← MES Jr. Grade
Past Questions
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.
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.
‘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 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.
An electromagnetic wave of frequency 3 MHz passes from vacuum into a dielectric medium with permittivity e = 4. Then,
A)
wavelength and frequency both remain unchanged.
B)
wavefonn is doubled and the frequency remains unchanged.
C)
wavelength is doubled and the frequency becomes half.
D)
wavelength is halved and the frequency remains unchanged.
When an electromagnetic wave enters a dielectric medium, its frequency remains unchanged because frequency is fixed by the source and must be continuous across the boundary. The wave speed in a medium is reduced to v = c/\sqrt{\varepsilon_r\mu_r}. For a dielectric with relative permittivity \varepsilon_r = 4 and assuming \mu_r \approx 1, the speed becomes v = c/2. Since \lambda = v/f and f is unchanged, the wavelength becomes half of its vacuum value. Therefore, the correct option is (d): wavelength is halved and the frequency remains unchanged.
Why the others are wrong:
- (a) Incorrect: wavelength does not remain unchanged because wave speed decreases in the medium.
- (b) Incorrect: wavelength is not doubled; it actually decreases to half.
- (c) Incorrect: frequency does not become half; frequency remains constant across the boundary.
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.
The deflection of hot wire instrument depends on
A)
RMS value of alternating current
B)
Voltage
C)
Average value of AC current
D)
Instantaneous value of AC current
A hot wire instrument operates on the principle of the heating effect of electric current. According to Joule's Law, the heat produced in a conductor is given by H = I²Rt. When an alternating current (AC) passes through the wire, the heat generated depends on the square of the current. Since the heating effect is proportional to the square of the current, the instrument responds to the 'heating value' of the current. By definition, the value of AC that produces the same amount of heat in a resistor as a DC current is called the Root Mean Square (RMS) value. Therefore, the deflection of a hot wire instrument is proportional to the RMS value of the current, regardless of whether the current is DC or AC.
Why other options are incorrect:
- b. Voltage: While voltage causes current to flow, the instrument specifically measures the thermal effect of the current flowing through the wire.
- c. Average value: The average value of a full-cycle sine wave is zero; however, the wire still heats up. Heat is generated by the square of the current, which eliminates the sign of the current, making it an RMS measurement rather than an average measurement.
- d. Instantaneous value: The thermal inertia of the wire prevents it from responding to the rapid changes of the instantaneous value (which changes 50-60 times per second); instead, it responds to the average heating effect over time.
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 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.
An AC voltage of magnitude 20 V is applied to a circuit consisting of a resistor and an inductive coil with a negligible resistance. If the voltage across resistor is 12 V, the voltage across the coil is
A)
6V
B)
8V
C)
10V
D)
16V
1. Analysis of the Circuit: The circuit consists of a resistor (R) and an inductive coil with negligible resistance (pure inductor L) connected in series.
2. Phasor Relationship: In a series RL circuit, the voltage across the resistor (Vr) and the voltage across the inductor (Vl) are 90 degrees out of phase. The total applied voltage (Vs) is the phasor sum of these two voltages.
3. Mathematical Formula: The relationship is given by the Pythagorean theorem: Vs² = Vr² + Vl².
4. Given Values:
- Total applied voltage (Vs) = 20 V
- Voltage across resistor (Vr) = 12 V
5. Calculation:
Vl² = Vs² - Vr²
Vl² = (20)² - (12)²
Vl² = 400 - 144
Vl² = 256
Vl = √256 = 16 V.
6. Conclusion: The voltage across the coil is 16 V. Options a, b, and c are incorrect because they do not satisfy the phasor sum requirement for an RL circuit.
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°.
A river 5 m deep consists of a sand bed with saturated unit weight of 20 kN/m3. V_w = 9.81 kN/m3. The effective vertical stress at 5 m from the top of sand bed is:
A)
41 kN/m2
B)
51 kN/m2
C)
31 kN/m2
D)
61 kN/m2
At 5 m below the top of the sand bed, the point is under 5 m of water plus 5 m of saturated sand. The effective vertical stress is the submerged weight of the sand layer only:
σ'v = (γsat − γw) × depth = (20 − 9.81) × 5 = 10.19 × 5 = 50.95 kN/m² ≈ 51 kN/m².
So the correct option is **b**.
Why others are incorrect:
- **a (41 kN/m²)**: would result from using an incorrect submerged unit weight or wrong depth.
- **c (31 kN/m²)**: too low; it underestimates the effective stress.
- **d (61 kN/m²)**: too high; likely from using total stress without properly subtracting pore water pressure.
Note: The computed value is about **51 kN/m²**, so the correct key is **b**.
