← MES Jr. Grade
Past Questions — 2024
In a network made up of linear resistors and ideal voltage sources, values of all resistors are doubled. Then the voltage across each resistor is
A)
doubled
B)
halved
C)
decreased four times
D)
not changes
To solve this problem, we analyze the network using the principles of linear circuit theory.
1. **Network Composition**: The network consists of linear resistors (R) and ideal voltage sources (V).
2. **Nodal Analysis/Voltage Division**: In any linear network, the voltage across a specific resistor is determined by the ratio of that resistor to the total equivalent resistance of the network. For example, in a simple series circuit with two resistors R1 and R2 and a source V, the voltage across R1 is: V1 = V * [R1 / (R1 + R2)].
3. **Scaling the Resistors**: If all resistors in the network are doubled, R1 becomes 2R1 and R2 becomes 2R2. The new voltage V1' becomes: V1' = V * [2R1 / (2R1 + 2R2)].
4. **Simplification**: Factoring out the 2 from the numerator and denominator: V1' = V * [2 * R1 / (2 * (R1 + R2))] = V * [R1 / (R1 + R2)].
5. **Conclusion**: The factor of 2 cancels out. Since all resistors are scaled by the same constant, the relative distribution of voltage (the voltage division ratio) remains identical. Therefore, the voltage across each resistor remains unchanged, although the current flowing through the network will be halved (since I = V/R and R is doubled).
- Option (a) is incorrect because voltage depends on the ratio of resistances, not the absolute value.
- Option (b) is incorrect; the current is halved, but the voltage remains the same.
- Option (c) is incorrect as there is no quadratic relationship affecting the voltage in this linear scenario.
At a temperature zero Kelvin, silicon material behaves as
A)
superconductor
B)
conductor
C)
insulator
D)
semiconductor
Step-by-step reasoning: 1. Silicon is an intrinsic semiconductor. In a semiconductor, the valence band is completely filled and the conduction band is empty at absolute zero (0 Kelvin). 2. For electrical conduction to occur, electrons must gain enough energy to jump across the energy gap (band gap) from the valence band to the conduction band. 3. At 0 K, there is no thermal energy available to excite electrons. Therefore, no electrons can move into the conduction band, and no holes are created in the valence band. 4. Since there are no free charge carriers (electrons or holes) available to move under the influence of an electric field, the material cannot conduct electricity. 5. A material that does not conduct electricity is defined as an insulator. Thus, at 0 K, silicon behaves as a perfect insulator. Other options: (a) Superconductors have zero electrical resistance, which is not the case for silicon; (b) Conductors have overlapping bands or partially filled bands; (d) While silicon is a semiconductor by nature, its 'behavior' at 0 K specifically is that of an insulator.
The breakdown voltage of air at normal temperature and pressure is
A)
20kV/cm
B)
30kV/cm
C)
60kV/cm
D)
90kV/cm
The dielectric strength (breakdown voltage) of a material is the maximum electric field that the material can withstand without experiencing electrical breakdown. For air at standard temperature and pressure (STP), the dielectric strength is widely accepted in electrical engineering textbooks and standards as approximately 30 kV per centimeter (or 3 kV/mm).
- Option (a) 20kV/cm is too low for standard dry air.
- Option (b) 30kV/cm is the standard theoretical and practical value used for calculations in high voltage engineering.
- Option (c) 60kV/cm and (d) 90kV/cm are significantly higher than the breakdown threshold of air at normal atmospheric pressure.
Which parameter represents the relationship between output current and output voltage in a two-port network?
A)
Z—Parameter
B)
Y—Parameter
C)
H—Parameter
D)
G—Parameter
In network theory, two-port network parameters are used to describe the relationship between the voltages and currents at the input and output ports.
1. **Y-Parameters (Admittance Parameters):** These are defined by expressing the currents ($I_1$ and $I_2$) as functions of the voltages ($V_1$ and $V_2$). The equations are:
$I_1 = Y_{11}V_1 + Y_{12}V_2$
$I_2 = Y_{21}V_1 + Y_{22}V_2$
Here, the output current ($I_2$) is expressed as a function of the input voltage ($V_1$) and output voltage ($V_2$). Specifically, $Y_{22}$ represents the relationship between the output current and output voltage when the input is open-circuited. Thus, Y-parameters fundamentally relate currents to voltages.
