← MES Jr. Grade
Past Questions
By properly coding and grouping longer sequence, it is possible to
A)
increase efficiency
B)
reduce delay in transmission
C)
increase delay
D)
make H = C
In information theory, proper coding and grouping of longer sequences helps reduce redundancy and improve the utilization of the channel. This increases coding efficiency. The other options are not the general result of coding longer sequences: delay is not necessarily reduced or increased, and H = C is not a standard consequence of grouping.
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).
If a line with Zo = 300 < 0° Ω is open circuited at far end, then VSWR is
A)
0
B)
1
C)
infinity
D)
2
For an open-circuited transmission line, the load impedance ZL = ∞. Hence the reflection coefficient magnitude |Γ| = 1. The VSWR is given by (1 + |Γ|) / (1 - |Γ|), which becomes (1 + 1) / (1 - 1) = ∞. Therefore, the VSWR is infinite. The characteristic impedance Zo = 300∠0° Ω does not change this result.
For a 50Ω resister for 3 GHz application, the stray capacitance should be less than
A)
1µF
B)
1nF
C)
1pF
D)
0.1pF
At 3 GHz, a resistor used in microwave applications must have extremely small stray capacitance so that its reactance remains very high compared with 50Ω. A capacitance of 1 pF would already have a very low reactance at 3 GHz, so it is too large. The required stray capacitance should therefore be much less than 1 pF, and among the given options, 0.1 pF is the correct choice.
In an optical fiber the refractive index of cladding material should be
A)
nearly 1
B)
very low
C)
less than that of core
D)
more than that of core
In an optical fiber, light is guided by total internal reflection at the core-cladding boundary. For total internal reflection to occur, the refractive index of the core must be higher than that of the cladding. Therefore, the cladding material should have a refractive index less than that of the core. Options (a), (b), and (d) do not satisfy this guiding condition.
In micro strip circuits, the dielectric material most commonly used is
A)
silicon
B)
rubber
C)
plastic
D)
bakelite
In microstrip circuits, the substrate is commonly made of a dielectric material such as plastic (or other low-loss insulating materials). Among the given options, plastic is the most suitable and commonly used dielectric. Silicon is a semiconductor, rubber is not used for microstrip substrates, and bakelite is an older insulating material but not the standard common choice here.
Microwave resonators are used in
A)
microwave oscillators
B)
microwave narrow band amplifiers
C)
microwave frequency meters
D)
All of the above
Microwave resonators are used in microwave oscillators to provide frequency-selective feedback, in microwave narrow band amplifiers to achieve selective amplification over a narrow frequency range, and in microwave frequency meters to measure frequency by resonance. Therefore, all the listed applications are correct.
Which of the following parameter is negligible in transmission lines?
A)
R
B)
L
C)
C
D)
G
In transmission line analysis, the four primary distributed parameters are resistance (R), inductance (L), capacitance (C), and conductance (G). For most practical overhead transmission lines, the shunt conductance of the line insulation is extremely small and is therefore usually neglected. Hence, G is the negligible parameter. R, L, and C are generally significant and must be considered.
Microwave are used for
A)
Telephony
B)
Radio broadcast
C)
TV systems
D)
All of the above
Microwaves are used in communication systems for telephony, radio broadcasting links, and TV transmission/reception systems. Since all three applications are valid, the correct answer is 'All of the above'.
Parity check bit for error detection is used in
A)
digital computer
B)
voice communication
C)
digital camera
D)
none of these
A parity check bit is an error-detection technique used in digital data transmission and storage, especially in digital computers. It adds one extra bit to make the number of 1s either even or odd, helping detect single-bit errors. It is not specifically used in voice communication or digital cameras.
What is the numerical aperture in an optical fiber when critical angle is 30°?
A)
0.5
B)
0.70
C)
0.886
D)
0.2
For an optical fiber, the numerical aperture is related to the critical angle by the common exam formula NA = \u221a(1 - cos^2\u03b8c) = sin\u03b8c when taken in the simplified form used in such questions. With critical angle \u03b8c = 30\u00b0, NA = sin 30\u00b0 = 0.5. Hence option (a) is correct. The other options do not match this standard result.
In VHF oscillator using butterfly capacitor
A)
only L
B)
only C
C)
both L and C
D)
none of these
A butterfly capacitor is a variable capacitor used for tuning in VHF oscillators. In VHF range, the oscillator frequency is generally controlled by the capacitance, while the inductance is usually fixed or less commonly varied. Therefore, the correct choice is only C. Options involving only L or both L and C are not the standard answer here.
Fibers have numerical aperture (NA) in the range of 0.15 to 0.4. Fibers with higher NA value would have
A)
reduced losses
B)
high bandwidth
C)
reduced losses and low bandwidth
D)
more losses and low bandwidth
Numerical aperture (NA) indicates the light-gathering ability of an optical fiber. A higher NA generally means a larger acceptance angle, but it also increases modal dispersion in multimode fibers. Greater modal dispersion reduces bandwidth, and the increased number of guided modes can contribute to higher losses. Therefore, fibers with higher NA value would have more losses and low bandwidth. Options (a) and (b) are incorrect because higher NA does not imply reduced losses or high bandwidth.
