← MES Jr. Grade
Past Questions — 2012
Phantom loading for testing of energy meters is used
A)
to isolate the current and potential circuits
B)
to reduce power loss during testing
C)
for meters having low current ratings
D)
to test meters having a large current rating for which loads may not be available in the laboratory
Phantom loading is used when an energy meter has to be tested at high current, but supplying the full load power would cause excessive power loss and require a very large load in the laboratory. In phantom loading, the current coil is supplied with the required current while the pressure coil is supplied separately, so the meter is tested under rated conditions without wasting large power. Hence, it is mainly used for meters having a large current rating for which suitable loads may not be available in the laboratory. Option (b) is a consequence, but the main purpose is as stated in (d).
A PMMC meter has an internal resistance 200Ω and the current required for its full-scale deflection is 50µA. The meter is capable of measuring, on its own, a maximum voltage of
A)
5mV
B)
10mV
C)
5µV
D)
10µV
For a PMMC meter, the maximum voltage it can measure on its own is given by V = I × R. Here, full-scale current I = 50 µA = 50 × 10^-6 A and internal resistance R = 200 Ω. So, V = 50 × 10^-6 × 200 = 0.01 V = 10 mV. Hence, the correct answer is 10 mV. The other options are incorrect because they do not match the product of the given current and resistance.
Two parallel conductors are carrying currents in the same direction. Which of the following statements is correct?
A)
There is a repelling force between the conductors
B)
The force increases when the current is decreased
C)
There is an attracting force between the conductors
D)
The force increases when the distance between the conductors is increased
Two parallel current-carrying conductors exert magnetic forces on each other. If the currents are in the same direction, the magnetic fields interact so that the conductors attract each other. If the currents were in opposite directions, they would repel. Also, the force is proportional to the product of currents and inversely proportional to the distance between them, so options (b) and (d) are incorrect.
Two parallel wires separated by a distance d are carrying a current I in the same direction. The magnetic field along a line running parallel to these wires and midway between them
A)
depends upon I
B)
is zero
C)
depends upon d
D)
depends upon the permeability of the medium between the wires
For a long straight current-carrying wire, the magnetic field at a point is proportional to I and inversely proportional to distance from the wire. Here the two parallel wires carry equal currents in the same direction, and the point on the line midway between them is at equal distance from both wires. By the right-hand rule, the magnetic fields produced by the two wires at the midpoint are equal in magnitude but opposite in direction, so they cancel out. Therefore, the net magnetic field along that midway line is zero.
The power consumption of a d.c. voltmeter using a direct coupling amplifier when measuring a voltage of 0.5V is of the order of
A)
a few watt
B)
a few milliwatt
C)
a few microwatt
D)
a few nanowatt
A d.c. voltmeter using a direct-coupled amplifier is designed to have very high input resistance, so it draws extremely small current from the circuit. For a low measured voltage like 0.5 V, the power consumed is therefore very small, typically of the order of microwatts. Watt or milliwatt levels would be far too high for a sensitive electronic voltmeter, and nanowatt is usually too low for this type of instrument.
With the increase in cross-sectional area of the conductor, the value of resistance
A)
increases
B)
remains same
C)
decreases
D)
none of these
Resistance of a conductor is given by R = ρL/A, where ρ is resistivity, L is length, and A is cross-sectional area. Since resistance is inversely proportional to area, increasing the cross-sectional area decreases the resistance. Hence, option (c) is correct. Options (a) and (b) are incorrect because resistance does not increase or remain unchanged when area increases.
A wave travelling in x-direction has Ey and Ex components at 45º phase apart. The wave is
A)
linearly polarized
B)
elliptically polarized
C)
circularly polarized
D)
horizontally polarized
For an electromagnetic wave, if two perpendicular electric field components are present with a phase difference other than 0° or 180°, the wave is generally elliptically polarized. Here, Ex and Ey are 45° phase apart, so the tip of the electric field vector traces an ellipse. Linear polarization requires zero phase difference, and circular polarization requires equal amplitudes with a 90° phase difference. Hence, the correct answer is elliptically polarized.
In a venturi meter the flow is 0.15 m3/s when the differential pressure is 30 KN/m2. The value of flow when the differential pressure is 60 KN/m2 is
A)
0.212 m3/s
B)
0.106 m3/s
C)
0.3 m3/s
D)
0.075 m3/s
For a venturi meter, discharge is proportional to the square root of differential pressure: Q ∝ √ΔP. Given Q1 = 0.15 m3/s at ΔP1 = 30 kN/m2, for ΔP2 = 60 kN/m2: Q2 = Q1 × √(ΔP2/ΔP1) = 0.15 × √(60/30) = 0.15 × √2 ≈ 0.212 m3/s. Hence, option (a) is correct.