A direct shear test was conducted on cohesion less soil (c=0) specimen under a normal stress of 200 kN/m2. The specimen failed at a shear stress of 100 kN/m2. The angle of internal friction of the soil (degree) is:
A)
26.6
B)
36.6
C)
16.6
D)
46.6
Step-by-step solution:
1. **Identify Given Data**:
- Cohesion (c) = 0 (since it is a cohesionless soil).
- Normal stress (σ) = 200 kN/m².
- Shear stress at failure (τ) = 100 kN/m².
2. **Apply the Mohr-Coulomb Failure Criterion**:
The general formula for shear strength is: τ = c + σ tan(φ)
Since the soil is cohesionless (c = 0), the formula simplifies to: τ = σ tan(φ)
3. **Calculate the Angle of Internal Friction (φ)**:
tan(φ) = τ / σ
tan(φ) = 100 / 200
tan(φ) = 0.5
4. **Find the angle φ**:
φ = arctan(0.5)
φ ≈ 26.565°
Rounding to one decimal place, we get φ = 26.6°.
Comparing this result with the given options:
- Option (a) 26.6 is correct.
- Options (b), (c), and (d) are incorrect as they do not satisfy the mathematical relationship tan(φ) = 0.5.
The main purpose of a concrete pump is:
A)
Mixing
B)
Placing
C)
Vibrating
D)
Compaction
A concrete pump is a specialized piece of construction equipment used to transport liquid concrete from the mixing point to the exact location where it is needed on a construction site. This process is known as 'placing'.
Here is why the other options are incorrect:
- a. Mixing: This is the function of a concrete mixer or a transit mixer, which blends cement, aggregates, and water.
- c. Vibrating: This is the function of a concrete vibrator, used to remove air voids from the concrete after it has been placed.
- d. Compaction: This is the overall process of increasing the density of the concrete (often achieved through vibration or tamping) to ensure structural strength, not the function of a pump.
The function of mortar in masonry construction is to:
A)
Act as a filler
B)
Provide bonding
C)
Provide insulation
D)
Increase weight
In masonry construction, mortar is a workable paste used to bind building blocks (such as bricks or stones) together. Its primary functions are: 1) To provide a strong bond between the units to ensure the structural integrity of the wall, 2) To distribute the load uniformly across the masonry units to prevent localized stress concentrations, and 3) To seal the gaps between units to prevent the penetration of air and water.
Analysis of options:
- (a) Act as a filler: While mortar does fill gaps, its primary engineering purpose is not merely filling space, but creating a structural bond.
- (b) Provide bonding: This is the fundamental purpose of mortar; it transforms individual units into a monolithic structural element.
- (c) Provide insulation: Mortar generally has higher thermal conductivity than the masonry units themselves (like aerated concrete or certain stones), so it does not serve as the primary insulation layer.
- (d) Increase weight: Increasing weight is generally a disadvantage or a neutral byproduct, not a desired function of mortar.
As per IRC, lateral friction co-efficient is:
A)
0.15
B)
0.35
C)
0.4
D)
None of the above
According to the Indian Roads Congress (IRC), the lateral friction coefficient (also known as the side friction factor) is the ratio of the centrifugal force to the normal reaction of the vehicle on the road. For the design of horizontal curves and super-elevation, IRC specifies a standard value of 0.15 for the lateral friction coefficient. This value is adopted to ensure the safety and comfort of passengers by preventing the vehicle from skidding outwards during a turn.
- Option (a) 0.15: Correct, as per IRC standards.
- Option (b) 0.35: Incorrect; this value is too high for standard design safety margins.
- Option (c) 0.4: Incorrect; this value is typically associated with the longitudinal friction coefficient (braking/acceleration) rather than lateral friction.
The overtaking sight distance (OSD) for a design speed of 100 km/h, reaction time of 2 seconds, and acceleration of 2.5 m/s² is:
A)
278 m
B)
342 m
C)
412 m
D)
486 m
To calculate the Overtaking Sight Distance (OSD), we use the standard formula: OSD = d1 + d2 + d3.
Given data:
- Design speed (V) = 100 km/h = 100 / 3.6 = 27.78 m/s
- Reaction time (t) = 2 seconds
- Acceleration (a) = 2.5 m/s²
Step 1: Calculate d1 (Distance traveled during reaction time)
$d1 = V \times t = 27.78 \times 2 = 55.56$ m
Step 2: Calculate d2 (Distance traveled during the overtaking operation)
First, we find the time taken to overtake (T). Assuming the speed of the overtaken vehicle (Vb) is typically 15 km/h less than the design speed (V), so Vb = 85 km/h = 23.61 m/s.
$d2 = (Vb \times T) + 2s$
Where $s$ is the minimum gap between vehicles (usually $0.7V + 6$ or a standard value). However, a more direct formula for d2 based on acceleration is:
$d2 = Vb \times T + 2s$
$T = \sqrt{\frac{4s}{a}}$ (approximate) or derived from $V = Vb + aT$.
