← MES Jr. Grade
Past Questions — 2012
An electron falls through a potential difference of 1V. If velocity is zero, the final velocity of electron is about
A)
1m/sec
B)
1Km/sec
C)
100Km/sec
D)
600Km/sec
An electron accelerated through a potential difference V gains kinetic energy eV. Here V = 1 V, so eV = 1.6 × 10^-19 J. Since initial velocity is zero,
(1/2)mv^2 = eV
v = sqrt(2eV/m)
= sqrt(2 × 1.6 × 10^-19 / 9.1 × 10^-31)
≈ sqrt(3.52 × 10^11)
≈ 5.9 × 10^5 m/s
= 590 km/s ≈ 600 km/s.
So the correct option is (d). Options (a), (b), and (c) are much too small.
The purpose of source coding is to
A)
increase the information transmission rate
B)
decrease the information transmission rate
C)
decrease the S/N rate
D)
decrease the probability error.
Source coding is used to remove redundancy from the source output so that the same information can be represented with fewer bits. This reduces the average number of bits needed per symbol, i.e., decreases the information transmission rate. Option (a) is the opposite of source coding. Option (c) refers to signal-to-noise ratio, which is not the purpose of source coding. Option (d) is more related to channel coding and error control, not source coding.
The magnitudes of OC and SC input impedances of a transmission line are 100Ω and 25Ω. The characteristic impedance is
A)
25Ω
B)
50Ω
C)
75Ω
D)
100Ω
For a transmission line, the characteristic impedance Z0 is the geometric mean of the open-circuit and short-circuit input impedances: Z0 = √(Zoc × Zsc). Here, Zoc = 100Ω and Zsc = 25Ω, so Z0 = √(100 × 25) = √2500 = 50Ω. Hence, the correct option is (b).
A lightning stroke has a current of 100KA and a duration of 200m/sec. If potential is 100KV the energy involved is
A)
2000 MJ
B)
200 MJ
C)
20 MJ
D)
0.2 MJ
Use the electrical energy formula: E = VIt. Here, V = 100 kV = 100,000 V, I = 100 kA = 100,000 A, and t = 200 ms = 0.2 s. So, E = 100,000 × 100,000 × 0.2 = 2 × 10^9 J = 2000 MJ. However, this value is not among the options, so we must check the intended interpretation. In such exam problems, the duration is often taken as 200 microseconds (μs) or the current/potential values may be intended differently. But with the given data and standard formula, the closest and likely intended option is 200 MJ only if one of the powers is misprinted. Since the question as written is inconsistent, the safest exam-key style answer is option (a) 2000 MJ if using the literal values. If the paper's intended unit was 200 microseconds, then E = 2000 MJ × 10^-3 = 2 MJ, still not matching. Therefore, the question appears misprinted.
Information content of a message has the characteristic that
A)
it is additive
B)
Increases linearly with time
C)
It increases monotonically
D)
All of the above
In information theory, the information content of a message is additive for independent events, and in many practical contexts it is treated as increasing monotonically with the amount of uncertainty or data. It also increases linearly with time when information is accumulated at a constant rate. Therefore, all the given statements are considered correct.
The first geostationary satellite was named
A)
Intelsat I
B)
Intelsat II
C)
Intelsat III
D)
Early bird
The first geostationary communications satellite was **Syncom 3** (first geostationary satellite overall), but the **first commercial geostationary satellite** was **Early Bird (Intelsat I)**, launched in 1965. In exam context, when asked “The first geostationary satellite was named,” the intended answer among the given options is **Early bird**. The other Intelsat satellites (II and III) were launched later, so they cannot be the first.
For a directional coupler the power is in the ratio of
A)
40dB
B)
30dB
C)
20dB
D)
10dB
A directional coupler is commonly specified by its coupling factor, and in standard microwave engineering the coupled power is often taken as 1/100 of the input power, i.e. a power ratio of 20 dB. Hence the correct option is 20 dB. The other values (10 dB, 30 dB, 40 dB) correspond to different coupling levels, but the standard expected answer here is 20 dB.
Number of side bands of an FM signal for fixed modulating voltage .............. with decrease in modulating frequency.
A)
increases
B)
decreases
C)
does not change
D)
none of these
In frequency modulation (FM), the modulation index is β = Δf / fm. For a fixed modulating voltage, the frequency deviation Δf remains constant. When the modulating frequency fm decreases, β increases. A larger modulation index produces more significant sidebands, so the number of sidebands increases. Hence, option (a) is correct.
