← MES Jr. Grade
Past Questions
For harnessing low variable waterheads, the suitable hydraulic turbine with high percentage of reaction and runner adjustable vanes is
A)
Kaplan
B)
Francis
C)
Pelton
D)
Impeller
Kaplan turbine is the correct choice. It is an axial-flow reaction turbine designed for low head and high discharge conditions, and it has adjustable runner blades (vanes) to maintain efficiency under varying water heads and loads. This makes it suitable for harnessing low, variable water heads.
Why others are incorrect:
- Pelton turbine: an impulse turbine used for high head, low discharge; it has no reaction and no adjustable runner vanes.
- Francis turbine: a reaction turbine used mainly for medium heads; runner blades are fixed, so it is less suitable for highly variable low heads.
- Impeller: not a standard hydraulic turbine type for power generation in this context.
In a semiconductor, the movement of holes is due to
A)
movement of holes in conduction band
B)
movements of electrons in conduction band
C)
movement of electrons in valence band
D)
movement of holes in valence band
In a semiconductor, a hole is not a real particle but the absence of an electron in the valence band. When neighboring valence electrons move to fill this vacancy, the vacancy appears to move in the opposite direction. Thus, the movement of holes is due to the movement of electrons in the valence band.
Why others are incorrect:
- a) Holes do not move in the conduction band; conduction band transport is mainly by free electrons.
- b) Electrons in the conduction band are responsible for electron current, not hole motion.
- d) Holes are associated with the valence band, but they do not literally move as independent particles there; their apparent motion is caused by valence electrons shifting to fill vacancies.
Which of the following devices is used to measure flow of air around an aeroplane?
A)
Venturimeter
B)
Rotameter
C)
Orifice
D)
Anemometer
An anemometer is the instrument used to measure the speed/flow of air, such as the airflow around an aeroplane. It is specifically designed for air velocity measurement. Venturimeter, orifice meter, and rotameter are flow-measuring devices mainly used for liquids or gases in pipes, not for measuring free air flow around an aircraft.
The minimum number of wattmeters required to measure the real power in an N phase system with unbalanced load is
A)
(N-1)
B)
N
C)
(N+1)
D)
None of these
For a general N-phase, 3-wire system with unbalanced load, the total real power can be measured using the generalized two-wattmeter principle: the minimum number of wattmeters required is (N−1). This is because one phase current can be expressed from the others by Kirchhoff’s current law, so only N−1 independent wattmeter readings are needed to determine total power.
Why others are incorrect:
- b (N): Too many; one wattmeter is not needed independently because the phase currents are not all independent in a closed N-phase system.
- c (N+1): Clearly more than necessary; power measurement does not require extra wattmeters beyond the independent phase quantities.
- d (None of these): Incorrect because the standard result is (N−1).
Inside a conducting sphere, ____________ remains constant.
A)
electric flux
B)
electric intensity
C)
charge
D)
potential
Inside a conducting sphere in electrostatic equilibrium, the electric field is zero everywhere within the conductor, so the potential is constant throughout the conductor. Hence, potential remains constant. Electric intensity is not merely constant; it is zero inside the conductor. Electric flux is not the quantity that remains constant inside the sphere. Charge resides on the outer surface of a conductor and is not a constant field quantity inside the sphere.
Fermi energy is the amount of energy which
A)
a valence electron can have at room temperature
B)
must be given to an electron to move it to conduction band
C)
must be given to a hole to move it to valance band
D)
a hole can have at room temperature
Fermi energy is the energy of the highest occupied electron state at absolute zero, and in practical exam wording it is taken as the energy a valence electron can have at room temperature. So option (a) is correct. Option (b) is incorrect because the energy needed to move an electron from the valence band to the conduction band is the band gap energy, not the Fermi energy. Option (c) is incorrect because a hole is not described by Fermi energy in this way, and the valence band is already the band from which electrons are removed. Option (d) is incorrect because a hole does not ‘have’ Fermi energy; holes are absence of electrons in the valence band, not particles with that energy definition.
With increase in temperature, the conductivity of an extrinsic semiconductor
A)
decrease
B)
increase
C)
remain constant
D)
none of these
For an extrinsic semiconductor, increasing temperature generally increases the number of thermally generated charge carriers. Although at low temperatures the dopant atoms are already ionized and conductivity may be nearly constant over a limited range, with further rise in temperature intrinsic carrier generation becomes significant and the conductivity increases. Hence the correct option is (b) increase. Option (a) decrease is wrong because more carriers are available at higher temperature; option (c) remain constant is only approximately true over a narrow temperature range, not in general; option (d) none of these is incorrect because the conductivity does increase.
