualized as shown:-

The approximate directivity is given by:

The correct answer is option 4): (f_T = f_βh_fe)

Concept:

In BJT, at the high–frequency end, there are two factors that define the – 3dB cutoff point:

The network capacitance ( parasitic and introduced) and the frequency dependence of hfe (β )

The frequency at which |hfe|dB = 0 dB is indicated by f_T.

f_T = f_βh_fe

The hybrid π model for high-frequency analysis is given as

Explanation:

Very High Frequency:

It has a range of  30 MHz to 300 MHz and are used in television broadcast.

Some ultra high frequencies are also used in television broadcasting.

Additional Information

The frequency spectrum for the complete range is:

Hence option (3) is the correct answer.

Norton's Theorem :

Any linear bilateral network consisting of multiple resistances and sources can be replaced by a resistor R_N in parallel of Current source I_N .

Calculation:

R_N : 

Disable all independent sources and find equivalent resistance.

R_N = ( 20 || 10 ) + 20 = 26.6 ohm

I_N :

Short circuited current

Let node voltage of 10 ohm is V

Apply nodal analysis

V = 2.5V

I_N = V /20 = 125 mA

The correct answer is Data packets sent over TCP/IP are private.

Key Points

• Transmission Control Protocol (TCP) is a communications standard that enables application programs and computing devices to exchange messages over a network.
• It is designed to send packets across the internet and ensure the successful delivery of data and messages over networks. 
• TCP guarantees the integrity of the data being communicated over a network. Before it transmits data, TCP establishes a connection between a source and its destination, which it ensures remains live until communication begins.
  • It then breaks large amounts of data into smaller packets, while ensuring data integrity is in place throughout the process.
• Data packets sent over TCP/IP are not private, which means they can be seen or intercepted. Hence option 2 is correct.
• So, it is recommended to avoid using public Wi-Fi networks to send private data and ensure information is encrypted. One way to encrypt data being shared through TCP/IP is through a virtual private network (VPN).
• TCP/IP communications provide end-to-end error checking of data transmission for reliable and sequential exchange of data, guarantees the sequential delivery of data packets and so it comes with high latency.

Additional Information

• Error control in TCP is mainly done through the use of three simple techniques: Checksum, Acknowledgement and Retransmission.
  • Every segment contains a checksum field which is used to find corrupted segments. If the segment is corrupted, then that segment is discarded by the destination TCP and is considered lost.
  • In acknowledgement affirm that the data segments have been delivered. Control segments that contain no data but have sequence numbers will be acknowledged as well but ACK segments are not acknowledged.
  • In retransmission, when a segment is missing, delayed in delivering to a receiver or corrupted when it is checked by the receiver then that segment is retransmitted again. 
  • Segments are retransmitted only during two events: when the sender receives three duplicate acknowledgements (ACK) or when a retransmission timer expires.
• ​ High-level protocols that need to transmit data like File Transfer Protocol (FTP), Secure Shell (SSH), and Telnet use TCP Protocol.  

Since output of system for any value of n is only depends on values of input at n or previous values the system, it is causal system

3y[n] = 2y [n - 2] – 2x[n] + x [n - 1]

Taking z – transform

3y[z] = 2y (z) z ^{- 2} – 2x[z] + x (z) z ^{- 1}

It has poles at

Since poles lie in unit circle |z_p| < 1 the system is stable

The Correct Answer is Option (1) i.e A program.

• An operating system is a software or a program that manages computer hardware, software resources, and provides common services for computer programs.
• An operating system is an interface between the computer user and computer hardware.
• The dominant desktop operating system is Microsoft Windows with a market of around 82.74%, followed by macOS by Apple Inc.
• Some popular operating system include Linux operating system, Windows operating system, VMS, OS/400, macOS, etc.
• Some of the important functions of an operating system are :
  • Memory Management
  • Processor Management
  • Device Management
  • File Management
  • Security
  • Job Accounting
  • Co-ordination between other software and users.

The transport layer can be either connectionless or connection oriented. A connectionless transport layer treats each segment as an independent packet and delivers it to the transport layer at the destination machine. A connection oriented transport layer makes a connection with the transport layer at the destination machine first before delivering the packets. After all the data are transferred, the connection is terminated.

• Complementary Metal-oxide-semiconductor (CMOS) uses complementary & symmetrical pair of P-type & n-type MOSFETS.

• The two important characteristics of CMOS devices are high noise immunity and low power dissipation.
• CMOS devices dissipate less power than NMOS devices because the CMOS dissipates power only when switching (“dynamic power), whereas N channel MOSFET dissipates power whenever the transistor is on because there is a current path from Vdd to Vss.
• In a CMOS, only one MOSFET is switched on at a time. Thus, there is no path from voltage source to ground so that a current can flow. Current flows in a MOSFET only during switching.
• Thus, compared to N-channel MOSFET has the advantage of lower drain current from the power supply, thereby causing less power dissipation.

Concept:

To avoid slope overload, we require

Calculations:

Given,

m(t) = A tanh(βt)

To avoid slope overload

Since,

It follows the maximum value of sec h(βt) is 1, which occurs at time t = 0.

