← JE Departmental
Past Questions — 2025
Megger is used to measure-
A)
breakdown voltage of insulation
B)
insulation resistance
C)
earth resistance
D)
none of above
A Megger (short for mega-ohmmeter) is a specialized electrical instrument designed specifically to measure very high resistances, typically in the range of megohms (MΩ). Its primary application is to measure the insulation resistance of electrical components such as wires, cables, and motor windings to ensure that the insulation is intact and preventing leakage current.
Why other options are incorrect:
- Option (a) Breakdown voltage: This is the voltage level at which insulation fails completely. Measuring this requires a High Potential (Hi-Pot) tester, not a Megger, which measures resistance at a specific voltage rather than the failure point.
- Option (c) Earth resistance: While some specialized earth testers look similar to meggers, earth resistance is typically measured using a dedicated Earth Tester (Ground Resistance Tester) which uses a different method (such as the fall-of-potential method) involving auxiliary spikes driven into the ground.
Sparking occurs when a load is switched off because the circuit has-
A)
high inductance
B)
high impedance
C)
high resistance
D)
high Capacitance
Sparking occurs during the switching off of a load due to the property of inductors to oppose sudden changes in current. According to Lenz's Law and the formula for induced voltage V = L(di/dt), when a switch is opened in a circuit with high inductance (L), the current (i) attempts to drop to zero almost instantaneously (dt is very small). This creates a very high rate of change of current, resulting in a large induced voltage (back EMF) across the switch contacts. If this voltage exceeds the dielectric strength of the air gap between the opening contacts, the air ionizes and becomes conductive, resulting in an electric arc or 'sparking'.
Why other options are incorrect:
- b. High impedance: Impedance is the total opposition to current; while it affects current flow, it is not the specific cause of the voltage spike during switching.
- c. High resistance: Purely resistive loads do not store energy in a magnetic field and therefore do not produce significant sparking when switched off.
- d. High Capacitance: While capacitors can cause sparking during switching (due to stored charge), the standard context for 'sparking when a load is switched off' in electrical engineering typically refers to inductive loads (like motors or transformers) because they generate a voltage spike to maintain current flow.
According to I.E. rules, the voltage fluctuation in residential buildings should not be more than
A)
2% of declared supply voltage for residential building
B)
3% of declared supply voltage for residential building
C)
6% of declared supply voltage for residential building
D)
none of above
According to the Indian Electricity (I.E.) Rules and the standards set by the Central Electricity Authority (CEA) regarding the quality of supply, the permissible variation in voltage for low-voltage (LV) supply to consumers is generally maintained within a specific tolerance band. For residential buildings and general domestic consumers, the standard permissible voltage fluctuation is ±6% of the declared nominal supply voltage. This ensures that electrical appliances operate efficiently without risking damage due to under-voltage or over-voltage conditions. Options a (2%) and b (3%) are too stringent and are typically not the standard for general residential distribution, while 6% is the industry and regulatory benchmark for these installations.
The main function of power transformer is
A)
to increase or decrease the d.c. voltage
B)
to increase or decrease the a.c. voltage
C)
to convert a.c. to d.c. vice-versa
D)
none of all
A power transformer is a static electrical device that operates on the principle of Faraday's Law of Electromagnetic Induction. Its primary purpose is to change the voltage level of an alternating current (AC) supply. Depending on the turns ratio of the primary and secondary windings, it can either increase the voltage (Step-up transformer) for efficient long-distance transmission or decrease the voltage (Step-down transformer) for safe distribution and utilization by consumers.
Analysis of other options:
- Option (a) is incorrect because transformers do not work with direct current (DC). DC does not create the changing magnetic flux required to induce a voltage in the secondary winding.
- Option (c) is incorrect because the conversion of AC to DC (or vice-versa) is the function of a Rectifier or an Inverter, not a transformer.
- Option (d) is incorrect as option (b) accurately describes the function.
Voltage drop allowed in the wiring from consumer’s board at any point inside the house should be
A)
2% of declared supply voltage plus one volt
B)
3% of declared supply voltage plus one volt
C)
4% of declared supply voltage plus one volt
D)
none of above
According to standard electrical installation guidelines and the Central Electricity Authority (CEA) regulations (which govern electrical standards in India), the permissible voltage drop from the consumer's distribution board to any point in the internal wiring of a residential or commercial building is limited to ensure the efficient operation of electrical appliances. The standard limit is defined as 3% of the declared supply voltage plus one volt. This margin accounts for the inherent resistance of the conductors and ensures that the voltage at the furthest point of the installation remains within the acceptable operating range of the equipment. Option (a) is too restrictive, and Option (c) exceeds the standard safety and efficiency limits.
