BEL Trainee Engineer Electronics Mock Test - 10 - Electronics and Communication Engineering (ECE) MCQ

The current gain β of the transistor is given by:

D_E: Diffusion constant of minority carriers in the emitter

D_B: Diffusion constant of minority carries in Base

N_E: emitter doping concentration

N_B: Base doping concentration

L_E: Length of diffusion of minority carriers in emitter

W_B: Base width.

 ratio of conductivity from emitter to base = 10 (Given)

LE_n (minority carries in emitter)  (hole diffusion length in base)

Taking the case of Si:

β = 2.1 × 10 × 5 = 105

Taking case of Si:

β = 2.6 × 10 × 5 = 130

None of the given option match, but if we check the option by satisfying the relation  

Only option 1 is satisfying

Hence option 1 is correct.

Concept:

The electric field for an isotropic antenna varies with the distance as given:

1/r term is the most dominant. 

Calculation:

Given,

E = 3 V/m for r = R

For r₂ = 3R,

Electric field E₂ will be 

Concept:

WBFM is given as:

Actual BW of WBFM is ∞ so to Bandlimit WBFM signal all of its significant lower-order sidebands retained and higher-order insignificant sidebands should be eliminated.

Calculation:

BW = 32 KHz

Given FM system has 8 sidebands (significant) i.e. 8 sidebands on the positive side of frequency & 8 sidebands on the negative side of the frequency

So, It has sidebands up to order ‘8’.

According to carson rule:

Significant sideband upto order of (β + 1) has B.W

BW = (β + 1) 2f­_m

So,

32 = 8 × 2 f_m

f_m = 2 kHz

Concept:

In an N-type semiconductor, the doping done is of pentavalent atoms.

As the doping density increases the Fermi level moves towards the conduction band.

If ND is the doping density, then the upward shift in the Fermi level is given by

From the expression

As ND increases  increases and the shift increases.

Calculation:

The electrostatic potential of n type semiconductor

E_F - E_i = qν_n

E_F - E_i = 0.359 eV

Concept:

If V₀ is the initial voltage across capacitor

Calculation:

Circuit diagram at laplace domain

As capacitor was initially relaxed: V₀ = 0 (Voltage source will be replaced by a short circuit)

Given;

C = 1μF

R= 5 Ω 

Use nodal analysis at node V_C

Determine the values of A and B using partial fraction

A_s=0 = (10/C)/(s + 1/RC) = (10/C)( 0 +1/RC) = 10R

B_{s= -1/RC} = (10/C)/s = -10R

⇒ V_C = (10R)/s + (-10R)/(s + 1/RC)

Using inverse laplace transform:

⇒ V_C = 10R -10Re^-t/RC

⇒ V_C = 10R(1 -e^-t/RC)

∴  V_C = 50(1 -e^-t/5)

Impulse function is exist only at t= 0 and the area under impulse function is unity

|_t=0 = 1

|_t=0 = 1

Discrete time Laplace transform is Z- transform.

40 GHz is the frequency limit of an optical fiber.

Additional Information

• Optical fibers are transparent fibers and act as a light pipe to transmit light between its two ends. They are made up of silicon dioxide.

• The total internal reflection occurs when the angle of incidence is greater than the critical angle.
• In optical fiber glasses of a high and lower index are assembled in the precise order.
• The light comes in from one end of the fiber and after thousands of successive internal reflections, the light reaches the opposite end of the fiber with almost zero loss. 

CONCEPT:

Gauss law:

• According to this law, the total flux linked with a closed surface called Gaussian surface is 1/ϵ_o times the charge enclosed by the closed surface.

• This law is applicable to all the closed surfaces.
• According to this law, the total flux linked with a closed surface called Gaussian surface is 1/ϵ_o times the charge enclosed by the closed surface.
• So this law is applicable to all the closed surfaces. Hence, option 2 is correct.

In satellite communication, uplink and downlink frequencies are kept different.

