BEL Trainee Engineer Electronics Mock Test - 5 Free MCQ Practice Test with Solutions - Electronics and Communication Engineering (ECE) (2024)

In ASK modulation each symbol is transmitted using a single bit.

Hence, the Bit rate = Baud rate

Baud rate = 300 baud

Note: QPSK transfers 2 bit for 1 symbol

∴ The baud rate for QPSK will be:

Baud rate = 300/2 = 150 baud

The current equation fora depletion type JFET is given by:

---(1)

I_DSS= Saturation Current of the JFET

V_GS= Gate to source voltage applied

V_P= Pinch off voltage at which I_D= I_DSS,which is also the voltage at which the channel ceases to exist. It is also denoted by V_GS(off).

The transconductance (g_m) is defined as the change in IDwith respect to a change in the gate to source voltage (V_GS), i.e.

Taking the differentiation of Equation (1), we get:

The maximum value of g_moccurs when V_gs= 0

With

g_m0 is the maximum transconductance at V_GS = 0

The characteristic graph for a change in V_GS is as shown:

The maximum radar range is given by:

Where

P_t is the transmitted power

A₀ is capture area

S = effective area

P_min = minimum power at the receiver which can be recognized

λ = wavelength of EM radiation

Concept:

The generator matrix is of the form G = [I₃ P^T] and parity check matrix is of the form H = [P I₃]

Calculation:

Given:

Concept:

Y parameter:

I₁ = V₁ Y₁₁ + V₂ Y₁₂

I₂ = V₁ Y₂₁ + V₂ Y₂₂

...1)

...2)

Calculation:

For the given question Z = 5Ω

Concept:

If we pass the input signal to a single T-flip flop, we will get half of the frequency at the output.

Similarly, when we pass the input signal to n-T flip-flops, the output frequency (f_out) will be:

Application:

Given Input frequency f = 200Hz

f_out= 25 Hz

The frequency spectrum of an Amplitude Modulated wave is given by:

Bandwidth in AM = 2f_m = 6 kHz

f_m = 3 kHz

Upper sideband = f_c + f_m = 600 kHz

∴ The carrier frequency is:

f_c = 600 – f_m = 600 - 3 kHz

f_c = 597 kHz

The amplitude Acis constant in a phase-modulated or a frequency modulated signal. RF power does not depend upon the modulation index.

A general expression for a phase or frequency modulated signal is:

m(t) =the modulating signal

ω_c= Carrier frequency

k_kbecomes k_cfor FM and k_pfor PM.

The average power (P_avg) is given by:

(Always)

We observe that the transmitted power is independent of the modulation index in the case of FM.

Important Point

Transmitted power in AM is given by:

P_c= Carrier Power

∴ We conclude that If the modulation indexincreases, transmitted power increases in the case of AM and is unchanged for FM.

The reflection coefficient at the load is defined as:

where Z_L = load impedance and Z₀ = characteristic impedance.

Also, VSWR

Clearly, VSWR is a function of Z_L and Z₀.

For the given transmission line, Z₀ = Z_t and Z_L = Z_L

So, VSWR is constant and it doesn’t change with the length of the line, In the smith chart, this is represented as a constant VSWR circle.

• In electronics, a super-heterodyne receiver uses frequency mixing to convert a received signal to a fixed intermediate frequency (IF) which can be more conveniently processed than the original carrier frequency and audio stage does not need alignment in a radio receiver.
• Audio mixing is one of the applications where two different audio frequencies are mixed using super-heterodyne receiver.

Norton's Theorem states thatit is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single current source and parallel resistance connected to a load.

Step 1

Calculate open voltage Circuit, V_AB.

V₁ = V_AB

Step 2

Short Circuit Current

7V₂ - 10 = 0

Concept:

Superposition theorem:

The superposition theorem states that the response of anycircuit with multiple voltage and current sources is equal to the algebraic sum of the response produced by individual sources.

Calculation:

Case 1: When 5 V is active:

Applying KVL in loop ABEA:

-5 + 2I₂+ 6I = 0

2I₂+ 6I = 5........(i)

Applying KVL in loop ABCEA:

-5 + 2I₂+ 4I₂- 4I= 0

6I₂- 4I = 5..........(ii)

Solving (i) and (ii), we get:

I₂= 1.13 A

I = 0.454 A

Case 2: When 10V is active:

Applying KVL in loop ABEA:

-10 + 4I₂+ 6I = 0

4I₂+ 6I = 10..........(i)

Applying KVL in loop ABCEA:

-10+ 4I₂+ 2I₂- 2I= 0

6I₂- 2I = 10..........(ii)

Solving (i) and (ii), we get:

I₂= 1.81A

I = 0.454 A

I_net=0.454 +0.454

Inet=0.908 A

The correct answer is QueueOrganization.

Key Points

• Computer organization deals with the components of the connections, and behaviorof a computer as seen by the user.
• It deals with low-level design issues. The important components of a computer organization are ALU. CPU, memory organization.
• There are three types of computer organization based on the address fields: Single Accumulator organization, General register organization, and Stack organization. Hence option 2 is correct.
• In a Single accumulator organization, the register is used implicitly to process the instructions of a program and store the results in the accumulator.
• The instruction format used in this organization is one address field and hence the Central Processing Unit (CPU) is called one address machine.
• In this, the first ALU operand is stored in the accumulator and the other is either in registers or memory.

