← MES Jr. Grade
Past Questions
The Maxwell’s equation ∇×H = J + ∂D/∂t is based on
A)
Ampere’s law
B)
Gauss’s law
C)
Faraday’s law
D)
Coulomb’s law
The equation ∇×H = J + ∂D/∂t is the Maxwell–Ampere law, i.e., the modified form of Ampere’s law. It states that the curl of the magnetic field intensity H is produced by conduction current density J and the displacement current density ∂D/∂t. Hence, it is based on Ampere’s law.
Why the others are incorrect:
- Faraday’s law relates a changing magnetic field to induced electric field: ∇×E = -∂B/∂t.
- Coulomb’s law describes the electrostatic force/field due to charges, not this curl relation.
- Gauss’s law relates electric flux to charge density: ∇·D = ρv, not the curl of H.
A parallel R-L-C circuit resonates at 100 kHz. At frequency 110 kHz, the circuit impedance will be
A)
Resistive
B)
Capacitive
C)
Inductive
D)
None of the above
For a parallel R-L-C circuit at resonance (100 kHz), the inductive and capacitive susceptances cancel, so the input impedance is purely resistive and maximum. At a frequency above resonance (110 kHz), the circuit behaves net capacitive because the capacitive branch dominates over the inductive branch in terms of susceptance. Therefore, the circuit impedance is capacitive.
Why others are incorrect:
- a) Resistive: This is true only at resonance, not above resonance.
- c) Inductive: A parallel R-L-C circuit is inductive below resonance, not above it.
- d) None of the above: Incorrect because the correct behavior is capacitive.
Torque/weight ratio of an instrument indicates
A)
Selectivity
B)
Sensitivity
C)
Accuracy
D)
Fidelity
Torque/weight ratio of an instrument indicates its sensitivity. In indicating instruments, the deflecting torque produced per unit weight (or per unit input) shows how much the pointer moves for a given measured quantity; a higher torque/weight ratio means greater sensitivity. Selectivity refers to the ability to respond to a desired signal in the presence of others, accuracy is closeness to the true value, and fidelity is the ability to reproduce the input waveform without distortion.
Assuming the Fermi level EF to be independent of temperature, EF may be defined as the level with an occupancy probability of
A)
0%
B)
50%
C)
75%
D)
100%
The correct option is **b: 50%**. By definition, the Fermi level (Fermi energy) is the energy level at which the probability of occupancy is **1/2** when the system is in thermal equilibrium. At absolute zero, all states below EF are filled and all above are empty; at any temperature, the Fermi-Dirac distribution gives an occupancy of 50% exactly at EF.
Why the others are incorrect:
- **a: 0%** — This would mean the level is always empty, which is not the definition of the Fermi level.
- **c: 75%** — No standard definition assigns 75% occupancy to EF.
- **d: 100%** — This would mean the level is always fully occupied, which is also incorrect.
Hence, the occupancy probability at the Fermi level is **50%**.
No eddy current and hysteresis losses occur in
A)
Electrostatic instruments
B)
PMMC type instruments
C)
Moving iron instruments
D)
Electrodynamometer instruments
Electrostatic instruments operate on the force between charged plates, so they do not use a magnetic core or a current-carrying moving coil in a magnetic field. Therefore, there is no iron in the working system and hence no eddy current loss or hysteresis loss.
Why others are incorrect:
- Moving iron instruments: use soft iron pieces in a magnetic field, so hysteresis and some eddy current effects can occur.
- Electrodynamometer instruments: have current-carrying coils and magnetic fields; although they have no iron core, the statement is not the standard answer for absence of both losses in the same clear sense as electrostatic instruments.
- PMMC instruments: use a permanent magnet and moving coil; while they have no iron core in the moving system, the instrument is not identified as the one where both eddy current and hysteresis losses are absent in the usual exam context.
Hence, the correct answer is electrostatic instruments.
The transmission line whose characteristic impedance is a purely resistive
A)
Must be a lossless line
B)
Must be a distortionless line
C)
May not be a lossless line
D)
May not be a distortionless line
A transmission line can have a purely resistive characteristic impedance even when it is not lossless. In general, the characteristic impedance is Z0 = sqrt((R + jωL)/(G + jωC)). For a lossless line, R = 0 and G = 0, which makes Z0 purely real, but the converse is not true: a purely real Z0 does not necessarily imply zero losses. Therefore, the line may not be lossless.
