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BEL Trainee Engineer Electrical Mock Test - 4 - Electrical Engineering (EE) MCQ


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30 Questions MCQ Test BEL Trainee Engineer Electrical Mock Test Series 2024 - BEL Trainee Engineer Electrical Mock Test - 4

BEL Trainee Engineer Electrical Mock Test - 4 for Electrical Engineering (EE) 2024 is part of BEL Trainee Engineer Electrical Mock Test Series 2024 preparation. The BEL Trainee Engineer Electrical Mock Test - 4 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The BEL Trainee Engineer Electrical Mock Test - 4 MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for BEL Trainee Engineer Electrical Mock Test - 4 below.
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BEL Trainee Engineer Electrical Mock Test - 4 - Question 1

According to Super position theorem, a Voltage source of 0 V can be replaced by a:

Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 1

Superposition Theorem:

  • It is state that in any linear, active, bilateral network having more than one source, the response across any element is the Algebraic sum of the response obtained from each source considered separately and all other sources are replaced by their internal resistance.
  • The principle of the superposition theorem is based on Linearity.
  • Voltage Source → short
  • Current source → open
  • Do not disturb the dependent source present in the network.

Limitations of SPT:

  • This theorem cannot be used to measure power.
  • This theorem is not applicable to unbalanced bridge circuits.
  • Applicable only to linear circuits not for nonlinear circuits.
  • Applicable only for the circuits having more than one source.

Application of SPT:

Superposition Theorem is applied to determine the current in one particular branch of a network containing several voltage source and/or current source.

BEL Trainee Engineer Electrical Mock Test - 4 - Question 2

A memory system has a total of 8 memory chips each with 12 address lines and 4 data lines. The total size of the memory system is

Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 2

Number of address lines, n = 12

No. of memory locations = m = 2n = 212

No. of data line = 4 = No. of bits/location

Total No. of chips given = 8

Total capacity = No. of chips given × No. of memory location (N) × No. of data line

= 8 × 212 × 4 bits

= 23 × 212 × 22 bits

= 212 × 22 bytes

∴ 1 byte = 23 = 8 bits

= 210 × 24 bytes

= 16 k bytes

∴ 210 = 1 k bytes

Total capacity of memory = 16 k bytes
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BEL Trainee Engineer Electrical Mock Test - 4 - Question 3

If I1 in the given circuit is 6 A, what will be the current I2 in the following circuit?

Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 3

Reciprocity Theorem:

In any passive linear bilateral network, if a single voltage source ‘V’ in the branch AB produces the current response ‘I’ in the branch CD, then the removal of voltage source from the branch AB and its insertion in the branch CD will produce same current ‘I’ in the branch AB.

In reciprocity theorem, if the position of excitation and responses are interchanged, then their ratio remains the same.

Application:

Circuit 1: V1 = 144 V, I1 = 6 A

Circuit 2: V2 = 144 V, I2 = ?

According to reciprocity theorem,

BEL Trainee Engineer Electrical Mock Test - 4 - Question 4

A 1000 × 1000 bus admittance matrix for an electric power system has 8000 non-zero elements. The minimum number of branches (transmission lines and transformers) in this system are:

Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 4

Concept:

  • Minimum number of branches in system = (Number of non-zero off diagonal elements) / 2
  • Number of diagonal elements = number of buses

Calculation:

Given -

Size of YBUS matrix = 1000 × 1000

Number of non-zero elements = 8000.

Number of diagonal elements = number of buses = 1000.

Number of non-zero off-diagonal elements = 8000 – 1000 = 7000

Minimum number of branches = 7000/2 = 3500

Alternate Method

Bus Admittance matrix = 1000 × 1000

Non-zero elements = 8000

Total number of elements = 1000 × 1000 = 1000000

Sparcity = (Total elements) - (Non-zero elements) = 1000000 - 8000 = 992000

The number of transmission line =

BEL Trainee Engineer Electrical Mock Test - 4 - Question 5

If the input and output of a system is related by the following differential equation, then find its transfer function.

Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 5

Concept:

A transfer function is defined as the ratio of Laplace transform of the output to the Laplace transform of the input by assuming initial conditions are zero.

TF = L[output]/L[input]

For unit impulse input i.e. r(t) = δ(t)

⇒ R(s) = δ(s) = 1

Now transfer function = C(s)

Therefore, transfer function is also known as impulse response of the system.