2. **Z-Parameters (Impedance Parameters):** These express voltages as functions of currents ($V = ZI$). They represent the relationship between output voltage and output current (the inverse of the question's requirement).
3. **H-Parameters (Hybrid Parameters):** These are a mix, expressing $V_1$ and $I_2$ as functions of $I_1$ and $V_2$. While they involve output current and voltage, they are specifically designed for transistor modeling rather than a general current-to-voltage relationship for the output port.
4. **G-Parameter:** This is not a standard primary two-port network parameter in the same category as Z, Y, H, and ABCD parameters.
Therefore, the parameter that represents the relationship where current is the dependent variable and voltage is the independent variable (specifically for the output) is the Y-parameter.
The potential at a point due to a charge is 9 V. If the distance from the charge is increased 3 times, the potential at that point will be
A)
3V
B)
9V
C)
18V
D)
27V
1. The electric potential (V) at a distance (r) from a point charge (q) is given by the formula: V = k * q / r, where k is the electrostatic constant.
2. From this formula, we can see that the potential is inversely proportional to the distance (V ∝ 1/r).
3. Let the initial potential be V1 = 9 V and the initial distance be r1.
4. According to the problem, the new distance r2 = 3 * r1.
5. The new potential V2 can be calculated using the ratio: V2 / V1 = r1 / r2.
6. Substituting the values: V2 / 9 = r1 / (3 * r1) = 1/3.
7. Therefore, V2 = 9 / 3 = 3 V.
Comparing this to the options:
- Option (a) 3V is correct.
- Option (b) 9V would imply the distance didn't change.
- Option (c) 18V and (d) 27V would imply the distance decreased, which contradicts the problem statement.
‘Creep’ in energy meters can be prevented by
A)
using extra turns on the voltage coil
B)
having two holes on opposite sides of the disc
C)
using a stronger brake magnet
D)
using steel laminations of high permeability
Creep in an induction-type energy meter is a phenomenon where the disc continues to rotate slowly even when there is no load current flowing (i.e., the current coil is not energized). This happens due to small stray magnetic fields, friction imbalances, or over-compensation of the brake magnet.
To prevent creep, two holes are drilled on opposite sides of the aluminum disc. When the disc is at rest, these holes create an asymmetry in the eddy current distribution and the resulting torque. This creates a 'dead band' or a stable equilibrium position where the driving torque is insufficient to overcome the frictional torque and the magnetic imbalance caused by the holes. Consequently, the disc will not rotate unless a minimum load current is applied.
Analysis of other options:
- Option (a): Extra turns on the voltage coil would increase the driving torque from the voltage coil, potentially worsening the creep.
- Option (c): While a stronger brake magnet increases the braking torque, it does not eliminate the possibility of creep and would affect the overall calibration/accuracy of the meter.
- Option (d): High permeability laminations are used to reduce hysteresis and eddy current losses in the core, but they do not address the mechanical/magnetic instability that causes creep.
Which of the following represents the Maxwell’s continuity equation for time varying electromagnetic fields
A)
∇·J + ∂ρ/∂t = 0
B)
∇·J = ρ
C)
∇×E = 0
D)
None of these
The continuity equation is a mathematical statement of the law of conservation of charge. It states that the current flowing out of a closed surface (divergence of current density J) must be equal to the rate of decrease of the charge enclosed within that surface.
Mathematically, the net current flowing out of a volume is given by the volume integral of the divergence of current density: ∫(∇·J)dV. According to the conservation of charge, this must equal the negative rate of change of total charge Q: -dQ/dt. Since Q = ∫ρdV, we have ∫(∇·J)dV = -∂/∂t ∫ρdV. For this to hold for any arbitrary volume, the integrands must be equal: ∇·J = -∂ρ/∂t, which rearranges to ∇·J + ∂ρ/∂t = 0.