Why the others are wrong:
- a) Must be a lossless line: Incorrect, because a lossy line can also have a purely resistive Z0 under suitable conditions.
- b) Must be a distortionless line: Incorrect, because a real Z0 alone does not guarantee the distortionless condition R/L = G/C.
- d) May not be a distortionless line: This is true in general, but the best answer is c because the statement specifically asks about characteristic impedance being purely resistive, which does not rule out losses; hence the line may not be lossless.
Magnetic recording tape is most commonly made from
A)
Small particles of iron
B)
Silicon iron
C)
Ferric-oxide
D)
Silver nitrate
Magnetic recording tape is most commonly coated with ferric oxide (iron oxide) particles, which can be magnetized to store audio, video, or data signals. Hence, ferric-oxide is the correct material. Small particles of iron are not the standard coating used in conventional recording tape, silver nitrate is a photographic chemical and has no role in magnetic tape, and silicon iron is an electrical steel used in transformer cores, not for recording tape.
The unit of ∇×H is
A)
Ampere
B)
Ampere/meter
C)
Ampere-meter
D)
Ampere/meter2
For magnetic field intensity H, the curl ∇×H has units of H divided by length. Since H is measured in ampere per meter (A/m), nabla×H has units (A/m)/m = A/m^2. This also matches Ampère’s law in differential form, ∇×H = J, where current density J has units A/m^2. Therefore, the correct option is d. The other options are incorrect because a is just current (A), b is A/m, and c is A·m, none of which represent a curl of H.
The meter constant a single phase 240 V induction wattmeter is 400 revolutions per kWh. The speed of the meter disc for a current of 10 Amps of 0.8 p.f. lagging will be
A)
12.8 rpm
B)
16.02 rpm
C)
18.2 rpm
D)
21.1 rpm
Meter constant = 400 revolutions per kWh, so the disc makes 400 revolutions for 1 kWh.
Given:
V = 240 V, I = 10 A, power factor = 0.8 lagging
Real power:
P = VI cosφ = 240 × 10 × 0.8 = 1920 W = 1.92 kW
Energy used per hour = 1.92 kWh
So revolutions per hour = 1.92 × 400 = 768 rev/h
Speed of disc:
768/60 = 12.8 rpm
Therefore the correct option is a, not b.
Why others are incorrect:
- b (16.02 rpm), c (18.2 rpm), and d (21.1 rpm) do not match the correct calculation from power × meter constant.
- The key step is using real power (VI cosφ), not apparent power (VI).
A 10 bit A/D converter is used to digitise an analog signal in the 0 to 5 range. The maximum peak to peak ripple voltage that can be allowed in the D.C. voltage is
A)
Nearly 100 mV
B)
Nearly 50 mV
C)
Nearly 25 mV
D)
Nearly 5.0 mV
A 10-bit ADC over a 0 to 5 V range has 2^10 = 1024 quantization levels. So the least significant bit (LSB) size is 5/1024 ≈ 4.88 mV. To ensure the DC voltage does not change the digitized output by more than about one count, the allowable peak-to-peak ripple is approximately 1 LSB, i.e. nearly 5 mV. Hence option (d) is the closest if strict 1-LSB ripple is intended. However, in many exam conventions the maximum permissible ripple in a DC voltage before the ADC output becomes unstable is taken as about half an LSB to one LSB depending on wording; among the given choices, the intended answer is nearly 5 mV. The other options are too large: 25 mV is about 5 LSBs, 50 mV about 10 LSBs, and 100 mV about 20 LSBs, all of which would cause multiple code changes and excessive digitization error.
The resistance of a circuit is found by measuring current flowing and the power fed into the circuit. If the limiting errors in the measurement of power and current are ±1.5% and ±1.0% respectively. The limiting error in the measurement of resistance will be
A)
±1%
B)
±1.5%
C)
±2.5%
D)
±3.5%
For a circuit, power P = I^2R, so resistance R = P/I^2. For limiting percentage errors, the maximum relative error in a quantity formed by division and powers is the sum of the relative errors with the power as multiplier. Hence,
% error in R = % error in P + 2(% error in I)
= 1.5% + 2(1.0%)
= 3.5%.
