← MES Jr. Grade
Past Questions — 2012
For proper functioning of a clamper, time constant of the circuit should be
A)
large
B)
small
C)
equal to signal time period
D)
greater than five times the signal time period
A clamper must hold the capacitor charge almost constant during the non-conducting interval, so the RC time constant should be much larger than the signal period. A common rule is RC ≥ 5T, where T is the signal time period. If the time constant is small, the capacitor discharges too quickly and proper clamping will not occur. Hence, the correct choice is greater than five times the signal time period.
A cycloconverter is a device which
A)
converts A.C. into D.C.
B)
converts D.C. into A.C.
C)
converts A.C. of one frequency into A.C. of other frequency
D)
none of these
A cycloconverter is an AC-to-AC power converter that directly changes AC power from one frequency to another frequency without an intermediate DC stage. Hence, option (c) is correct. Option (a) describes a rectifier, and option (b) describes an inverter.
A class B push-pull amplifier has the main advantage of being free from
A)
any circuit imbalances
B)
unwanted noise
C)
even-order harmonic distortion
D)
d.c. magnetic saturation effects
In a class B push-pull amplifier, the two transistors conduct on alternate half-cycles. This causes the even-order harmonics generated in one half to cancel with those from the other half, so the amplifier is free from even-order harmonic distortion. The other options are incorrect because push-pull amplifiers may still have circuit imbalance, noise is not eliminated, and d.c. magnetic saturation is mainly related to transformer-coupled circuits, not the core advantage of class B push-pull operation.
The basic function of an electronic filter circuit in radio communication is to
A)
replace tuning circuits
B)
eliminate ripples in D.C. power supplies
C)
separate different frequencies in a composite signal
D)
divert RF currents out of a given circuit
An electronic filter in radio communication is used to pass desired frequency components and reject unwanted ones, i.e., to separate different frequencies in a composite signal. Option (b) refers to power-supply filtering, which is a specific application, not the basic function in radio communication. Option (a) is incorrect because filters do not simply replace tuning circuits. Option (d) describes a special RF bypassing function, not the general basic purpose of a filter.
The channel of JFET
A)
consists of p-type material only
B)
consists of n-type material only
C)
may consist of p- type or n-type material
D)
none of these
A JFET (Junction Field Effect Transistor) can be of two types: n-channel JFET and p-channel JFET. Therefore, its channel may be made of either n-type material or p-type material. Option (a) and (b) are incomplete because they mention only one type, and (d) is incorrect.
A communication channel has a bandwidth of 4 KHz and a signal-to-noise ratio is 25, the channel capacity will be nearly
A)
18000 bits/sec.
B)
4000 bits/sec.
C)
1500 bits/sec.
D)
none of these
Use Shannon’s channel capacity formula: C = B log2(1 + SNR). Here B = 4 kHz = 4000 Hz and SNR = 25. So C = 4000 × log2(26). Since log2(26) is about 4.7, capacity ≈ 4000 × 4.7 ≈ 18800 bps, which is nearly 18000 bps. Hence option (a) is correct. The other options are far from the computed value.
Number of operators required to build a 5-bit Analog to Digital converter is
A)
5
B)
11
C)
21
D)
31
For an n-bit flash ADC, the number of comparators required is 2^n - 1. Here n = 5, so required comparators = 2^5 - 1 = 32 - 1 = 31. Hence the correct answer is 31. The other options do not match the standard flash ADC comparator count.
CMOS circuits are extensively used for one-chip computers mainly because of their extremely
A)
low power dissipation
B)
large packing density
C)
high noise immunity
D)
low cost
CMOS circuits are widely used in one-chip computers because they consume very little power. This low power dissipation allows more components to be packed on a single chip without excessive heating, making CMOS ideal for integrated circuits. While CMOS also offers high packing density and good noise immunity, the main reason asked here is its extremely low power dissipation.