Transfer function = L[IR]

IR = L-1 [TF]

Calculation:

Given differential equation is,

⇒ s2 Y(s) + 3s Y(s) + 2 Y(s) = U(s) + s U(s)

BEL Trainee Engineer Electrical Mock Test - 4 - Question 6

If RST 7.5 interrupt is invoked in 8085, it will execute the ISR from ________ address

Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 6

For 8085 the interrupt service routine address is given as:

RST 4.5 (TRAP)

ISR address: 4.5 × 8 = (36)10 = (0024) H

RST 5.5

ISR address: 5.5 × 8 = (44)10 = (002C) H

RST 6.5

ISR address: 6.5 × 8 = (52)10 = (0034) H

RST 7.5

ISR address: 7.5 × 8 = (60)10 = (003C) H

Important Points:

BEL Trainee Engineer Electrical Mock Test - 4 - Question 7

An overhead transmission line has a span of 220 meters, conductor weighing 804 kg/km. Calculate the maximum SAG if the ultimate tensile strength of the conductor is 5758 kg. Assume safety factor as 2. Also find maximum tension T.

Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 7

Concept:

Sag s:

The distance between the highest point of electric poles or towers and the lowest point of a conductor connected between two poles or towers.

Span length:

It is the shortest distance between two towers or poles.

Sag

Where,

S is the sag of the conductor

W is the weight of the conductor

l is the span length of the conductor

T is the working tension on the conductor

Calculation:

Span length = 220 m

Breaking strength (Ultimate strength) = 5758 kg

Weight of conductor (W) = 804 kg/km

= 804 kg/1000m (1km = 1000m)

W = 0.804 kg/m

working tension, T = ultimate strength/safety factor

= 5758/2

T = 2879 kg

Sag

=

= 1.69 m

The maximum sag of an overhead transmission line is 1.69 m

BEL Trainee Engineer Electrical Mock Test - 4 - Question 8

Identify the function of logic circuit given below

Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 8

The given circuit is redrawn as:

The Output expression is:

Output = A̅B + AB̅

This is the expression for an XOR gate.

The Truth Table for an XOR gate is as shown:

Important Point

XNOR Gate Output: A̅B̅ + AB

AND Gate Output: AB

OR Gate Output: A + B

BEL Trainee Engineer Electrical Mock Test - 4 - Question 9
8085 microprocessors has ________ individual flags during arithmetic and logic operations.
Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 9

There is an 8-bit flag register out of which only 5 bits are used

BEL Trainee Engineer Electrical Mock Test - 4 - Question 10

Which one of the following statements is not correct for the use of bundled conductors in transmission lines?

Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 10

Bundled conductor are those conductors which form from two or more stranded conductors, bundled together to get more current carrying capacity.

By using bundle conductors instead of the single conductor in the transmission line increases the GMR of the conductors.

So, inductance reduces according to the following equation

H/m/ph

Hence, the inductance per phase of the conductor decreases

So, capacitance increases according to the following equation

F/m/ph

Hence, by using bundle conductors GMR is increased, so inductance L decrease and capacitance C increases.

Advantages of Bundled Conductors:

  • Bundling of conductors leads to a reduction in line inductance.
  • Bundling of conductors leads to increases in capacitance.
  • An important advantage of bundled conductors is its ability to reduce corona discharge
  • Reduction in the formation of corona discharge leads to less power loss and hence improved transmission efficiency of the line.
  • Reduction in communication line interference.
BEL Trainee Engineer Electrical Mock Test - 4 - Question 11

A 30 kV, 50 Hz, 50 MVA generator has the positive, negative, and zero sequence reactance’s of 0.25 pu, 0.15 pu, and 0.05 pu, respectively. The neutral of the generator is grounded with a reactance so that the fault current for a bolted LG fault and that of a bolted three-phase fault at the generator terminal are equal. The value of grounding reactance is:

Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 11

Concept:

Fault current for LG fault

Fault current for three phase fault =

where,

X1 = Positive sequence reactance

X2 = Negative sequence reactance

X0 = Zero sequence reactance

Xn = Grounding reactance

Calculation:

Given that,

X1 = 0.25 pu

X2 = 0.15 pu

X0 = 0.05 pu

Fault current for LG fault is

Fault current for three phase fault,

Given that both the fault currents are equal.

