← MES Jr. Grade
Past Questions — 2012
In microwave resistor the wall thickness is made as small as possible to
A)
have high R
B)
have low R
C)
reduce capacitor
D)
reduce inductance
In a microwave resistor, the wall thickness is kept as small as possible so that the current path length is minimized and the lead/wall inductance becomes very low. At microwave frequencies, even a small inductance can cause significant impedance, so reducing inductance is important. Hence, the correct choice is (d) reduce inductance. Options (a) and (b) are about resistance, but wall thickness is not reduced for that purpose here, and (c) capacitor is not the relevant effect.
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.
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.
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).
Taking the earth’s radius as 6370Km and electrosphere height as 25Km, the capacitance of earth electrosphere combination is about
A)
1 F
B)
0.2F
C)
0.01F
D)
1µF
Treat the earth and the electrosphere as a spherical capacitor. For a spherical capacitor, C = 4πϵ0ab/(b−a), where a is Earth’s radius and b is the electrosphere radius. Here, a = 6370 km and b = 6395 km, so the separation (b−a) = 25 km = 2.5×10^4 m. Since the gap is very small compared to Earth’s radius, we can use the approximation C ≈ 4πϵ0a^2/d, which gives a value of the order of 10^-6 F. Numerically this comes out close to 1 µF. Hence the correct option is (d).
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.
Which mode has the lowest cutoff frequency in circular wave guides?
A)
TE01
B)
TE10
C)
TE11
D)
TE20
In a circular waveguide, the cutoff frequency depends on the mode indices. Among the given modes, TE11 has the lowest cutoff frequency and is the dominant mode of a circular waveguide. TE01, TE10, and TE20 all have higher cutoff frequencies than TE11.
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.
If a 240 watt carrier is simultaneously modulated by two sine waves with modulation indices of 0.3 and 0.4, its total modulation index is
A)
0.5
B)
0.7
C)
0.12
D)
0.75
For simultaneous modulation by two independent sine waves, the total modulation index is the root-sum-square of the individual indices: m = √(m1² + m2²). Here, m1 = 0.3 and m2 = 0.4, so m = √(0.3² + 0.4²) = √(0.09 + 0.16) = √0.25 = 0.5. Hence, the correct option is (a). The carrier power value 240 W is not needed to find the modulation index.
An FM signal with a modulation index mf is passed through a frequency tripler. The wave in the output of the tripler will have a modulation index of
A)
mf /3
B)
mf
C)
3mf
D)
9mf
In frequency modulation, the modulation index is proportional to the frequency deviation. A frequency tripler multiplies both the carrier frequency and the frequency deviation by 3. Therefore, the new modulation index becomes 3 times the original: mf' = 3mf. Hence, the correct option is (c).
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.
Two spherical drops of water each having a charge Q and surface potential V are combined to form a bigger drop. The surface potential of the bigger drop will be
A)
V
B)
2V
C)
< V but > 2V
D)
none of these
For a spherical drop, surface potential is V = kQ/R. When two identical drops combine, the total charge becomes 2Q and the volume doubles, so the new radius becomes R' = 2^(1/3)R. The new potential is V' = k(2Q)/R' = k(2Q)/(2^(1/3)R) = 2^(2/3)V. However, this is not among the given options. In many exam settings, the intended result is that the potential remains the same only if charge and radius scale proportionally, but for two identical charged drops the exact value is 2^(2/3)V, which is greater than V. Since that exact value is not listed, the correct choice from the options is 'none of these'.
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.
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.
As wavelength decreases, the size of the high directivity antenna
A)
decreases
B)
increases
C)
not affected
D)
either (b) or (c)
For a high-directivity antenna, the physical size is generally related to the wavelength. As wavelength decreases, the same electrical size can be achieved with a smaller physical antenna. Therefore, the size of a high-directivity antenna decreases. Options (b) and (c) are incorrect because the antenna does not need to increase in size or remain unaffected when wavelength decreases.
In an oscillator crystal, the parallel resonant frequency is .............. series resonant frequency
A)
equal to
B)
higher than
C)
lower than
D)
none of these
In a crystal oscillator, the series resonant frequency is the lower resonant point, where the crystal impedance is minimum. The parallel resonant frequency occurs at a slightly higher frequency due to the effect of the crystal’s shunt capacitance. Therefore, the parallel resonant frequency is higher than the series resonant frequency. So option (b) is correct.
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.
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.
The main advantage of using microwaves for communication is
A)
Large bandwidth
B)
Small bandwidth
C)
Low power
D)
High power
Microwaves are widely used in communication because they offer a very large bandwidth, which allows transmission of a high amount of information. They also support highly directional antennas, reducing interference and enabling point-to-point links. The other options are incorrect because microwaves do not have small bandwidth as an advantage, and low power or high power are not the main communication benefit.
The noise which assumes great importance at high frequencies is
A)
flicker noise
B)
transit time noise
C)
Johnson noise
D)
shot noise
Transit time noise becomes significant at high frequencies because the finite time taken by carriers to travel through a device is no longer negligible compared with the signal period. This causes phase lag and fluctuations that dominate at high frequency. Flicker noise is mainly important at low frequencies, while Johnson noise and shot noise are broadband noises not specifically associated with high-frequency dominance.
The skin depth at 1000 MHz is compared to that at 500 MHz is
A)
2
B)
√2
C)
0.707
D)
0.5
Skin depth in a good conductor is inversely proportional to the square root of frequency: δ ∝ 1/√f. Therefore, when frequency increases from 500 MHz to 1000 MHz, the ratio is δ1000/δ500 = √(500/1000) = √(1/2) = 1/√2 ≈ 0.707. So the skin depth at 1000 MHz is 0.707 times that at 500 MHz.
Divergence of D is a
A)
scalar point function
B)
vector point function
C)
phasor
D)
may be scalar or vector point function
The divergence of the electric flux density D is a scalar quantity at each point in space, so it is a scalar point function. Divergence converts a vector field into a scalar field. Hence option (a) is correct. Option (b) is wrong because divergence is not a vector. Option (c) is unrelated. Option (d) is incorrect because divergence is specifically scalar, not sometimes vector.
Power spectral density of white noise .............. with frequency.
A)
is constant
B)
decreases
C)
increases
D)
none of above
White noise has equal power at all frequencies, so its power spectral density does not vary with frequency. Therefore, the PSD of white noise is constant. Options (b) and (c) are incorrect because white noise does not decrease or increase with frequency.
Impedance of a half wave folded dipole antenna is
A)
75 ohms
B)
150ohms
C)
300ohms
D)
none of these
A half-wave folded dipole has an input impedance about 4 times that of a simple half-wave dipole. Since a half-wave dipole has an impedance of about 73–75 ohms, the folded dipole impedance is about 300 ohms. Hence, option (c) is correct. Options (a) and (b) are too low, and (d) is incorrect.
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.