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RRB JE Electrical (CBT II) Mock Test- 9 - Electrical Engineering (EE) MCQ


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30 Questions MCQ Test - RRB JE Electrical (CBT II) Mock Test- 9

RRB JE Electrical (CBT II) Mock Test- 9 for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The RRB JE Electrical (CBT II) Mock Test- 9 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The RRB JE Electrical (CBT II) Mock Test- 9 MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for RRB JE Electrical (CBT II) Mock Test- 9 below.
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RRB JE Electrical (CBT II) Mock Test- 9 - Question 1

In the figure, D in an ideal diode. If the rms value of the input voltage is 50V, then the rms current through is

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 1

Output will be rectified half wave,then

RRB JE Electrical (CBT II) Mock Test- 9 - Question 2

If the P.F of load decreases, the line losses-

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 2

Taking that the pf of the load is improved ( say from 0.7 lag to 0.85 lag) for the same real power P, then the magnitude of the line current will decrease, decreasing the line losses. So transmission efficiency will improve.

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RRB JE Electrical (CBT II) Mock Test- 9 - Question 3

The maximum power in the shown load is

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 3

Then by using the following Ohm’s Law equations:

RRB JE Electrical (CBT II) Mock Test- 9 - Question 4

What is the unit of relative permeability?

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 4

The relative permeability (μr) of a substance is the ratio of magnetic permeability (μ) of the substance to
the permeability of free space (μ0)
μr = μ/μ0
Hence, it is a dimensionless quantity and is equal to 1 for vacuum

RRB JE Electrical (CBT II) Mock Test- 9 - Question 5

The wattmeter measures________ power.

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 5

The wattmeter measures average power

RRB JE Electrical (CBT II) Mock Test- 9 - Question 6
What is the process of slow cooling of hot glass called?
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 6
  • Annealing is one of the major processes in the manufacture of glass.
  • It is a controlled slow cooling process of heat treatment of glass.
  • It helps in strengthening the glass and relieving the stress.
RRB JE Electrical (CBT II) Mock Test- 9 - Question 7
What is the typical value of VBE set in PNP Ge transistor?
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 7
RRB JE Electrical (CBT II) Mock Test- 9 - Question 8
Where is the largest population of Bhil Tribe found in India?
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 8
The Bhil are an indigenous tribe in the Kolarian group and were the earliest settlers in India. According to the 2011 Census of India, Bhil is the most populous tribe in Madhya Pradesh with a total population of 4,618,068, constituting 37.7 per cent of the total ST population.
RRB JE Electrical (CBT II) Mock Test- 9 - Question 9
Which one of the following is not a fractional horse power motor?
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 9

A fractional-horsepower motor (FHP) is an electric motor with a rated output power of 746 Watts or less.

There is no defined minimum output. However, it is generally accepted that a motor with a frame size of less than 35 mm square can be referred to as a micro-motor.

The synchronous motor is not a fractional horsepower motor.
RRB JE Electrical (CBT II) Mock Test- 9 - Question 10

In an Induction type meter, maximum torque is produced when the phase angle between two fluxes is

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 10

An induction type meter consists of two electromagnets, called shunt magnet and series magnet of laminated construction.

A coil having a large number of turns of fine wire is wound on the middle limb of shunt magnet. This coil is known as voltage coil or pressure coil and it is connected across the supply mains. It has many turns and is arranged to be as highly inductive as possible. This causes the current and therefore the flux, to lag the supply voltage by nearly 90°.
An adjustable copper shading rings are provided on the central limb of the shunt magnet to make the phase angle displacement between the magnetic field set up by shunt magnet and the supply voltage is approximately 90°.
The series electromagnet is energized by a coil which is known as current coil and it is connected in series with the load so that it carries the load current. The flux produced by this magnet proportional to and in
phase with the load current.
The moving system essentially consists of a light rotating aluminium disk mounted on a vertical spindle or shaft. The fluxes produced by shunt and series magnet induce eddy currents in the aluminium disc. The interaction between these two magnetic fields and eddy currents set up a driving torque in the disc.
As in energy meter instrument, we have two fluxes and two eddy currents and therefore two torques are produced by
(i) First flux interacting with eddy current produced by second flux and
(ii) Second flux interacting with eddy currents generated by the first flux
Maximum torque is produced when the phase angle between these two fluxes is 90°

RRB JE Electrical (CBT II) Mock Test- 9 - Question 11
Single phase induction motor is also known as:
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 11

A fractional-horsepower motor (FHP) is an electric motor with a rated output power of 1 HP (746.9 or 746 Watts) or less.

Single phase induction motors have very less rating hence these are also known as fractional horsepower motors.
RRB JE Electrical (CBT II) Mock Test- 9 - Question 12

The rms value of the current in a wire which carries a DC current of 10 A and a sinusoidal alternating current of peak value 20 A is

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 12

RRB JE Electrical (CBT II) Mock Test- 9 - Question 13

Assertion (A): When a series RLC circuit is in resonance, the current flowing in the circuit is maximum.

Reason (R): The inductive reactance and the capacitive reactance are equal in magnitude at resonance.