Analysis of other options:
- Option (b) ∇·J = ρ is incorrect; it does not represent a standard physical law in electromagnetics.
- Option (c) ∇×E = 0 represents the electrostatic condition (where the electric field is conservative), but in time-varying fields, ∇×E = -∂B/∂t (Faraday's Law).
A wire conductor of resistance 1 Ω is doubled in length. Its resistance becomes
A)
the same as that when cold
B)
higher than cold resistance
C)
lower than cold resistance
D)
higher or lower than the cold resistance depending on the supply frequency
The resistance (R) of a conductor is governed by the formula R = ρ(L/A), where ρ is the resistivity of the material, L is the length, and A is the cross-sectional area.
1. According to the formula, resistance is directly proportional to length (R ∝ L).
2. If the length of the wire is doubled (L becomes 2L), the resistance will also double (R becomes 2Ω), assuming the cross-sectional area and material remain constant.
3. Since 2 Ω is greater than the original 1 Ω (which represents the initial or 'cold' state of the wire), the resistance becomes higher.
Regarding the options:
- Option (a) is incorrect because resistance changes with length.
- Option (b) is correct because doubling the length increases the resistance.
- Option (c) is incorrect because resistance increases, not decreases, with length.
- Option (d) is incorrect because DC resistance of a wire is independent of supply frequency.
If a dynamometer type wattmeter is connected in an AC circuit, the power indicated by the wattmeter will be
A)
volt-ampere product
B)
average power
C)
peak power
D)
instantaneous power
A dynamometer-type wattmeter consists of two coils: a fixed current coil (connected in series with the load) and a movable pressure coil (connected in parallel with the load). The torque produced in the instrument is proportional to the product of the current flowing through the current coil and the voltage across the pressure coil. In an AC circuit, the instantaneous torque varies sinusoidally; however, due to the inertia of the moving system, the pointer cannot follow these rapid fluctuations and instead settles at a position representing the time-average of the instantaneous power. Mathematically, the power indicated is P = Vrms * Irms * cos(φ), where cos(φ) is the power factor. This value is the 'average power' (also known as real power or active power).
Why other options are incorrect:
- (a) Volt-ampere product: This refers to Apparent Power (S = V * I), which is measured by a power analyzer or calculated, not by a standard wattmeter.
- (c) Peak power: This is the maximum instantaneous value of power, which the instrument's inertia filters out.
- (d) Instantaneous power: This varies 100/120 times per second in a 50/60Hz system; a mechanical pointer cannot track this.
The Q factor of a series RLC circuit increase if
A)
R increases
B)
R decreases
C)
applied voltage increases
D)
current decreases
The Quality factor (Q factor) of a series RLC circuit is a dimensionless parameter that describes how under-damped an oscillator is. It is defined as the ratio of the resonant frequency to the bandwidth, or more fundamentally, the ratio of the energy stored to the energy dissipated per cycle.
Mathematically, for a series RLC circuit, the Q factor is given by the formula:
Q = (1/R) * √(L/C)
From this formula, we can analyze the relationship between Q and the circuit components:
1. Resistance (R): Q is inversely proportional to R (Q ∝ 1/R). Therefore, if the resistance R decreases, the Q factor increases. Conversely, if R increases, the Q factor decreases.
2. Inductance (L) and Capacitance (C): Q increases if L increases or if C decreases.
Evaluating the options:
- Option (a) R increases: This would decrease the Q factor.
- Option (b) R decreases: This increases the Q factor (Correct).
- Option (c) applied voltage increases: The Q factor is a property of the circuit components (L, C, R) and is independent of the input voltage.
- Option (d) current decreases: The Q factor is a structural property of the circuit and does not depend on the instantaneous current flowing through it.
Which of the following magnetic materials have small area of hysteresis loop?
A)
Saturable magnetic material
B)
Soft magnetic material
C)
Hard magnetic material
D)
Diamagnetic material
The area of a hysteresis loop represents the energy loss (hysteresis loss) per unit volume per cycle of magnetization.
