RRB JE Stage II exam is scheduled to be conducted from 28th August to 1st September2019. The overall merit for the selection will be made on the basis of the marks scored in stage II, so candidates need to prepare thoroughly for the second stage. To help the aspirants, Adda247 has come up with a study plan in which we will provide you daily quizzes of all technical subjects. The below quiz will be helpful for the candidates who opted Electronics as the exam group in RRB JE stage II.
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Q1. Two waves of the same frequency have opposite phase when the phase angle between them is
Sol. When the phase difference between two waves is 180°(it may be = + 180° or – 180°), then the waves are said to be in ‘Phase opposition.
Q2. Maxwell’s curl equation for the static electric field E can be expressed as
(a) Δ × E=0
(b) Δ × E=constant
(c) Δ . E= constant
(d) Δ . E= 0
Sol. Maxwell’s curl equation for the static electric field E is
Δ × E=0
Q3. The type 0 system has ………… at the origin
(a) no pole
(b) net pole
(c) simple pole
(d) two poles
Sol. The type of the system is defined as the property of the system which has pole at the origin.
Q4. In case of type- 1 system steady state acceleration is
Sol. In case of type-1 system steady state acceleration is infinity as for the type less than 3 acceleration is not defined it is infinity.
Q5. When derivative action is included in a proportional controller, the proportional band………
(c) remains unaffected
(d) none of the above
Sol. When derivative action is included in a proportional controller, the proportional band remains unaffected.
Q6. Signal diodes are called ………… diodes.
(a) general purpose
(b) high power
(c) special purpose
(d) none of the above
Q7. Which region of a transistor is lightly doped?
(d) All regions are equally doped.
Sol. In most transistors, emitter is heavily doped. Its job is to emit or inject electrons into the base. These bases are lightly doped and very thin, it passes most of the emitter-injected electrons on to the collector.
Q8. CMRR (Common Mode Rejection Ratio) for a differential amplifier should be
Sol. CMRR = 20 log (Ad/Acm)
For a perfect differential amplifier, the CMRR is equal to as Acm is zero.
Q9. In a monolithic IC, resistors are formed from
(a) aluminium ribbon
(b) ceramic material
(c) manganin wire
(d) p-type semiconductor
Sol. In a monolithic IC, resistors are formed from p-type semiconductor.
You may also like to read:
- Check RRB JE STAGE 1 result
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- Check RRB JE Stage II Exam Pattern & Syllabus
- Click here for RRB JE Salary & Job Profile Details
- Click Here to view Latest Govt. Jobs 2019