F.M. systems as compared to A.M. systems
A)
are equally effected by noise
B)
are less effected by noise
C)
are more effected by noise
D)
are highly effected by noise as they operate at VHF and UHF
FM (frequency modulation) is less affected by noise than AM (amplitude modulation) because in FM the information is carried by frequency variations, while most noise mainly disturbs amplitude. FM receivers can also use amplitude limiters to remove noise-induced amplitude changes. Hence, compared to AM systems, FM systems are less affected by noise. Options (a), (c), and (d) are incorrect.
A current measuring instrument can be used to measure voltage by
A)
adding resistance in series
B)
adding resistance in parallel
C)
using larger wire in the coil
D)
using a high resistance shunt
A current-measuring instrument (ammeter) can be converted into a voltmeter by connecting a high resistance in series with it. This series resistance is called a multiplier and limits the current so the instrument can measure voltage safely. Adding resistance in parallel or using a shunt is used for extending current measurement, not voltage measurement.
The velocity of a travelling electromagnetic wave in free space is given by
A)
µ0ε0
B)
√ µ0ε0
C)
1/ µ0ε0
D)
1/√µ0ε0
For an electromagnetic wave in free space, the speed is given by c = 1/√(μ0ε0). This comes from Maxwell’s equations. Hence the correct option is (d). Options (a), (b), and (c) are dimensionally incorrect for velocity.
The direction of the magnetic lines of force is
A)
from positive to negative charges
B)
from south to north poles
C)
from one end of the magnet to the other
D)
from north pole to the south pole
Magnetic field lines outside a magnet emerge from the north pole and enter the south pole. So the direction of magnetic lines of force is from north pole to the south pole. Option (a) refers to electric field lines, (b) is the opposite of the correct magnetic direction, and (c) is too vague and not the standard scientific description.
In wire wound strain gauges, the change in resistance on application of strain is mainly due to
A)
change in length of wire
B)
change in diameter of wire
C)
change in both length and diameter of wire
D)
change in resistivity
In a wire-wound strain gauge, when strain is applied, the wire elongates and its cross-sectional area decreases. Since resistance R = ρL/A, the resistance changes mainly because both the length increases and the diameter (hence area) decreases. The resistivity is assumed nearly constant for the mechanical strain effect, so option (d) is not the main reason.
A differential transformer is a
A)
variable inductance transducer
B)
variable pressure transducer
C)
constant displacement transducer
D)
constant pressure transducer
A differential transformer, commonly known as an LVDT (Linear Variable Differential Transformer), works on the principle of mutual induction and changes its output voltage with the displacement of the core. Since its inductance varies with position, it is a variable inductance transducer. The other options are incorrect because it is not related to pressure measurement, and it is not a constant displacement or constant pressure device.
Piezo-electric crystals are used for the measurement of
A)
temperature
B)
velocity
C)
sound
D)
none of these
Piezo-electric crystals generate an electrical signal when subjected to mechanical stress and are widely used in microphones, ultrasonic transducers, and sound/vibration measurement. Hence they are used for measurement of sound. Temperature is measured by thermocouples/RTDs, and velocity is not the standard direct use here.
Relative permeability of vacuum is
A)
4π x 10-7/h/m
B)
1 H/m
C)
1
D)
1/4 π
Relative permeability (μr) is defined as the ratio of the permeability of a medium to the permeability of free space: μr = μ/μ0. For vacuum, μ = μ0, so μr = 1. Option (a) is the permeability of free space (μ0 = 4π × 10^-7 H/m), not relative permeability. Option (b) is incorrect because 1 H/m is not the permeability of vacuum. Option (d) is also incorrect.
A uniform plane wave travelling in a perfect dielectric is incident normally on the surface of a perfect conductor. Then
A)
the wave is transmitted into the conductor without attenuation
B)
50% of the incident wave is transmitted and 50% is reflected
C)
a standing wave is set up in the conducting medium
D)
a standing wave is set up in the dielectric
For a perfect conductor, the tangential electric field at the surface must be zero. So when a uniform plane wave in a perfect dielectric is incident normally on it, the wave cannot be transmitted into the conductor. The reflected wave has equal amplitude and opposite phase, and the incident and reflected waves superpose in the dielectric to form a standing wave. Hence, option (d) is correct. Options (a) and (b) are wrong because a perfect conductor does not allow transmission. Option (c) is wrong because the standing wave is formed in the dielectric region, not inside the conductor.
