← MES Jr. Grade
Past Questions — 2012
The basic elements of a microprocessor are
A)
ALU and memory
B)
ALU and control unit
C)
ALU, memory and I/O devices
D)
ALU, control unit and memory
A microprocessor is fundamentally built around two main parts: the Arithmetic Logic Unit (ALU), which performs arithmetic and logical operations, and the Control Unit, which directs and coordinates all operations. Memory and I/O devices are not basic internal elements of the microprocessor; they are external components interfaced with it. Therefore, the correct answer is ALU and control unit.
If the carrier of a 100 percent modulated AM wave is suppressed, the percentage power saving will be
A)
50
B)
33.33
C)
100
D)
66.66
In standard AM, total power is P_t = P_c(1 + m^2/2), where m is the modulation index. For 100% modulation, m = 1, so P_t = 1.5 P_c. If the carrier is suppressed, only the sideband power remains: P_s = P_c(m^2/2) = 0.5 P_c. Hence power saved = (1.5P_c - 0.5P_c) / 1.5P_c × 100 = 1/1.5 × 100 = 66.66%. So the correct option is (d). Other options do not match the standard AM power relations.
SMPS stands for
A)
service monitored power supply
B)
self monitored power supply
C)
switched mode power supply
D)
single mode power supply
SMPS is the standard abbreviation for Switched Mode Power Supply. It is a power supply that regulates output by switching elements on and off at high frequency. The other options are not standard expansions of SMPS.
A half wave rectifier has a load resistance R and a capacitance filter connected across the load. The capacitance of the filter is C. It is desired that the output voltage should have small ripple. Then,
A)
RC should be very large
B)
RC should be very small
C)
C should be reciprocal of R
D)
none of these
In a half-wave rectifier with a capacitor filter, the capacitor charges near the peak of the rectified voltage and discharges slowly through the load between peaks. For small ripple, the discharge should be very slow compared to the time between charging pulses. That happens when the time constant RC is much larger than the period of the ripple, i.e., RC should be very large. If RC is small, the capacitor discharges quickly and ripple increases. The option that best ensures small ripple is therefore (a).
When a reverse biased is applied to a p-n junction, the width of the depletion layer
A)
remains the same
B)
is increased
C)
is decreased
D)
may increase or decrease
In a p-n junction, applying reverse bias pulls majority carriers away from the junction. This exposes more fixed ions near the junction, so the depletion region becomes wider. Hence, the depletion layer width increases. Option (a) is wrong because the width does not remain unchanged, (c) is wrong because it does not decrease, and (d) is wrong because under reverse bias the effect is specifically an increase.
A single instruction to clear the lower four bits of the accumulator in 8085 assembly language is
A)
XRIOFH
B)
ANIFOH
C)
XRIFOH
D)
ANOFH
To clear the lower four bits of the accumulator in 8085, we use the logical AND operation with the mask F0H (11110000B). This preserves the upper four bits and resets the lower four bits to 0. In 8085 assembly, the instruction is written as ANI F0H. Among the given options, ANIFOH corresponds to ANI F0H. XOR would not selectively clear only the lower four bits, and the other options are not the correct masking instruction.
The interface chip used for data transmission between 8086 and a 16-bit ADC is
A)
8259
B)
8255
C)
8253
D)
8251
The 8255 Programmable Peripheral Interface (PPI) is used to interface 8086 with peripheral devices, including ADCs, for parallel data transmission. A 16-bit ADC needs parallel input/output control, which is provided by 8255. 8259 is an interrupt controller, 8253 is a timer/counter, and 8251 is a serial communication interface, so they are not suitable here.
A bi-stable multi-vibrator (BMV) is a .............. oscillator
A)
triggered
B)
free-running
C)
sine-wave
D)
sawtooth
A bi-stable multivibrator has two stable states and changes state only when an external trigger pulse is applied, so it is a triggered oscillator. A free-running oscillator operates without external triggering, while sine-wave and sawtooth refer to waveform types, not the operating mode of a bi-stable multivibrator.