1. **Soft Magnetic Materials**: These materials are characterized by easy magnetization and demagnetization. They have low coercivity (the intensity of the magnetic field required to reduce magnetization to zero) and low retentivity. Consequently, the hysteresis loop is narrow, meaning the area is small, resulting in low energy loss. Examples include soft iron and permalloy. These are ideal for transformer cores and electromagnets.
2. **Hard Magnetic Materials**: These materials are difficult to magnetize and demagnetize. They have high coercivity and high retentivity, resulting in a wide hysteresis loop with a large area. This indicates high energy loss per cycle. Examples include Alnico and ferrite. These are used to make permanent magnets.
3. **Saturable Magnetic Materials**: This refers to a state of magnetization rather than a specific class of material defined by loop area. All ferromagnetic materials eventually saturate.
4. **Diamagnetic Materials**: These materials do not exhibit a hysteresis loop because they do not retain magnetization after the external field is removed; they show a very weak, negative response to magnetic fields.
At what temperature does the resistivity of mercury completely disappear?
A)
4.2°K
B)
12.5°K
C)
22.2°K
D)
55.2°K
Superconductivity is a phenomenon where the electrical resistivity of a material drops to zero when cooled below a characteristic temperature known as the critical temperature (Tc). Mercury (Hg) was the first material in which superconductivity was discovered by Heike Kamerlingh Onnes in 1911. He observed that the resistivity of mercury completely disappeared at a critical temperature of approximately 4.15 K to 4.2 K.
Analysis of options:
- Option (a) 4.2°K: This is the scientifically accepted critical temperature for mercury.
- Option (b) 12.5°K: This does not correspond to the Tc of mercury.
- Option (c) 22.2°K: This does not correspond to the Tc of mercury.
- Option (d) 55.2°K: This is far above the critical temperature of mercury; mercury is a Type-I superconductor with a very low Tc.
When two 2-port network are connected in parallel, it is convenient to use
A)
z-parameters
B)
y-parameters
C)
h-parameters
D)
inverse-h parameters
In network theory, different 2-port parameters are used depending on how networks are interconnected.
1. **Parallel Connection**: When two networks are connected in parallel, the voltage across the input ports is the same, and the total current is the sum of the individual currents. The Admittance parameters (y-parameters) are defined by the equations: I1 = y11V1 + y12V2 and I2 = y21V1 + y22V2. Because currents add algebraically when voltages are common, the equivalent y-matrix of two parallel networks is simply the sum of their individual y-matrices ([Y]total = [Y]1 + [Y]2). This makes y-parameters the most convenient for parallel connections.
2. **Series Connection**: When networks are connected in series, the currents are the same and voltages add. In this case, Impedance parameters (z-parameters) are most convenient ([Z]total = [Z]1 + [Z]2).
3. **Cascade Connection**: When networks are connected in cascade (output of one to input of another), Transmission parameters (ABCD parameters) are used, as the overall matrix is the product of individual matrices.
4. **h-parameters and inverse-h parameters**: These are typically used for transistor modeling (hybrid configurations) and are not the primary choice for simple parallel or series combinations of networks.
Inductance is measured in terms of capacitance and resistance by using
A)
Schering bridge
B)
Anderson bridge
C)
Maxwell-Wien bridge
D)
Wien bridge
The question asks for a bridge used to measure inductance in terms of capacitance and resistance.
1. **Anderson Bridge**: This is a modification of the Maxwell-Wien bridge. It is specifically designed to measure self-inductance in terms of a standard capacitor (C) and resistors (R). It is highly accurate because it allows the use of a fixed capacitor and variable resistors to balance the bridge, making it the standard choice for measuring inductance using capacitance.
2. **Schering Bridge**: This bridge is primarily used for the measurement of capacitance and the dissipation factor (dielectric loss) of a capacitor, not for measuring inductance.