Additional Information

• A register is a high-speed temporary storage found in the CPU which speeds up the implementation of programs.
• In General register organization, multiple general-purpose registers are used instead of a single accumulator.
  • Here the computer uses two or three address fields in their instruction format.
• The Stack organization is also called Last In First Out (LIFO) which means the last stored element will be retrieved first.
• The ALU operations are performed on stack data. Both operands are always required on the stack, after the operation, the result is placed in the stack.

Concept:

Span:It is defined as the difference between the largest and smallest reading of the instrument.

If voltmeter scale is -V1to V2, then the span is given by (V2+ V1)

Explanation:

The voltmeter scale is -5 V to 5 V

If voltmeter scale is -V1to V2, then the span is given by (V2+ V1)

Now, span = 10+ 10= 20 V

Concept:

The data flow between the two devices can be in simplex, half duplex and full duplex modes.

1. Simplex

• In Simplex, communication is unidirectional. Only one of the devices sends the data and the other one only receives the data.
• The channel capacity is used by only one device all the time.

Example: A CPU sends data while a monitor only receives data.

2. Half Duplex

• In half duplex both the stations can transmit as well as receive but not at the same time. When one device is sending other can only receive and vice-versa
• The channel capacity is shared by both the devices but at a time only once device uses the full capacity.

Example: A walkie-talkie

3. Full Duplex

• In Full duplex mode, both stations can transmit and receive at the same time.
• The channel capacity is shared by both communicating devices at all times.

Example: mobile phones

The modulation index,for a Frequency modulated wave, is given by:

Δf = Frequency deviation

f_m = Modulating frequency

Calculation:

Given,Δf = 50 kHz and f_m = 100 Hz

β = 500

NAND Gate:

• NAND gate represents the complement of the AND operation.
• The graphic symbol for the NAND gate consists of an AND symbol with a bubble on the output, denoting that a complement operation is performed on the output of the AND gate.
• The logic NAND function can be expressed by the Boolean expression of, A.B.

​

Analysis:

Given A = 1100 and B = 1010

∴ The output will be 0111.

Concept:

The characteristic equation of the JK flip flop is:

where, J and K are the inputs

Q_n= Present output

Q_n+1= Future output

Calculation:

If an inverter is connected between J and K inputs, then

This equation is similar to thecharacteristic equation of the Dflip flop which is:

Hence, option 4 is correct

Concept:

Average Value:

The average value can be calculated by dividing the area by time period for one cycle

Rms value:

V_rms= V_m/ √3 = 0.577 V_m

Note: The result of RMS value for standard triangular waveform and sawtooth waveform should be remembered asV_m/√3

Application:

Given,

Peak voltage V_m = 200 V

Time period T = 2 sec

Concept

Zener Diode

• Zener diodes are specially designed to give accurate and stable reverse breakdown voltage.
• When it is forward-biased, it behaves like a standard diode.
• When reverse-biased, a small leakage current flows through it.
• If the reverse voltage across Zener diode is increased, a value of voltage will be attained at which reverse breakdown occurs.
• As the breakdown occurs, a sudden increase in current takes place called as Zener Current.
• Application: Voltage Regulator.

Zener Breakdown:

• Zener diode is a highly doped PN junction diode, due to which the depletion region between P and N becomes very narrow and the electric field is quite high.
• The breakdown easily occurs in reverse biased conditions because of the narrow junction in Zener Diode.
• Even though the electric field is very high, the electrons cantunnel through duenarrow depletion region.
• Incase, we narrow the space charge region, It isobvious that with the decreasedreverse voltage applied reverse breakdown will occur.
• This causes a high reverse current to flow even in the reverse bias. This can be used as a normal PN junction diode in forward bias because the junction is not destroyed.

Fig:Circuit Diagram or Layer Structure of Zener Diode

Additional Information
Difference between Zener and Avalanche Breakdown:

Laser is the abbreviation ofLightAmplification by theStimulatedEmission ofRadiation. It is a device that creates a narrow and low-divergent beamof coherent light, while most other light sources emit incoherent light, which has a phase that varies randomly with time and position.The principle of a laser is based on three separate features:

a) stimulated emission within an amplifying medium,

b) population inversion of electronics and

c) an optical resonator.

These are based on the principle of population inversion at a junction as explained in below figure

PWM :

It stands for pulse width modulation.

In this narrow or wide pulse is sent which means pulse width is varied.

For the generationof PWM

• Comparator
• MonoStable Vibrator

Hence correct option is "4"

The correct answer is 40 W

Concept:

Let's take the given below direction of voltage and current as a reference.

If any direction other than the reference was given, then don't forget to take the negative sign in the final answer.

Solution:

The graph shows that the current is zero, which means an open circuit condition. In this case, V is nothing but the Thevenin voltage across that point.

V_th= - 40 V

But the given direction is opposite to the reference direction.

V_th = - (-40 V) = 40 V

From the graph, as we take output voltage zero means short circuit then short circuit current would be I_SC= 4 A

= 10

According to the maximum power transfer theorem

= 40 W

The correct answer is option 1):(Timer)

Concept:

• This is a control system that manages a process defined by a series of tasks to be performed i.e. a sequence of operations, one after the other. Each operation in the sequence is performed either for a certain period of time and is called asequential control system
• Sequential timer circuits play a significant role in all sorts of timer applications such as switching devices in sequence for a given amount and switch appliances off after desired time.
• Relays are electrically operated switches that open and close the circuits by receiving electrical signals from outside sources.
• A contactor is a special type of relay used for switching an electrical circuit on or off. They are most commonly used with electric motors and lighting applications.
• A control transformer is a transformer that is used to step down the voltage to power the control devices of a circuit or machine. The reduced voltage provides a much safer environment for technicians working on the equipment.