OP-AMPs have become very popular in industry mainly because
A)
they are very cheap
B)
their external characteristics can be changed to suit any application
C)
of their extremely small size
D)
they are available in different packages
Operational amplifiers are popular in industry because their behavior can be easily tailored using external feedback components. By changing the external resistors/capacitors, the same op-amp can be configured as an amplifier, comparator, integrator, differentiator, filter, etc. This flexibility is the main reason for their wide use. The other options are not the primary reason: op-amps are not mainly chosen because they are cheap, small, or available in different packages.
The number of memory chips of size 1K × 4 bits required to build a memory bank of size 16 K × 8 bits is
A)
64
B)
32
C)
16
D)
8
A 1K × 4-bit chip stores 1024 words of 4 bits each. The required memory bank is 16K × 8 bits, i.e. 16384 words of 8 bits each.
Step 1: Match the word size.
- Each chip gives 4 bits per word.
- Required word size is 8 bits.
- So, 2 chips are needed in parallel to make 8 bits.
Step 2: Match the memory depth.
- One chip provides 1K words.
- Required depth is 16K words.
- So, 16 chips are needed in series for the address space.
Step 3: Total chips required.
- 16 × 2 = 32 chips.
Therefore, the correct answer is 32.
Feedback in an amplifier always helps to
A)
control its output
B)
increase its gain
C)
decrease its input impedance
D)
stabilize its gain
Feedback in an amplifier is mainly used to make the amplifier performance less sensitive to variations in transistor parameters, temperature, and supply changes. Negative feedback especially helps stabilize the gain. It does not generally increase gain; in fact, it usually reduces gain but improves stability, bandwidth, and linearity. Hence, the correct answer is 'stabilize its gain'.
The reason of preferring FETs to bipolar transistors in integrated circuits is that
A)
they occupy very small space
B)
it has become a tradition
C)
fabrication of circuit is easy
D)
it is cheaper
FETs are preferred in integrated circuits mainly because they are very small in size and can be packed densely on a chip. This makes IC fabrication more suitable and efficient. Option (b) is irrelevant, (c) is not the main reason compared to density and miniaturization, and (d) is not the primary advantage in this context.
The chief advantage of Scottky TTL logic family is its least
A)
power dissipation
B)
propagation delay
C)
fan-in
D)
noise immunity
Schottky TTL uses Schottky diodes to prevent transistor saturation, which makes switching faster. Therefore its chief advantage is the least propagation delay. It is not known for least power dissipation, fan-in, or noise immunity compared with other logic families.
An ideal OP-AMP has
A)
infinite open-loop gain
B)
infinite input resistance
C)
Zero output resistance
D)
all of these
An ideal op-amp is defined by several ideal characteristics: infinite open-loop gain, infinite input resistance, and zero output resistance. Since options (a), (b), and (c) are all true, the correct answer is (d) all of these.
The noise figure of an amplifier is 6dB. If the input S/N ratio is 38 dB, then the output S/N ratio
A)
44 dB
B)
40 dB
C)
36 dB
D)
32 dB
Noise figure in dB is the degradation in signal-to-noise ratio from input to output. So,
Output S/N (dB) = Input S/N (dB) - Noise figure (dB) = 38 - 6 = 32 dB.
However, the options given do not include 32 dB as the result of this standard relation. If the question intended noise factor addition in terms of noise figure affecting S/N directly, the correct computed output is 32 dB, which matches option (d), not (c). Therefore the correct option is (d) 32 dB.
In a single-phase full converter, if the output voltage has peak and average values of 325 V and 133 V respectively, then the firing angle is
A)
40°
B)
140°
C)
50°
D)
130°
For a single-phase full converter, the average output voltage is given by Vdc = (2Vm/π) cos α, where Vm is the peak value of the supply/output waveform. Here the peak value is 325 V, so the maximum average (at α = 0°) is 2Vm/π ≈ 2×325/π ≈ 206.9 V. Given average output voltage is 133 V, so cos α = 133 / 206.9 ≈ 0.643. Therefore α ≈ cos⁻¹(0.643) ≈ 50°. Hence the correct option is (c) 50°.
