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RRB JE Electrical (CBT II) Mock Test- 6 - Electrical Engineering (EE) MCQ


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30 Questions MCQ Test - RRB JE Electrical (CBT II) Mock Test- 6

RRB JE Electrical (CBT II) Mock Test- 6 for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The RRB JE Electrical (CBT II) Mock Test- 6 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The RRB JE Electrical (CBT II) Mock Test- 6 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- 6 below.
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RRB JE Electrical (CBT II) Mock Test- 6 - Question 1

The z-transorm of z(an)=

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 1

RRB JE Electrical (CBT II) Mock Test- 6 - Question 2

The drive motor used in a mixer-grinder is a

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 2

Universal motor: It is used in high-speed vacuum cleaners, vacuum cleaners, drink and food mixers, domestic sewing machine, mixer-grinder.

D.C. shunt motor: It is a constant speed motor and is used in centrifugal pumps, fans, blowers

3-phase synchronous motor: It is used for constant speed applications where speed is independent of the load over the operating range of the motor

Induction motor: It is used in a drilling machine, fan, blower printing machines etc.

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RRB JE Electrical (CBT II) Mock Test- 6 - Question 3

Which of the following comes under the secondary air pollutant?

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 3

Peroxy Acetyl Nitrate is a secondary air pollutant whereas, Suspended Particulate Matter, SO2 and NO2 are primary air pollutants.

RRB JE Electrical (CBT II) Mock Test- 6 - Question 4

In which of the following during the emission of radioactive rays, no changes occur in the mass or charge?

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 4

In γ-emission rays ( Gamma rays), no changes occur in the mass/charge.

Gamma rays are electromagnetic waves of short wavelength and high frequency.

They are emitted by most radioactive sources along with alpha or beta particles

Through releasing a gamma photon it reduces to a lower energy state. Hence, Gamma rays have no electrical charge connected with them.

Alpha rays are released by high mass whereas beta has high energy electrons.

RRB JE Electrical (CBT II) Mock Test- 6 - Question 5

What is the chemical symbol of Cobalt?

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 5

Cobalt is a chemical element with symbol Co and atomic number 27.

RRB JE Electrical (CBT II) Mock Test- 6 - Question 6
Which one of the following is not alkaline?
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 6
Alkaline refers to a nutrient solution or growing media with a pH greater than 7. Alkaline soils are normally composed of various types of materials such as calcium carbonate. Highly alkaline soils are more prevalent in dry regions. Because soil pH is directly linked to nutrient availability, alkaline soils tend to have a deficiency in phosphorus, zinc, and iron. So sulphur is not an alkaline .
RRB JE Electrical (CBT II) Mock Test- 6 - Question 7
Hysteresis is the phenomenon of_________ in a magnetic circuits.
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 7
Hysteresis is the phenomenon of lagging of B behind H in a magnetic circuits.
The phenomenon of hysteresis in ferromagnetic materials is the result of two effects: rotation of magnetization and changes in size or number of magnetic domains. In general, the magnetization varies (in direction but not magnitude) across a magnet, but in sufficiently small magnets, it does not. In these single-domain magnets, the magnetization responds to a magnetic field by rotating. Single-domain magnets are used wherever a strong, stable magnetization is needed (for example, magnetic recording).
RRB JE Electrical (CBT II) Mock Test- 6 - Question 8
A reluctance motor runs at___________.
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 8
A reluctance motor runs at synchronous speed.
Synchronous motor is act as a reluctance motor ,if field winding excition is zero.
It run same as synchronous motor.
It receive both active power and reactive power for motoring oparation.
RRB JE Electrical (CBT II) Mock Test- 6 - Question 9
__________ is used as a programming language in first generation computers?
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 9
A first generation (programming) language (1GL) is a grouping of programming languages that are machine level languages used to program first-generation computers.
RRB JE Electrical (CBT II) Mock Test- 6 - Question 10

When excitation of the synchronous motor is increased up to normal excitation from under excitation, the armature current

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 10

By observing V curves,
When excitation of the synchronous motor is increased up to normal excitation from under excitation, armature current decreases
When excitation of the synchronous motor is increased to overexcitation from normal excitation, armature current increases

RRB JE Electrical (CBT II) Mock Test- 6 - Question 11

Three resistances of 1 ohm, 2 ohms and 3 ohms are connected in delta. These resistances are to replace by star connection as shown in the figure below, maintaining the same terminal conditions. The value of the highest resistance in the star will be

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 11

To convert delta to star,

RRB JE Electrical (CBT II) Mock Test- 6 - Question 12
Electroplating is done for:
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 12

Electroplating is defined as the deposition of a metal over any metallic or non-metallic surfaces. Electroplating is usually employed

1. To protect the metals from corrosion by atmospheric air, moisture and CO2

2. To give the reflecting properties to reflectors

3. To replace worn out metals

4. To give a shiny appearance to articles
RRB JE Electrical (CBT II) Mock Test- 6 - Question 13

A current source i(t) = 1.25 cos (5t – 15°), a driving load = 12∠53°Ω. What is the value of the complex power delivered by the source to the load?