⇒ Xn = 0.1 pu

Base kV = 30 kV

Base MVA = 50 MVA

BEL Trainee Engineer Electrical Mock Test - 4 - Question 12

Determine the ac dynamic resistance of diode if current varies from 2 to 17 mA and voltage changes from 0.65 to 0.725 V?

Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 12

Concept:

The dynamic resistance can be defined from the I-V characteristic of a diode in forward bias. It is defined as the ratio of a small change to voltage to a small change in current, i.e.

VT = Thermal voltage

I = Bias current

VT ∝ T

The dynamic resistance of the diode is directly proportional to the temperature.

The dynamic resistance is given by the inverse of the slope of i-v characteristics as shown:

Calculation:

Dynamic resistance of diode

ΔVd = 0.725 – 0.65 = 0.075 V

ΔID = 17 – 2 = 15 mA

Dynamic resistance of diode

BEL Trainee Engineer Electrical Mock Test - 4 - Question 13

Polar plot of sinusoidal transfer function is a plot of:

Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 13

Polar plot:

  • The polar plot of a transfer function G(jω) is the plot of the magnitude of G(jω) versus the phase angle of G(jω) as ω is varied from 0 to positive infinity.
  • For all pole systems, type indicates the starting point of the polar plot and order indicates the ending point of the polar plot.

The sinusoidal transfer function G(jω) is a complex function and it is given by

G(jω) = Re [G(jω)] + j Im [G(jω)]

G(jω) =

= M ∠ϕ Polar form

From the above equation,

G(jω) may be represented as a phasor of magnitude M and phase angle ϕ

The starting points (ω = 0) of a polar plot for different types of minimum phase systems is given below:

  • As seen from the above figure, when a zero is added the type decreases, and the end of the polar plot shifts by +90°.
  • When a pole is added, the type of the system increases, and hence the end of the polar plot shifts by -90°.

Therefore, the Polar plot of the sinusoidal transfer function is a plot of magnitude and phase angle.

BEL Trainee Engineer Electrical Mock Test - 4 - Question 14
If A = 5t2 I + t J – t3 K, B = sin t I – cos t J, find
Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 14

A = 5t2 I + t J – t3 K

B = sin t I – cos t J

The dot product of the vectors A and B is given by

A.B = (5t2 I + t J – t3 K) . (sin t I – cos t J)

= 5t2 sin t – t cos t

= 5t2 cos t + 11t sin t – cos t

BEL Trainee Engineer Electrical Mock Test - 4 - Question 15

In the given Data were conversion, Find the value of R1, R2 & R3.

Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 15

Concept:

1) Delta to Star conversion:

2) Star to Delta conversion:

Calculation:

RAB = 4 Ω , RBC = 3 Ω , RAC = 2 Ω (Given)

BEL Trainee Engineer Electrical Mock Test - 4 - Question 16

Consider the following statements regarding capacitance

  1. Voltage of capacitance cannot change suddenly
  2. Current through a capacitance cannot change suddenly.

Tick the correct answer:

Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 16

Concept:

Capacitance:

  • Capacitance is defined as it is the capability of an element to store electric charge within it. It is also defined the amount of charge
    required to create a unit p.d. between its plates.
  • Capacitance is the property that possesses the capacitor. Its unit is Farad.
  • It opposes the sudden change in voltage.

It is given by,

Inductance:

  • Inductance is the property of a material by virtue of which it opposes any changes of magnitude or direction of current passing through the conductor.
  • Inductance is said to be one henry when the current through a coil of the conductor changes at the rate of one ampere per second induction one volt across the coil.
  • The unit of inductance is Henry.
BEL Trainee Engineer Electrical Mock Test - 4 - Question 17
The causal system represented by is
Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 17

Concept:

The characteristic equation of standard second order system is given by

s2 + 2ξ ωn s + ω2n = 0

The system is said to be

  • undamped if ξ = 0
  • critically damped if ξ = 1
  • underdamped if ξ < />
  • overdamped if ξ > 1

Calculation:

Characteristic equation: s2 + 6s + 9 = 0

By comparing with standard second order system,

2ζ ωn = 6 ⇒ 2 × ζ × 3 = 6

⇒ ζ = 1

So, the system is critically damped.
BEL Trainee Engineer Electrical Mock Test - 4 - Question 18

The purpose of emitter bypass capacitor in a CE BJT amplifier is to

Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 18
  • The emitter bypass capacitor is connected across emitter resistance RE in a CE BJT amplifier.

