← MES Jr. Grade
Past Questions — 2012
A class A transformer coupled power amplifier is to deliver 10W output. The power rating of transistor should not be less than
A)
5W
B)
10W
C)
20W
D)
40W
In a class A transformer-coupled power amplifier, the maximum theoretical efficiency is 50%. So, to deliver 10 W output power, the transistor must be able to handle at least twice that power as input dissipation/handling capacity. Therefore, required transistor power rating should not be less than 20 W. Hence, option (c) is correct.
Which amplifier has the highest power gain?
A)
common collector
B)
common base
C)
common emitter
D)
both CC and CE
In transistor amplifier configurations, the common emitter (CE) amplifier provides the highest power gain. It offers both significant current gain and voltage gain, so the product of the two gives the largest power gain. The common collector (CC) has high current gain but voltage gain is about 1, so its power gain is lower. The common base (CB) has current gain less than 1, so its power gain is also lower than CE.
Generally a servo motor operates at
A)
high torque and high speed
B)
low torque and low speed
C)
low torque and high speed
D)
Not provided
A servo motor is designed for precise control of position, speed, and acceleration. In general, it operates with low torque and high speed, especially in control applications where quick response is needed. Hence, option (c) is correct. Option (a) is incorrect because servo motors are not generally characterized by high torque and high speed together in basic exam context. Option (b) is incorrect because servo motors are not typically low speed. Option (d) is not applicable.
The force required to initiate motion between two contacting surfaces is called
A)
Coulomb force
B)
viscous force
C)
stiction
D)
none of these
The force needed to start motion between two surfaces in contact is the static friction force, often called stiction. It acts before relative motion begins. Coulomb force refers to friction during motion in an idealized model, and viscous force is resistance due to fluid-like damping, not contact initiation force.
The maximum theoretical efficiency for class A amplifier is
A)
12.5%
B)
25 %
C)
50%
D)
75%
For a class A amplifier, the maximum theoretical efficiency is 25% when using a resistive load. This is because the transistor conducts for the full 360° of the input cycle, causing significant power loss. Option (a) 12.5% is too low, while (c) 50% and (d) 75% are efficiencies associated with other amplifier classes or idealized push-pull arrangements, not class A.
The open loop gain of an amplifier is 50 but is likely to decrease by 20% due to various factors. If negative feedback with β=0.1 is used the change in gain will be about
A)
20%
B)
10%
C)
3.33%
D)
0.5%
For a feedback amplifier, the closed-loop gain is A_f = A / (1 + Aβ). Here A = 50 and β = 0.1, so the feedback factor is Aβ = 5. Thus the gain reduction sensitivity is 1/(1+Aβ) = 1/6.
If the open-loop gain decreases by 20%, the fractional change in closed-loop gain is approximately:
20% × 1/6 = 3.33%.
So negative feedback greatly stabilizes the gain, reducing the effect of a 20% open-loop variation to about 3.33%.
Hence, the correct answer is (c) 3.33%.
Why others are incorrect:
- 20%: this would be the change without effective feedback stabilization.
- 10%: too large for Aβ = 5.
- 0.5%: too small; would require much larger loop gain.
Which of the following schemes has unique representation of zero?
A)
Sign magnitude
B)
1’s complement
C)
2’s complement
D)
both (b) & (c)
In number representation, a scheme has a unique representation of zero if it has only one zero value. In 2’s complement, zero is represented only as 0000...0, so it is unique. In sign magnitude and 1’s complement, both +0 and -0 exist, so zero is not unique. Therefore, the correct answer is 2’s complement.
When the ac base voltage in a CE amplifier circuit is too high, the ac emitter current is
A)
zero
B)
constant
C)
alternating
D)
distorted
In a common-emitter (CE) amplifier, the emitter current should ideally follow the input signal linearly. If the AC base voltage becomes too high, the transistor is driven beyond its linear operating region, causing non-linear operation. As a result, the AC emitter current becomes distorted. The other options are incorrect because the current does not become zero, constant, or merely alternating in the ideal sense; the key effect of excessive input is distortion.
Which amplifier has a voltage again of less than unity?
A)
common base
B)
common collector
C)
common emitter
D)
both CB and CC
A common collector amplifier (emitter follower) has a voltage gain slightly less than 1, i.e., less than unity. In contrast, a common base amplifier has high voltage gain, and a common emitter amplifier also provides voltage gain greater than 1. Therefore, the correct answer is common collector.
Which OP-Amp circuit uses a resistance in series with input and capacitor in feedback path?
A)
differentiating amplifier
B)
integrator
C)
logarithmic amplifier
D)
exponential amplifier
In an op-amp integrator, the input element is a resistor in series with the input signal, and the feedback element is a capacitor. This makes the output proportional to the time integral of the input. A differentiating amplifier has the capacitor at the input and resistor in feedback, so it is the opposite. Logarithmic and exponential amplifiers use nonlinear elements like diodes/transistors, not this RC arrangement.
A potential of 7V is applied to a silicon diode in series with a 1KΩ load resistor. If the diode is forward biased, then the current in the circuit is
A)
7mA
B)
6.3mA
C)
0.7mA
D)
zero
For a silicon diode in forward bias, the forward voltage drop is about 0.7 V. So the voltage across the 1 kΩ resistor is 7.0 - 0.7 = 6.3 V. Using Ohm’s law, I = V/R = 6.3/1000 = 6.3 mA. Hence the correct option is (b).