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 13

Given that, (t) = 1.25 cos (5t – 15°)

Complex power delivered by the source to the load is

RRB JE Electrical (CBT II) Mock Test- 6 - Question 14
The area of the hysteresis loop for a specimen is found to be large. This means that the hysteresis loss in this specimen would be _______
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 14
The energy loss associated with hysteresis is proportional to the area of the hysteresis loop. As the area of the hysteresis loop for a specimen is found to be large, the hysteresis loss in this specimen is also large.
RRB JE Electrical (CBT II) Mock Test- 6 - Question 15
Who has assumed the charge as Vice Chief of the Naval Staff on 30 Jan 2019?
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 15
  • Vice Admiral G Ashok Kumar, AVSM, VSM assumed charge as Vice Chief of the Naval Staff on 30 Jan 2019.
  • Kumar has held various Staff and Command assignments during his distinguished naval career spread over more than three decades.
  • He is an alumnus of National Defence Academy, Khadakvasla, Pune and was commissioned into the Executive Branch of the Indian Navy on 01 Jul 1982.
RRB JE Electrical (CBT II) Mock Test- 6 - Question 16
To calculate the power factor which of the following is odd one:
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 16

In AC circuits, the power factor is defined as the ratio of the real power flowing to the load to the apparent power in the circuit.

Hence power factor can be defined as watts to volt amperes

Power factor = cosϕ

Where, ϕ is the angle between voltage and current.

If current lags the voltage, power factor will be lagging.

If current leads the voltage, power factor will be leading.

From the options,

1) Ratio of real power to apparent power

2) P/VI represents the ratio of the real power to apparent power, it is known as power factor

3) R/Z also represents power factor from the impedance triangle

4) V/I represents resistance but not power factor
RRB JE Electrical (CBT II) Mock Test- 6 - Question 17
During the operation of a connecting a busbar in parallel with an incoming alternator the:
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 17

During the operation of a connecting a busbar in parallel with an incoming alternator, the following condition should be satisfied:

  • The phase sequence of the busbar voltages and the incoming machine voltage must be the same
  • The busbar voltages and the incoming machine terminal voltage must be in phase
  • The terminal voltage of the incoming machine and the alternator which is to be connected in parallel or with the busbar voltage should be equal
  • The frequency of the generated voltage of the incoming machine and the frequency of the voltage of the busbar should be equal
RRB JE Electrical (CBT II) Mock Test- 6 - Question 18

An RLC circuit in series resonance at the frequency of f Hz. If the value of all the components doubled, the new frequency will be

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 18

The frequency of a series RLC circuit at resonance is,

If the value of all the components doubled, the new frequency will be

RRB JE Electrical (CBT II) Mock Test- 6 - Question 19
In an AC circuit, power is consumed only in
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 19

In an AC circuit, power is consumed only in resistor whereas inductor and capacitor act as power sources.

Resistor dissipates the energy whereas inductor and capacitor stores the energy.

Inductor stores energy in the form of magnetic field and capacitor stores energy in the form of the electrostatic field.
RRB JE Electrical (CBT II) Mock Test- 6 - Question 20
Which of the following is not an example of a biomass energy source?
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 20

Energy resources can be classified into two categories:

  1. Renewable source (Non-conventional energy)
  2. Non-Renewable source (Conventional energy)

Renewable energy resources are natural resources which can be regenerated continuously and are inexhaustible. They can be used again and again in an endless manner. Examples are Wood, Solar energy, wind energy, hydropower, tidal energy, geothermal energy, Biomass Energy etc.

  • Biomass Energy: Energy produced by organic matter like plants and animals are called Biomass Energy
  • Examples of biomass are wood, crop residues, seeds, cattle dung, sewage, agricultural wastes etc
  • Biogas and biofuel are examples of biomass energy

Non-Renewable energy resources are natural resources which cannot be regenerated once they are exhausted. They cannot be used again. Examples are Oil, Coal, petroleum, natural gas, nuclear fuels etc.

RRB JE Electrical (CBT II) Mock Test- 6 - Question 21

If ‘t’ be the thickness of the lamination, then eddy current loss in a generator will vary to:

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 21

Eddy current losses are given by
W e = Kef2B2mt2v
Where, Ke = eddy current constant.
f is frequency
Bm is flux density
t is thickness
v is volume
Here eddy current losses are directly proportional to square of thickness

RRB JE Electrical (CBT II) Mock Test- 6 - Question 22

A series RL circuit with R = 100 ohms and L = 50 H is supplied by a D.C. source of 100 V. The time taken by the current to rise to 70% of its steady state value is

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 22

i(0+) = 0 A
i(∞) = V/R = 100/100 = 1
Time constant τ = L/R = 50/100 = 0.5
i(t) = i(∞)+[i(0+)-i(∞)]e-t/τ
i(t) = 1 - e-t/0.5
⇒ 0.7 = 1 - e-t/0.5
⇒ t = 0.6 sec

RRB JE Electrical (CBT II) Mock Test- 6 - Question 23

Who won her second international gold in women's 200m with a top finish at the Kutno Athletics Meet in Poland in July 2019?