Analyzing each statement in the question:

Statement-1:

• The term transient response and steady-state response arise naturally in the context of sinewave analysis.
• When the sinewave is switched on, the filter takes a while to “settle down” to a perfect sinewave at the same frequency.
• The filter response during this settling period is called the transient response of the filter. The amplitude and phase of the output are determined by the filter’s frequency response.

Now, length N FIR filters only remember N – 1 sample into the past.

Thus, for N filters, the duration of the transient response is N – 1 sample.

Example: Consider a length N = 1 (zero or less) FIR filter (a simple gain) will have no memory state at all. ∴ It is will be in steady-state immediately when the input sinewave is switched on.

A length N = 2 FIR filter reaches steady-state, one sample after the input sinewave is switched on as it has one sample of delay.

∴ We can observe that the transients are depending upon the number of memory elements, i.e. When the number of memory elements is finite, the transient is also of limited duration.

Hence statement 1 is correct.

Statement 2:

The general input-output relation for an FIR filter is given by:

y(n) = b_o x(n) + b₁ x(n - 1) + …. + b_N x(n - N)

x(n) = Input signal

y(n) = Output signal

b_i are the filter coefficients

Since b_i is finite, so if x(n) is bounded, y(n) is also bounded.

∴ The FIR filter is Bounded Input Bounded Output.

Hence statement 2 is correct.

Statement 3:

The transfer function of an FIR filter is not a rational function in z^-1. It is actually a polynomial function in z^-1, i.e. H(z) of FIR filter is:

Hence, Statement 3 is not correct.

Omni directional antennas radiate EM waves in all directions.

If the radiation pattern for this type of antenna is plotted, the pattern is a constant signifying that the radiated power is constant measured at any point around the antenna.

The Azimuth plane pattern is non-directional ,that means the antenna radiates its energy equally in all directions in the azimuth plane.

CONCEPT:

• The skin effect is the tendency of analternating electric current to become distributed within a conductorsuch that the current density is largest near the surface of the conductor and decreases with greater depths in the conductor.
• The electric current flows mainly at the skin of the conductor, between the outer surface and a level called theskin depth.
• Skin depth is defined as thereciprocal ofattenuation constanti.e.,

• Theattenuation constant (Real Part)isgiven by:

• For an idealconductor withσ≈∞,

Important Point

α is the attenuation constant given by:

ω = Operating frequency

μ = Permeability of the material

σ = Conductivity of the material

For a good conductor σ>> 1. The above expression for the attenuation constant can be approximated as:

∴ The attenuation constant becomes:

Thus, the skin depth becomes:

Note:Theskin depth is inversely proportional to the square root of frequency.

The correct answer is Memory Address Register.

Key Points

• Memory Address Register or MAR is a temporary storage in CPU or a parallel loading register that contains the next memory address to be read or written.
• Contents of the MAR are directly connected to the address bus.
• It is used to handle the address transferred to the memory unit. Hence option 2 is correct.
• When reading from memory, data addressed by MAR is stored into the Memory Data Register (MDR) and is used by the CPU.
• When writing to memory, the CPU writes data from MDR to the memory location whose address is stored in MAR.

Additional Information

• A memory Data Register or MDR is a register used for holding the information that is in the process of being transferred from the memory to the central processor.
  • It is also calledMemory Buffer Register.
• It contains the copy of the value in the memory location specified by MAR.
• It acts as a buffer that allows the processor and memory to act independently without being affected by minor differences in operation.

Concept:

The total memory can be calculated from the number of address lines and date-lines, i.e.

Total Memory = 2address lines× Data Lines

Also, 1 Kilo = 1 K = 210

1 Mega = 1 M = 220

1 Giga = 1 G = 230and

1 Tera = 1 T = 240

Calculation:

Since 1 Byte = 8 bits

8 KB = 8× 210× 8 bits

8 KB = 213× 8 bits

So, the Number of address lines = 13, and

The Number of data lines = 8

Explanation:

Mechatronics is the synergistic integration of mechanical engineering with electronics and intelligent control algorithms in the design and manufacture of products processes.

Mechatronicshas evolved through several stages that are primarily defined in terms of

• Primary level mechatronics
• Secondary level mechatronics
• Tertiary level mechatronics
• Quaternary level mechatronics

Primary Level Mechatronics:

• This level encompasses input/output (I/O) devices such as sensors and actuators that integrate electrical signalling with mechanical action at the basic control level. Electrically controlled fluid valves and relay switches are two examples.

Secondary Level Mechatronics:

• Itintegrates microelectronics into electrically controlled devices. Sometimes these products are stand-alone. An example is a cassette tape player.

Tertiary Level Mechatronics:

• This level further enhances the quality in terms of sophistication by incorporating advanced feedback functions into the control strategy.
• The mechatronic systems at this level are called smart systems.
• The control strategy uses microelectronics, microprocessors, and other application-specific integrated circuits (ASIC) as bits and pieces for control realization.
• A microprocessor-based electrical motor used for actuation purposes in typical industrial robots can be considered as an example of a tertiary level mechatronic system.

Quaternary Level Mechatronics: Intelligent control is very much a part of mechatronics now.