Hence |sec h(βt)|_max = 1

∴ 

Δ ≥ Aβ T_s.

Hence option (1) is correct.

Scattering:

• Scattering is the loss of signal caused by the diffusion of a light beam, where the diffusion itself is caused by microscopic variations in the transmission medium
• Scattering typically happens when a light signal hits an impurity in the fibre

Dispersion:

• Dispersion is defined as pulse spreading in an optical fibre
• As a pulse of light propagates through a fibre, elements such as numerical aperture, core diameter, refractive index profile, wavelength, and laser linewidth cause the pulse to broaden

Attenuation:

• The attenuation of an optical fibre measures the amount of light lost between input and output
• Total attenuation is the sum of all losses
• For a given fibre, these losses are wavelength-dependent

Inter-symbol Interference:

• This is the effect of dispersion on optical fibre
• Intersymbol interference occurs when the pulse spreading caused by dispersion causes the output pulses of a system to overlap

Moving Iron Instruments are the most common type of ammeter and voltmeter used at power frequencies in laboratories.

These instruments are very accurate, cheap, and rugged as compared to other AC instruments.

Working Principle of Moving Iron Instruments:

In Moving Iron Instruments, a plate or van of soft iron or of high permeability steel forms the moving element of the system.

The iron van is so situated that it can move in the magnetic field produced by a stationary coil.

The below figure shows a simple moving iron instrument.

The stationary coil is excited by the current or voltage under measurement.

When the coil is excited, it becomes an electromagnet, and the iron van moves in direction of offering a low reluctance path.

Thus the force of attraction is always produced in a direction to increase the inductance of the coil.

Mind that as the van follows the low reluctance path, the net flux in the air gap will increase which means increased flux linkage of the coil, and hence inductance of coil will increase.

It shall also be noticed that the inductance of the coil is variable and depends on the position of the iron van.

Torque Equation of Moving Iron Instruments:

Deflecting torque in Moving iron Instruments is given as

Td =  (1/2)I²(dL/dƟ)

In moving iron instruments, the controlling torque is provided by spring. Controlling torque due to spring is given as

Tc = KƟ

Where K = Spring constant

Ɵ = Deflection in the needle

In equilibrium state, deflecting and controlling torque shall be equal as below.

Deflecting Torque = Controlling Torque

⇒ Td = Tc

⇒ (1/2)I²(dL/dƟ) = KƟ

⇒ Ɵ = (1/2)(I²/K)(dL/dƟ)

PSK(Phase Shift Keying):

In PSK (phase shift keying), binary 1 is represented with a carrier signal and binary 0 is represented with 180° phase shift of a carrier, i.e. the phase of the carrier signal is changed by varying the sine and cosine inputs at a particular time.

For binary ‘1’ → S₁ (A) = Acos 2π f_Ct

For binary ‘0’ → S₂ (t) = A cos (2πf_Ct + 180°) = - A cos 2π f_Ct

The Constellation Diagram Representation is as shown:

Important Point

ASK, PSK and FSK are signaling schemes used to transmit binary sequences through free space. In these schemes, bit-by-bit transmission through free space occurs.

FSK (Frequency Shift Keying):

In FSK (Frequency Shift Keying) binary 1 is represented with a high-frequency carrier signal and binary 0 is represented with a low-frequency carrier, i.e. In FSK, the carrier frequency is switched between 2 extremes.

For binary ‘1’ → S₁ (A) = Acos 2π f_Ht

For binary ‘0’ → S₂ (t) = A cos 2π f_Lt . The constellation diagram is as shown:

ASK(Amplitude Shift Keying):

In ASK (Amplitude shift keying) binary ‘1’ is represented with the presence of a carrier and binary ‘0’ is represented with the absence of a carrier:

For binary ‘1’ → S₁ (t) = Acos 2π f_ct

For binary ‘0’ → S₂ (t) = 0

The Constellation Diagram Representation is as shown:

where ‘I’ is the in-phase Component and ‘Q’ is the Quadrature phase.

As we can say that ,
Nebulous is indistinct something that does not have a clear outline, Concrete is solid and of definite shape.
Hence , Concrete is antonym ( opposite ) of Nebulous . 

The correct answer is 1024.

Key Points

• One gigabyte in the memory context of data storage comprises 1024 MB (megabyte) or 1 million bytes.
• In SI units, it is written as 109. This is the most commonly used unit of storage in modern times.

Additional Information

•  Unit Conversion Table:
  • 4 bit = 1 Nibble
  • 8 bit = 1 byte
  • 1 KB (kilobyte) = 1024 byte
  • 1 MB (megabyte) = 1024 KB
  • 1 GB (gigabyte) = 1024 MB
  • 1 TB (terabyte) = 1024 GB
  • 1 PB (petabyte) = 1024 TB
  • 1 XB (Exabyte) = 1024 PB
  • 1 ZB (zettabyte) = 1024 XB
  • 1 YB (yottabyte) = 1024 ZB

The correct answer is ​Program Unit.