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 23
  • Hima Das won her second international gold in women's 200m with a top finish at the Kutno Athletics Meet in Poland.
  • Hima took 23.97 seconds to clinch the gold while VK Vismaya bagged the silver in 24.06 seconds.
  • This was Hima's second competitive 200m race of 2019. She has a personal best of 23.10s, which she clocked in 2018.
RRB JE Electrical (CBT II) Mock Test- 6 - Question 24

Varley loop tests are preferred over Murray loop tests because

Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 24
  • Varley loop test is used for locating short-circuit and earth faults in underground cables.
  • This test also employs the principle of the Wheatstone bridge. However, the difference between Murray loop test and Varley loop test is that, in Varley loop test resistances R1 and R2 are fixed, and a variable resistor is inserted in the faulted leg.
  • If the fault resistance is high, the sensitivity of Murray loop test is reduced, and Varley loop test may be more suitable.

RRB JE Electrical (CBT II) Mock Test- 6 - Question 25
Which among these is a demerit of underground service mains?
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 25

Appearance: The general appearance of an underground system is better as all the distribution lines are invisible.

Fault location and repairs: In general, there are little chances of faults in an underground system. However, if a fault does occur, it is difficult to locate and repair this system. On an overhead system, the conductors are visible and easily accessible so that fault locations and repairs can be easily made.

Initial cost: The underground system is more expensive due to the high cost of trenching, conduits, cables, manholes and other special equipment. The initial cost of an underground system may be five to ten times than that of an overhead system.
RRB JE Electrical (CBT II) Mock Test- 6 - Question 26
The carbon arc welding has the advantages of:
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 26

Advantages of carbon arc welding:

  • The heat developed during the welding can be easily controlled by adjusting the length of the arc
  • It is quite clean, simple, and less expensive when compared to other welding processes
  • Both the ferrous and non-ferrous metals can be welded
  • Easily adaptable to automation

Disadvantages of carbon arc welding:

  • Input current required in this welding, for the workpiece to raise its temperature to welding temperature, is more
  • In case of ferrous metal, there is a chance of disintegrating the carbon at high temperature and transfer to the weld, which causes weld deposit to be harder and brittle
  • A separate filler rod has to be used if any filler metal is required
RRB JE Electrical (CBT II) Mock Test- 6 - Question 27
The critical temperature above which ferromagnetic materials lose their magnetic property is known as
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 27

Curie temperature: It is the temperature above which ferromagnetic materials lose their permanent magnetic field and the magnetism completely disappears.

The magnetic susceptibility decreases with an increase in temperature. So, the ferromagnetism decreases with rising temperature. It is maximum at absolute zero temperature and becomes zero at Curie temperature. Above this temperature, the ferromagnetic material behaves as paramagnetic substance.
RRB JE Electrical (CBT II) Mock Test- 6 - Question 28
If four 90-ohm resistors are connected in series across an 18 V source, the current equals
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 28

When four 90-ohm resistors are connected in series, the equivalent resistance becomes:

Req = 4 × 90 = 360 Ω

Voltage (V) = 18 V

Current (I) = V/R = 18/360 = 0.05 A = 50 mA
RRB JE Electrical (CBT II) Mock Test- 6 - Question 29
Different types of line supports used for transmission lines include
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 29

Wooden poles, steel poles, RCC and PCC poles and steel towers are used for transmission line.

Wooden poles:

Wooden poles are used for temporary electrical line. The length of wooden pole is kept 6-9 meters out of which 1-2 meter length buried in the ground. The diameter of the pole is kept 15-25 cm so as to have safety factor of 3.5. Maximum permissible span for wooden poles is 75 meter.

Steel poles:

  • Steel poles possess greater mechanical strength and thus permit the use of longer spans (60 to 80 metres).
  • These poles have the longer life (more than 40 years) which can further be increased by regular painting.
  • These poles need protection against corrosion. Hence at the bottom (the portion which is buried underground), these poles are set in concrete muffs in order to protect them from chemical reactions.

RCC and PCC poles:

In a service line of a residential colony normally RCC poles are preferred. RCC means reinforced cement concrete.
RRB JE Electrical (CBT II) Mock Test- 6 - Question 30
The electric flux and field intensity inside a conducting sphere is _____.
Detailed Solution for RRB JE Electrical (CBT II) Mock Test- 6 - Question 30

According to Gauss law, the electric field inside a conducting sphere is always zero irrespective of how much charge resides on its surface. Due to zero electric field inside a sphere, the electric flux inside a sphere is zero.

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