• Decision-making is done an influence is needed to perform the decided upon action.
• Machine or plant control intelligence is moving into the realm of human-like intelligence.
• This level attempts to improve smartness a step ahead by introducing intelligence and FDI (Fault Detection and Isolation) capability into the systems.
• At this level, mechatronics means many things to many people. Artificial Neural Networks and Fuzzy Logic try to capture some of the intellectual capabilities of intelligence.

Concept:

The maximum range of a radar system is given as

Where:

R_max= Maximum range of the radar

S_min= Minimum Detectable Signal Power

P_t= Peak transmitted power

A_e= Aperture Area

D = Antenna diameter

We observe that

Rmax∝ D^1/2 xP_t^1/4

Calculation:

P(new) = 16 P(old)

D(new) = 2D(old)

R_max∝D^1/2 xP_t^1/4

R(new) = √2D^1/2 x 2P^1/4

R(new) = 2√2R(old)

R(new) = √8R(old)

Hence, If thepeak transmitted poweris increased by a factor of 16,and the antenna diameter is increased by a factor of 2, the radar range will be increased by a factor of √8.

IMPATT diodeis used as a power generating device in semiconductor microwave oscillators and delivers a relatively high RF power to produce carriersignals for microwave transmission systems, particularly airborne and ground-based low-powerradarsystems.

IMPATT diode:

IMPATT diode consists of 4 layers according to the construction.

• An IMPATT diode hasn+-p-i-p+structure and is used with reverse bias.
• It exhibits negative resistance and operates on the principle of avalanche breakdown. I
• IMPATT diode circuits are classified as a broadly tunable circuit, low Q circuit, and high Q circuit. The impedance of the IMPATT diode is a few ohms.
• The word IMPATT stands for Impact Avalanche Transit Time diode.

The features of IMPATT diode oscillator are:

1) Frequency 1 to 300 GHz, Power output (0.5 W to 5 W for single diode circuit and up to 40 W for the combination of several diodes),

2) It has an efficiencyof about 20%.

3) Its applications include police radar systems for detection of speed, low power microwave transmitters, etc.

Concept:

The conductivity (σ) of a semi-conductor is given by:

---(1)

Where,

n₀ and p₀ are the concentration of electrons and holes in doped semiconductor

μ_nis electron mobility

μ_pis hole mobility

Analysis:

For a doped semiconductor, one of the charge concentration will dominate and the conductivity will be:

(For semiconductors doped with donor impurities only)

(For semiconductors doped with acceptor impurities only)

Since Boron is a trivalent impurity, the acceptor concentration will be:

N_a =10¹⁸atom/cm³

The conductivity of sample A doped with Boron impurity will be:

Similarly, since phosphorus is a pentavalent impurity, the donor concentration will be:

N_d =10¹⁸atom/cm³

And the conductivity of sample Bdoped with Phosphorus impurity will be:

The required ratio will be:

Given N_a = N_d = 10¹⁸ /cm³

Concept:

R – 2R ladder DAC:

3 bit R – 2R DAC by using Inverting op-amp is shown below:

Output voltage is defined as:

V₀= - I₁(R_F) ⋯ (i)

Current is calculated as:

⋯ (ii)

Current drawn from the source is

I = V / R ⋯ (iii)

Current is calculated as:

The final output voltage is

For n bit DAC output voltage is

Calculation:

Given digital input is 1010 and the reference voltage is 5V

a₃= 1, a₂= 0, a₁= 1, a₀= 0

considering the value of the feedback resistor for Op amp is R Ω

Output voltage is

V₀= -3.125 V

• A mobile identification number (MIN) is a serial number that uniquely identifies a cell phone services subscriber within a mobile carrier network.
• It is used by mobile phone services providers to identify subscribers within its database, specifically when routing calls.
• MIN can also be called a mobile subscriber identification number (MSIN).

Opticalfibersaretransparentfibersand act as a light pipe totransmit light between its two ends. They are made up ofsilicon dioxide.

Refractive index:The ratio of thespeed of light in a vacuum to the speed of light in a mediumis calledthe refractive indexof that medium.

It is also called anabsolute refractive index.

Critical angle:When a ray of light is going from a denser medium to a rare medium then theangle of incidenceat which the refraction angle is 90° is called thecritical angle.

Thecritical angleis given by:

Θ_C= critical angle = sin^-1(n₂/n₁)

Where n₂is therefractive indexof second medium in which light ray is going and n₁is therefractive indexof first medium from which light is going to second medium.

Advantages and disadvantages of step-index mono-mode optical fiber.

A waveguide with conducting ridges protruding into the center of the waveguide from the top wall or bottom wall or both walls, is called as a Ridged Waveguide.

Single Ridged Waveguides:

A rectangular waveguide with a single protruding ridge from the top or bottom wall is called a Single Ridged Waveguides.

Different types of waveguides are shown in the figure below:

Concept:

1. Optical fibres can be used to transmit light and thus information over long distances. The electromagnetic spectrum is composed of visible and near-infrared light like that transmitted by fibre, and all other wavelengths used to transmit signals such as AM and FM radio and television.

For fibre optics with glass fibres, we use light in the infrared region which has wavelengths longer than visible light, typically around 850, 1300 and 1550 nm.

2. Radar is a detection system that uses radio waves to determine the range, angle, or velocity of objects.

Radio waves (pulsed or continuous) from the transmitter reflect off the object and return to the receiver, giving information about the object's location and speed have frequencies as high as 300 gigahertz (GHz) to as low as 30 hertz (Hz). X-band is a segment of the microwave radio region of the electromagnetic spectrum, within the frequency range 8–12 GHz.