Uplink frequency > Downlink frequency

Hence, option C is correct.

Explanation:

• The transmission of a signal from the first earth station to satellite through a channel is called uplink​.
• The transmission of a signal from the satellite to the second earth station through a channel is called the downlink.
• Uplink frequency is the frequency at which, the first earth station is communicating with the satellite.
• The satellite transponder converts this signal into another frequency and sends it down to the second earth station.
• This frequency is called Downlink frequency. In a similar way, the second earth station can also communicate with the first one

In the atmosphere, the attenuation is directly proportional to frequency.

Since the satellite cannot afford to have high powers downlink is sent at lower frequencies, while the station at earth can afford to transmit at high frequencies.

• A frame check sequence (FCS) refers to an error-detecting code added to a frame in a communications protocol

Layer 2 Frame Format:

Ethernet Frame Format consists of Cyclic Redundancy Check (CRC) in Frame check sequence Field used for error-detection, but CRC can also be used for error correction.

The principle of operation of fibre-optic cable is Reflection.

Key Points

• In an optical fiber, the information is passed through light, which must not escape outside of it.
• This phenomenon of confining the light inside the optical fiber is termed as Total internal reflection.
• For this, the construction and material used to ensure that the total internal reflection of light takes place to prevent the escape of it.

Principle:

• When light travels from a high refractive index medium to a low-refractive-index medium, it is refracted away from the normal as shown:

• At a certain angle θ_i, there is no refracted wave and the wave is totally internally reflected  (θ_r = 90°).
• This angle is called a critical angle.
• Inside an optical fiber, we have a high refractive index core (n₁) and low refractive index cladding (n₂).
• This results in the propagation of waves inside a fiber through total internal reflection phenomenon.

The correct answer is 'option 2'

Solution:

In moving-coil instruments, the deflection is directly proportional to the current. i.e. θ ∝ I. 

∴ The scale used in the moving coil instruments is linear in nature.

Additional Information

• In thermocouple instruments, the deflection is directly proportional to the square of the current. i.e. θ ∝ I²
• In moving iron instruments, the deflection is directly proportional to the square of the current. i.e. θ ∝ I²
• In hotwire instruments, the deflection is directly proportional to the square of the current. i.e. θ ∝ I²

The resistances are classified depending upon the values as:

• The resistances of the order of 1 Ω (or) less than 1 Ω are classified as low resistances.
• The resistances from 1 Ω to 100 KΩ are classified as medium resistances.
• The resistances of the order of 100 KΩ (or) higher are classified as high resistances.

The methods used for the measurement of low resistances are:

Kelvin's Double Bridge Method: Kelvin's Double Bridge Method is a modification of the Wheatstone bridge method. 

The circuit diagram is shown below:

R is the unknown resistance and S is the standard resistance.

Ammeter Voltmeter Method:

Connecting voltmeter across the load is suitable for low resistance measurements.

Ammeter reading is:

Voltmeter reading = E

The calculated resistance is:

Potentiometer Method:

• It works by comparing the unknown resistance with the standard resistance.
• The voltage drop across the known and unknown resistance is measured and by comparison the value of known resistance is determined.

Important Points

Given: 

R_b = 4 kbps

In FSK, the best possible interval between the carriers is given as:

ΔT = 0.5 msec

Important Point

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.

PSK(Phase Shift Keying):

In PSK (phase shift keying) binary 1 is represented with a carrier signal and binary O is represented with 180° phase shift of a carrier

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:

Straw in the wind: something believed to be a sign or warning of a future event.
Shot in the arm: something that has a sudden and positive effect on something, providing encouragement and new activity.
Stroke of luck: a fortunate occurrence that could not have been predicted or expected.
Plain sailing: smooth and easy progress in a process or activity.
Hence, the correct answer is option A. 

The author says that government has launched a number of schemes. But no solid step taken.

The correct answer is Option 3. 