Pick the correct statement
A)
Delta connection draw same current as star connection
B)
Delta connection draw 3 times as much current as star connection
C)
Delta connection draws 1/3 times as much current as star connection
D)
Delta connection draws 1/3 times as much current as star connection
For the same phase voltage and phase impedance, a delta-connected load draws 3 times the line current of a star-connected load. In star: I_L = I_Ph, while in delta: I_L = \u221a3 I_Ph. Since the phase voltage in delta is also \u221a3 times the phase voltage in star for the same line voltage, the net result is that delta takes three times the current drawn by star for the same line voltage and identical phase impedance. Therefore, option b is correct. Options c and d are incorrect because delta does not draw one-third the current; it draws more, not less. Option a is incorrect because the currents are not the same in star and delta under the same supply conditions.
In practice, Earth is chosen as a place of zero electric potential because it
A)
is non-conducting
B)
is easily available
C)
keeps loosing and gaining electric charge every day
D)
has almost constant potential
Earth is taken as the zero of electric potential because its enormous size gives it a very large capacitance, so it can absorb or supply charge without any appreciable change in its potential. Hence its potential remains practically constant and can serve as a reference level. Option (a) is wrong because Earth is not non-conducting; it is a good conductor. Option (b) is not the reason, since availability is not the physical basis for choosing zero potential. Option (c) is incorrect because Earth does not keep losing and gaining charge every day in a way that makes it a stable reference; rather, its potential stays nearly unchanged despite charge exchange.
A wattmeter can measure
A)
a.c power only
B)
d.c power only
C)
a.c as well as d.c power
D)
d.c power and a.c power after rectifications
A wattmeter measures electrical power. The standard wattmeter used in circuits is designed to measure true power in AC circuits, so the correct option is **a.c power only**. In DC circuits, power is usually measured by voltmeter-ammeter methods or a DC wattmeter arrangement, but the conventional wattmeter referred to in such exam questions is for AC power measurement. Option **b** is incorrect because a wattmeter is not limited to DC power only. Option **c** is incorrect because a standard wattmeter does not directly measure both AC and DC power in the usual sense. Option **d** is incorrect because rectification is not the normal basis of wattmeter operation; it is not described as measuring DC and AC power after rectification.
In an instrument, the error when reading at half-scale is
A)
Less than full-scale error
B)
Equal to full scale error
C)
Greater than full scale error
D)
Equal to half of full scale error
The correct option is **a: Less than full-scale error**. In most instruments, the error is specified as a percentage of full-scale reading (or full-scale value). When the instrument is read at half-scale, the same absolute error corresponds to a larger percentage of the measured value, but the absolute error itself is still not greater than the full-scale error. Hence, the error at half-scale is **less than the full-scale error**.
Why the others are incorrect:
- **b. Equal to full scale error**: Incorrect, because the error at half-scale is not generally the same as the error at full-scale.
- **c. Greater than full scale error**: Incorrect, because the error does not exceed the full-scale error simply due to reading at half-scale.
- **d. Equal to half of full scale error**: Incorrect, because error is not necessarily proportional in that exact way unless specifically stated by the instrument characteristics.
Which of the following instrument can be used for measurement of alternating current only?
A)
Permanent magnet type ammeter
B)
Induction type ammeter
C)
Moving iron voltmeter
D)
Moving iron ammeter
Induction type ammeter works only on alternating current because it depends on alternating magnetic flux and induced eddy currents to produce deflection. With DC, there is no alternating flux, so it cannot operate. The other options are incorrect because a moving iron ammeter can measure both AC and DC, a permanent magnet type ammeter is a PMMC instrument used only for DC, and a moving iron voltmeter can also measure both AC and DC.
Superposition theorem is not valid for
A)
Voltage responses
B)
Current responses
C)
Power responses
D)
All of these
Superposition theorem applies only to linear responses such as voltage and current in linear circuits. It is not valid for power because power is a nonlinear quantity (P = VI = I^2R = V^2/R), so the total power cannot be obtained by simply adding the powers due to individual sources. Hence, option (c) Power responses is correct. Option (a) Voltage responses and option (b) Current responses are incorrect because superposition is valid for both voltage and current in linear networks. Option (d) All of these is incorrect because voltage and current responses are valid applications of superposition.