3. Sonar is a system used for the detection of objects under water by emitting sound pulses and detecting or measuring their return after being reflected.

Ultrasonic waves (sounds having frequency greater than 20,000 Hz) are used in sonar because Ultrasonic waves can penetrate water to long distances (because of their high frequency and very short wavelength), but ordinary sound waves or infrasonic waves cannot penetrate water to such long distances.

4. A mobile phone receives microwave signals from a nearby phone (or transmitter) and sends microwave signals back. These short bursts of microwave radiation can transmit a lot of information in a very short time with frequencies between 300 MHz (1 m) and 300 GHz (1 mm).

Therefore, the correct matched pairs are

Thus, the correct answer is A-Q, B-S, C-P, D-R.

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

Uplink frequency > Downlink frequency

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.
• Since thesatellite cannot afford to deliver high powers because of battery power constraints, the downlink is sent at lower frequencies, while the station at earth can afford to transmit at high frequencies, so it transmits at higher frequencies.
• This helps in avoiding interference between its powerful transmitted signal and the weak incoming signal.
• Thesatellite transponderconverts this signal into another frequency and sends it down to the second earth station.
• This frequency is calledDownlink frequency.In a similar way, the second earth station can also communicate with the first one.

The block diagram of the transponder system is as shown:

A repeater is a device that operates only in the physical layer. A repeater receives a signal and, before it becomes too weak or corrupted, regenerates the original bit pattern.

A repeater does not actually connect two LANs; it connects two segments of the same LAN. The segments connected are still part of one single LAN. A repeater is not a device that can connect two LANs of different protocols.

Concept:

• Packet filter firewall can investigate up to network layer
• Packet filter firewall cannot investigate in transport layer section hence cannot block user with a specified port number

Explanation:

A packet filter network firewall can investigate and block network traffic based on several fields in a network packet header.
These fields include:

I. the source and destination port numbers.

II. the source and destination IP addresses.

Additional Information
MAC addresses are included in the packet header, they are typically used for local network communication and are not used for filtering traffic by a packet filter firewall.

Note: However, some advanced firewalls may be able to filter traffic based on the MAC address in certain scenarios, such as in virtualized environments or for wireless access points.

Hence, the correct answer is option 2.

Calculation:

Given equation is

y[n] = x[n] - x[n-1]

Applying DTFT,

Y(Ω) = X(Ω) -e-jΩX(Ω)

Y(Ω) = X(Ω)[1 -e-jΩ]

H(Ω) = e-jΩ/2[ejΩ/2-e-jΩ/2]

H(Ω) = 2 e-jΩ/2sin[Ω/2]

AtΩ = 0

|H(Ω)| = 0

Hence, it is a highpass filter.

Various frequency bands associated with the Radar are:

Baud Rate:

• The baud rate is the rate at which information is transferred to a communication channel.
• The baud rate indicates the number of electrical oscillations per second that occurs within a data transmission.
• Baud rate is commonly used when discussing electronics that use serial communication.
• The higher the band rate, the more bits per second that are transferred.
• In the serial part content, ‘9600 band’ means that the serial port is capable of transferring a maximum of 9600 bits per second.
• At baud rates above 76,800, the cable length will need to be reduced.
• Higher the baud rate, the mere sensitive the able becomes to the quality of installation, due to how much of the wire is untwisted around each device.
• Standard baud rates include 100, 300, 600, 1200, 2400, 4800, 9600, 14400, 19200, 38400, 57600, 115200, 128000 and 256000 bits per second.

Note:

• Bit rate is the transmission of the number of bits per second whereas baud rate is the number of signal units transmitted per second.

TCP/IP model, RFC 1122

• The application layer in TCP/IP is equivalent to the combined session, presentation, and application layers in the OSI model.
• The application layer allows a user to access the services of our private internet or the global Internet.
• Many protocols are defined at this layer to provide services such as electronic mail, file transfer, accessing the World Wide Web, and so on.

Sensitivity of measuring instrument

• All calibrations and specifications of an instrument are only valid under controlled conditions of temperature, pressure, and so on. These standard ambient conditions are usually defined in the instrument specification.
• As variations occur in the ambient temperature, certain static instrument characteristics change, and the sensitivity to disturbance is a measure of the magnitude of this change.
• Such environmental changes affect instruments in two main ways, known as zero drift and sensitivity drift.
• Zero drift or bias describes the effect where the zero reading of an instrument is modified by a change in ambient conditions. This causes a constant error that exists over the full range of measurement of the instrument.
• Sensitivity driftdefines the amount by which an instrument's sensitivity of measurement varies as ambient conditions change. It is quantified by sensitivity drift coefficients that define how much drift there is for a unit change in each environmental parameter that the instrument characteristics are sensitive to.

Concept:

Thestandard equation of Frequency Modulation(FM)is given as:

whereβ= modulation index

The Bandwidth of the FM signal is given by

B.W = 2(1 + β) fm

Calculation:

β = 15, f_m = 100 Hz,

BW = 2(1 + β)f_m

= 2(1 + 15)100 = 3.2 kHz

Important Points

• In FM transmission,the waveform has infinite sidebands occurring at frequencies fc±nfm, where n is an integer.
• Hence theoretical bandwidth requirement is infinite.
• But, the amplitude (thereby energy content) of the sidebands is influenced by modulation index and frequency deviation. Generally in FM communication, the modulation index is chosen as small values for which the first two sidebands are significant and the rest are ignored.