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 13

We know that, in a series RLC circuit impedance is given by

Current flowing in the circuit is given by I=V/Z

Where, Z is impedance

R is resistance

XL is inductive reactance

XC­ ­is capacitive reactance

At resonance condition, the inductive resistance is equal to capacitive reactance.

At this condition, Z = R which is minimum value of impedance

At minimum impedance, the current flowing in the circuit is maximum

Hence both A and R are true, and R is the correct explanation of A.

RRB JE Electrical (CBT II) Mock Test- 9 - Question 14

Assertion (A): Inductors carrying steady direct currents act as effective short circuits with zero voltage across it.

Reason (R): The voltage induced across an inductance is proportional to the rate of change of current di/dt.

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 14

The voltage induced across an inductance is proportional to the rate of change of current di/dt.
V L = L di/dt
V L  ∝ di/dt
For DC supply, di/dt = 0 ⇒ VL = 0 V
For DC supply Inductors carrying steady direct currents act as effective short circuits with zero voltage across it.

RRB JE Electrical (CBT II) Mock Test- 9 - Question 15

What are the advantages of Waste to energy?

i) It is economical

ii) Reduce volume of waste

iii) Recover useful energy

iv) High degree of sophistication is required

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 15

Concept:

Waste to Energy (WtE) or Energy from Waste (EfW) is the process of generating energy in the form of electricity and/or heat from the primary treatment of waste, or the processing of waste into a fuel source. WtE is a form of energy recovery.

The methods by which energy can be recovered from Waste to Energy are:

  • Heat
  • Electricity
  • Co-generation

Advantages of Waste to Energy:

  • Reduction volume of waste
  • Reduction of waste going to landfill sites
  • Reduction of carbon emissions
  • Reduction of the use of fossil fuels
  • Electricity and heat can be generated from waste which provides an alternative and more environment-friendly source of energy
  • The local community around EfW facilities benefits from the creation of jobs to cost effective energy
RRB JE Electrical (CBT II) Mock Test- 9 - Question 16
A series circuit has 100 resistors, each having a resistance of 1 Ω and 1 A current is flowing through this circuit. What will be the current in the circuit when these 100 resistors are connected in parallel?
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 16

We know that,

In parallel circuit sum of the currents through each path is equal to the total current that flows from the source.

Total resistor in the circuit = 100

The value of each resistor = 1 Ω

The current in each resistor when connected in series = 1 A

Source voltage = 1 × 100 = 100 V

The current in the circuit when these 100 resistors are connected in parallel = 100 × 100 = 10000 A
RRB JE Electrical (CBT II) Mock Test- 9 - Question 17
In operating a 400 Hz transformer at 50 Hz:
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 17
  • If frequency of the transformer is decreased, magnetizing current in the primary of the transformer increases as it is inversely proportional to frequency and directly proportional to the applied voltage.
  • This magnetizing current is used to set up flux in core of the transformer.
  • If magnetizing current goes beyond certain limit, the transformer core may saturate. Therefore, it is required to reduce applied voltage along with frequency in same proportion to keep the magnetizing current same.
  • This will reduce applied voltage (400/50 = 8) by 8 times and thus its KVA rating too will reduce 8 times.
RRB JE Electrical (CBT II) Mock Test- 9 - Question 18
How does deforestation cause an increase in greenhouse gases in the atmosphere?
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 18
  • Tropical forest trees, like all green plants, take in carbon dioxide and release oxygen during photosynthesis
  • Plants also carry out the opposite process known as respiration in which they emit carbon dioxide, but generally in smaller amounts than they take in during photosynthesis; The surplus carbon is stored in the plant, helping it to grow
  • When trees are cut down and burned or allowed to rot, their stored carbon is released into the air as carbon dioxide
  • So, the levels of carbon dioxide will increase and cause global warming
  • And this is how deforestation and forest degradation contribute to the increase in greenhouse gases and hence global warming
RRB JE Electrical (CBT II) Mock Test- 9 - Question 19

A ball of mass 8 kg is dropped from a height of 20 m. What is the velocity with which it strikes the ground?

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 19

PE lost = KE gained

KE gained = mgh

KE gained = 8*10*20

KE gained = 1600 J

KE = (1/2)mv2

1600 = (1/2)*8*v2

Solve for v

V = Square root of 400

V = 20 m/s2

RRB JE Electrical (CBT II) Mock Test- 9 - Question 20

The expansion of RAM is _______.

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 20

Primary Memory (Main Memory): Primary memory is also known as main memory. It holds only those data and instructions on which computer is currently working. It has limited capacity and data is lost when power is switched off.
Primary memory can be divided into RAM (Random Access Memory) and ROM (Read Only Memory)

RRB JE Electrical (CBT II) Mock Test- 9 - Question 21
As there is no _______ on conductors, therefore an entire cross-section of conductor is usefully utilized in DC transmission.
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 21

Skin effect is the tendency of an alternating electric current to become distributed within a conductor such 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 the skin depth. The skin effect causes the effective resistance of the conductor to increase at higher frequencies where the skin depth is smaller, thus reducing the effective cross-section of the conductor.