Key Points

• Secondary memory, also known as auxiliary memory or non-volatile memory, is a type of computer memory that retains data even when the power is turned off.
• It is non-volatile, meaning it does not lose its data when the power supply is disconnected.
• Secondary memory is used for long-term storage of data and files, such as hard disk drives (HDDs), solid-state drives (SSDs), magnetic tapes, and optical discs like CDs and DVDs.
• In contrast, volatile memory, such as random access memory (RAM), is temporary and loses its data when the power is removed.

Additional Information

•  Memory refers to the electronic storage space where data and instructions are stored for processing by the computer's central processing unit (CPU). Computer memory is an essential component that allows the computer to perform tasks and store information.
• Here are the main types of memory in a computer system:
  • Random Access Memory (RAM): RAM is the primary type of computer memory used for temporary data storage while the computer is running. It provides fast access to data and instructions that are actively used by the CPU. RAM is volatile, meaning its contents are lost when the computer is powered off or restarted.
  • Read-Only Memory (ROM): ROM is a type of memory that stores permanent instructions or data that cannot be modified or erased. It contains firmware or software instructions that are necessary for booting up the computer and initializing essential components. ROM retains its contents even when the computer is powered off.
  • Cache Memory: Cache memory is a small, high-speed memory that resides closer to the CPU. It serves as a buffer between the CPU and main memory (RAM). Cache memory stores frequently accessed data and instructions, allowing the CPU to quickly retrieve them without accessing the slower main memory. It helps improve overall system performance.
  • Virtual Memory: Virtual memory is a technique used by operating systems to extend the available memory beyond the physical RAM capacity. It uses a portion of the computer's hard drive as an extension of RAM. Virtual memory allows running more programs and handling larger datasets than what can fit in physical memory alone.
  • Flash Memory: Flash memory is a non-volatile memory used for long-term data storage in devices like USB drives, solid-state drives (SSDs), and memory cards. It retains data even when the power is turned off. Flash memory is popular

• Definition: A bootstrap is a small initialization program that starts up a computer system. It is responsible for loading and executing the operating system or other software components necessary for the system to function.


• Function: The bootstrap program resides in the computer's firmware or read-only memory (ROM) and is automatically executed when the system is powered on or restarted.


• Initialization: The bootstrap program initializes the hardware components of the computer, such as the processor, memory, and input/output devices.


• Bootloader: The bootstrap program then loads and executes a more complex bootloader, which further initializes the system and loads the operating system or other software from storage devices, such as hard drives or solid-state drives.


• Chain of Execution: The bootstrap process involves a chain of execution, where each component in the chain is responsible for loading and executing the next component until the operating system is fully loaded and running.


• Importance: The bootstrap program is crucial for the proper functioning of a computer system as it ensures that the necessary software is loaded and executed in the correct order.

The correct answer is Tab

Important Points

• Tab key should be pressed continuously while moving a column in MS Excel.
• Select the column that you want to move. Hold the Shift Key from your keyboard. Move your cursor to the edge of the selection.
• It would display the move icon (a four-directional arrow icon). Click on the edge (with the left mouse button) while still holding the shift key.
• Move it to the column where you want this row to be shifted.

Key Points

• Open a blank workbook on Excel and press the TAB key and your active cell should move to the column on the right. Hold in Shift and then press the TAB key and it reverses direction moving your active cell to the column on the left. You can also use this shortcut when tabbing through options in a dialogue box.
• Press the Enter key and your active cell should move to the row below. Hold in Shift and then press the Enter key and it reverses direction moving your active cell to the row above.
• 
• Using Shift and the arrow keys will select cells and using Shift when clicking on cells will select a contiguous cell range.

The correct answer is Ctrl + 3.

Key Points

The correct answer is a - ii, b - iii, c - iv, d - i.