Schottky Diode:

It is not a typical diode because it does not have a p-n junction.

Instead, it consists a doped semiconductor (usually n-type) and metal, bound together as shown:

• When the metal comes into contact with n-type semiconductor the majority carriers, electrons, are infused into metal commonly called "Hot carrier" since injected electrons from semi-conductor have very high kinetic energy compared to electrons in a metal.
• ​The Schottky diode’s significant characteristic is itsfast switching speedas it doesn't allow the diode to reach saturation.
• This makes theShottky diode useful for high-frequencies and digital applications.
• Schottky diode is also known as the Schottky barrier diode, surface barrier diode,majority carrier device,hot-electron diode, or hot carrier diode.
• Its symbolic representation is as shown:-

1

• The Schottky diode is a semiconductor formed by the junction of a semiconductor with metal. It has a low forward voltage drop and near-zero reverse recovery time hence used in very fast switching circuits.
• Schottky barrier diode can be used as a low noise amplifier due to metal-semiconductor junction.

Differential pulse-code modulators:

• The DPCM module implements one of the simplest compression algorithms.
• It combines a spatial prediction loop with a non-uniform quantizer of which the representation levels are Huffman encoded.
• It is a signal encoder that uses PCM but adds some functionalities based on the prediction of the samples of the signal.
• The input to the DPCM can be an analog or digital signal.
• The number of encoded bits is smaller than in the PCM system.
• It needs far fewer bits per error sample than what would have been needed for the original samples themselves.
• This results in fewer bandwidth requirements by the DPCM system and hence has a lowerdynamic range than the original message itself.​

Hence, options 1 and 2 are correct.

Unconnected not joined together or to something else.
Chaotic : in a state of complete confusion and disorder.
Unaccountable : unable to be explained; Inexplicable.
Confusing : bewildering or perplexing.
Inexplicable : unable to be explained or accounted for.

Synonym of Inexplicable is Unaccountable

Cardinal means of the greatest importance; fundamental

Complacency would be India's cardinal sin

Porous means (of a rock or other material) having minute interstices through which liquid or air may pass

Synonyms: permeable, penetrable, pervious
For instance, the new moonlet might be porous, like an orbiting icy rubble pile
Very high capacity supercapacitors use a porous carbon-based electrode material

The given sentence is in active form of simple past tense. The structures for active/passive voices are:

Active: Subject + verb (IInd form) + object...

Passive: Object + was/were + verb (IIIrd form) + by + subject...

The question statement is in simple past tense, so the converted statement too should be in the same.

This eliminates options (b) & (c).

Out of options (a) and (d), the latter has ambiguities in the meaning.

So, the correct response is option (a).

The typical elements of process image are:

• User data


• System data


• User program


• System stack

Explanation:

• User data: This includes the data that is specific to the user and is used by the user program.


• System data: This includes the data that is used by the operating system to manage the process, such as process control block (PCB), process state, priority, etc.


• User program: This is the actual program code that is executed by the processor.


• System stack: This is the stack used by the system for managing function calls and local variables.

Therefore, the typical elements of a process image include user data, system data, user program, and system stack.

1. Seek time: Seek time is the time taken by the disk drive to position the read/write head to the desired track on the disk. It involves moving the head arm across the platter to access the required data. Seek time is crucial because it directly affects the overall time required to access data on the disk.


2. Rotational time: Rotational time, also known as latency or rotational delay, is the time taken for the disk platter to rotate and bring the desired data under the read/write head. It depends on the rotational speed of the disk and the position of the data on the platter. It is an important factor as it contributes to the overall access time.


3. Transmission time: Transmission time refers to the time taken to transfer the data from the disk to the computer's memory or vice versa. It depends on the transfer rate of the disk and the speed of the data bus. While important, it is not specific to accessing data on the disk.


4. Waiting time: Waiting time refers to the time spent by the disk drive in handling other requests or performing internal operations before it can process the data access request. It includes time spent in the disk queue, waiting for other tasks to complete, or performing internal tasks like error correction. While it can affect the overall system performance, it is not directly related to accessing data on the disk.

In conclusion, the most crucial time while accessing data on the disk is the seek time, as it involves positioning the read/write head to the desired track. However, it is important to consider other factors like rotational time, transmission time, and waiting time to fully understand the overall process of accessing data on the disk.

Material consisting of text and numbers is best represented as a table slide.

Table Slide:

- A table slide is a type of presentation slide that is specifically designed to display information in a tabular format.
- It is ideal for presenting material that consists of text and numbers, such as data, statistics, or comparisons.
- Tables allow for clear and organized representation of information, making it easier for the audience to understand and interpret.
- Columns and rows can be used to categorize and structure the data, providing a visual hierarchy.
- Tables can also be formatted to include additional features such as color-coding, highlighting, or data sorting.

Bullet Slide:

- A bullet slide is a presentation slide that typically uses bullet points to list and emphasize key information.
- While bullet slides can be effective for presenting concise text or ideas, they may not be the best choice for displaying extensive textual or numerical content.
- Bullet points are usually used for brief summaries, key points, or to highlight important information.

Title Slide:

- A title slide is the opening slide of a presentation that typically includes the title, subtitle, and presenter's name.
- It is used to introduce the topic and set the tone for the presentation.
- While a title slide can be useful for providing an overview or introduction, it is not suitable for presenting detailed text and numbers.
Conclusion:
- When it comes to representing material consisting of text and numbers, a table slide is the most appropriate choice.
- It allows for clear organization and presentation of data, making it easier for the audience to understand and interpret the information.