  • When a bypass capacitor is connected in parallel with an emitter resistance, its input impedance reduces while the mid band voltage gain of the amplifier increase.
  • Bypass capacitors also reduce both the power supply noise and the result of spikes on the supply line.
  • The bypass capacitor is a capacitor that shorts AC signals to the ground in a way that any AC noise that present on a DC signal is removed producing a much cleaner and pure DC signal.
  • It bypasses AC noise that may be on a DC signal, filtering out of AC so that a clean, pure DC signal goes through without AC ripples.
BEL Trainee Engineer Electrical Mock Test - 4 - Question 19
When port 1 of a two port cirucit is short circuited, I1 = 4I2 and V2 = 0.25I2, which of the following is true?
Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 19

Concept:

I1 = Y11 V1 + Y12 V2

I2 = Y21 V1 + Y22 V2

When port 1 is short cirucited, i.e. V1 = 0,

I1 = Y12 V2 and I2 = Y22 V2

Calculation:

The given equations are:

I1 = 4I2 and V2 = 0.25 I2

⇒ I2 = 4V2 ⇒ Y22 = 4

I1 = 4 (4V2) = 16 V2

Y12 = 16
BEL Trainee Engineer Electrical Mock Test - 4 - Question 20
_________in a computer controls, coordinates and supervises the operations of the computer.
Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 20

Option 3 is the correct answer: Central Processing Unit (CPU) in a computer controls, coordinates and supervises the operations of the computer.

  • Central Processing Unit (CPU) is also known as the brain of the computer.
  • It is the key component that processes the data to give useful information.
  • Commonly referred to as Processor it is made up of a number of micro transistors mounted on a chip.
  • It consists of two basic components:
    • Arithmetic and Logical Unit (ALU): This unit of CPU performs all the arithmetic and logic operations.
    • Control unit: It directs the flow of data and information.
  • Random Access Memory (RAM):
    • It is the volatile memory of the computer where information is temporarily stored.
    • The data is stored until the system is working and loses it as soon as power is switched off.
  • Read-Only Memory (ROM):
    • It is a non-volatile memory in which the information is stored permanently.
    • The content does not get erased off even when the system is switched off.
BEL Trainee Engineer Electrical Mock Test - 4 - Question 21

A 3-phase semi-converter works as a six-pulse converter when the firing angle α is:

Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 21

3-phase semi-converter:

Circuit diagram:

Where,

Va Vb Vc are the input voltages

ia ib ic are the input currents

T1 T3 T5 are the thyristors

D4 D6 D2 are the diodes

V0 i0 are output voltage and current respectively

R is the load resistance

  • This is obtained by a series connection of a 3 pulse controlled converter and a 3 pulse uncontrolled one.
  • The three arms of the former consist of thyristors and the three arms of the latter comprise diodes.
  • The thyristors commutate at the phase angle at which they are fired. The diodes commu­tate at the natural firing instant α = 0.
  • The thyristors conduct for 120° and are fired at intervals of 120°.
  • A thyristor and diode conduct at any given time the diode is forward biased at the natural firing instant.
  • A thyristor conducts, even if it is reverse biased until the next thyristor in the sequence is fired.
  • Thus there is a natural free wheeling of the load current via the incoming diode and outgoing thyristor.
  • The load voltage is zero during the freewheeling period.
  • Freewheeling due to the conduction of the diode does not allow negative excursions of load voltage.
  • This reduces the ripple content in the output voltage.
  • The ripple frequency of the output voltage at α = 0 is 6f.
  • For α < 60°,="" free-wheeling="" does="" not="" take="" place="" since="" the="" voltage="" is="" always="" positive="" on="" the="" dc="" />
  • A 3-phase semi-converter works as a six-pulse converter when the firing angle α is between 0° to 60°
BEL Trainee Engineer Electrical Mock Test - 4 - Question 22

I believe that if you lower taxes it will be an incentive for them to work harder as they can save more. What is the meaning of “incentive”?

Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 22

Incentive means a thing that motivates or encourages someone to do something
Some of its synonyms are inducement, motivation, motive, reason, stimulus, stimulant, spur, impetus, encouragement
He had no incentive to work after he was refused a promotion
Companies provide various incentives to attract and retain talent

BEL Trainee Engineer Electrical Mock Test - 4 - Question 23

Which of the following is an input device?

Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 23

The correct answer is "Joystick".

