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RSMSSB JE Electrical Mock Test - 4 - Electrical Engineering (EE) MCQ


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30 Questions MCQ Test - RSMSSB JE Electrical Mock Test - 4

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RSMSSB JE Electrical Mock Test - 4 - Question 1

Who among the following was the founder of Niranjani sect?

Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 1

The correct answer is Saint Haridas.

Key Points

  • Saint Haridas was the founder of the Niranjani sect a bhakti religious community primarily based in central Rajasthan.
  • He was believed to be lived from the mid-sixteenth to early seventeenth century.
  • He was born in Kapadod village which is located near the town of Didwana in Rajasthan.
  • He was a dacoit before becoming a saint and he is called Valmiki of Kaliyuga.

Additional Information

  • Saint Ramcharan
    • ​He is the Rajasthani Hindu guru, inspirator of a religious tradition called Ramsnehi Sampradaya or Ramdwara.
    • He initiated and illustrated Nirguna Bhakti.
    • His works are collected in Vaani Ji consisting of 36250 compositions.
  • Saint Ramdas
    • ​Samarth Ramdas is also known as Sant Ramdas was an Indian Marathi Hindu saint, philosopher, poet, writer, and spiritual master.
    • Some of his notable literary works are Manache shlok, Dasbodh, Shree Maruti Stotra, Aatmaaram, 11-Laghu Kavita, Shadripu Nirupan.
  • Saint Niranjandas
    • ​He is a spiritual leader at Dera Sach Khand a socio-religious social organization based in the village of Ballan near Jalandhar, Punjab.
    • Dera Sach Khand was founded by devotees of Guru Ravidas.
RSMSSB JE Electrical Mock Test - 4 - Question 2

"Ras Rahasya" text was written by-

Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 2

The correct answer is Kulpati Mishra.

Key Points

  • Ras Rahsya text was written by Kulpati Mishra.
  • Sangram-Sar and Nakhsikh were other texts written by Kulpati Mishra.
  • Kulpati Mishra, Shankar Bhatt, Ganesh Deoal, and Dalpati Ram were writers in Maharaja Ramsingh I reign.
  • Ramsingh I (1667-1682 AD) was son of Mirza Raja Jai Singh.


Additional Information

  • Chagta Padshahi text was written by Dalpati Ram.
  • Vaidh Vinod text was written by Shankar Bhatt.
  • Muhrat Tatv text was written by Ganesh Deoal.
RSMSSB JE Electrical Mock Test - 4 - Question 3

Who is the author of "Mutinies in Rajputana"?

Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 3

The correct answer is I. T. Prichard.

Key Points

  • He was an officer in the Bengal army.
  • The same mutinied in 1857.
  • Its scope is contained in the detail that appears on the title page after the main heading.
  • A personal narrative of the mutiny at Nasirabad, with subsequent residence at Jodhpur, and journey across the desert into Sindh, together with an account of the outbreak at Neemuch, and the mutiny of the Jodhpur Legion at Erinpoora, and attack on Mount Abu.

Additional Information

RSMSSB JE Electrical Mock Test - 4 - Question 4
Who was the author of the book 'Shok Shatak'?
Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 4

The correct answer is Balabakhsh.

  • Balabakhsh was the author of the book Shok Shatak.

Key Points

  • Some of the works of Balabakhsh:
    • Ashva Vidhan Soochna
    • Bhupal Sujsavarn
    • Shata Shastra Saransh
    • Sandhopasana Atyanika
    • Kshatriya Education Panchashika
    • Shok Shatak
    • Narukul Suyash
    • Assis Ashtak
    • Shastra Prakash
    • Maan Mahotsav Mahima

Additional Information

  • Balabakhsh Palawat
    • The Palawats are considered under the "Barhat" branch.
    • In this Palawat branch, there were great devotees, donors, philanthropists like Balabakhsh Ji, who donated several thousand rupees to the Kashi Nagari Pracharini Sabha, with the help of which "Balbakhsh, Rajput barn, book garland" created the history of Rajputs and barons.
    • He was born in Hanutia village under Jaipur state.
RSMSSB JE Electrical Mock Test - 4 - Question 5
'Grameen Gaurav Path Yojana' in Rajasthan was started in which of the following years?
Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 5
The correct answer is 2014-15 Key Points
  • Gramin Gaurav Path is a major road construction program started by the state Government of Rajasthan and is implemented by Public Works Department (PWD).
  • Under the Gramin Gaurav Path project, the government aims to construct main Gram Panchayat roads in 33 districts in the Rajasthan state.
  • In the first phase, the government planned to construct 1720 km of roads in the 1972 gram panchayat.
  • In the second phase, the government plan to construct 2098 km of roads in 2086 gram panchayat.
  • The government has aims to construct roads in all villages that come under the main gram panchayat for hassle-free transport for villagers.

Important Points

  • The main objective of the Gramin Gaurav Path Scheme is to construct a Cement Concrete road and proper drainage at each Gram Panchayat headquarters with the purpose of creating a neat and clean environment.
  • The program also aims to provide damage-free roads for commuters at each Gram Panchayat Headquarters.
  • Rural villagers now easily travel to schools, hospitals, temples, farms, nearby villages, and towns because of the construction of good roads.
  • The Rajasthan government has approved Rs 1,115 crore for the Gramin Gaurav Path program
Additional Information
  • Rajasthan is a state in northern India.
  • It covers 342,239 square kilometres or 10.4 per cent of India's total geographical area.
  • It is the largest Indian state by area and the seventh-largest by population
  • Area: 342,239 km²
  • Governor: Kalraj Mishra
  • Capital: Jaipur
  • Population: 6.89 crores
  • Chief minister: Ashok Gehlot
RSMSSB JE Electrical Mock Test - 4 - Question 6
Who was the last official ruler of Princely state of Bundi?
Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 6

The correct answer is Maharao Raja Bahadur Singh.

Key Points

  • Maharao Raja Bahadur Singh was the last king of Bundi.

Additional Information

  • Bundi, Kota, Jhalawar and Baran regions are included in Hadauti.
    • In ancient times this area was ruled by the Meenas.
    • This name of Bundi was named after Bunda Meena.
    • Bundi was earlier called "Bunda-ka-Nal", Nal meaning "narrow passage".
    • The region was later ruled by Rao Deo Hada, who took over Bundi from Jaita Meena in 1342 and established a princely state called Bundi, naming the surrounding area Hadauti, the land of the great Hada Rajputs.
    • In 1818 AD, Vishnusinh, the ruler of Bundi concluded a subsidiary alliance with the East India Company.
    • On March 25, 1948, Bundi merged with Rajasthan.
RSMSSB JE Electrical Mock Test - 4 - Question 7
The main objective of RIICO is -
Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 7

The correct answer is All of the above.

Key Points

  • Rajasthan State Industrial Development and Investment Corporation (RIICO):
    • RIICO is an apex organization engaged in promoting the growth of industrialization in the state.
    • RIICO's mission is to catalyze the planned and rapid industrialization of Rajasthan.
    • RIICO has set up 28 regional offices across Rajasthan to conduct the development and management of industrial areas.
    • RIICO has pioneered the industrialization of the state of Rajasthan by setting up industrial areas.
    • RIICO also acts as a financial institution by providing loans to large, medium, and small-scale projects.
  • RIICO​'s Role:
    • Identifying investment opportunities.
    • Providing access to important information for setting up projects.
    • To assist in the translation of investment possibilities into concrete investment propositions.
    • The nodal agency for single-window clearance.
    • Ensuring prompt attention to proposals on behalf of investors along with all government departments/agencies.
    • Assisting in site selection and procurement of infrastructure facilities.
    • Intervention with government departments for necessary clearances.
RSMSSB JE Electrical Mock Test - 4 - Question 8

In order to control the speed of a 3-phase slip ring induction motor through injected voltage in its rotor circuit, this voltage and the rotor voltage should essentially be

Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 8

To control the speed of a 3-phase slip ring induction motor, the injected voltage in its rotor circuit must have the same frequency as the rotor voltage. This matching frequency allows for the adjustment of slip and enables speed control.

Concept:

Rotor emf injection method:
For below-rated speeds: In this method, injected emf has the same frequency of rotor slip frequency and that emf is 180° out of phase with rotor emf.

E2R is resultant emf in the rotor

E2R = E2 – E1

R2 is rotor resistance

T is the torque

s is the slip

Here, the value of rotor emf becomes less. To maintain constant torque, the value of slip will increase. Therefore, the speed will be decreased.

At this condition, effective rotor resistance increases.

For above-rated speeds: In this method, injected emf has the same frequency of rotor slip frequency and that emf is in phase with rotor emf.

E2R is resultant emf in the rotor

E2R = E2 + E1

R2 is rotor resistance
T is the torque
s is the slip

Here, the value of rotor emf becomes more. To maintain constant torque, the value of slip will decrease. Therefore, the speed will be increased.

RSMSSB JE Electrical Mock Test - 4 - Question 9
Most common application of UJT is –
Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 9

The applications of a unijunction transistor are:

  • As a triggering device for SCR’s and Triacs
  • Saw-toothed generators
  • Simple oscillators
  • Phase control
  • Timing circuits

The simplest and most common application of all UJT circuits is the Relaxation Oscillator producing non-sinusoidal waveforms.

RSMSSB JE Electrical Mock Test - 4 - Question 10
One of the following can acts as an Inverse Transducer?
Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 10

The inverse transducer is defined as a device that converts an electrical quantity into a non-electrical quantity.

A piezoelectric crystal acts as an inverse transducer because when a voltage is applied across its surfaces, it changes its dimensions causing a mechanical displacement.

Note:

A piezoelectric sensor is a device that uses the piezoelectric effect to measure changes in pressure, acceleration, temperature, strain, or force by converting them to an electrical charge. It is used to measure displacement.

RSMSSB JE Electrical Mock Test - 4 - Question 11

The final value of  will be

Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 11

Concept:

1. Final value theorem:
A final value theorem allows the time domain behavior to be directly calculated by taking a limit of a frequency domain expression
The final value theorem states that the final value of a system can be calculated by

Where F(s) is the Laplace transform of the function.
For the final value theorem to be applicable system should be stable in a steady-state and for that real part of the poles should lie on the left side of s plane.

2. Initial value theorem:


It is applicable only when the number of poles of C(s) is more than the number of zeros of C(s).

Calculation:
Given that,
Poles lies at s = 0, ±j 1
Roots lies on the imaginary axis, so it is marginally stable.
So the Final value theorem is not applicable as the system is oscillatory in nature.

∴ The correct answer is option C 'cannot be determined'.

RSMSSB JE Electrical Mock Test - 4 - Question 12
Voltage applied to a load is 100√2 sin500t. Current through the load is 10√2sin(500t + π/3). The power consumed by the load is:
Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 12

Concept:

Real Power transmitted to the load can be given as,

P = Vrms × Irms × cos ϕ

Where,

Vrms = load terminal voltage in volt

Irms = load current in amper

cos ϕ is the power factor of the load

ϕ is the power factor angle

Calculation:

v = 100√2 sin (500t) V

I = 10√2 sin (500t + π/3) A

Power factor angle

Active power consumed by load = Vrms× Irms × cos ϕ

=

= 500 W

RSMSSB JE Electrical Mock Test - 4 - Question 13

A series RLC circuit has a resonance frequency of 1 KHz and a quality factor Q = 100. If each of R, L, C doubled from its original value. The new quality factor Q of the circuit is

Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 13

Concept:

The quality factor is defined as the ratio of the maximum energy stored to maximum energy dissipated in a cycle

The quality factor in a series RLC circuit is given by:

Calculation:

If R, L, and C double from its original value, the quality factor becomes half of the previous value.

New quality factor (Q) will be:
Q = 100/2 = 50

RSMSSB JE Electrical Mock Test - 4 - Question 14

Class A commutation is often referred to as :

Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 14

The correct answer is option 1):(Load commutation)

Commutation of SCR:

  • Commutation is nothing but the turn-OFF method of an SCR. It is one method used to bring an SCR or thyristor from the ON state to the OFF state. We know that an SCR can be activated by using a gate signal toward an SCR when it is in forwarding bias. But SCR needs to turn off when required for power controlling otherwise power conditioning.
  • The commutation techniques of thyristors are classified into two types: Natural Commutation and Forced Commutation
  • Class A commutation of the thyristor is a type of forced commutation and is also referred to as Load Commutation. It is one of the widely used commutation techniques that mainly find applications in high-frequency operations. Hence option 1 is correct.
  • Class B commutation is also called as Resonant-Pulse Commutation. It is one of the types of forced commutation
  • Class-C commutation is also known as complementary commutation. Complementary commutation is also called voltage commutation, forced commutation, complimentary impulse commutation, or Class-C commutation. It is used in dc choppers and inverters.
  • The auxiliary voltage commutation is also called “Impulse commutation”, or class D commutation. it uses an auxiliary SCR to switch the charged capacitor.

RSMSSB JE Electrical Mock Test - 4 - Question 15

The magnetic flux through a 150 turns coil increases at the rate of 0.08 wb/s. What is the induced EMF between the ends of the coil?

Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 15

Concept:

  • Magnetic flux is total magnetic field that passes through the given area.
  • If we choose a simple flat surface with area A and there is an angle θ between the normal to the surface and a magnetic field vector (magnitude B) then the magnetic flux is ϕ = BA cosθ
  • Faraday's Law: Any change in the magnetic flux of wire will cause a voltage (induced emf) to be "induced" in the coil.
  • This change can be produced by changing the magnetic field strength, moving the coil into or out of the magnetic field, moving a magnet toward or away from the coil,rotating the coil relative to the magnet, etc.
  • Induced emf

Calculation:

Given that N = 150 turns
Δϕ/Δt = 0.08 Wb/s

induced emf

= 150 × 0.08 = 12 Volt

Important Point

  • Lenz's law says that the direction of the induced current in coil will be always in such a way as to oppose the change which produces the current.
  • It is just a small addition to Faraday's law. See negative sign in the formula. Negative sign shows opposement.
RSMSSB JE Electrical Mock Test - 4 - Question 16

Which of the following transductions is done by the Hall effect device as a transducer?

Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 16

Concept:

Hall Effect Transducer :

  • It is a device that is used for the measurement of magnetic field strength.
  • This transducer uses a conducting strip to convert the magnetic field into proportional potential difference across the opposite faces of the strip using the Hall Effect./

Working Principle:

  • The working principle of the Hall Effect Transducer is based on the Hall Effect.
  • The Hall Effect is basically the process of the development of potential differences across the two faces of a current-carrying strip when the strip is kept in a magnetic field.
  • The magnitude of voltage depends upon the current, strength of the magnetic field, and the property of conducting material.
  • The Hall Effect is found in conducting material and semiconductors in varying amounts depending upon the density and mobility of the current carrier.
  • The figure given below explains the working principle of the Hall effect transducer.

  • In the above figure, a conducting strip is kept in a transverse magnetic field. Note that the current through the strip and the magnetic field are perpendicular to each other.
  • The flow of current means the flow of positive charges in the direction of the current.
  • This means that the magnetic field will exert a force on the moving positive charges as per F = q(v × B).
  • Where v & B are the velocity and strength of the magnetic field respectively.
  • v & B are in vector form.
  • Since v and B are perpendicular to each other, the magnitude of the force on the moving positive charges will be F = qvB
  • The expression given below shows the voltage developed due to the Hall Effect.
  • This voltage is called the Hall effect emf (EH) and is used to either measure the magnitude of current or magnetic field strength.
  • Where KH is a constant of proportionality called the Hall Effect Coefficient and 't' is the thickness of the strip.
  • The magnitude of Hall Effect emf is very small in conductors and hence very difficult to measure.

  • However, its value is quite sufficient in semiconductors and can easily be measured by sensitive moving coil instruments.

Major applications:

  • Measurement of Magnetic Field Strength.
  • Measurement of Displacement.
  • Measurement of Current.
RSMSSB JE Electrical Mock Test - 4 - Question 17

The number of roots of s3 + 5s2 + 7s + 3 = 0 in the left half of the s-plane is

Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 17

Concept:

The characteristic equation for a given open-loop transfer function G(s) is

1 + G(s) H(s) = 0

According to the Routh tabulation method,

The system is said to be stable if there are no sign changes in the first column of the Routh array

The number of poles lies on the right half of s plane = number of sign changes

Calculation:

Characteristic equation: s3 + 5s2 + 7s + 3 = 0

By applying the Routh tabulation method,

Number of roots lie on the right side of the s-plane = Number of sign changes = 0

Number of roots lie on the left side of the s-plane = Total number of roots of the characteristic equation - number of roots lie on the right side of the s-plane

As the characteristic equation is a cubic equation

∴ Total number of roots of the characteristic equation = 3

Hence, the number of roots lie on the left side of the s-plane = 3 - 0

Number of roots that lie on the left side of the s-plane = 3

RSMSSB JE Electrical Mock Test - 4 - Question 18
The voltage at the two ends of a line are 132 kV and its reactance is 40 ohms. The capacity of the line is
Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 18

Concept:

The maximum power transfer capacity of the transmission line is given by

Where,

|Vs| is sending end voltage of transmission line

|Vr| is receiving end voltage of transmission line

X is the series reactance of transmission line

Calculation:

Given

Vs = Vr = 132 kV

X = 40 Ω

The capacity of the line P

P = (132 × 103)2 / 40 = 435.6 MW

Therefore, the capacity of the line is 435.6 MW

RSMSSB JE Electrical Mock Test - 4 - Question 19
Resistance switching is normally resorted in case of-
Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 19

Resistance switching is normally resorted in the Air blast Circuit breaker.

Resistance switching:

Circuit diagram for the resistance switching of circuit breaker is shown below.

  • Resistance Switching in Circuit Breaker refers to a method adopted for dampening the over voltage transients due to current chopping, capacitive current breaking etc.
  • In this method, a shunt resistance is connected across the contacts of circuit breaker.
  • The main role of shunt resistance R is to limit the growth of re-strike voltage and cause it to grow exponentially up to recovery voltage.
  • R is so selected that the circuit becomes critically damped then re-strike voltage rises exponentially till recovery voltage is reached.
  • The value of R for critical damping is
  • Resistance Switching also reduces the oscillatory growth of Re-striking Voltage
  • Voltage appearing across the electrodes of circuit breaker or switching voltage

RSMSSB JE Electrical Mock Test - 4 - Question 20

A transformer can have regulation closer to zero on __________.

Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 20

Voltage regulation:

Voltage regulation is the change in secondary terminal voltage from no load to full load at a specific power factor of load and the change is expressed in percentage.
E2 = no-load secondary voltage
V2 = full load secondary voltage

Voltage regulation for the transformer is given by the ratio of change in secondary terminal voltage from no load to full load to no load secondary voltage.

Voltage regulation

It can also be expressed as,

Regulation
+ sign is used for lagging loads and
- ve sign is used for leading loads

Hence voltage regulation can be negative only for capacitive loads

In a transformer, minimum voltage regulation occurs when the power factor of the load is leading.

The voltage regulation of the transformer is zero at a leading power factor load such as a capacitive load.

For zero voltage regulation, E2 = V2 

⇒ IR cosϕ = IX sinϕ (negative sign represents leading power factor loads)

This is the leading power factor at which voltage regulation becomes zero while supplying the load.

RSMSSB JE Electrical Mock Test - 4 - Question 21

A damper winding is used in a synchronous motor for-

Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 21

In large alternators, damper windings are used to

  • Reduce the vibrations (hunting)
  • To increase the stability i.e. the voltage fluctuations will get reduced

The damper winding in a synchronous motor performs two functions:

  • Prevents hunting (damped out oscillations)
  • Provides starting torque (made self-starting)

Note: Under normal running conditions, damper winding does not carry any current.

Important Points

Hunting is a phenomenon that occurs in synchronous motors due to varying load or supply frequency.

Causes of hunting

  • Periodic variation of load
  • Sudden changes in load
  • Faults occur in the system when supplied by the generator
  • Sudden change in the field current
  • Cyclic variations of the load torque

Effects of hunting

  • It can lead to loss of synchronism
  • It can cause variations of the supply voltage producing undesirable lamp flicker
  • The possibility of resonance condition increases. If the frequency of the torque component becomes equal to that of the transient oscillations of the synchronous machine, resonance may take place
  • Large mechanical stresses may develop in the rotor shaft
  • The machine losses increases and the temperature of the machine rises
RSMSSB JE Electrical Mock Test - 4 - Question 22

What are the materials which exhibit electric polarization even in the absence of an applied electrical field?

Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 22

Properties of Antiferroelectric Material:

  • Antiferroelectric material can exhibit electric polarization even in absence of an applied electric field (spontaneous polarization).
  • By varying, the direction of the field applied we can invert the direction of polarization.
  • It is analogous to the antiferromagnetic material.
  • Examples: PbZrO3 (Lead Zirconate), NH4H2PO4 (ADP: Ammonium dihydrogen Phosphate), NaNbO3 (Sodium Niobate)
  • Application: Supercapacitors, Used in integration with ferromagnetic materials, High energy storage devices, etc

Properties of Ferroelectric material:

  • It has a parallel arrangement of the dipole and all dipoles become parallel after the applied electric field.
  • It can also maintain the polarisation even once the field is removed.
  • Polarization can be reversed by reversing the applied electric field.
  • Above Curie temperature, it acts as a piezoelectric material.
  • It is analogous to the ferromagnetic material.
  • Example: BaTiO3, PbTiO3, Lead Zirconate Titanate (PZT), Triglycine Sulphate, PVDF, Lithium tantalite, etc.
  • Application: Thermistors, Oscillators, Non-volatile memory, Filters, Capacitors, Light deflectors, Trans chargers, Electro-optic materials, Modulators, Piezoelectric, Display, etc

Properties of Paramagnetic Material:

  • Atomic dipole exists.
  • The net magnetic dipole moment is zero without an external magnetic field.
  • On applying an external magnetic field, the atomic dipole aligns in the direction of the applied magnetic field.
  • Atomic dipole alignment increased with an increase in external field strength and decreased with temperature.
  • They are weakly attracted to the external magnetic field.
  • The value of susceptibility is positive and very small in the order of 10-3 to 10-6.
  • The value of relative permeability is slightly greater than 1.
  • Example: Sodium, Aluminium, Gold, Copper, etc

Properties of Ferromagnetic Material:

  • Atomic dipole exists just like Paramagnetic material.
  • On applying an external magnetic field, the atomic dipole aligns in the same direction as the applied magnetic field.
  • They are strongly attracted to the external magnetic field.
  • The value of susceptibility is positive and large.
  • The value of relative permeability is much greater than 1.
  • Above Curie temperature, it acts as paramagnetic material.
  • Example: Iron, Nickel, Cobalt, etc
RSMSSB JE Electrical Mock Test - 4 - Question 23
Amorphous materials are
Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 23

Amorphous material:

  • Amorphous materials are in which there is no definite atomic structure and atoms exist in a random pattern just as in a liquid.
  • An amorphous material (AM) has a non-crystalline structure that differs from that of its iso-chemical liquid and does not undergo structural relaxation and the glass transition when heated. Examples are Glass, Gels, plastics, various polymers, wax, thin films.
  • The terms amorphous and non-crystalline are synonymous under this definition. The term glassy has the same structural meaning, but besides it also usually implies that the material exhibits a ‘glass transition’. Although the presence of dynamic disorder in the case of liquids complicates matters, the average atomic structure of liquids can be described in similar ways to that of amorphous solids.
  • Amorphous alloys are a class of metal alloy that, unlike ordinary metals do not have a long-range crystal structure.
  • Amorphous alloys have no grain boundaries.

Properties of Amorphous Solids:

Amorphous solids are sometimes described as supercooled liquids because their molecules are arranged randomly somewhat as in a liquid state.

  1. Lack of long-range order: Amorphous Solid does not have a long-range order of arrangement of their constituent particles. However, they may possess small regions of orderly arrangement. These crystalline parts of an otherwise amorphous solid are known as crystallites.
  2. No sharp melting point: An amorphous solid does not have a sharp melting point but melts over a range of temperatures. For example, glass on heating first softens and then melts over a temperature range. Glass, therefore, can be molded or blown into various shapes. Amorphous solid does not possess the characteristic heat of fusion.
  3. Conversion into a crystalline form: Amorphous solid, when heated and then cooled slowly by annealing, becomes crystalline at some temperature. That is why glass objects of ancient times look milky because of some crystallization having taken place.

Additional Information

Difference between Crystalline and Amorphous Solid :

RSMSSB JE Electrical Mock Test - 4 - Question 24

Rajasthan is endowed with a solar radiation intensity of around_______Kwh/sq-m/day.

Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 24

The correct answer is 6 to 7.

Key Points

  • Rajasthan has a lot of solar energy potential.
  • Rajasthan has a semi-arid climate, and the Thar Desert covers 66.66% of the state's total land area.
  • With these climatic characteristics, it may receive 6-6.4 kwh/m2 of solar radiation per day.
  • This is the second-highest amount in the globe and around 300–325 sunny days annually.
  • The western cities of Rajasthan often experience temperatures between 35 and 40 degrees, with summertime highs of above 45 degrees.
  • Rajasthan has a solar energy availability of 6 to 7 kw/km2.
  • It has the capacity to produce 100000 MW of power year, albeit only 442.25 MW is currently being produced.

Additional Information

  • Solar radiation intensity:
    • It is the density of solar radiation coming per unit area of the photoelectric module.
    • Above the earth's atmosphere, solar radiation has an intensity of approximately 1380 watts per square meter (W/m2).
RSMSSB JE Electrical Mock Test - 4 - Question 25
In the application of electrical circuits, the nichrome that is used to make the heating element in an electric cooker has ________.
Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 25

The correct composition of nichrome used in the heating element of an electric cooker is typically around 80% nickel and 20% chromium.

Nichrome is chosen for its ability to resist oxidation and maintain its structural integrity at high temperatures, making it suitable for heating elements in various electrical appliances, including electric cookers.

Nichrome alloy is used for making the heating elements of electrical appliances because:

  • It has a high melting point such that it can be heated till red hot without melting.
  • Even at high temperatures, it does not get oxidized easily.
  • It can easily drawn in wires.
  • It has high value of specific resistance.
RSMSSB JE Electrical Mock Test - 4 - Question 26

UJT is used for which of the following applications?

Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 26
  • The uni-junction transistor is a solid-state three-terminal device that can be used in gate pulse, timing circuits and trigger generator applications to switch and control both thyristors and triacs for AC power control type applications.
  • When a UJT is used for triggering an SCR, the wave shape of the voltage obtained from the UJT circuit is a saw-tooth wave.
  • The UJT is a three-terminal, semiconductor device which exhibits negative resistance and switching characteristics for use as a relaxation oscillator in phase control applications.

  • The device has only one junction, so it is called the unijunction device
  • The device, because of one P-N junction, is quite similar to a diode but it differs from an ordinary diode as it has three terminals
  • The structure of a UJT is quite similar to that of an N-channel JFET
  • The main difference is that P-type (gate) material surrounds the N-type (channel) material in the case of JFET and the gate surface of the JFE T is much larger than emitter junction of UJT
  • In a unijunction transistor, the emitter is heavily doped while the N-region is lightly doped, so the resistance between the base terminals is relatively high, typically 4 to 10 kilo Ohm when the emitter is open
  • The N-type silicon bar has a high resistance and the resistance between emitter and base1 is larger than that between emitter and base2; It is because emitter is closer to base2 than base1
  • UJT is operated with emitter junction forward- biased while the JFET is normally operated with the gate junction reverse-biased
  • UJT does not have the ability to amplify but it has the ability to control a large ac power with a small signal; It exhibits a negative resistance characteristic and so it can be employed as an oscillator
RSMSSB JE Electrical Mock Test - 4 - Question 27
A 3-phase 208 V 60 Hz 6 pole star connected induction motor delivers 15 kW. Find the slip if rotor speed 1116 RPM.
Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 27
Concept:

The synchronous speed is given by

Where,

f is the supply frequency in Hz or C/s

P is the number of poles

The induction motor rotates at a speed (Nr) close but less than the synchronous speed.

Slip of an induction motor is given by,

Where,

Ns is the synchronous speed

and Nr is the rotor speed

Rotor current frequency (fr) = (s)(f)

The slip is negative when the rotor speed is more than the synchronous speed of the rotor field and is in the same direction.

Calculation:

Given -

Number of poles (P) = 6

Frequency (f) = 60 Hz

Nr = 1116 rpm

Now rotor speed is

Nr = Ns (1 – s)

⇒ 1116 = 1200 (1 - s)

= 0.93 = 1 - s

Hence, s = 0.07

RSMSSB JE Electrical Mock Test - 4 - Question 28

Torque produced in shaded pole structure induction type relay is:

Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 28

The correct answer is option 1):Proportional to square of the current

Concept:

  • The construction of shaded pole induction type relay is given as
  • The main flux is split into two fluxes displaced in time and space with the help of a shading ring.
  • A shading ring or coil is made of copper (Cu) and encircles the part of the pole face of each pole at air-gap.
  • Thus, two alternating fluxes are produced and are displaced in time and space.
  • The air-gap flux of the shaded poles lags behind that of the non-shaded poles.
  • The torque is produced by the interaction of one flux with the eddy current produced by another flux.
  • This resultant torque rotates the disc, which in turn operates the relay contact mounted on it.
  • A permanent magnet is provided to produce eddy current braking in the disc.
  • Torque produced in shaded pole structure induction type relay is Proportional to square of the current.
RSMSSB JE Electrical Mock Test - 4 - Question 29
The instrument used to measure the frequency of AC signals is called a/an:
Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 29

The instrument used to measure the frequency of AC signals is called a Frequency Meter (option 3).

A brief explanation of the other instruments:

DC Meter: This is used to measure the current or voltage in a Direct Current (DC) circuit.

AC Meter: This is used to measure the current or voltage in an Alternating Current (AC) circuit.

Dynamometer: This is a device that measures the torque or rotational speed (RPM) of a motor or other rotating system. In some cases, it can be used to calculate power.

RSMSSB JE Electrical Mock Test - 4 - Question 30

Two synchronous generators G1 and G2 rated 200 MW and 400 MW respectively are operated in parallel to supply a total load of 300 MW. If the governors in both the machines are set to a droop of 4%, what would be the individual power supplied by each generator?

Detailed Solution for RSMSSB JE Electrical Mock Test - 4 - Question 30

Concept:

Load frequency characteristics of two Generator:

Considered two generators are running in parallel. The load-frequency characteristics of this machine can be drawn as,

Let,
W1 = full load power rating of the generator 1.
W2 = full load power rating of the generator 2.
P1 = power shared by the generator 1.
P2 = power shared by the generator 2.
P = Total power shared by both generators.
f01 = no load frequency of generator 1.
f02 = no load frequency of generator 2.
f1 = full load frequency of generator 1.
f2 = full load frequency of generator 2.
f = common operating frequency of both generators.

For Generator 1:
Drop-in frequency from no load to full load = f01 - f1
Drop-in frequency per unit rating 
Drop-in frequency for load 

Operating frequency = no load frequency - drop in frequency

.... (1)

For Generator 2:

Drop-in frequency from no load to full load = f02 - f2

Drop-in frequency per unit rating 
Drop-in frequency for load 
Operating frequency = no load frequency - drop in frequency

.... (2)

Both the generator shared the load P,

∴ P = P1 + P2 .... (3)

Therefore,

System frequency or common frequency of two generators is given by


f0 = No-load frequency of both generators
SR = Speed regulation of generator
PR = Rated loading of the generator
PL = Loading of Generator
From the equation (1), (2), and (3), we can find the load shared by generator and frequency.
The load supplied by the generator is the function of operating frequency.
The amount of power generated by the generator is determined by the speed of the prime mover or frequency.

Application:
Given:
PA rated = 200 MW, PB rated = 400 MW, Speed regulation of A and B = 4% = 0.04
PA + PB = 300 MW ..............(1)

.............(2)

.............(3)
By equating (2) and (3) we get

..................(4)
Solving (1) & (4) we get
PA = 100 MW, PB = 200 MW

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