The correct answer is Fire fox.

Key Points

• AnOperating System
  • It is software that acts as an interface between computer hardware components and the user.
  • Every computer system must have at least one operating system to run other programs. Applications like Browsers, MS Office, Notepad Games, etc., need some environment to run and perform their tasks.

• The OS helps you to communicate with the computer without knowing how to speak the computer's language. The user can't use any computer or mobile device without having an operating system.

Additional Information

• Symbian
  • It is a discontinued mobile operating system and computing platform designed for smartphones.
  • Symbian was originally developed as a proprietary software OS for PDAs in 1998 by the Symbian Ltd. consortium.
• Android
  • It is a mobile operating system based on a modified version of the Linux kernel and other open-source software, designed primarily for touchscreen mobile devices such as smartphones and tablets.
• Firefox
  • It is a Browser also known as Mozilla Firefox or simply Firefox is a free and open-source web browser developed by the Mozilla Foundation and its subsidiary, the Mozilla Corporation.
  • Firefox uses the Gecko layout engine to render web pages, which implements current and anticipated web standards.
• iOS
  • This is the world's most advanced mobile operating system.

The correct answer isQuick Access Toolbar

Key Points
Quick Access Toolbar -

• Positioned just above the Ribbon, the Quick Access Toolbar in MS-PowerPoint 2016 provides commands you are likely to use frequently. It's customizable and by default, it includes commands such as Save, Undo, and Redo. It remains constant no matter which tab/menu in the Ribbon you are currently accessing.

Additional Information

Status Bar - Located at the bottom of the PowerPoint window, the status bar displays information about the current presentation such as the number of slides, slide number being viewed, theme, etc. It can also have view buttons and a zoom slider.

Title Bar - This is the uppermost part of the application window where the name of your presentation is displayed. It also contains buttons to close, minimize, or maximize the window.

Scroll Bar - It's situated on the right and sometimes bottom side of the PowerPoint window, the scroll bar allows you to navigate through your slides or if zoomed in, throughout different parts of a single slide, by clicking and dragging the scroll bar or clicking the arrows at its end.

The correct answer isMicrosoft Outlook.

Key Points:

• Microsoft SQL Server is a relational database management system developed by Microsoft.
• MySQL Database Service is a fully managed database service to deploy cloud-native applications.
• Microsoft Outlook is the email service provided by Microsoft.
  • Hence this is the answer.
• Microsoft Access is the only application software you need to create apps, utilize databases, and store data in SQL Server and Microsoft Azure SQL.

Hardware that calculates CRC(Cyclic Redundancy Check) uses:

The hardware that calculates CRC (Cyclic Redundancy Check) uses the following components:

Shift Register:

• One of the primary components used in CRC calculation hardware is the shift register.


• A shift register is a digital circuit that can shift its stored bits either to the left or right, based on clock pulses.


• It is used to perform the polynomial division required in CRC calculations.

XOR Unit:

• Another important component used in CRC calculation hardware is the XOR (Exclusive OR) unit.


• The XOR unit performs the XOR operation on the input data and the feedback data in each stage of the shift register.


• This operation is necessary to generate the remainder bits in the CRC calculation process.

Both (a) and (b):

• The hardware used for CRC calculation combines the shift register and XOR unit to perform the required polynomial division and generate the CRC remainder bits.


• Both the shift register and XOR unit work together to calculate the CRC value.


• Hence, the correct answer is option C: Both (a) and (b).

Instruction Register:

• The instruction register is not directly involved in the CRC calculation process.


• It is a component used in a processor to hold the current instruction being executed.


• It is not related to the hardware used specifically for CRC calculation.

In conclusion, the hardware that calculates CRC (Cyclic Redundancy Check) uses a shift register and XOR unit in combination to perform the necessary polynomial division and generate the CRC remainder bits. The correct answer is option C: Both (a) and (b).

The correct answer isCtrl + Z.

Key Points

• Undois atechnique or a commandwhich is implemented in manycomputer programs.
• It erases the last change done to the document.
• We canundo, redo, or repeatmany actions inMicrosoft Word, PowerPoint, and Excel.
• With the help ofundo command, the users can explore and work without fear of making mistakes, because changes can be made whenever required.
• The opposite ofundo is Redo.
• The redo command reverses the undo command.
• In Microsoft Windows applications, the keyboard shortcut for the Undo command is Ctrl+Z or Alt+Backspace, and the shortcut for Redo is Ctrl+Y or Ctrl+Shift+Z.

The answer to the question is B: Resolution.

Resolution refers to the pixel spacing or the level of detail that can be displayed or captured by a digital device. It is an important factor in determining the clarity and sharpness of images.

Here are some key points to understand about resolution:

• Resolution is measured in pixels per inch (PPI) or dots per inch (DPI). It indicates the number of pixels or dots that can be displayed in one inch of a digital image.


• Higher resolution means more pixels or dots per inch, resulting in greater detail and sharper images.


• Lower resolution means fewer pixels or dots per inch, resulting in less detail and more pixelation.


• Resolution is a crucial factor in various digital devices, such as computer monitors, cameras, printers, and smartphones.


• For example, a higher resolution in a camera allows for capturing more details in an image, while a higher resolution in a monitor provides a clearer and more detailed display.


