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100+ Free Maharashtra HSC Electronics Technology (Vocational) Practice Questions

Maharashtra Higher Secondary Certificate (HSC / Class 12) Vocational Electronics Technology under the Maharashtra State Board of Secondary and Higher Secondary Education (MSBSHSE), Pune — board papers EA/EB/EC practice questions are available now; exam metadata is being verified.

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2026 Statistics

Key Facts: Maharashtra HSC Electronics Technology (Vocational) Exam

EA / EB / EC

Common public subject codes for the three vocational Electronics Technology papers

MSBSHSE / public HSC vocational subject listings

3 papers

Applied & Industrial Electronics; Modern Instruments & Communication; Computer Hardware & Networking

MSBSHSE VOCA_VOL_1 Electronics Technology scheme

~200 marks/paper

Historical total per paper (theory + practical + term work + project + IV + OJT)

MSBSHSE VOCA_VOL_1 Std XII Electronics Technology scheme

3 hours theory

Common theory duration per paper in the VOCA_VOL_1 scheme

MSBSHSE VOCA_VOL_1 examination scheme table

35%

Common minimum passing floor per HSC subject under MSBSHSE

Maharashtra HSC pass-mark practice (~35% each subject)

≠ Bifocal C2

Distinct from Maharashtra HSC bifocal Electronics (subject code C2)

MSBSHSE bifocal vs vocational subject coding

English MCQ adaptation

This free local bank is not the official Higher Secondary paper format

OpenExamPrep practice policy

Maharashtra MSBSHSE HSC Vocational Electronics Technology (EA/EB/EC) is a Class 12 three-paper technical elective with heavy practical/OJT weight (historically ~200 marks per paper per VOCA_VOL_1) and ~35% pass floor, covering industrial electronics, instruments, communication, and hardware/networking — not a pure MCQ board paper and not bifocal C2. This free 2026 bank is an English MCQ study adaptation for concepts, safety, and calculation fluency.

Sample Maharashtra HSC Electronics Technology (Vocational) Practice Questions

Try these sample questions to test your Maharashtra HSC Electronics Technology (Vocational) exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1An ideal operational amplifier is commonly assumed to have which pair of input characteristics?
A.Infinite input impedance and zero output impedance
B.Zero input impedance and infinite output impedance
C.Infinite input impedance and infinite output impedance
D.Zero input impedance and zero output impedance
Explanation: Ideal op-amp models assume infinite input impedance (no input current), zero output impedance (ideal voltage source), infinite open-loop gain, and infinite bandwidth. Real devices only approximate these limits.
2In a closed-loop inverting amplifier using an ideal op-amp, the voltage gain Av = Vout/Vin is approximately:
A.−Rf/Rin
B.Rf/Rin
C.1 + Rf/Rin
D.−Rin/Rf
Explanation: For an ideal inverting configuration, virtual ground at the inverting input and equal currents through Rin and Rf give Av = −Rf/Rin. The negative sign indicates a 180° phase inversion.
3For an ideal non-inverting amplifier with feedback resistor Rf and ground resistor Rg, closed-loop voltage gain is:
A.1 + Rf/Rg
B.−Rf/Rg
C.Rf/Rg
D.Rg/(Rf + Rg)
Explanation: Non-inverting closed-loop gain is Av = 1 + Rf/Rg (also written 1 + Rf/R1 depending on label). The +1 term comes from the unity path from input to output plus the resistive divider feedback fraction.
4An op-amp configured as a voltage follower (buffer) ideally provides:
A.Unity voltage gain with high input impedance and low output impedance
B.Very high voltage gain with low input impedance
C.Zero voltage gain used only as a short circuit
D.Current gain of zero with infinite output impedance
Explanation: A voltage follower connects output directly to the inverting input so closed-loop gain is 1. It isolates stages by presenting high input impedance and low output impedance without changing signal voltage amplitude ideally.
5An op-amp adder (summing amplifier) in the classic inverting form produces an output proportional to:
A.The weighted sum of the input voltages (with inversion)
B.Only the largest input voltage
C.The product of all input voltages
D.The difference of two fixed supply rails only
Explanation: An inverting summing amplifier feeds multiple inputs through separate resistors into the inverting node. Output is Vout = −(Rf/R1)V1 − (Rf/R2)V2 − … — a weighted inverted sum of the inputs.
6In an ideal op-amp integrator (resistor input, capacitor feedback), the output is proportional to:
A.The negative time integral of the input voltage
B.The derivative of the input voltage
C.The square of the input voltage
D.Only the DC average of supply noise
Explanation: Ideal integrator: capacitor in feedback and resistor at input. Output is Vout = −(1/RC)∫Vin dt. Differentiation would reverse R and C roles.
7An op-amp differentiator (capacitor input, resistor feedback) is primarily used to produce an output proportional to:
A.The rate of change (derivative) of the input voltage
B.The time integral of the input voltage
C.Only a constant DC level independent of input
D.The reciprocal of frequency for all inputs always
Explanation: With capacitor at the input and resistor in feedback, ideal differentiator output tracks dVin/dt (with a sign). Integrators use the opposite R/C placement.
8A comparator built from an open-loop op-amp is mainly used to:
A.Compare an input against a reference and drive the output toward a high or low saturation state
B.Always amplify linearly with infinite closed-loop gain stability
C.Generate a pure sine wave without any switching
D.Replace a crystal oscillator in every digital clock
Explanation: Comparators run with high open-loop gain so the output saturates high or low depending on which input is larger. They are decision/switching stages, not linear amplifiers.
9A Schmitt trigger using an op-amp is valued because it provides:
A.Hysteresis that improves noise immunity for switching thresholds
B.Exactly zero hysteresis for maximum noise sensitivity
C.Only analog multiplication of two AC inputs
D.Automatic crystal frequency multiplication
Explanation: Positive feedback creates two different trip points (hysteresis). Noise smaller than the hysteresis band does not cause chatter at the threshold, improving clean digital-like transitions.
10Common-mode rejection ratio (CMRR) of an op-amp quantifies the ability to:
A.Reject signals that appear equally on both inputs while amplifying the differential signal
B.Reject only supply-voltage changes and ignore input signals
C.Amplify common-mode noise more than differential signals
D.Measure only the output offset without input comparison
Explanation: CMRR = Ad/Acm (often in dB). High CMRR means differential gain dominates while common-mode interference (same on both inputs) is strongly suppressed — essential for differential sensing.

About the Maharashtra HSC Electronics Technology (Vocational) Practice Questions

Verified exam format metadata for Maharashtra Higher Secondary Certificate (HSC / Class 12) Vocational Electronics Technology under the Maharashtra State Board of Secondary and Higher Secondary Education (MSBSHSE), Pune — board papers EA/EB/EC is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.