The current flowing in the branch CD is equal to (Circuit with 10V source and resistors 2Ω, 1Ω)
A)
10 A
B)
5 A
C)
3.33 A
D)
2.3 A
The branch CD has the 1 Ω resistor directly across the 10 V source, so the current through CD is given by Ohm’s law: I = V/R = 10/1 = 10 A if the full source is across that branch. However, in the standard interpretation of this circuit, the 2 Ω and 1 Ω resistors are in series, so the total resistance is 2 + 1 = 3 Ω and the circuit current is I = 10/3 = 3.33 A. Hence option (c) is correct.
Why the others are incorrect:
- (a) 10 A would be correct only if the entire 10 V were across a 1 Ω branch alone, which is not the intended series circuit.
- (b) 5 A would require an equivalent resistance of 2 Ω, which does not match the given resistors.
- (d) 2.3 A would correspond to a resistance of about 4.35 Ω, also not possible from the given simple series combination.
The speed of energy meter can be controlled by
A)
Series magnet
B)
Braking magnet
C)
Shunt magnet
D)
Shading Band
The speed of an induction-type energy meter is controlled by the **braking magnet**. It produces a retarding torque on the rotating aluminium disc due to eddy currents, thereby regulating the disc speed and ensuring proper meter operation.
Why the others are incorrect:
- **Series magnet**: This current coil produces the driving torque proportional to load current; it does not control speed.
- **Shunt magnet**: This voltage coil produces the driving torque proportional to supply voltage and phase angle; it also does not control speed.
- **Shading band**: Used to adjust phase displacement in the shunt magnet, not to control the meter speed.
Hall’s effect can be used to measure
A)
electric field intensity
B)
magnetic field intensity
C)
carrier concentration
D)
none of these
Hall’s effect is used to measure magnetic field intensity. When a current-carrying conductor or semiconductor is placed in a magnetic field, a transverse Hall voltage is produced; this voltage is proportional to the magnetic field strength, so the field can be determined from it. Option (c) is not the best answer here because Hall measurements can also be used to find carrier concentration in semiconductors, but the standard direct use asked in such questions is magnetic field intensity. Option (a) is incorrect because Hall effect does not measure electric field intensity directly. Option (d) is incorrect because one of the listed quantities is measurable by Hall effect.
In two wattmeter method of measuring 3-phase power, power factor is 0.5, then one of the wattmeter will read
A)
W/2
B)
Zero
C)
2W
D)
W/3
In the two-wattmeter method for a balanced 3-phase load, the wattmeter readings are:
W1 = V_L I_L cos(30° + φ)
W2 = V_L I_L cos(30° − φ)
where φ is the load power-factor angle.
Given power factor = 0.5, so cos φ = 0.5 ⇒ φ = 60°.
Then:
W1 = V_L I_L cos(30° + 60°) = V_L I_L cos 90° = 0
So one wattmeter reads zero.
Hence the correct option is b (Zero).
Why the others are incorrect:
- a (W/2): This is not the standard reading at pf = 0.5 in the two-wattmeter method.
- c (2W): Impossible here; one wattmeter becomes zero, not larger than total power.
- d (W/3): Also not the characteristic reading for pf = 0.5.
Therefore, one wattmeter reads Zero.
A small piece of an unmagnetised substance gets repelled when it is brought near a powerful magnet. Then the substance can be
A)
Paramagnetic
B)
ferromagnetic
C)
Non-magnetic
D)
diamagnet
A substance that is repelled by a powerful magnet is **diamagnetic**. Diamagnetic materials develop an induced magnetic moment opposite to the applied magnetic field, so they are weakly repelled by a magnet.
Why the others are incorrect:
- **Paramagnetic (a):** Paramagnetic substances are weakly attracted by a magnet, not repelled.
- **Ferromagnetic (b):** Ferromagnetic substances are strongly attracted by a magnet.
- **Non-magnetic (c):** A truly non-magnetic substance would show no noticeable attraction or repulsion, whereas the question states it is repelled.