Key Points

• Optical Mouse
  • It is a computer mouse that uses a light source, typically a light-emitting diode (LED), and a light detector, such as an array of photodiodes, to detect movement relative to a surface.
  • It uses special-purpose image processing chips, as the mouse takes over 1,000 images/ps below the surface level to detect movement through reflected light changes.
  • It uses camera technology and digital processing to compare and track the position of the mouse.
• Track Ball
  • ​It is an input device used to enter motion data into computers or other electronic devices.
  • It is designed with a moveable ball on the top, which can be rolled in any direction.
  • A moving ball is mounted on a stationary device that can be rotated mechanically (manually) using the fingers
• Digitizing(Graphic) Tablet
  • ​It is a computer input device that enables a user to hand-draw images, animations and graphics, with a special pen-like stylus.
  • Alternatively, it is a tool used to convert hand-drawn images into a format suitable for computer processing.
• Joystick
  • ​It is an input device that can be used for controlling the movement of the cursor or a pointer in a computer device.
  • It was first invented by C. B. Mirick at the U.S. Naval Research Laboratory, and patented in 1926.
  • It includes a base and a stick that can be move to the different direction.
  • It is used to indicate position in video game play.

Fragmentation is a common issue in memory management, where available memory becomes divided into small, non-contiguous blocks. This can lead to inefficient memory utilization and can negatively impact the performance of a system. The problem of fragmentation occurs in heap allocation due to the following reasons:

• Internal Fragmentation: This type of fragmentation occurs when allocated memory blocks have unused space within them. It happens because memory is allocated in fixed-size chunks, and if the requested memory size is smaller than the chunk size, the remaining space is wasted.



• External Fragmentation: External fragmentation occurs when free memory blocks are scattered throughout the heap, making it difficult to find contiguous blocks of memory for allocation. This can happen when memory blocks are allocated and deallocated in a non-sequential order, leaving gaps between allocated blocks.



• Compaction: Compaction is a technique used to reduce external fragmentation. It involves moving allocated blocks of memory closer together and filling the gaps left by deallocated blocks. However, compaction can be time-consuming and may require updating memory references, which can be challenging in systems with dynamic memory allocation.



• Allocation Policies: The choice of allocation policies can also contribute to fragmentation. For example, a first-fit allocation policy may result in more external fragmentation compared to a best-fit policy.

Overall, fragmentation in heap allocation can lead to inefficient memory usage, decreased performance, and increased memory overhead. It is important to consider fragmentation issues and employ appropriate memory management techniques to mitigate these problems.

The correct answer is .xlsx.

• .xlsx is the default XML-based file format for Excel 2010 and Excel 2007.
• Tar, Zip, rar are archive formats.

Additional Information

• An archive format is the file format of an archive file.
• Some file types and file extensions are as follows.

Given:

6% of √(3x) = 0.18

Concept:

Concept of percentage. It is calculated on the basis of 100

For example x% means x out of 100

Calculation:

According to the question,

⇒ 6% of √(3x) = 0.18

⇒ (6/100) × √(3x) = 18/100

⇒ √(3x) = 3

⇒ 3x = 3²

⇒ 3x = 9

⇒ x = 3

∴ Required value of x is 3.

As 'Table' is made from 'Wood' in the same way 'Shirt' is made from 'Cloth'.

In the figure the simplest Triangles are GLK, DLJ, DJM,HMN,QRE,IRA,IPA, and FPO is the 8 triangle

and another is BDO,CDQ,DLM, PRA, KFI, NEI, HJI, GJI,DKI, and DNI is the 10 triangle

and another is DIE, DFI, DOA, DQA, and GHI is the 5 triangle

and DCA, DBA is the 2 triangle and DEF and ABC so that according to the figure the total no. of triangle is
8 + 10 + 5 + 2 + 1 + 1
= 27

The logic followed here is:

The figure in image 2 is water image of the figure in image 1. The figure in image C of answer figure will be the water image of the figure in image 3.

Hence, image C of answer figure is the correct answer.