100+ Free BSNL JTO Practice Questions

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2026 Statistics

Key Facts: BSNL JTO Exam

120

Total MCQs

Exam pattern

-1 / +4

Marking Scheme

Negative marking applies

3 Hours

Time Limit

180 minutes duration

₹2000

General Fee

UR/OBC candidates

₹1000

Reserved Fee

SC/ST/PWD candidates

40%

Min. Qualifying

General sectional cutoff

The BSNL JTO exam consists of 120 MCQs divided into three sections: Engineering Stream I (50 questions), Engineering Stream II (50 questions), and General Ability (20 questions). The exam has a duration of 3 hours and is worth 480 marks. Each correct answer awards 4 marks, while each incorrect response results in a deduction of 1 mark.

Sample BSNL JTO Practice Questions

Try these sample questions to test your BSNL JTO exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1A step-index optical fiber has a core refractive index of 1.50 and a cladding refractive index of 1.47. What is the numerical aperture (NA) of this fiber?

A.0.300

B.0.225

C.0.150

D.0.090

Explanation: The numerical aperture (NA) is calculated using the formula NA = sqrt(n1^2 - n2^2), where n1 is the core index and n2 is the cladding index. Substituting the values gives NA = sqrt(1.50^2 - 1.47^2) = sqrt(2.25 - 2.1609) = sqrt(0.0891) ≈ 0.2985, which rounds to 0.300.

2In optical fiber communications, what is the standard engineering unit used to express the attenuation coefficient of a signal over a long distance?

A.decibels per kilometer (dB/km)

B.decibels per milliwatt (dBm)

C.nepers per meter (Np/m)

D.decibels per microvolt (dBμV)

Explanation: The attenuation coefficient in fiber optics represents the rate of signal loss per unit length of fiber and is standardly expressed in decibels per kilometer (dB/km). Higher values indicate greater signal loss over distance.

3A lossless transmission line has an inductance of 250 nH/m and a capacitance of 100 pF/m. What is the characteristic impedance (Z0) of this transmission line?

A.50 ohms

B.75 ohms

C.100 ohms

D.250 ohms

Explanation: The characteristic impedance of a lossless transmission line is calculated as Z0 = sqrt(L/C). Substituting the values, Z0 = sqrt((250 * 10^-9 H) / (100 * 10^-12 F)) = sqrt(2500) = 50 ohms.

4An optical fiber has a core refractive index of 1.48 and a cladding refractive index of 1.46. Calculate the critical angle at the core-cladding interface for total internal reflection.

A.80.6 degrees

B.70.5 degrees

C.9.4 degrees

D.45.0 degrees

Explanation: The critical angle (θc) is given by θc = arcsin(n2/n1), where n1 is the core index and n2 is the cladding index. Here, θc = arcsin(1.46 / 1.48) = arcsin(0.9865) ≈ 80.57 degrees, which rounds to 80.6 degrees.

5A transmission line has a voltage reflection coefficient of 0.2. What is the Voltage Standing Wave Ratio (VSWR) on this transmission line?

A.1.50

B.1.20

C.1.25

D.2.00

Explanation: The Voltage Standing Wave Ratio is calculated using the formula VSWR = (1 + |Γ|) / (1 - |Γ|), where Γ is the reflection coefficient. Substituting the values gives VSWR = (1 + 0.2) / (1 - 0.2) = 1.2 / 0.8 = 1.5.

6Determine the propagation velocity of an electromagnetic wave travelling through a coaxial cable with a relative permittivity (dielectric constant) of 2.25.

A.2.00 * 10^8 m/s

B.1.50 * 10^8 m/s

C.3.00 * 10^8 m/s

D.1.33 * 10^8 m/s

Explanation: The velocity of propagation in a medium is given by v = c / sqrt(εr), where c is the speed of light in vacuum (3 * 10^8 m/s) and εr is the relative permittivity. Here, v = (3 * 10^8) / sqrt(2.25) = (3 * 10^8) / 1.5 = 2.00 * 10^8 m/s.

7A step-index single-mode fiber has a core radius of 4.5 micrometers. If the core index is 1.46 and the relative refractive index difference is 0.25%, what is the cutoff wavelength of the fiber?

A.1.21 micrometers

B.1.55 micrometers

C.1.31 micrometers

D.0.85 micrometers

Explanation: The cutoff wavelength (λc) is given by λc = (2 * pi * a * NA) / Vc, where Vc = 2.405 for step-index fiber. NA ≈ n1 * sqrt(2 * Δ) = 1.46 * sqrt(2 * 0.0025) = 1.46 * 0.0707 = 0.1032. Thus, λc = (2 * pi * 4.5 * 0.1032) / 2.405 ≈ 1.21 micrometers.

8A laser diode transmitter launches 2 mW of optical power into a fiber. If the signal suffers a total loss of 15 dB before reaching the photodetector, calculate the received power in microwatts.

A.63.2 microwatts

B.20.0 microwatts

C.6.32 microwatts

D.2.00 microwatts

Explanation: Input power Pin = 2 mW = 3 dBm. The received power in dBm is Pin(dBm) - Loss(dB) = 3 dBm - 15 dB = -12 dBm. Convert -12 dBm back to power: P = 10^(-1.2) mW ≈ 0.0631 mW = 63.2 microwatts.