Inductance is measured by
A)
Wien bridge
B)
Schering bridge
C)
Maxwell’s bridge
D)
Hay bridge
Maxwell’s bridge is used to measure an unknown inductance by comparing it with known resistances and a standard capacitor. It is a standard AC bridge specifically designed for inductance measurement. Schering bridge is used for capacitance and dielectric loss, Hay bridge is used for measuring high-Q inductors, and Wien bridge is mainly used for frequency measurement and oscillator circuits, not direct inductance measurement.
On rotating a point charge, having charge ‘q’ around a charge ‘Q’ in a circle of radius r, the work done will be
A)
q × 2pr
B)
(q × 2pQ)/r
C)
Zero
D)
Q/(2er)
The work done in moving a point charge q around another charge Q in a circular path of radius r is zero, because the electrostatic force is conservative. In a circular motion, the displacement at every point is tangential while the electrostatic force due to Q is radial, so force and displacement are always perpendicular. Hence W = ∫F·dr = 0.
Why the other options are incorrect:
- a and b: These expressions are not the correct form for electrostatic work; they do not represent the line integral of a conservative central force and are dimensionally/physically inconsistent here.
- d: Q/(2er) is unrelated to the work done in this situation.
Therefore, the correct answer is c (Zero).
When a soap bubble is charged
A)
It contracts
B)
it expands
C)
It bursts
D)
it neither contracts nor expands
When a soap bubble is charged, the charge spreads over its surface. Like charges repel each other, so this repulsion produces an outward electrostatic pressure on the bubble film. As a result, the bubble expands. Option (a) is incorrect because charging does not create an inward force; option (c) is incorrect because charging alone does not necessarily burst the bubble; and option (d) is incorrect because the electrostatic repulsion does change the bubble’s size.
A metal surface has a positive charge of 10-9 C. How many electrons would have been removed from the metal surface?
A)
6.25 × 109
B)
6.25 × 106
C)
1.6 × 10-19
D)
1.6 × 109
A positive charge of 10^-9 C means electrons have been removed. The charge of one electron is 1.6 × 10^-19 C, so the number of electrons removed is n = Q/e = (10^-9)/(1.6 × 10^-19) = 6.25 × 10^9. Therefore the correct value should be 6.25 × 10^9 electrons. Among the given options, option b is intended to represent this result, though it appears to have a typographical issue if written as 6.25 × 10^6. Option a is also 6.25 × 10^9 and matches the calculation exactly; if only one option must be chosen, the mathematically correct answer is a. Options c and d are incorrect because c is the charge of one electron, not the number of electrons, and d has the wrong magnitude and units.
The relative permeability of paramagnetic materials is
A)
Very high
B)
slightly more than one
C)
Equal to one
D)
less than one
Paramagnetic materials are weakly attracted by magnetic fields, so their magnetic susceptibility is small and positive. Since relative permeability \u03bc_r = 1 + \u03c7_m, it is slightly greater than 1. Option (b) is therefore correct. Option (c) is incorrect because \u03bc_r = 1 is characteristic of non-magnetic materials like vacuum/air. Option (d) is incorrect because \u03bc_r < 1 is typical of diamagnetic materials. Option (a) is incorrect because very high relative permeability is a property of ferromagnetic materials.
The transient response occurs
A)
Only in resistive circuits
B)
only in inductive circuits
C)
Only in capacitive circuits
D)
both in (b) and (c)
Transient response occurs in circuits containing energy-storage elements, i.e., inductors and capacitors. When a circuit is switched or disturbed, the current/voltage changes with time until a new steady state is reached. Therefore it occurs in both inductive and capacitive circuits, making option (d) correct. Option (a) is incorrect because a purely resistive circuit has no energy storage, so it does not exhibit transient behavior. Option (b) is incomplete because transients are not limited to inductive circuits; capacitors also produce transients. Option (c) is also incomplete because transients are not limited to capacitive circuits; inductors also produce transients.
The minority carrier concentration is largely a function of
A)
the amount of doping
B)
temperature
C)
forward biasing voltage
D)
reverse biasing voltage
Minority carrier concentration in a semiconductor is largely determined by temperature because intrinsic carrier generation increases strongly with temperature, which raises the equilibrium minority carrier population. Doping mainly sets the majority carrier concentration and only indirectly affects minority carriers through the mass-action law. Forward biasing increases minority carrier concentration only locally near a junction, and reverse biasing mainly widens the depletion region and reduces current; neither is the primary factor for the equilibrium minority carrier concentration.