A push pull amplifier is a
A)
power amplifier
B)
voltage amplifier
C)
current amplifier
D)
both (a) and (c)
A push-pull amplifier is mainly used as a power amplifier because it delivers higher output power with better efficiency. It also acts as a current amplifier since it can supply a larger load current than the input stage. Therefore, the correct choice is both power amplifier and current amplifier.
Which of the following 8085 instruction does not belong to arithmetic group?
A)
SHLD
B)
SUI
C)
ADC
D)
INR
In 8085, arithmetic instructions include operations like SUI (subtract immediate), ADC (add with carry), and INR (increment). SHLD is not an arithmetic instruction; it is a data transfer instruction used to store the HL pair directly into memory. Therefore, SHLD does not belong to the arithmetic group.
What does the operator ( ) mean in C?
A)
function expression
B)
array expression
C)
structure operator
D)
increment/decrement
In C, the operator ( ) is the function call operator. It is used to call a function, so among the given options the closest and correct answer is 'function expression'. The other options are incorrect: array expression uses [ ], structure operator uses . or ->, and increment/decrement uses ++ or --.
The statement !( b < 5 ) in C means
A)
B = 5
B)
b > 5
C)
b >= 5
D)
b <= 5
In C, the expression `b < 5` is true when `b` is less than 5. The `!` operator negates this condition, so `!(b < 5)` means `b` is not less than 5, i.e. `b >= 5`. Hence the correct option is (c).
Most of the memory chips with static RAM need
A)
2V supply
B)
5V supply
C)
12V supply
D)
none of above
Static RAM (SRAM) chips traditionally operate with a 5V power supply in standard computer architecture questions. 2V is too low for most conventional SRAM chips, and 12V is not the normal operating voltage for SRAM. Therefore, the correct answer is 5V supply.
Which of the following is not a library string function in C?
A)
strlen
B)
strcpy
C)
strdup
D)
strbig
In C, `strlen`, `strcpy`, and `strdup` are known string-related library functions (though `strdup` is POSIX, not part of the C standard library). `strbig` is not a standard C library string function. Therefore, the correct answer is `strbig`.
Which of the following is not a valid real constant in C?
A)
426
B)
321.0
C)
– 627.431
D)
+ 201.84
In C, a real constant must contain a decimal point or exponent part. Option (a) 426 is an integer constant, not a real constant. Options (b) 321.0, (c) – 627.431, and (d) + 201.84 are real constants because they include decimal points.
The number of storage classes in C is
A)
2
B)
3
C)
4
D)
6
In C, the commonly taught storage classes are `auto`, `register`, `static`, and `extern`. That makes the total number 4. Hence option (c) is correct. Options (a), (b), and (d) do not match the standard set of C storage classes.
In 8085 the pins for + 5V input and ground are
A)
20 & 40
B)
40 & 20
C)
1 & 2
D)
2 & 1 respectively
In the 8085 microprocessor, pin 40 is VCC (+5V supply) and pin 20 is GND (ground). Therefore, the correct pair for +5V input and ground is 40 & 20. The other options do not match the standard 8085 pin configuration.
An OP-Amp uses
A)
RC coupling
B)
direct coupling
C)
transformer coupling
D)
either (b) or (c)
An op-amp is a DC amplifier, so its internal stages are directly coupled to amplify both AC and DC signals. RC coupling is not used inside op-amps, and transformer coupling is also not used because it is bulky and unsuitable for integrated circuits. Therefore, the correct answer is direct coupling.
Which of the following identifiers is invalid in PASCAL?
A)
Beta
B)
MAX40
C)
2.Nd
D)
A Max
In Pascal, an identifier must begin with a letter and may contain only letters, digits, and sometimes underscores, depending on the compiler. It cannot start with a digit, and it cannot contain a period or space.
- Beta: valid identifier
- MAX40: valid identifier
- 2.Nd: invalid because it starts with a digit and contains a period
- A Max: invalid because it contains a space, but the clearly invalid Pascal identifier among the options is 2.Nd
So the correct answer is 2.Nd.
In a class C power amplifier the input frequency of ac signal is 1MHz. If tank circuit has C=1000pF, the value of L is
A)
25µH
B)
50µH
C)
100µH
D)
200µH
For a class C power amplifier, the tank circuit resonates at the input frequency: f = 1 / (2π√(LC)).
Given:
- f = 1 MHz = 10^6 Hz
- C = 1000 pF = 10^-9 F
Rearranging:
L = 1 / (4π^2 f^2 C)
Substitute values:
L = 1 / (4π^2 × (10^6)^2 × 10^-9)
≈ 1 / (39.48 × 10^3)
≈ 2.53 × 10^-5 H
≈ 25 µH
However, checking the standard resonance relation carefully, the correct calculation gives approximately 25 µH, which matches option (a), not (b).
6 bytes means
A)
6 bits
B)
24 bits
C)
48 bits
D)
96 bits
1 byte = 8 bits. Therefore, 6 bytes = 6 × 8 = 48 bits. So the correct option is (c). The other options are incorrect because they do not match the standard byte-to-bit conversion.
In which of the following circuit is OP-Amp used in open loop configuration?
A)
comparator
B)
Summing amplifier
C)
Integrator
D)
multiplier
An op-amp used in open-loop configuration means there is no negative feedback. In this mode, even a very small difference between the input voltages drives the output to saturation, which is exactly how a comparator works. Summing amplifier and integrator both require negative feedback, so they are closed-loop circuits. Multiplier circuits are not the standard open-loop op-amp application here.