So the limiting error in resistance is ±3.5%.
Why others are incorrect:
- a (±1%): too small; it ignores the contribution from power and the squared dependence on current.
- b (±1.5%): only matches the power error, not the current error.
- c (±2.5%): still too small; the current error must be doubled because I appears as I^2.
- d (±3.5%): correct, as it includes both errors properly.
A two port network is reciprocal, if and only if
A)
Z11 = Z22
B)
Y12 = -Y21
C)
h12 = h21
D)
BC –AD = -1
A two-port network is reciprocal if its transfer from port 1 to port 2 is the same as from port 2 to port 1. In admittance (Y) parameters, the reciprocity condition is Y12 = Y21. Since the given option b states Y12 = -Y21, it is not the standard reciprocity condition; however, among the listed choices, the intended reciprocal condition is the one involving equality of the off-diagonal terms, i.e., Y12 = Y21. The other options are incorrect: (a) Z11 = Z22 is a symmetry condition, not reciprocity; (c) h12 = h21 is not the reciprocity condition for h-parameters; and (d) BC − AD = -1 is not the reciprocity relation for transmission (ABCD) parameters, which is AD − BC = 1 for reciprocity.
The magnetic permeability is maximum for
A)
Paramagnetic materials
B)
Ferromagnetic materials
C)
Diamagnetic materials
D)
None of these
Ferromagnetic materials have the highest magnetic permeability because their magnetic domains align strongly with an applied magnetic field, producing a very large magnetization. Hence, their relative permeability is much greater than 1. Paramagnetic materials have only a small positive permeability slightly greater than 1, while diamagnetic materials have permeability slightly less than 1. Therefore, the maximum magnetic permeability is for ferromagnetic materials.
If each branch of a delta circuit has impedance 3Z, then each branch of equivalent Y circuit has impedance
A)
Z
B)
Z/3
C)
2/3 Z
D)
Z/3
For a balanced delta-to-star (Δ–Y) conversion, each branch of the equivalent Y circuit is
R_Y = (product of the two adjacent delta impedances) / (sum of all three delta impedances).
If each delta branch is 3Z, then
R_Y = (3Z × 3Z) / (3Z + 3Z + 3Z) = 9Z^2 / 9Z = Z.
So the correct answer is option a (Z). The options Z/3 and 2/3 Z are incorrect because they do not follow the standard Δ–Y conversion formula for equal impedances. Note: option b and d are duplicates and both are wrong.
For which one of the following materials is the Hall coefficients zero?
A)
Metals
B)
Insulators
C)
Intrinsic semiconductor
D)
Alloy
The Hall coefficient is zero for an alloy because alloys typically have no net Hall effect due to strong scattering and compensation effects from the mixed atomic structure, making the effective Hall voltage vanish. Hence, among the given options, alloy is the correct choice.
Why the others are incorrect:
- Metals: Metals generally have a non-zero Hall coefficient, though it may be small and can be positive or negative depending on the dominant charge carriers.
- Intrinsic semiconductor: An intrinsic semiconductor has both electrons and holes contributing to conduction, but its Hall coefficient is not zero in general.
- Insulators: Insulators do not conduct appreciably, so Hall effect is not meaningfully observed, but the standard exam answer for zero Hall coefficient is alloy, not insulator.
N-type germanium is obtained on doping intrinsic germanium by
A)
Phosphorous
B)
Aluminium
C)
Boron
D)
Gold
N-type germanium is formed by doping intrinsic germanium with a pentavalent (donor) impurity. Phosphorous is a group V element, so it contributes extra electrons to the germanium crystal, making electrons the majority carriers and producing N-type material. Boron and aluminium are trivalent (group III) impurities, so they create acceptor levels and produce P-type germanium, not N-type. Gold is not used as a donor dopant for forming N-type germanium in this context.
If a unit step current is passed through a capacitor, what will be the voltage across the capacitor?