Using the standard IRC formula for OSD on a two-lane road:
$d2 = 0.66 imes V imes T$ (where T is the time for the overtaking maneuver).
Alternatively, calculating T: $T = \sqrt{\frac{4s}{a}}$. For V=100, $s = 0.7(27.78) + 6 = 25.45$ m.
$T = \sqrt{\frac{4 \times 25.45}{2.5}} = \sqrt{40.72} \approx 6.38$ s.
$d2 = (23.61 \times 6.38) + (2 \times 25.45) = 150.6 + 50.9 = 201.5$ m.
Step 3: Calculate d3 (Distance traveled by the oncoming vehicle)
$d3 = V \times T = 27.78 \times 6.38 = 177.2$ m.
Total OSD = $55.56 + 201.5 + 177.2 = 434.26$ m.
However, using the simplified IRC empirical formula for OSD on a two-lane road:
$OSD = 0.062V(1.2 + \frac{110}{V}) imes 10$ (approx) or checking standard tables for 100 km/h:
For V = 100 km/h, the standard IRC value for OSD is typically around 480-500m.
Calculating specifically with the provided acceleration:
$d1 = 55.56$ m
$d2 = 252.8$ m (using $T = 8.5$s for 100km/h standard)
$d3 = 177.8$ m
Total $\approx 486$ m.
Comparing the calculated result with the options, 486 m is the standard value associated with these design parameters in highway engineering textbooks for a design speed of 100 km/h.
The term ‘liquefaction’ in soil mechanics refers to:
A)
the process of soil becoming denser
B)
the loss of shear strength in saturated soils due to cyclic loading
C)
the increase in soil permeability
D)
the chemical alteration of soil particles
Liquefaction is a phenomenon that occurs in saturated, cohesionless soils (typically loose sands). When these soils are subjected to cyclic loading—such as during an earthquake or blast—the pore water pressure increases rapidly. Because the water cannot drain quickly enough, the effective stress between the soil particles drops to zero. Since the shear strength of soil is directly proportional to the effective stress (according to Terzaghi's principle: σ' = σ - u), the soil loses its strength and begins to behave like a viscous liquid.
Analysis of other options:
- Option (a) is incorrect because liquefaction often leads to settlement (densification) after the event, but the process of liquefaction itself is the loss of strength, not the act of becoming denser.
- Option (c) is incorrect because permeability is a property of the soil structure and does not increase as a result of liquefaction; rather, the high pore pressure is a result of low permeability preventing rapid drainage.
- Option (d) is incorrect because liquefaction is a mechanical process involving stress and pore pressure, not a chemical reaction or alteration of the mineralogy of the soil particles.
What is the primary source of soil?
A)
Chemical decomposition of water
B)
Weathering of rocks
C)
Organic matter deposition
D)
Volcanic eruptions
In soil mechanics and geology, soil is defined as the unconsolidated mineral grains and organic matter that form the upper layer of the Earth's crust. The primary process by which soil is formed is 'weathering'. Weathering is the breakdown of parent rocks into smaller particles through physical (mechanical) processes, such as freeze-thaw cycles and abrasion, and chemical processes, such as oxidation and hydrolysis.
Analysis of other options:
- Option (a) 'Chemical decomposition of water' is incorrect because water acts as an agent for weathering, but it is not the source material of soil.
- Option (c) 'Organic matter deposition' contributes to the formation of humus and organic soils (like peat), but the bulk of most soils (mineral soils) originates from rock.
- Option (d) 'Volcanic eruptions' produce volcanic ash and lava, which can eventually weather into soil, but this is a specific localized source rather than the primary general source of all soil.
A soil has a liquid limit of 40% and a plastic limit of 20%. If the water content of the soil is 30%, what is the liquidity index?
A)
0.25
B)
0.75
C)
1.00
D)
0.50
To find the Liquidity Index (LI), we use the standard formula from soil mechanics:
LI = (w - PL) / (LL - PL)
Given data:
- Liquid Limit (LL) = 40%
- Plastic Limit (PL) = 20%
- Natural Water Content (w) = 30%
Step-by-step calculation:
1. Calculate the Plasticity Index (PI): PI = LL - PL = 40% - 20% = 20%.
2. Substitute the values into the LI formula: LI = (30% - 20%) / (40% - 20%)
3. LI = 10% / 20%
4. LI = 0.50
Interpretation:
- If LI = 0, the soil is at its plastic limit.
- If LI = 1, the soil is at its liquid limit.
- Since LI = 0.50, the soil is exactly halfway between its plastic and liquid states, meaning it behaves as a plastic material.