Key Points

  • Plotter: A plotter is referred to as a type of printer. It is an output device. The main job of the plotter is to convert the user commands from a computer into line drawings on paper. It has one or more automated pens which help to draw the lines on the paper. 
  • Speaker: The speaker is an output peripheral of the computer. If the user plays some audio or video files on the system, the sound comes out from the speaker.
  • Projector: The projector is an output device. The projector is used for projecting images or video on the screen.
  • Joystick: The joystick is referred to as an input peripheral. The joystick is mainly used for playing games in the computing system.

Thus the correct answer is "Joystick".

Additional Information

  • The word computer is derived from the word compute. In the year 1833, English mechanical engineer Charles Babbage designed the first analytical engine. He is known as the father of the computer.
  • In the beginning, computers are used for calculating large numbers and can only store a small amount of data.
  • The software and the hardware parts of the computers are running together to execute the operations smoothly.
  • Software is a set of instructions or programs given to the computers' hardware to perform a task. For example, system software (OS), application software( MS Office, Browser), and programming software(compiler).
  • The hardware parts of the computer are tangible parts of the computer which collect the input (e.g. keyboard, mouse, scanner etc.), and process (e.g. C.P.U). It and display it to the respective output devices(e.g. monitor, printer etc.). Sometimes it stores the processed data in the memory.
BEL Trainee Engineer Electrical Mock Test - 4 - Question 24
IBM 360 was developed in which of the following generations of computers?
Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 24

The correct answer is Third Generation.

Key Points

  • The IBM System/360 was announced and developed in the mid-1965s, during the period associated with third-generation computers.
  • Third-generation computers were defined by the use of integrated circuits (ICs) or semiconductor chips, which represented a significant advancement over the previous generation's technology.
  • The System/360 was a groundbreaking series of compatible mainframe computers, allowing customers to upgrade from lower-cost models to larger systems without having to rewrite their applications.
  • It also introduced the 8-bit byte to the commercial world, which remains a fundamental data management unit today.
  • The S/360 was not just a computer but a whole series of computers that helped to establish the concept of a computer "family".

Additional Information

  • First Generation:
    • These computers (late 1946s to mid-1959s) were based on vacuum tubes.
    • They were large, expensive, and generated a lot of heat.
    • Examples include the ENIAC and UNIVAC.
  • Second Generation:
    • Second generation computers (mid-1959s to mid-1965s) moved from vacuum tubes to transistors, which were smaller, faster, cheaper, and more reliable.
    • IBM's 7000 series belongs to this generation.
  • Fourth Generation:
    • These computers (1971 to the mid-1980s) are characterized by microprocessors which contain many, sometimes millions, of ICs.
    • Personal computers, such as the IBM PC, fall into this category.
BEL Trainee Engineer Electrical Mock Test - 4 - Question 25

Sheela recalls using transistors as part of her computer.

Which generation of computers used it?

Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 25

The correct answer is Second.

Key Points

  • Second-generation computers (1955-1964) were made of transistors.
  • The second generation of computers, which used transistors instead of vacuum tubes, were released in the late 1950s and early 1960s as interest in computer technology grew quickly.
  • The second generation of computers, however, was entirely built using transistors rather than vacuum tubes.

Additional Information

BEL Trainee Engineer Electrical Mock Test - 4 - Question 26
The Linux OS was developed by Linus Torvalds in __________.
Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 26

The correct answer is "1991".


Key Points

  • Linux: Linux is a good example of an open-source operating system invented by Linus Torvalds in the year 1991.
  • Like any other kind of operating system, Linux also builds communication between the hardware and the software.
  • It comes with both Graphical User Interface and Command Line Interface. It is a very secure operating system.
  • So that malware and viruses cannot attack the operating system. It is very flexible and can be installed in desktop applications as well as server applications.
  • It is also multiprogramming and multitasking like other kinds of operating systems and as well as it supports multi-user.

Thus the correct answer is 1991.

Additional Information

  • MS-DOS: The full form of MS-DOS is Microsoft Disk Operating System. It was developed by Microsoft in the year 1980. The MS-DOS operating system uses  ​a command-based interface to interact with the users. It is referred to as a single-tasking operating system. The main job of the MS-DOS is to help the user to navigate open and modify files on their own computer system.
  • Windows: Microsoft first introduced its first GUI (Graphical User Interface) based operating system, named Microsoft Windows in the year 1983. Microsoft Windows can perform all the tasks just like an operating system does. It can execute multiple instructions at a time. It also made communication between the hardware and software. It is also multiprogramming and multitasking like other kinds of operating systems.
BEL Trainee Engineer Electrical Mock Test - 4 - Question 27

Which data type is NOT in M.S. Excel?

Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 27

The correct option is (2) Label

Key Points

  • When you wish to enter data manually into Excel, you have a few choices. You can enter data into a single cell, multiple cells at once, or across multiple worksheets.
  • You can enter text, numbers, dates, or times as your data. The data can be formatted in many different ways.
  • Text, number, logical, and error data are the four categories of data. Each type can be used for a variety of tasks, so knowing which to use and when to use it is crucial.
  • You might also keep in mind that while exporting data into a spreadsheet, some data types might change.

Additional Information
Label:- Text that you have entered into a cell is typically referred to as a label. Excel interprets the label as a "tag" that designates that range of numbers if it occurs next to a continuous list of values. In formulas, the label can then be used in place of specific cell references.

BEL Trainee Engineer Electrical Mock Test - 4 - Question 28
What is used for holding program instructions that can't be changed throughout the life of the computer?
Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 28

The correct answer is ROM.

Key Points

  • ROM
    • Read-Only Memory is a ROM in a computer.
    • Read-only memory is a storage medium that permanently stores data on computers and electronic devices.
    • The data stored in these chips is nonvolatile i.e. it is not lost when power is removed.
    • There are five basic ROM types:
      • ROM: Read-Only Memory.
      • PROM: Programmable read-only memory.
      • EPROM: Erasable programmable read-only memory.
      • EEPROM: Electrically erasable programmable read-only memory.
      • Flash memory: Flash memory, a type of EEPROM that uses in-circuit wiring. 
    • BIOS uses Flash memory.

Additional Information

  • Register
    • The smallest and fastest memory of a computer is registered memory located in the computer's central processing unit(CPU).
    • It generally stores the most used instructions, data, etc. which are to be used by CPU.
    • Control Unit is part of the CPU.
    • It advises the input and output devices, and computer memory on how to respond to the instructions sent to the processor.
  • RAM
    • It stores data that the computer needs to use temporarily. So it is faster than the secondary storage devices. It is a volatile memory i.e. the data disappear from it when the power is off.
    • It is known as the Read-Write Memory. i.e. the user can write information into RAM and read information from it.
    • It is called random access since any memory location can be accessed randomly for reading and writing. The access time is the same for each memory location.
    • It is usually called “temporary” memory, which means that when the system is shut down, the memory is lost.
  • Cache
    • A cache memory is used by the central processing unit of a computer to reduce the average cost time or energy to access data from the main memory.
    • Generally Cache memory is logically positioned between CPU and main memory.
BEL Trainee Engineer Electrical Mock Test - 4 - Question 29

Directions: Each of the following consists of a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statements are sufficient to answer the question:

How is ‘HEALTHY’ written in that code language?
Statement I:
In a certain code language, EATING is written as HDRLLE.
Statement II: In a certain code language, REPLACE is written as PHNJDAH.

Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 29

From Statement I: From this statement, we get the following inference:

All the vowels are coded as 3rd next letter as per English alphabetical series and the consonants are coded as 2nd previous letter as per English alphabetical series.

So, the code of ‘HEALTHY’ is ‘FHDJRFW’.

Clearly, Statement I alone is sufficient to answer the question.

From Statement II: From this statement, we get the following inference:

All the vowels are coded as 3rd next letter as per English alphabetical series and the consonants are coded as 2nd previous letter as per English alphabetical series.

So, the code of ‘HEALTHY’ is ‘FHDJRFW’.

Clearly, Statement II alone is sufficient to answer the question.

Therefore, the question can be answered using either statement I alone or statement II alone.

Hence, option C is the correct answer.

BEL Trainee Engineer Electrical Mock Test - 4 - Question 30

P, Q, R, S and T are sitting in a straight line facing the North. P sits next to S but not to T. Q is sitting next to R, who sits on the extreme left corner. Who sits to the immediate left of S if T does not sit next to Q?

Detailed Solution for BEL Trainee Engineer Electrical Mock Test - 4 - Question 30


P sits to the immediate left of S.

Hence, option A is correct.

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Top Courses for Electrical Engineering (EE)

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Top Courses for Electrical Engineering (EE)