3. **Maxwell-Wien Bridge**: While this bridge measures inductance, it does so by comparing the unknown inductance with a known standard inductor (not primarily in terms of a capacitor) or using a capacitor in a way that is less precise for a wide range of values compared to the Anderson bridge. The Anderson bridge was developed specifically to improve upon the Maxwell-Wien bridge by allowing the use of a smaller, fixed capacitor.
4. **Wien Bridge**: This bridge is used for the measurement of frequency (often used in audio oscillators) and the measurement of capacitance/resistance in specific RC circuits, but not for measuring inductance.
How many junction points are there in the circuit shown below?
A)
2
B)
3
C)
4
D)
7
A junction point is a point in a circuit where three or more branches meet. In the given circuit, there are 3 such junction points. Therefore, the correct answer is option (b) 3. The other options are incorrect because 2 and 4 do not match the actual count of multi-branch meeting points, and 7 is too high since not every connection or terminal in the circuit is a junction point.
The driving point impedance function Z(s)= s/(s^2+1) can be realized as a
A)
R-C network
B)
R-L network
C)
L-C network
D)
R-L-C network
To determine the type of network that can realize the driving point impedance function Z(s) = s / (s^2 + 1), we analyze the properties of the function based on network synthesis theory:
1. **Analysis of the Function**: The given function is Z(s) = s / (s^2 + 1).
2. **Lossless Network Condition**: A network is lossless (consisting only of Inductors (L) and Capacitors (C)) if its driving point impedance Z(s) is a lossless function. A function is lossless if:
- It is a rational function of s.
- All its poles and zeros are simple and lie on the imaginary axis (jω axis).
- The poles and zeros alternate between the origin and infinity along the imaginary axis.
3. **Checking the Poles and Zeros**:
- **Zeros**: Set the numerator to zero: s = 0. There is a simple zero at the origin.
- **Poles**: Set the denominator to zero: s^2 + 1 = 0 => s = ±j1. There are simple poles on the imaginary axis at s = j and s = -j.
4. **Alternation Property**:
- Zero at s = 0
- Pole at s = ±j1
- (The next zero would be at infinity)
Since the poles and zeros are simple, lie on the imaginary axis, and alternate, Z(s) is a lossless function.
5. **Conclusion**: A lossless driving point impedance function can be realized using only L and C elements. If the function had real parts in the poles/zeros or coefficients that implied energy dissipation, it would require resistors (R). Since this function is purely imaginary for s = jω (Z(jω) = jω / (1 - ω^2)), it represents a purely reactive network.
Therefore, the network is an L-C network.
A superconducting material when placed in a magnetic field will
A)
not influence the magnetic field
B)
repel all the magnetic lines of force passing through it
C)
attract the magnetic field towards its centre
D)
attract the magnetic field but transfer it into a concentrated zone
The phenomenon described in the question is known as the Meissner Effect. When a material transitions into the superconducting state (below its critical temperature Tc), it doesn't just become a perfect conductor; it becomes a perfect diamagnet. According to the Meissner Effect, a superconductor expels all magnetic flux from its interior. This is achieved by the spontaneous creation of surface currents that generate a magnetic field exactly equal and opposite to the applied external magnetic field, resulting in a net magnetic field of zero (B = 0) inside the bulk of the material. Therefore, the material repels all magnetic lines of force passing through it.
Analysis of other options:
- Option (a) is incorrect because the superconductor actively alters the field by expelling it.
- Option (c) is incorrect because attracting the field would imply paramagnetism or ferromagnetism, which is the opposite of the Meissner effect.
- Option (d) is incorrect because the field is expelled from the interior, not concentrated within it.
Strain gauge LVDT and thermocouple may be classified as
A)
Active transducers
B)
Analogue transducers
C)
Primary transducers
D)
None of the above
To determine the correct classification, let's analyze the three devices mentioned:
1. **Strain Gauge**: It converts mechanical strain into a change in electrical resistance. The output is a continuous electrical signal (voltage/current) proportional to the strain.
2. **LVDT (Linear Variable Differential Transformer)**: It converts linear motion into a continuous AC voltage signal.
3. **Thermocouple**: It converts temperature differences into a continuous DC voltage (Seebeck effect).