In a wave guide the dielectric region is generally
A)
air
B)
magnetic material
C)
brass
D)
mica
In a waveguide, the dielectric filling is generally air because it has very low loss and supports efficient propagation of electromagnetic waves. Magnetic materials are not used as the dielectric region, brass is the conducting wall material of the waveguide, and mica is not the usual filling medium.
A wave guide section in a microwave circuit acts as
A)
LPF
B)
BPF
C)
HPF
D)
BSF
A waveguide supports propagation only above its cutoff frequency, so it blocks low-frequency signals and passes higher-frequency signals. Therefore, in a microwave circuit, a waveguide section acts as a high-pass filter (HPF). The other options are incorrect because a low-pass filter passes low frequencies, a band-pass filter passes only a band, and a band-stop filter rejects a band.
A measure of the ability of a radio receiver to receive weak signal is
A)
fidelity
B)
selectivity
C)
sensitivity
D)
ductility
The ability of a radio receiver to receive weak signals is called its sensitivity. Sensitivity indicates the minimum input signal strength required to produce a usable output. Fidelity refers to faithful reproduction of the input signal, selectivity refers to the ability to separate one desired signal from others, and ductility is unrelated to communication systems.
A reflex klystron oscillator uses
A)
one cavity resonator
B)
two cavity resonator
C)
three cavity resonator
D)
none of the above
A reflex klystron oscillator uses a single cavity resonator. The electron beam passes through the cavity gap, is reflected back by the repeller electrode, and interacts again with the same cavity to sustain oscillations. A two-cavity klystron uses two cavities, so option (b) is incorrect. Three-cavity resonators are not used in a reflex klystron, so (c) is incorrect. Therefore, the correct answer is (a) one cavity resonator.
Power monitor and reflectometers use the principle of
A)
directional computer
B)
Klystron amplifier
C)
Gunn diode
D)
TWT
Power monitors and reflectometers in microwave measurements work on the principle of a directional coupler, which samples forward and reflected power separately. Among the given options, 'directional computer' is clearly intended to mean 'directional coupler'. The other options—klystron amplifier, Gunn diode, and TWT—are microwave devices used for generation or amplification, not for power monitoring or reflection measurement.
The semiconductor diode which can be used in switching circuit in microwave range is
A)
PIN code
B)
varactor diode
C)
Tunnel diode
D)
Gunn diode
A PIN diode is widely used as a microwave switch because its intrinsic layer gives it low capacitance and good high-frequency performance. It can act as an effective RF/microwave switch by changing its resistance with bias. Varactor diodes are mainly used for voltage-controlled capacitance/tuning, tunnel diodes for negative resistance oscillators, and Gunn diodes for microwave oscillators, not switching circuits.
The width of a radio beam from a 1m diameter parabolic antenna at 1GHz is about
A)
100°
B)
50°
C)
10°
D)
2°
For a parabolic antenna, the beamwidth is approximately inversely proportional to the electrical size of the aperture. Here, wavelength at 1 GHz is about 0.3 m, so D/λ = 1/0.3 ≈ 3.33. A 1 m parabolic antenna at 1 GHz therefore has a narrow beam, on the order of a few degrees. The commonly used approximation gives a beamwidth around 2° (about 2–3°), so the closest option is 2°. The other options are far too wide for a 1 m dish at 1 GHz.
Decibel is a logarithmic unit denoting
A)
noise level
B)
current
C)
voltage
D)
power ratio
A decibel (dB) is a logarithmic unit used to express a ratio, most commonly a power ratio. For power, dB = 10 log10(P2/P1). It is also used for voltage or current ratios only when the impedance is specified, but the basic definition is for power ratio. Hence, the correct option is power ratio.
If the output current wave shape of class C circuit has a period 1µsec and pulse width of 0.006µsec, the duty cycle is
A)
6%
B)
0.6%
C)
0.06%
D)
0.006%
Duty cycle = (pulse width / period) × 100. Here, pulse width = 0.006 µs and period = 1 µs. So duty cycle = (0.006 / 1) × 100 = 0.6%. However, since the options include 0.06%, we must check the decimal carefully: 0.006 × 100 = 0.6, not 0.06. Therefore the correct computed value is 0.6%, which corresponds to option (b).
The 8085 instruction LDA 0FFAH is an example of
A)
direct addressing
B)
register addressing
C)
immediate addressing
D)
implicit addressing
In the 8085 instruction `LDA 0FFAH`, the 16-bit memory address `0FFAH` is given explicitly in the instruction itself. The accumulator loads data from that memory location, so this is **direct addressing**.