9If the Voltage Standing Wave Ratio (VSWR) on an RF transmission line is measured to be 3.0, what is the return loss in decibels?

A.6.02 dB

B.9.54 dB

C.12.04 dB

D.3.01 dB

Explanation: First, find the reflection coefficient Γ = (VSWR - 1) / (VSWR + 1) = (3 - 1) / (3 + 1) = 2/4 = 0.5. Return loss (RL) in dB is RL = 20 * log10(1/|Γ|) = -20 * log10(0.5) ≈ -20 * (-0.301) = 6.02 dB.

10A fiber optic link spanning 50 km consists of fiber with attenuation of 0.3 dB/km. It includes 4 splices with 0.1 dB loss each, and 2 connectors with 0.5 dB loss each. If the transmitter launches 0 dBm, what is the received power?

A.-16.4 dBm

B.-15.0 dBm

C.-17.5 dBm

D.-13.6 dBm

Explanation: Calculate the total link loss: Fiber attenuation loss = 50 km * 0.3 dB/km = 15 dB. Splice losses = 4 * 0.1 dB = 0.4 dB. Connector losses = 2 * 0.5 dB = 1.0 dB. Total loss = 15 + 0.4 + 1.0 = 16.4 dB. Received power = Transmitted power - Total loss = 0 dBm - 16.4 dB = -16.4 dBm.

About the BSNL JTO Exam

The BSNL Junior Telecom Officer (JTO) examination is a national competitive recruitment test. It evaluates engineering graduates on electrical, electronic, telecommunication, computer engineering principles, and general reasoning abilities to staff technical operations at BSNL.

Questions

120 scored questions

Time Limit

3 hours

Passing Score

40% for General (35% for SC/ST/PWD)

Exam Fee

₹2000 (UR/OBC), ₹1000 (SC/ST/PWD) (Bharat Sanchar Nigam Limited (BSNL))

BSNL JTO Exam Content Outline

42%

Engineering Stream I

Covers telecom basics (optical fiber, transmission lines, propagation), network theory (RLC resonance, AC transients, two-port networks), and digital electronics (combinational/sequential circuits, ADC/DAC).

42%

Engineering Stream II

Covers analog electronics (op-amps, BJT/MOSFET amplifiers, feedback), microprocessors (8085 architecture, programming, interfacing), communication systems (AM/FM, PCM, digital keying), and control systems (stability, transient response, state-space).

16%

General Ability

Covers English vocabulary, grammar, analytical reasoning, and general general knowledge on Indian telecom developments.

How to Pass the BSNL JTO Exam

What You Need to Know

Passing score: 40% for General (35% for SC/ST/PWD)
Exam length: 120 questions
Time limit: 3 hours
Exam fee: ₹2000 (UR/OBC), ₹1000 (SC/ST/PWD)

Keys to Passing

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

BSNL JTO Study Tips from Top Performers

1Dedicate adequate time to circuit calculation problems, particularly involving op-amps, RLC resonance, and Thevenin equivalent parameters.

2Thoroughly practice 8085 assembly instruction execution tracking, as register status questions are common.

3Learn standard digital communication formulas, including Shannon's channel capacity and PCM signal-to-quantization-noise ratio (SQNR).

4Do not ignore the General Ability section; its 20 questions can significantly raise your overall rank and help satisfy sectional cutoffs.

5Practice negative marking management: since every incorrect answer loses 1 mark, avoid blind guessing on complex calculations.

Frequently Asked Questions

What is the BSNL JTO exam pattern?

The BSNL JTO exam is a single paper of 3 hours duration containing 120 multiple-choice questions. It is divided into three sections: Engineering Stream I (50 questions), Engineering Stream II (50 questions), and General Ability (20 questions). Each correct response scores 4 marks, and there is a negative marking of 1 mark for each incorrect response.

What is the eligibility criteria for the BSNL JTO recruitment?

Candidates must possess a B.E./B.Tech. degree (or equivalent) in Telecommunications, Electronics, Radio, Computer Science, IT, or Electrical Engineering. Alternatively, candidates with an M.Sc. in Electronics or Computer Science from a recognized Indian university are also eligible.

Are there section-wise cutoff marks in the BSNL JTO exam?

Yes, BSNL typically mandates minimum qualifying marks in each of the three sections separately, as well as an aggregate score requirement. Usually, general and OBC candidates need at least 40% in each section, whereas SC/ST/PWD candidates need at least 35%.

What is the age limit to apply for BSNL JTO?

The standard age limit is between 18 and 30 years as of the date specified in the official recruitment advertisement. Relaxations in the upper age limit are provided for reserved categories: 5 years for SC/ST, 3 years for OBC, and up to 10 years for PWD candidates.

What is the syllabus of Engineering Stream I and Stream II?

Engineering Stream I contains Telecom Basics, Network Theory, and Digital Electronics. Engineering Stream II contains Analog Electronics, Microprocessors (specifically 8085 and 8051), Communication Systems (analog and digital), and Control Systems.