The bus used to transfer data from main memory to a peripheral device is
A)
data bus
B)
input bus
C)
DMA bus
D)
output bus
A peripheral device that receives data from main memory is on the output side of the system, so the bus used for transferring data from main memory to a peripheral device is the output bus. The data bus is a general-purpose bus for data transfer between CPU, memory, and I/O, while input bus is for devices sending data into the system. DMA bus is not the standard term for this direction of transfer in such MCQs.
The only function of a NOT gate is to
A)
stop a signal
B)
complement a signal
C)
invert an input signal
D)
act as a universal gate
A NOT gate performs logical inversion: if the input is 1, the output is 0, and if the input is 0, the output is 1. This is the same as complementing the signal. Option (b) is also conceptually correct, but in MCQ wording the standard answer is (c) because a NOT gate inverts the input signal. Option (a) is incorrect because it does not stop a signal, and (d) is incorrect because a NOT gate is not a universal gate.
The main purpose of modulation is to
A)
Combine two waves of different frequencies for
B)
achieve wave-shaping of the carrier wave
C)
transmit low-frequency information over long distances efficiently
D)
produce sidebands
Modulation is used to shift a low-frequency information signal onto a high-frequency carrier so that it can be transmitted efficiently over long distances. This improves antenna size, radiation efficiency, and reduces interference. Option (d) is a result of modulation, not the main purpose. Options (a) and (b) do not describe the primary objective.
One of the serious disadvantages of FM transmission is its
A)
high static noise
B)
limited line-of-sight range
C)
expensive equipment
D)
adjacent channel interference
FM transmission operates at VHF frequencies, so it generally travels by line of sight and does not follow the earth’s curvature well. Hence its coverage range is limited compared with lower-frequency systems.
Why others are incorrect:
- (a) High static noise: FM is actually less affected by noise and static than AM.
- (c) Expensive equipment: this may be a practical drawback, but it is not the main serious disadvantage asked here.
- (d) Adjacent channel interference: FM has better noise immunity and channel separation; this is not its primary disadvantage.
Schottky barrier diodes can be used as
A)
low noise amplifier
B)
variable capacitance device
C)
power supply rectifier
D)
low level detector
A Schottky barrier diode has very fast switching and a low forward voltage drop, so it is commonly used in high-speed and low-level signal detection applications. It is not a variable capacitance device (that is a varactor diode), and while it can be used in rectification, the most appropriate option here is low level detector. Low noise amplifier is not a diode application.
In FM, when frequency deviation is doubled,
A)
modulation is doubled
B)
modulation is halved
C)
carrier swing is halved
D)
modulation index is decreased
In frequency modulation (FM), the modulation index is directly proportional to the frequency deviation. If the frequency deviation is doubled, the modulation index also doubles. Hence, the correct option is (a) modulation is doubled. The other options are incorrect because halving the modulation or carrier swing would be the opposite effect, and the modulation index does not decrease when deviation increases.
An address is the number used by the CPU to specify
A)
a location in the memory
B)
a location in flags
C)
a location in accumulator
D)
a location in stack pointer
An address is a unique number used by the CPU to identify a specific memory location. The CPU uses addresses to read from or write to memory. Flags, accumulator, and stack pointer are CPU registers, not the general meaning of an address.
PROM is
A)
Permanent read only memory
B)
Polarised read only memory
C)
Positive read only memory
D)
Programmable read only memory
PROM stands for Programmable Read Only Memory. It is a type of memory that can be programmed once after manufacturing. The other options are incorrect because PROM does not mean permanent, polarised, or positive read only memory.
In AM transmission, the frequency which is not transmitted is
A)
upper side frequency
B)
lower side frequency
C)
audio frequency
D)
carrier frequency
In amplitude modulation (AM), the transmitted signal contains the carrier frequency and two sidebands: the upper side frequency and the lower side frequency. The original audio frequency itself is not transmitted as a separate component; it is used to modulate the carrier. Therefore, the frequency which is not transmitted is the audio frequency. Options (a), (b), and (d) are all present in an AM signal.