- **Direct addressing**: the address of the operand is specified in the instruction. ✅
- **Register addressing**: operand is in a register, not a memory address. ❌
- **Immediate addressing**: the data itself is part of the instruction, not a memory address. ❌
- **Implicit addressing**: operand is implied by the opcode. ❌
In an Op-Amp comparator, the positive or negative output is
A)
saturated
B)
amplified
C)
un-saturated
D)
any of the above
In an op-amp used as a comparator, the output is driven to one of its extreme levels depending on the input polarity. Therefore the positive or negative output is in the saturated state. It is not operating in the linear amplification region, so 'amplified' and 'un-saturated' are incorrect. 'Any of the above' is also incorrect because only saturation describes comparator output behavior.
If input frequency is 50Hz, the frequency of output wave in a full wave diode rectifier circuit is
A)
25Hz
B)
50Hz
C)
100Hz
D)
200Hz
In a full-wave rectifier, both the positive and negative half-cycles of the AC input are converted into output pulses. So the output frequency becomes twice the input frequency. For 50 Hz input, output frequency = 2 × 50 = 100 Hz. Hence, option (c) is correct. Option (a) is half-wave related, while (b) is the same as input frequency and (d) is too high.
In a self bias circuit for CE amplifier, the base voltage is
A)
equal to supply voltage
B)
more than supply voltage
C)
less than supply voltage
D)
none of these
In a self-bias circuit for a common-emitter (CE) amplifier, the base is biased through a resistor network so that the base voltage is kept below the supply voltage. This allows the transistor to operate in the active region with proper base-emitter forward bias. Hence, the base voltage is less than the supply voltage. Options (a) and (b) are incorrect because the base is not directly at or above the supply voltage in a proper self-bias arrangement.
The radius of constant N circle for N = 1 is
A)
√2
B)
2
C)
1
D)
1/√2
For the Nichols chart, the constant-N circle for a given closed-loop magnitude ratio N has center at \( \frac{N^2}{1-N^2} \) on the real axis and radius \( \frac{N}{|1-N^2|} \). For \(N=1\), this special case becomes the unity-gain circle, whose radius is \(1/\sqrt{2}\). Hence the correct option is (d).
In a CE amplifier the ac cut-off voltage is 9V and slope of ac load line is –0.5mA/V, the ac saturation current is
A)
18mA
B)
9mA
C)
4.5mA
D)
3mA
For an AC load line, the slope is given by ΔI/ΔV. Here the slope is –0.5 mA/V, so the magnitude is 0.5 mA/V. The AC cut-off voltage is 9 V, meaning at cut-off the collector current is zero and the voltage swing is 9 V. Therefore, the AC saturation current is:
I_sat = slope × voltage = 0.5 × 9 = 4.5 mA
But note that in a CE amplifier load line, the current intercept at saturation is found from the full line relation using the voltage intercept and slope. Since the slope is negative, the current at zero voltage is 0.5 × 9 = 4.5 mA only if the intercept is taken directly from the given cut-off voltage. However, the standard interpretation in such exam questions is that the AC cut-off voltage corresponds to the voltage axis intercept and the AC saturation current is the current axis intercept, which is:
I_sat = V_cutoff / |slope| = 9 / 0.5 = 18 mA.
Hence, the correct answer is 18 mA.
The number of interrupt lines in 8085 processor is
A)
2
B)
3
C)
4
D)
5
The 8085 microprocessor has 5 interrupt inputs in total: TRAP, RST 7.5, RST 6.5, RST 5.5, and INTR. However, if the question asks for the number of interrupt lines, standard exam convention usually counts the hardware interrupt lines as 5. Since the options do not include 5 as a separate correct choice? Actually option (d) is 5, so the correct answer is 5. The earlier count of 4 would be incorrect because 8085 has one non-maskable interrupt (TRAP), three maskable vectored interrupts (RST 7.5, 6.5, 5.5), and one general interrupt request line (INTR).
The open loop gain of an amplifier is 200. If negative feedback with β=0.2 is used, the closed loop gain will be
A)
200
B)
40.12
C)
4.878
D)
2.2
For a negative feedback amplifier, the closed-loop gain is
A_f = A / (1 + Aβ)
Given open-loop gain A = 200 and feedback factor β = 0.2:
Aβ = 200 × 0.2 = 40
So,
A_f = 200 / (1 + 40) = 200 / 41 = 4.878
Hence, the correct answer is 4.878. The other options do not match the standard negative feedback gain formula.