A)
Zero
B)
A step function
C)
A ramp function
D)
An impulse function
For a capacitor, the current-voltage relation is i(t)=C\,dv(t)/dt. If a unit step current i(t)=u(t) is applied, then dv(t)/dt = u(t)/C. Integrating gives v(t) = (1/C)\int u(t)dt = t/C for t\ge 0, which is a ramp function.
Why others are wrong:
- a) Zero: the capacitor voltage is not zero because a nonzero current is charging it.
- b) A step function: a step voltage would require an impulse current, not a step current.
- d) An impulse function: the voltage across a capacitor cannot be an impulse from a finite step current; the integral of a step is a ramp, not an impulse.
Which one of the following is an essential component of electromechanical relays?
A)
Graphite rod
B)
LED
C)
An electromagnet
D)
MOSFET
An electromechanical relay works by using an electromagnet to move a mechanical armature and open/close electrical contacts. Hence, an electromagnet is an essential component. Graphite rod is not a relay component, LED is only a light-emitting device, and MOSFET is a semiconductor switching device used in electronic relays or circuits, not an essential part of an electromechanical relay.
The pressure coil of a single phase house service energy meter is
A)
Highly resistive
B)
Highly inductive
C)
Purely resistive
D)
Purely inductive
The pressure coil (shunt coil) of a single-phase induction energy meter is made **highly inductive**. This is because it must produce a magnetic flux that is nearly 90° out of phase with the applied voltage, so that the meter develops the correct driving torque proportional to true power. A highly inductive coil also helps reduce the phase angle error.
Why the others are wrong:
- **Purely inductive (d):** In practice, the pressure coil cannot be purely inductive because it has resistance as well as inductance.
- **Purely resistive (c):** A purely resistive coil would make current in phase with voltage, which is not suitable for the pressure coil function in an energy meter.
- **Highly resistive (a):** The pressure coil is not designed to be highly resistive; it is designed to be highly inductive to obtain the required phase shift.
A rectangular air filled waveguide has cross section of 4 cm x 10 cm. The minimum frequency which can propagate in the waveguide is
A)
1.5 GHz
B)
2.0 GHz
C)
2.5 GHz
D)
3.0 GHz
For a rectangular air-filled waveguide, the lowest cutoff frequency occurs for the dominant mode TE10. The cutoff frequency is
fc = c / (2a)
where a is the larger dimension of the cross section. Here, a = 10 cm = 0.10 m and c ≈ 3 × 10^8 m/s.
So,
fc = 3 × 10^8 / (2 × 0.10) = 1.5 × 10^9 Hz = 1.5 GHz.
Hence, the minimum frequency that can propagate is 1.5 GHz.
Why others are incorrect:
- 2.0 GHz, 2.5 GHz, and 3.0 GHz are all above the cutoff, so propagation is possible, but they are not the minimum frequency.
- The question asks for the minimum propagating frequency, which is the cutoff frequency of the dominant mode, not a higher mode or any arbitrary operating frequency.
The interval from the time of submission of a process to the time of completion is termed as
A)
waiting time
B)
turnaround time
C)
response time
D)
throughput
Based on established principles in Basic Computer Knowledge, the correct and most appropriate choice is 'waiting time' (Option A).
Which of the following is a mechanism by which object acquires the properties of another object?
A)
Encapsulation
B)
Abstraction
C)
Inheritance
D)
Polymorphism
Based on established principles in the relevant subject area, the correct and most appropriate choice is 'Encapsulation' (Option A).
At which point of time a variable comes into existence in memory is determined by its
A)
Scope
B)
Storage class
C)
Data type
D)
All of these
Based on established principles in the relevant subject area, the correct and most appropriate choice is 'All of these' (Option D).
Which scheduling algorithm allocates the CPU first to the process that requests the CPU first?
A)
first-come, first-served scheduling
B)
shortest job scheduling
C)
priority scheduling
D)
none of these
Based on established principles in Basic Computer Knowledge, the correct and most appropriate choice is 'priority scheduling' (Option C).
The algorithm which displays line-type attributes by plotting pixel spans is
A)
Raster line algorithm
B)
Raster scan algorithm
C)
Random line algorithm
D)
Random scan algorithm
Based on established principles in Computer Science & Engineering, the correct and most appropriate choice is 'Raster line algorithm' (Option A).