Analysis of Options:
- **Active Transducers (Option a)**: An active transducer generates its own electrical output without an external power source (e.g., Thermocouple). However, a Strain Gauge and LVDT are **passive transducers** because they require an external excitation voltage to operate. Since not all three are active, this option is incorrect.
- **Analogue Transducers (Option b)**: All three devices produce a continuous output signal that varies linearly or proportionally with the input physical quantity. They do not produce discrete pulses or digital bits. Therefore, they are all classified as analogue transducers. This is the correct option.
- **Primary Transducers (Option c)**: A primary transducer is the first element in a measurement chain that senses the physical quantity. While a thermocouple is a primary transducer, an LVDT is often used as a secondary transducer (converting a displacement provided by another element into voltage). This classification is not universally applicable to all three in the same way 'analogue' is.
Conclusion: The common characteristic shared by the strain gauge, LVDT, and thermocouple is that they all provide an analogue output signal.
A two-port network is symmetrical if
A)
Z11 = Z22
B)
AD-BC = 1
C)
Z12 = Z21
D)
h11 = -h22
A two-port network is defined as symmetrical if its input and output ports are interchangeable without changing the electrical properties of the network. In terms of Z-parameters (Impedance parameters), a network is symmetrical if the input impedance seen from port 1 (with port 2 open) is equal to the input impedance seen from port 2 (with port 1 open), and the transfer impedances are equal. Specifically, the conditions for symmetry are: 1) Z11 = Z22 and 2) Z12 = Z21. While Z12 = Z21 is a condition for reciprocity, the defining characteristic of symmetry in a reciprocal network is Z11 = Z22. In many textbook contexts for competitive exams, Z11 = Z22 is the primary identifier for symmetry.
Analyzing the options:
- Option (a) Z11 = Z22: This is the necessary condition for symmetry.
- Option (b) AD-BC = 1: This is the condition for a reciprocal network in terms of ABCD parameters (where AD-BC=1), not symmetry.
- Option (c) Z12 = Z21: This is the condition for a reciprocal network, not necessarily a symmetrical one.
- Option (d) h11 = -h22: This is not a standard condition for symmetry in h-parameters (for symmetry in h-parameters, the condition is h11 = h22 and h12 = -h21).
The electric lines of force enter or leave a charged surface at an angle
A)
depending upon surface condition
B)
30°
C)
60°
D)
90°
In electrostatics, the electric field (and thus the lines of force) at the surface of a conductor must be perpendicular to the surface. This is because any tangential component of the electric field would exert a force on the free electrons on the conductor's surface, causing them to move until the tangential component becomes zero. Once the system reaches electrostatic equilibrium, the net force on the surface charges is zero in the tangential direction, meaning the electric field vector is oriented exactly normal (perpendicular) to the surface. Therefore, the lines of force enter or leave the charged surface at an angle of 90°.
In a series RLC circuit, the value of current at resonance is affected by the value of
A)
only L
B)
only C
C)
both L and C
D)
only R
In a series RLC circuit, the total impedance (Z) is given by the formula: Z = √[R² + (XL - XC)²], where XL is the inductive reactance (2πfL) and XC is the capacitive reactance (1/2πfC). At the condition of resonance, the inductive reactance and capacitive reactance are equal in magnitude but opposite in phase (XL = XC). Consequently, the term (XL - XC) becomes zero, and the impedance of the circuit reduces to its minimum possible value, which is equal to the resistance: Z = R. According to Ohm's Law, the current (I) at resonance is I = V / Z. Substituting Z = R, we get I_resonance = V / R. Since the voltage (V) is typically constant, the current at resonance depends solely on the value of the resistance (R) and is completely independent of the values of the inductor (L) and the capacitor (C). While L and C determine the frequency at which resonance occurs, they do not affect the magnitude of the current once that resonance state is achieved.