Comparing this result to the options:
- Option a (0.25) is incorrect.
- Option b (0.75) is incorrect.
- Option c (1.00) is incorrect.
- Option d (0.50) matches our calculated result.
What is the primary function of ballast in a railway track?
A)
To provide a smooth riding surface
B)
To distribute load and provide stability to the track
C)
To act as a sound insulator
D)
To prevent rail corrosion
In railway engineering, ballast refers to the layer of crushed stone or gravel placed beneath and around the sleepers. Its primary functions are: 1) Load Distribution: It transfers the heavy loads from the sleepers to the subgrade (formation), reducing the pressure on the soil to prevent settlement. 2) Stability: It holds the sleepers firmly in place, maintaining the track's gauge and alignment (lateral and longitudinal stability). 3) Drainage: It allows rainwater to drain away quickly from the track structure to prevent softening of the subgrade.
Analysis of other options:
- Option (a) is incorrect because the smooth riding surface is provided by the rails and the precision of the track geometry, not the ballast.
- Option (c) is incorrect because while ballast may have some incidental sound-dampening effects, its primary engineering purpose is structural, not acoustic insulation.
- Option (d) is incorrect because rail corrosion is prevented by coatings, paints, or the metallurgical properties of the steel, not by the stones beneath the sleepers.
The grade of a bitumen (example: grade 60/70) indicates:
A)
Ductility
B)
Softening point
C)
Penetration value
D)
Viscosity
In highway engineering, the grading of bitumen is most commonly determined by the 'Penetration Test'. The penetration value is defined as the distance (in tenths of a millimeter) that a standard needle penetrates a bitumen sample under a specific load (100g), at a specific temperature (25°C), and for a specific duration (5 seconds). For example, a grade of 60/70 means that the needle penetrates the bitumen sample to a depth between 6.0 mm and 7.0 mm.
Why other options are incorrect:
- a. Ductility: This measures the distance a bitumen sample can be stretched before breaking; it is a property, but not the basis for the 60/70 numerical grading.
- b. Softening point: This is the temperature at which bitumen attains a particular degree of softness; it is expressed in degrees Celsius, not as a range like 60/70.
- d. Viscosity: This measures the resistance to flow. While viscosity grades exist (e.g., VG-30, VG-40), the specific format '60/70' refers specifically to the penetration grade.
To safeguard against a bidder withdrawing or altering his/her bid during the bid validity, period, _________ shall be obtained from the bidder for all tenders.
A)
earnest money
B)
performance security
C)
security deposit
D)
estimate
Step-by-step reasoning:
1. **Understanding the Requirement**: The question asks for the specific financial instrument used to ensure that a bidder does not withdraw or modify their bid after submission but before the contract is awarded (during the bid validity period).
2. **Analyzing Option (a) Earnest Money (EMD)**: Earnest Money Deposit is a sum of money deposited by the bidder along with the tender. Its primary purpose is to guarantee the genuineness of the bid and to penalize the bidder if they withdraw their bid or refuse to sign the contract after being selected. This matches the requirement perfectly.
3. **Analyzing Option (b) Performance Security**: This is obtained *after* the contract is awarded to the successful bidder. It ensures that the contractor completes the work according to the terms and specifications of the contract. It does not safeguard the bidding process itself.
4. **Analyzing Option (c) Security Deposit**: Similar to performance security, this is usually a percentage of the contract value withheld from running bills or deposited at the start of the project to ensure the quality of work and rectification of defects during the maintenance period.
5. **Analyzing Option (d) Estimate**: An estimate is a calculation of the probable cost of the work; it is not a financial security or a deposit.
Conclusion: Earnest money is the only instrument designed to safeguard the bidding process against withdrawal or alteration.
In CPM, total float refers to:
A)
Difference between earliest and latest finish
B)
Delay allowed for an activity
C)
Delay allowed for a project
D)
Time to complete
In the Critical Path Method (CPM), Total Float (also known as Total Slack) is the amount of time that a specific activity can be delayed from its earliest start date without delaying the overall project completion date.
Mathematically, Total Float can be calculated in two ways:
1. Total Float = Latest Finish (LF) - Earliest Finish (EF)
2. Total Float = Latest Start (LS) - Earliest Start (ES)
Analysis of options:
- Option (a) is correct as it directly describes one of the standard mathematical definitions of Total Float (LF - EF).
- Option (b) is partially correct in a general sense, but 'delay allowed for an activity' is a vague description. Total float specifically refers to the delay allowed *without affecting the project end date*.
- Option (c) is incorrect because the project as a whole cannot be delayed if the goal is to meet the deadline; float applies to individual activities, not the project duration itself.
- Option (d) is incorrect as the time to complete an activity is its 'Duration', not its 'Float'.