Hence resistance in AC is greater than resistance in DC
RRB JE Electrical (CBT II) Mock Test- 9 - Question 22
Major aerosol pollutant in jet plane emission is
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 22

Aerosols are chloroform-hydrocarbon compounds released into air with force in the form of vapour. Main source of aerosols is the emission of jet planes, where fluorocarbons are used. These chlorofluorocarbons deplete the ozone layer in the higher atmosphere. These CFC ‘s has produced a hole in the ozone layer.

RRB JE Electrical (CBT II) Mock Test- 9 - Question 23

If an induction type energy meter runs fast, it can be slowed by:

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 23

The breaking torque is given by

∴ N ∝ 1/d i.e speed of disc is inverse relation with the position of spindle from the permanent magnet.
If an induction type energy meter runs fast, it can be slowed by adjusting the position of braking magnet and making it move away from the centre of the disc.

RRB JE Electrical (CBT II) Mock Test- 9 - Question 24
Major contributing activity towards Global Warming by Greenhouse gases is
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 24

Greenhouse gases are released by many different types of activities, not just the burning of fossil fuels, but also farming, deforestation and some industrial processes.

Energy sector contributes majorly towards global Warming by releasing greenhouse gases. The percentages of greenhouse gas emissions for different activities are given below.

Energy sector: Electricity and heat (24.9%), Industry (14.7%), Transportation (14.3%), Other fuel combustion (8.6%), Fugitive emissions (4%).

Agriculture sector (13.8%)

Deforestation (12.2%)

Industrial processes (4.3%)

Waste (3.2%)
RRB JE Electrical (CBT II) Mock Test- 9 - Question 25
A 3-phase synchronous generator connected to infinite bus is operating at full load at lagging power factor. Suddenly, the prime mover fails, the machine will
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 25
In case of failure of the prime mover of the generator and the excitation is present then generator draws power from the other parallel generator and starts working as a motor with the same direction and for this motor, the turbine is working as a load. Because of the huge mass of the turbine the motor draws a huge amount of current which damages the winding. It will absorb power from the bus bar i.e working at lagging power factor
RRB JE Electrical (CBT II) Mock Test- 9 - Question 26

The soil pollutants that affect the food chain and food web by killing micro-organisms and plants are

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 26

The soil pollutants that affect the food chain and food web by killing micro-organisms and plants are pesticides.

Agricultural waste is waste produced as a result of various agricultural operations. It includes manure and other wastes from farms, poultry houses and slaughterhouses; harvest waste; fertilizer run off from fields; pesticides that enter into the water, air or soils; and salt and silt drained from fields.

A pathogen or infectious agent is a biological agent that causes disease or illness to its host.

RRB JE Electrical (CBT II) Mock Test- 9 - Question 27

Work sheets are part of

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 27
  • Microsoft Excel is the electronic spreadsheet program within the Microsoft Office suite
  • An electronic spreadsheet is an application used to perform numeric calculations and to analyse and present numeric data
  • In Excel, the electronic spreadsheet is called a worksheet and it is contained in a file called workbook, which has the file extension .xlsx
  • Cells are the basic unit for storing data
  • In any spreadsheet program such as Excel, each rectangular box is referred to as cell
  • A cell is the intersection point of a column and a row
  • Data entered into an Excel spreadsheet is entered into a cell
  • Each cell can hold only one piece of data at a time

RRB JE Electrical (CBT II) Mock Test- 9 - Question 28
The mass of the earth is 6 × 1024 kg and that of the moon is 7.4 × 1022 kg. If the distance between the earth and the moon is 3.84×105 km, calculate the force exerted by the earth on the moon. (Take G = 6.7 × 10-11 N m2 kg-2)
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 28

The mass of the earth, M = 6 × 1024 kg

The mass of the moon, m = 7.4 × 1022 kg

The distance between the earth and the moon, d = 3.84 × 105 km = 3.84 × 105 × 1000 m = 3.84 × 108 m

G = 6.7 × 10-11 N m2 kg-2

The force exerted by the earth on the moon is

F = G x (M m)/d2

F = 2.02 × 1020 N

Thus, the force exerted by the earth on the moon is 2.02 × 1020 N

RRB JE Electrical (CBT II) Mock Test- 9 - Question 29

What is the peak factor value for full-wave rectified sine wave?

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 29

For Half Wave rectifier:

  • Average Vaule = Vm/r =π
  • rms value = Vm /2
  • Ripple factor = 1.21
  • Form factor = 1.58
  • Efficiency = 40.6 %

For Full Wave rectifier:

  • Average Vaule = 2V m/π
  • rms value = Vm /√2
  • Ripple factor = 0.48
  • Form factor = 1.11
  • Peak factor = 1.41
  • Efficiency= 81.2%
RRB JE Electrical (CBT II) Mock Test- 9 - Question 30

Which of the following will need the highest level of illumination?

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 9 - Question 30

Among the given options, proofreading need the highest level of illumination. The illumination levels for different areas are given

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