Static RAM is preferred over dynamic RAM when the requirement is
A)
slow speed of operation
B)
larger storage capacity
C)
lower access time
D)
lower power consumption
Static RAM (SRAM) is faster than dynamic RAM (DRAM), so it is preferred when low access time is required. DRAM is used for larger storage capacity and lower cost, but it is slower and needs refreshing. Hence, the correct choice is lower access time. The other options are incorrect because SRAM is not chosen for slow speed, larger capacity, or lower power consumption.
Silicon is invariably used in the manufacture of photodiodes because
A)
more electron-hole pairs are generated in it
B)
its thermally-generated minority current is extremely small
C)
it is more rugged than Ge
D)
it needs less reverse bias
Silicon is preferred for photodiodes because it has a much lower intrinsic carrier concentration than germanium, so the thermally generated minority current is very small. This reduces dark current and improves sensitivity and stability. Option (a) is incorrect because the key advantage is not generating more electron-hole pairs, but lower unwanted thermal generation. Option (c) may be true in a general sense, but it is not the main electrical reason for photodiode use. Option (d) is incorrect because the required reverse bias depends on device design, not the material choice in this way.
The primary function of a filter is to
A)
minimize a.c. input variations
B)
suppress odd harmonics in the rectifier output
C)
stabilize d.c. level of the output voltage
D)
remove ripples from the rectified output
A filter in a rectifier circuit is used to smooth the pulsating DC output by reducing the AC ripple component. It does not primarily stabilize the DC level or specifically suppress only odd harmonics. Therefore, the correct function is to remove ripples from the rectified output.
In a certain telemetry system, the measured values are converted to digital form. The digital values can then be transmitted via FSK (binary or quaternary) or BPSK or QPSK systems. Out of these the best noise immunity can be obtained with
A)
binary FSK
B)
quaternary FSK
C)
BPSK
D)
QPSK
Noise immunity is best when the signal constellation has the maximum separation for a given bit energy. Among the given schemes, BPSK is a 2-phase system and is more robust than QPSK and FSK forms at the same bit rate/power conditions. QPSK carries 2 bits per symbol, so its symbol points are closer together than BPSK, making it less immune to noise. Binary FSK is generally better than quaternary FSK, but among all the listed options, BPSK gives the best noise performance.
The main reason why electrons can tunnel through a P-N junction is that
A)
they have high energy
B)
barrier potential is very low
C)
depletion layer is extremely thin
D)
impurity level is low
In a P-N junction, tunneling occurs when the depletion region becomes extremely thin, so electrons can pass through the potential barrier by quantum tunneling. This is the principle behind tunnel diodes. The other options are incorrect because tunneling is not mainly due to high electron energy, low barrier potential alone, or low impurity level.
The clipping level is primarily determined by
A)
the shape of the input waveform
B)
battery voltage
C)
value of the resistor
D)
knee voltage of the diode
In a diode clipper circuit, clipping starts when the input exceeds the diode’s forward conduction threshold. Therefore, the clipping level is mainly set by the diode’s knee voltage (and any biasing, if present). The input waveform shape does not determine the clipping level, and the resistor mainly limits current rather than fixing the clip point. Battery voltage can shift the clipping level in biased clippers, but in the basic case the diode knee voltage is the primary factor.
In microprocessors, flag is a character for
A)
identification of a word
B)
occurrence of some condition
C)
marking a tagging
D)
all of these
In a microprocessor, a flag is a status bit used to indicate the occurrence of a particular condition, such as zero result, carry, sign, overflow, etc. Hence, the correct meaning is 'occurrence of some condition'. Option (a) is incorrect because a flag does not identify a word, and option (c) is too vague. Therefore, (d) cannot be correct.
Ripple frequency of the output waveform of a bridge rectifier when fed with a 50Hz sine wave is
A)
100Hz
B)
25Hz
C)
50Hz
D)
200Hz
In a bridge rectifier, both half-cycles of the AC input are converted into pulsating DC. Therefore, the ripple frequency is twice the input frequency. For a 50 Hz supply, ripple frequency = 2 × 50 = 100 Hz. Hence, option (a) is correct. Option (b) 25 Hz is too low, option (c) 50 Hz is the input frequency, and option (d) 200 Hz would correspond to a 100 Hz input.