An inverter is a device that converts
A)
D.C. into A.C.
B)
A.C. into D.C.
C)
A.C. into A.C. of higher frequency
D)
A.C. into A.C. of lower frequency
An inverter converts direct current (D.C.) into alternating current (A.C.). Option (b) describes a rectifier, while options (c) and (d) describe frequency changers/cycloconverters, not an inverter.
Negative feedback in an amplifier
A)
lowers its lower 3 dB frequency
B)
raises its upper 3 dB frequency
C)
increases its bandwidth
D)
all of these
Negative feedback in an amplifier reduces gain but improves frequency response. It lowers the lower cutoff frequency, raises the upper cutoff frequency, and therefore increases the overall bandwidth. Hence all the given statements are correct.
FET has offset voltage of approximately
A)
0.2 volt
B)
0.6 volt
C)
1.1 volt
D)
zero
A FET (Field Effect Transistor) is a voltage-controlled device and its input gate current is ideally zero. Therefore, the offset voltage is approximately zero. The other values (0.2 V, 0.6 V, 1.1 V) are associated with diode or BJT junction forward voltages, not FET offset voltage.
When used in a circuit, the Zener diode is always
A)
forwarded biased
B)
connected in series
C)
troubled by overheating
D)
reverse biased
A Zener diode is designed to operate in the reverse-bias region, where it maintains an almost constant voltage after breakdown. That is why it is used for voltage regulation. In forward bias, it behaves like an ordinary diode, so that is not its normal circuit use. The other options are incorrect because a Zener diode is not defined by being connected in series, and overheating is not its intended operating condition.
The highest voltage gain can be obtained from a .............. configuration
A)
CC
B)
CE
C)
CB
D)
both CE and CB
In a BJT, the common-emitter (CE) configuration provides the highest voltage gain among the three basic configurations. It offers both significant current gain and voltage gain, making it the most widely used amplifier configuration. Common-collector (CC) has voltage gain less than 1, and common-base (CB) has good voltage gain but generally less practical overall than CE for this purpose. Therefore, the correct answer is CE.
The thyristor is turned off
A)
by switching off the voltage applied to the principal circuit and changing into reverse blocking mode
B)
by switching off the voltage applied to the gate circuit only
C)
by changing the thyristor into forward blocking state
D)
none of these
A thyristor (SCR) is turned off by reducing its anode current below the holding current and applying reverse bias or removing the main circuit voltage so it returns to the reverse blocking state. The gate terminal cannot turn it off; it only triggers turn-on. Hence option (a) is correct. Option (b) is wrong because gate voltage alone cannot turn off an SCR. Option (c) is incorrect because forward blocking is not the turn-off state.
According to Boolean algebra, (A + Ā) equals
A)
A
B)
1
C)
0
D)
Ā
By the complement law in Boolean algebra, a variable ORed with its complement equals 1: A + Ā = 1. So the correct option is (b). The other options are incorrect because A + Ā is not A, not 0, and not Ā.
In microprocessors, return from a subroutine is affected by
A)
a jump instruction
B)
an RST instruction
C)
a RET instruction
D)
a hardware interrupt signal
In microprocessors, a subroutine is entered using a CALL instruction, and it is returned from using a RET (return) instruction. RET pops the saved return address from the stack and transfers control back to the instruction after the CALL. A jump instruction only transfers control unconditionally, RST is a restart/call-like instruction, and a hardware interrupt signal invokes an interrupt service routine rather than directly returning from a subroutine.
Non-sinusoidal waveforms
A)
are departures from sine waveform
B)
have low mark-to-space ratio
C)
are much easier to generate
D)
are unfit for digital operation
A non-sinusoidal waveform is any waveform that is not a pure sine wave, so it is a departure from sine waveform. Option (b) is not a defining property, (c) is not generally true for all non-sinusoidal waveforms, and (d) is incorrect because many digital signals are non-sinusoidal and are widely used in digital operation.