• A program unit is a sequence of one or more lines, organized as statements, comments, and directives. A subprogram is either a function or a subroutine and is either an internal, external, or module subprogram.
• A program unit is usually a sequence of statements that define the data environment and the steps necessary to perform calculations; it is terminated by an END statement.

Additional Information 

• A Fortran program consists of one or more program units.
• Alternately referred to as a processor, central processor, or microprocessor, the CPU (pronounced sea-pea-you) is the central processing unit of the computer.
• A computer's CPU handles all instructions it receives from hardware and software running on the computer. It is often referred to as the brain of the computer.
• Components of CPU:
  • ALU (arithmetic logic unit) - performs mathematical, logical, and decision operations.
  • CU (control unit) - directs all the processor's operations.
  • Storage Unit.

User-level threads are very effective because a mode switch is not required to switch from one thread to another. Here is a detailed explanation:

• Definition: User-level threads are managed by the application without the intervention of the operating system. They are created and scheduled by the application itself.


• Advantages: User-level threads offer several advantages, including:




• No mode switch: User-level threads do not require a mode switch from user mode to kernel mode when switching between threads. This makes the context switch faster and more efficient.


• Thread management control: The application has full control over thread management, including scheduling, synchronization, and resource allocation.


• Portability: User-level threads can be easily ported across different operating systems, as they are independent of the underlying kernel.




• Disadvantages: User-level threads also have some limitations:




• Blocking system calls: If a user-level thread performs a blocking system call, it will block the entire process, as the kernel is unaware of the individual threads.


• Limited parallelism: User-level threads may not fully utilize the available hardware resources, as the operating system schedules threads at the process level rather than at the thread level.

In conclusion, user-level threads are very effective because they eliminate the need for a mode switch and provide greater control over thread management. However, they also have some limitations that need to be considered in certain scenarios.

Internet:

1. The Internet (or internet) is the global system of interconnected computer networks that uses the Internet protocol suite (TCP/IP) to communicate between networks and devices.
2. It is a network of networks that consists of private, public, academic, business, and government networks of local to global scope, linked by a broad array of electronic, wireless, and optical networking technologies.
3. The Internet carries a vast range of information resources and services, such as the inter-linked hypertext documents and applications of the World Wide Web (WWW), electronic mail, telephony, and file sharing.

Therefore, The word 'Internet' is a composition of Inter-connection and Network.

Confusion Points 

Internet is an interconnected computer network. That's why it is Inter-connection and Network. 

Dos

The correct option is (4) Alt + K

Key Points

• Add transitions between slides by opening the Transitions tab. tap Alt+K.
• Choose the slide to which you wish to apply the transition. Choose a transition from the Transitions tab. To get a preview, choose a transition.
• The keyboard shortcut Alt+R is frequently used to open the Review tab in the Ribbon of Office products.
• To open the slide show settings in PowerPoint, use Alt+S on the keyboard.

Additional Information

Slide transitions are a great way to liven up your presentation.

• Choose the slide to which you wish to apply the transition.
• Choose a transition from the Transitions tab.
• To determine the kind and direction of the transition, select Effect Options.
• To get a preview of the transition, select Preview.

According to question,
A's weight is twice as much as of B.
1. A = 2B.
B's weight is four and half times as much as of C.
2. B = 4.5C
C's weight is half times as much as of D.
3. C = D/2.
D's weight is half as much as of E.
4. D = E/2.
E weight is less than A but more than C.
5. A > E > C
from 1 and 2
6. A = 2B = 9C
from 6 and 3
7. A = 2B = 9C = 4.5D
from 7 and 4
8. A = 2B = 9C = 4.5D = 2.25E
so from 8 and 5
A > B > E > D > C

So A is the heaviest in the weight.

The logic followed is:

Therefore, the number of students in the group are 20 + 30 + 70 = 120

The number of students = 120

Hence, the correct answer is "120".

Logic: We have to make a family tree as per the above conditions.

Consider the below symbols.

The family tree for the given problem is as follows:

In the question, Her father is the only son of my father means woman's father and the man's father's only son is the same person means he is talking about himself. 

As the man is the father of that woman.

Therefore, the correct answer should be "Option 1: Father".  

The main triangle shown is in the given figure and this the total no. of triangle is 15. remaing triangle we can find out in the drawing the triangle in the image.

From the final arrangement, we get S sits exactly opposite to U.
Hence, Option D is correct.
References:
Q sits second to the right of P.
Inferences:
From the above reference, we get two different cases:
In case 1, P sits at the corner or in the middle of the side of the table.

References:
R is not an immediate neighbor of P.
R faces inside.
R sits third to the left of W.
Inferences:
From the above references, we get different cases for R, in case 1(a) in which R faces P and in case 1(b) R faces Q.

References:
T sits third to the left of U.
V sits second to the left of U.
Inferences:
So, from this case 1(b) and case 2 will be eliminated as there is no possible place for V in these cases after placing U and T. Hence, we get our final arrangement in which S sits on the immediate left of P and V sits on the immediate right of R.
Final arrangement: