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100+ Free RRB JE CBT 2 - Electronics Practice Questions

Pass your RRB Junior Engineer CBT 2 — Electronics & Allied Engineering exam on the first try — instant access, no signup required.

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2026 Statistics

Key Facts: RRB JE CBT 2 - Electronics Exam

150

CBT 2 Questions

Railway Recruitment Boards

120 min

CBT 2 Duration

RRB CEN

100 Marks

Technical Section

Electronics & Allied Eng.

1/3

Negative Marking

Per Wrong Answer

40% / 30% / 25%

Min Qualifying (UR / OBC-SC / ST)

RRB CEN

Rs 500 / Rs 250

Application Fee (Refundable)

RRB CEN

The RRB Junior Engineer CBT 2 for Electronics is the final selection stage, consisting of 150 questions in 120 minutes. It features 100 technical questions on Electronics Engineering and 50 non-technical questions on General Awareness (15), Physics & Chemistry (15), Computer Applications (10), and Environment & Pollution Control (10). A negative marking of 1/3 applies for every wrong response.

Sample RRB JE CBT 2 - Electronics Practice Questions

Try these sample questions to test your RRB JE CBT 2 - Electronics exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1Which of the following metals exhibits the highest electrical conductivity at 20°C?
A.Silver
B.Copper
C.Gold
D.Aluminum
Explanation: Silver has the highest electrical conductivity of all metals (6.30 x 10^7 S/m at 20°C), slightly higher than copper (5.96 x 10^7 S/m) and gold (4.11 x 10^7 S/m).
2Which of the following magnetic materials exhibits spontaneous magnetization below a specific temperature known as the Curie temperature?
A.Paramagnetic materials
B.Diamagnetic materials
C.Ferromagnetic materials
D.Antiferromagnetic materials
Explanation: Ferromagnetic materials exhibit spontaneous parallel alignment of magnetic dipoles (spontaneous magnetization) below the Curie temperature. Above this temperature, they behave as paramagnetic materials.
3In optical fiber cables (OFC), what is the primary purpose of applying index-matching gel during mechanical splicing?
A.To lubricate the fiber for easier alignment inside the splice tray
B.To minimize reflection losses by matching the refractive index of the fiber core
C.To protect the exposed silica cladding from moisture and mechanical stress
D.To act as a permanent adhesive to bond the two fiber ends together
Explanation: Index-matching gel has a refractive index very close to that of the silica glass fiber core. It fills the air gap between the two fiber ends to eliminate Fresnel reflection and minimize insertion loss.
4What is the nominal terminal voltage of a single fully-charged lead-acid battery cell under normal open-circuit conditions?
A.1.2 V
B.1.5 V
C.2.1 V
D.3.7 V
Explanation: A standard fully-charged lead-acid cell has a nominal open-circuit terminal voltage of approximately 2.1 V (typically ranging from 2.1 V to 2.15 V). A standard 12V lead-acid battery consists of 6 of these cells in series.
5Which component combination constitutes a standard snubber circuit connected across an electromagnetic relay contact to suppress inductive arcing?
A.A resistor and capacitor in parallel
B.A resistor and capacitor in series
C.An inductor and capacitor in series
D.A resistor and inductor in parallel
Explanation: A snubber circuit connected across relay contacts typically consists of a resistor and capacitor in series (RC snubber). The capacitor absorbs the high voltage transient when the contact opens, while the resistor limits the discharge current when the contact closes.
6What physical phenomenon describes the complete expulsion of magnetic flux lines from the interior of a superconducting material when it is cooled below its critical temperature?
A.Hall Effect
B.Seebeck Effect
C.Meissner Effect
D.Peltier Effect
Explanation: The Meissner effect is the complete expulsion of magnetic fields from the interior of a superconductor when it transitions into the superconducting state below its transition temperature, establishing perfect diamagnetism.
7A semiconductor material has a Hall coefficient (RH) of -2.5 x 10^-4 m^3/C. What can be concluded about the type and majority carrier concentration of this semiconductor?
A.It is p-type with a majority carrier concentration of 2.5 x 10^22 m^-3
B.It is n-type with a majority carrier concentration of 2.5 x 10^22 m^-3
C.It is p-type with a majority carrier concentration of 4.0 x 10^22 m^-3
D.It is n-type with a majority carrier concentration of 2.5 x 10^18 m^-3
Explanation: A negative Hall coefficient (RH) indicates that the majority carriers are electrons, meaning it is an n-type semiconductor. The majority carrier concentration (n) is calculated as n = -1 / (q * RH) = -1 / (-1.6 x 10^-19 * 2.5 x 10^-4) = 2.5 x 10^22 m^-3.
8What is the value and SI unit of the permeability of free space (μ0)?
A.8.854 x 10^-12 F/m
B.4π x 10^-7 H/m
C.4π x 10^-7 F/m
D.1.257 x 10^-6 H
Explanation: The permeability of free space (μ0) is a physical constant defined as 4π x 10^-7 Henries per meter (H/m), which is approximately 1.257 x 10^-6 H/m.
9What is the full form of MCB in the context of electrical switchgear and circuit protection?
A.Main Circuit Breaker
B.Miniature Circuit Breaker
C.Molded Circuit Board
D.Magnetic Current Breaker
Explanation: MCB stands for Miniature Circuit Breaker. It is an electromagnetic device that automatically interrupts an electrical circuit during overcurrent (overload or short circuit) conditions.
10What is the theoretical ripple factor of a full-wave rectifier operating with a pure resistive load and no filter circuit?
A.1.21
B.0.48
C.0.82
D.0.30
Explanation: The ripple factor (γ) of a rectifier is the ratio of the RMS value of the AC component of the output voltage to the DC value. For a full-wave rectifier, γ = √((Vrms/Vdc)^2 - 1) = √(( (Vm/√2) / (2Vm/π) )^2 - 1) = √((π^2 / 8) - 1) ≈ 0.482 (or 48.2%).

About the RRB JE CBT 2 - Electronics Exam

The RRB Junior Engineer CBT 2 Electronics & Allied Engineering exam is the merit-deciding second stage of the selection process. It consists of 150 multiple-choice questions to be completed in 120 minutes. The paper is split into 100 marks of technical questions specific to Electronics & Allied Engineering, and 50 marks of non-technical questions. Uniquely, the exam includes one-third (1/3) negative marking for every incorrect answer. Candidates must qualify CBT 1 first to be shortlisted for CBT 2.

Questions

150 scored questions

Time Limit

120 minutes

Passing Score

Minimum qualifying: 40% (UR), 30% (OBC/SC), 25% (ST)

Exam Fee

Rs 500 / Rs 250 (Railway Recruitment Boards (RRBs), Ministry of Railways, Government of India)

RRB JE CBT 2 - Electronics Exam Content Outline

66.7%

Technical Abilities (Electronics & Allied Engineering)

Electronic components, devices and circuits, digital electronics, linear integrated circuits, microprocessors and microcontrollers, electronic measurements, communication, data communication and networks.

10%

General Awareness

Current affairs, Indian history, geography, polity, economics, sports, awards, and general knowledge of national importance.

10%

Physics & Chemistry

General physics and chemistry topics up to CBSE 10+2 standard.

6.7%

Basics of Computers & Applications

Computer architecture, input/output devices, storage, operating systems, networking, internet, web browsers, and email.

6.7%

Basics of Environment & Pollution Control

Ecosystem, air, water, and noise pollution, waste management, global warming, acid rain, and ozone layer depletion.

How to Pass the RRB JE CBT 2 - Electronics Exam

What You Need to Know

  • Passing score: Minimum qualifying: 40% (UR), 30% (OBC/SC), 25% (ST)
  • Exam length: 150 questions
  • Time limit: 120 minutes
  • Exam fee: Rs 500 / Rs 250

Keys to Passing

  • Complete 500+ practice questions
  • Score 80%+ consistently before scheduling
  • Focus on highest-weighted sections
  • Use our AI tutor for tough concepts

RRB JE CBT 2 - Electronics Study Tips from Top Performers

1Prioritize the technical section (100 marks) as it represents two-thirds of the total weight. Focus on electronic devices, digital systems, and communication engineering.
2Strengthen your physics and chemistry concepts up to 10+2 standard, which provides a high-yielding 15 marks.
3Practise selective answering to combat the 1/3 negative marking. Skip questions you are not reasonably confident about.
4Solve quantitative and calculation-based questions on operational amplifiers, network theory, and modulation parameters daily.
5Allocate dedicated preparation time for the computer applications and environmental science sections, which contribute a total of 20 marks.

Frequently Asked Questions

What is the RRB JE CBT 2 Electronics exam pattern?

The exam consists of 150 multiple-choice questions to be completed in 120 minutes. It includes 100 questions from the technical abilities core (Electronics & Allied Engineering) and 50 questions from non-technical areas including General Awareness (15), Physics & Chemistry (15), Basics of Computers (10), and Basics of Environment (10).

Is there negative marking in the RRB JE CBT 2 exam?

Yes. There is a negative marking of one-third (1/3) mark for every incorrect answer. Unanswered questions carry no penalty. It is important to avoid guessing to maximize your score.

What are the minimum qualifying marks for RRB JE CBT 2?

The minimum qualifying marks are 40% for General/UR, 30% for OBC and SC, and 25% for ST candidates. However, final selection depends on securing a position within the zone-wise merit list based on cut-offs and available vacancies.

Who is eligible to appear for the CBT 2 Electronics exam?

Only candidates who qualify in the screening stage (CBT 1) and possess a three-year Diploma or a Bachelor's degree (B.E./B.Tech) in Electronics Engineering or allied disciplines from a recognized institution are eligible.

What is the application fee for the RRB JE exam?

The fee is Rs 500 for General/OBC candidates, of which Rs 400 is refunded after appearing in CBT 1. For SC/ST/Ex-Servicemen/PwBD/Female/Transgender/Minorities/EBC candidates, the fee is Rs 250, which is fully refunded upon appearing in CBT 1.

How is the final merit list prepared for the Junior Engineer post?

The final merit list is prepared based solely on the normalized marks obtained in CBT 2. Candidates who qualify CBT 2 are called for Document Verification (DV) and a Medical Examination.