The unit of electric field intensity is
A)
NC-1
B)
Fm-1
C)
Nm-1
D)
FC-1
Electric field intensity is defined as force per unit charge: E = F/q. So its unit is newton per coulomb (N/C), which is written as NC^-1. Option (a) matches this. Option (b) Fm^-1 is the unit of permittivity, option (c) Nm^-1 is not the correct unit for electric field, and option (d) FC^-1 is farad per coulomb, which is incorrect.
If the carrier of a 100% modulated AM transmitter is suppressed, the percentage power saving will be
A)
50%
B)
66.67%
C)
78%
D)
100%
For a conventional AM transmitter at 100% modulation, the total power is P_t = P_c(1 + m^2/2). With m = 1, P_t = 1.5P_c. The sideband power is P_s = P_t - P_c = 0.5P_c. If the carrier is suppressed, only sideband power is transmitted, so the new power is 0.5P_c. Power saving = (1.5P_c - 0.5P_c) / 1.5P_c × 100 = 66.67%. Hence the correct option is (b).
A branched duplexer requires
A)
TR tube
B)
ATR tube
C)
Both TR and ATR tubes
D)
None of the above
A branched duplexer uses both a TR (Transmit-Receive) tube and an ATR (Anti-Transmit-Receive) tube. The TR tube protects the receiver during transmission, while the ATR tube helps direct the transmitted pulse to the antenna and isolate the receiver. Hence, the correct answer is both TR and ATR tubes.
A source with memory is known to be
A)
Markov source
B)
stochastic source
C)
consists of long sequence of symbols
D)
none of these
A source with memory means the present symbol depends on previous symbols. Such a source is modeled as a Markov source. A stochastic source is a broader term and does not specifically imply memory. Option (c) is unrelated, and (d) is incorrect because (a) is the standard answer.
Which mode has the minimum cutoff frequency in rectangular wave guides?
A)
TE11
B)
TE10
C)
TE01
D)
TE20
In a rectangular waveguide, the cutoff frequency is lowest for the dominant mode. For the usual rectangular guide, the dominant mode is TE10, because it has the smallest cutoff frequency among the listed modes. TE11, TE01, and TE20 all have higher cutoff frequencies than TE10.
Data rate R and information rate Rt are different because of
A)
channel capacity C, being high
B)
equivocation
C)
entropy h(x)
D)
none of these
In information theory, the data rate R and information rate Rt differ due to the presence of uncertainty in the received message, called equivocation. Equivocation measures the ambiguity remaining after reception, so the effective information rate is reduced. Hence R and Rt are different because of equivocation. Option (a) is not the reason, and entropy h(x) is only a measure of source uncertainty, not the direct cause of the difference between data rate and information rate.
Sometimes microwave signals follow the earth’s curvature. This is due to
A)
ionospheric reflection
B)
faraday rotation
C)
tropospheric scatter
D)
ducting
Microwave signals normally travel in straight lines, but under certain atmospheric conditions they can be trapped in a layer of the atmosphere and guided along the earth’s curvature. This phenomenon is called ducting. Ionospheric reflection is mainly for HF radio waves, Faraday rotation is the rotation of polarization in the ionosphere, and tropospheric scatter is a different long-distance propagation mechanism but not the one that makes microwaves follow the earth’s curvature.
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 charge on an isolated conductor resides
A)
at the surface of the conductor
B)
inside the conductor
C)
partly at surface and partly inside the conductor
D)
with uniform distribution over the volume of the conductor.
For an isolated conductor in electrostatic equilibrium, the electric field inside the conductor is zero. Therefore, any excess charge cannot remain inside the conductor and moves to the outer surface. Hence, the charge resides at the surface. The other options are incorrect because charge does not stay inside, does not split partly inside and partly outside, and is not uniformly distributed throughout the volume.
A particle having charge e is released in an electric field E. If there are no collisions than the velocity of the particle will be
A)
constant
B)
increasing
C)
decreasing
D)
none of these
A charged particle in an electric field experiences a force F = eE. Since the particle is released from rest and there are no collisions, this force continuously accelerates it in the direction of the field (for a positive charge). Therefore its speed increases with time. So the correct option is (b) increasing. Option (a) is wrong because the force is not zero, and option (c) is wrong because the particle is not slowing down.
In low level AM system, the amplifier following the modulator stage must be
A)
harmonic device
B)
class-C amplifier
C)
linear device
D)
class-A amplifier
In low-level AM, the modulated signal is still weak after the modulator, so the following amplifier must preserve the amplitude variations exactly. Therefore it must be a linear amplifier. A class-C amplifier is non-linear and is used in high-level AM power stages, not in low-level AM. Class-A is linear, but the general correct requirement here is a linear device, which includes class-A operation.
The force between two charges in 120N. If the distance between the charges is doubled, the force will be
A)
120N
B)
60N
C)
30N
D)
15N
By Coulomb’s law, electrostatic force is inversely proportional to the square of the distance between two charges: F ∝ 1/r². If the distance is doubled, the force becomes 1/4 of the original. So, new force = 120/4 = 30 N. Therefore the correct option is (c) 30N.