• Resolution is often expressed as the width and height of an image in pixels, such as 1920x1080 for Full HD or 3840x2160 for 4K Ultra HD.

Therefore, in the context of pixel spacing, the correct answer is B: Resolution.

The correct answer is thePresentation program.

Key Points

• Microsoft PowerPoint
  • It is a powerfulslide show presentation program. Hence, Option 4 is correct.
  • It is a standard component of the company's Microsoft Office suite software and is bundled together with Word, Excel, and other office productivity tools.
  • The program uses slides to convey information rich in multimedia.

Additional Information

The correct answer isControl Unit.

• A synchronization signal is a specific signal used in various communication and electronics systems to ensure different parts of a system or multiple systems work together coherently and with proper timing.
• The primary purpose of synchronization signals is to establish a common reference time or clock for all the components involved.
• Synchronization signals can come in various forms, depending on the type of system they are used in.

• A control unit or CU is circuitry that directs operations within a computer's processor. ..
• Examples of devices that utilize control units include CPUs and GPUs. A control unit works by receiving input information that it converts into control signals, which are then sent to the central processor.
• A memory unit is the amount of data that the memory can hold. Besides, we measure this storage capacity in terms of bytes. Moreover, there are different units of memory as per the requirement.

Name Box: This is a small rectangular box located in the top left corner of the Excel window, to the left of the formula bar. It displays the address of the currently selected cell (for example, "A1" if the first cell is selected). Apart from displaying cell references, you can use it to define names to cells or ranges.

Formula Bar: This is located just below the ribbon and to the right of the name box. It is used to enter or edit a formula, a function, or data in a cell. It displays the contents of the currently active cell and allows you to create and edit formulas.

Status Bar: This is located at the bottom of the Excel window. It provides various types of status information, including the average, count, and sum of selected numbers, whether the scroll lock is on, and other status messages. It doesn't display the address of the selected cells.

Title Bar: This is at the very top of the Excel window, displaying the name of the currently open workbook (Excel file). The title bar is largely used to move the application window around and doesn't show the address of the selected cells.

Among these options, only "Name Box" displays the reference of the currently selected cell or cells.

The Correct answer isHigh Definition Multimedia Interface

Key Points

• HDMI stands forHigh Definition Multimedia Interface.
• It carries bothvideo and audio signals; you don't need separate audio cables.
• It allows you to turn your TV into a computer monitor as the latest computers and laptops come with an HDMI connection.

Additional Information

• Display Port (DP)is a digital display interface developed by a consortium of PC and chip manufacturers and standardized by the Video Electronics Standards Association (VESA).
• There are two main types of cables:analog and digital.
• Analog cables work by sending information via a stream of electricity (an electrical signal) from one point to another.
• Digital cables send information via abinary code stream(1s and 0s) from one point to another.

From the common explanation, we get C sits opposite to F.

Hence, Option C is correct.

Common Explanation:

References:

One person sits between A and C, who sits second to the right of H.

G sits opposite to B.

Both G and B do not sit immediate left of H.

Inferences:

From the above reference, we get two different cases:

References:

D sits second to the left of B.

A doesn’t sit beside D.

E doesn’t face C.

Inferences:

From the above reference, case 1 will be eliminated because A doesn’t sit beside of D.

So, the final arrangement:

According to the information given in the question, we can draw the following diagram.

Thus we can see if a man has to travel to point E from point A(through these points by the shortest distance), he has to follow a path point A→ B→ F→ E.

Thus travelling from point A, he will pass the firstpoint B.

Hence, the correct answer is "Point B".

The logic is:

Column 1: 9× 5× 3 = 135

Column 2: 5× 7× 4 = 140

Similarly,

Let the missing number in column 3 be 'a'.

So, 5× a× 5 = 150

⇒ 25× a = 150

⇒ a = 150÷ 25

⇒ a = 6

Hence, ‘6’ is the correct answer.

Given number: 7 8 3 4 3 2 9 5 1 3

After interchanging the digits,
New number: 8 7 4 3 2 3 5 9 3 1

6th digit from right end

27 → 3³; 125 → 5³; 64 → 4³

Therefore ? = 6³ = 216.

From Statement 1:
H sits to the immediate right of D.
Two persons sit between D and F.
A is an immediate neighbour of M, who sits to the immediate right of F.

So, H sits to the immediate left of N.
The data in Statement I alone is sufficient to answer the question.
From Statement 2:
F and N are immediate neighbours and M sits immediate right of F.
A sits third to the left of N.
Here, we can not determine who sits immediate left of N.

Hence, option A is the correct answer.

P, Q, R, S, T, V, W, and Z are sitting around a circle facing at the centre not necessarily in this same order.

i)T is second to the right ofR whois third to the right of P.

Mistake Points

Whois used for thesecond personin a connecting statement.

For example:

T is second to the right of R who is third to the right of P.

Here,whois used for Rand not for T.

ii)S is second to the left of P and fourth to the right of Q.

iii)V whois not an immediate neighbour of P. (This means V cannot sit immediate left or right of P).

iv)Z is third to the right ofV

Thus we can see R is second to the right of W

Hence, the correct answer is"Option 1".

The sequence is a combination of two series:
I. 3, 5, 7, 9
II. 6, 20, 42, (____)

Consider II sequence
1st term = 6
2nd term = 6 + 14= 20
3rd term = 20 + 22 = 42

Thus, the missing term is 4th term = 42 + 30 = 72

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