When the load impedance is equal to the characteristic impedance of the transmission lines, then the reflection coefficient and standing wave ratio are respectively
A)
0 and 0
B)
1 and 0
C)
0 and 1
D)
1 and 1
When the load impedance equals the characteristic impedance (ZL = Z0), the line is perfectly matched. Therefore, no wave is reflected, so the reflection coefficient Γ = 0. Also, the standing wave ratio (SWR) becomes 1, which is the minimum possible value. Hence the correct answer is 0 and 1.
Electromagnetic waves may be diffracted
A)
by reflection from the ground
B)
when encountered with a spherical wave front
C)
around the edge of a sharp obstacle
D)
while passing through a large slot
Diffraction is the bending/spreading of waves when they pass around an obstacle or through an aperture. Electromagnetic waves can therefore be diffracted around the edge of a sharp obstacle. Reflection from the ground is reflection, not diffraction. A spherical wave front is just a wave shape, not a cause of diffraction. Passing through a large slot may produce only negligible diffraction because diffraction is significant when the opening is comparable to the wavelength.
Creeping error in single-phase energy meter can be minimized by
A)
adjusting the braking magnet
B)
using short circuited loops on the outer limbs of the shunt magnet
C)
drilling two holes in the disc on the opposite side of the spindle
D)
adjusting the shading band
Creeping is the slow continuous rotation of the energy meter disc at no-load or very light load. It is commonly minimized by making two opposite holes/slots in the disc near the rim, so the disc tends to stop when one of these holes comes under the driving system. Hence option (c) is correct. Option (a) relates to braking torque adjustment, (b) is used in induction wattmeters/meters for phase compensation, and (d) is for shading band adjustment, not creeping prevention.
The order of minimum displacement that can be measured with capacitive transducers is
A)
1cm
B)
1mm
C)
1µm
D)
1 × 10-12 m
Capacitive transducers are highly sensitive and are commonly used for very small displacement measurements. Their minimum measurable displacement is typically of the order of micrometres. Hence, 1 µm is the correct order. 1 cm and 1 mm are too large, while 1 × 10^-12 m is far beyond the practical range of ordinary capacitive transducers.
Which of the following transducers is used for transmitting as well as receiving the acoustic energy in an ultrasonic flow meter?
A)
LVDT
B)
RTD
C)
Piezo-electric crystals
D)
Strain gauge
In an ultrasonic flow meter, the same transducer is used to generate ultrasonic waves and to detect the received acoustic energy. Piezo-electric crystals are ideal for this because they convert electrical energy into mechanical vibrations and vice versa. LVDT measures displacement, RTD measures temperature, and strain gauge measures strain, so they are not used for transmitting/receiving ultrasonic energy.
In a reed type frequency meter, all the reeds
A)
have the same natural frequency
B)
have different natural frequency
C)
have different natural frequency with a specific difference of the frequencies of adjacent reeds
D)
none of the above
A reed type frequency meter contains a set of reeds, each tuned to a slightly different natural frequency. The reeds are arranged so that adjacent reeds have a specific small frequency difference, allowing the reed whose natural frequency is closest to the supply frequency to vibrate the most. Hence, the correct choice is (c). Option (a) is wrong because all reeds cannot have the same natural frequency, and option (b) is incomplete because the reeds are not just different randomly; they are spaced by a specific frequency interval.
“In all cases of electromagnetic induction, an induced voltage will cause a current to flow in a closed circuit in such a direction that the magnetic field which is caused by the current will oppose the change that produces the current” is the original statement of
A)
Lenz’s law
B)
Faraday’s law of magnetic induction
C)
Fleming’s law
D)
Ampere’s law
The statement describes the direction of induced current: it always flows so that the magnetic field produced by it opposes the change in magnetic flux that caused the induction. This is exactly Lenz’s law. Faraday’s law gives the magnitude of induced emf, Fleming’s law is used for direction in motors/generators, and Ampere’s law relates magnetic field to current.
The magnitude of statically induced e.m.f. depends on
A)
the magnitude of flux
B)
the rate of change of flux
C)
the resistance of the coil
D)
none of these
A statically induced e.m.f. is produced when the magnetic flux linking a stationary coil changes with time. By Faraday’s law, the induced e.m.f. is proportional to the rate of change of flux linkage, not to the flux magnitude itself or the coil resistance. Therefore, the correct answer is (b).