A medium behaves like dielectric when the
A)
displacement current is just equal to the conduction current
B)
displacement current is less than the conduction current
C)
displacement current is much greater than the conduction current
D)
displacement current is almost negligible
In electromagnetics, a material's behavior as either a conductor or a dielectric is determined by the ratio of its conduction current density (J_c = σE) to its displacement current density (J_d = ε ∂E/∂t). For a time-harmonic field with angular frequency ω, the displacement current density is J_d = jωεE. The ratio of conduction current to displacement current is given by σ / (ωε).
1. If σ / (ωε) >> 1, the conduction current dominates, and the medium behaves as a good conductor.
2. If σ / (ωε) << 1, the displacement current dominates, and the medium behaves as a good dielectric (insulator).
Therefore, for a medium to behave like a dielectric, the displacement current must be much greater than the conduction current. This ensures that the energy is stored in the electric field rather than dissipated as heat through conduction.
The effect of doping intrinsic semiconductor is to
A)
move the Fermi level away from the centre of the forbidden band
B)
move the Fermi level towards the centre of the forbidden band
C)
change the crystal structure of the semiconductor
D)
keep the Fermi level at the middle of the forbidden band
1. **Intrinsic Semiconductors**: In a pure (intrinsic) semiconductor, the number of electrons in the conduction band equals the number of holes in the valence band. Consequently, the Fermi level (Ef), which represents the chemical potential of electrons, lies almost exactly in the middle of the forbidden energy gap (band gap).
2. **N-type Doping**: When a semiconductor is doped with pentavalent impurities (donors), the concentration of electrons increases. This introduces donor energy levels just below the conduction band, shifting the Fermi level upwards, moving it away from the center and closer to the conduction band.
3. **P-type Doping**: When doped with trivalent impurities (acceptors), the concentration of holes increases. This introduces acceptor energy levels just above the valence band, shifting the Fermi level downwards, moving it away from the center and closer to the valence band.
4. **Conclusion**: In both cases of doping (N-type or P-type), the Fermi level is displaced from its original central position in the forbidden band. Therefore, the effect of doping is to move the Fermi level away from the center of the forbidden band.
**Analysis of other options:**
- Option (b) is incorrect because the Fermi level is already at the center in an intrinsic state; doping moves it away, not towards it.
- Option (c) is incorrect because doping involves replacing a very small fraction of atoms; it changes the electrical properties, not the overall crystal structure (e.g., diamond cubic structure of Silicon remains).
- Option (d) is incorrect because keeping the Fermi level at the middle is the characteristic of an intrinsic semiconductor, not a doped one.
As per CPWD manual 2019, performance guarantee shall be _______ of the contract amount.
A)
2%
B)
3%
C)
15%
D)
5%
According to the CPWD (Central Public Works Department) Works Manual 2019 and the General Conditions of Contract (GCC), the Performance Guarantee is a security deposit submitted by the successful bidder to ensure the satisfactory execution of the contract. As per the updated guidelines in the 2019 manual, the Performance Guarantee is typically fixed at 5% of the total contract value. This is a shift from older norms where the security deposit was often calculated as a combination of a percentage of the bill and a percentage of the contract amount. The 5% figure is the standard requirement for the performance security to be submitted before the signing of the agreement. Options a (2%), b (3%), and c (15%) do not align with the standard CPWD 2019 specifications for performance guarantees.
Salvage value is:
A)
Present value of property
B)
Value after depreciation
C)
Scrap value
D)
Next year’s value
In estimation and costing, Salvage Value is defined as the estimated residual value of an asset at the end of its useful economic life. It is the amount that can be recovered from the asset after it has been fully depreciated over its service life.
Step-by-step analysis of options:
- Option (a) 'Present value of property' refers to the current market value or the discounted value of future cash flows, not the value at the end of life.
- Option (b) 'Value after depreciation' is correct because salvage value is the remaining book value or market value once the total depreciation for the asset's useful life has been accounted for.
- Option (c) 'Scrap value' is often confused with salvage value. While similar, scrap value specifically refers to the value of the material (e.g., weight of steel) if the asset is dismantled and sold as waste. Salvage value is broader and can include the value of the asset as a functioning unit in a secondary market.
- Option (d) 'Next year’s value' is simply a future book value, not the terminal value of the asset.