The speed for a residential elevator is between
A)
1.5-2.5 m/s
B)
0.5-2.5 m/s
C)
1.0-2.0 m/s
D)
none of above
In elevator engineering, the rated speed is determined by the height of the building and the intended use of the lift. Residential elevators (used in low to mid-rise apartment buildings) typically operate at lower speeds to ensure passenger comfort and safety, and because the travel distance is relatively short. Standard industry ranges for residential/low-rise elevators generally fall between 0.5 m/s (for very low-rise/home lifts) and 2.5 m/s (for mid-rise residential complexes). Speeds above 2.5 m/s are typically reserved for high-rise commercial buildings or skyscrapers to reduce travel time. Therefore, option (b) provides the most comprehensive range that covers both basic residential lifts and those in larger residential complexes.
The maximum permissible voltage drop from supply terminals to any point in the installation for power wiring is
A)
7.5%
B)
5%
C)
4%
D)
none of above
According to standard electrical installation codes (such as the National Electrical Code and the Indian Standard IS 732), the maximum permissible voltage drop from the point of supply (supply terminals) to any point in the installation is generally limited to ensure the efficient operation of electrical equipment. For power wiring and general installations, the standard limit is 5%. A voltage drop exceeding this limit can lead to overheating of conductors, malfunction of electronic equipment, and reduced efficiency of motors (causing them to draw more current and potentially overheat). Therefore, option (b) 5% is the correct technical standard.
Standard frequency of supply in India is-
A)
25Hz
B)
100 Hz
C)
50 Hz
D)
none of above
The standard frequency of the Alternating Current (AC) power supply in India is 50 Hz (Hertz). This means that the current changes its direction 50 times per second. This standard was adopted during the British era and is consistent with the power grid standards used across most of Europe and Asia.
Analysis of options:
- Option (a) 25Hz: This is an obsolete frequency used in some early railway systems but is not the national standard.
- Option (b) 100 Hz: This is not a standard utility frequency for power distribution.
- Option (c) 50 Hz: This is the correct national standard for India.
- Option (d) none of above: Incorrect, as option (c) is correct.
In two W meter method, if the power factor is unity, then,
A)
W1 > W2
B)
W1 < W2
C)
W1 = W2
D)
none of above
In the two-wattmeter method for measuring three-phase power, the readings of the two wattmeters (W1 and W2) are given by the formulas: W1 = V_L I_L cos(30° + φ) and W2 = V_L I_L cos(30° - φ), where φ is the phase angle of the load.
1. When the power factor is unity (cos φ = 1), the phase angle φ = 0°.
2. Substituting φ = 0° into the formulas:
W1 = V_L I_L cos(30° + 0°) = V_L I_L cos(30°)
W2 = V_L I_L cos(30° - 0°) = V_L I_L cos(30°)
3. Therefore, W1 = W2.
Additional context for other options:
- If the power factor is between 0.5 and 1, both W1 and W2 are positive, but not necessarily equal.
- If the power factor is 0.5, one wattmeter reads zero (W2 = 0).
- If the power factor is less than 0.5, one wattmeter will give a negative reading.
In electrical instruments, electrical energy is converted to-
A)
light energy
B)
mechanical energy
C)
chemical energy
D)
heat energy
Electrical measuring instruments (such as ammeters, voltmeters, and wattmeters) are designed to provide a visual reading of an electrical quantity. To achieve this, the electrical energy from the circuit being measured is converted into mechanical energy (displacement). This process typically involves three torques: the deflecting torque (which moves the pointer), the controlling torque (which opposes the deflection to ensure a linear scale), and the damping torque (which prevents oscillation). For example, in a Permanent Magnet Moving Coil (PMMC) instrument, the interaction between the magnetic field and the current-carrying coil produces a mechanical force that rotates the pointer across a scale. While heat energy (option d) is often a byproduct (I²R loss), it is not the intended conversion for the purpose of measurement in standard electrical instruments.
E.m.f. equation of a generator is (Where is flax per pole, N is speed in r.p.m., Z is number of conductor, P is number of pole and A is number of parallel paths)
A)
e.m.f = NA P / 60 z volts
B)
e.m.f. = AZ P / 60 N volts
C)
e.m.f. = NZ P / 60 A volts
D)
none of the above
The electromotive force (e.m.f.) generated in a DC generator is derived from Faraday's Law of Electromagnetic Induction. The standard formula for the generated e.m.f. (E) is: E = (ΦZN P) / (60A).
Breaking down the variables:
- Φ (Phi): Flux per pole (Webers)
- Z: Total number of armature conductors
- N: Speed of armature in revolutions per minute (r.p.m.)
- P: Number of poles
- A: Number of parallel paths (A = P for lap winding, A = 2 for wave winding)
- 60: Conversion factor to change r.p.m. to revolutions per second.
Comparing this to the options:
- Option (a) incorrectly places P in the numerator and Z in the denominator.
- Option (b) incorrectly places A in the numerator and N in the denominator.
- Option (c) correctly follows the formula: e.m.f. = (Φ * N * Z * P) / (60 * A).
Efficiency of a power transformer is of the order of-
A)
75%
B)
50%
C)
98%
D)
100%
A power transformer is highly efficient because it has no moving parts and its losses are relatively small (mainly copper loss and core/iron loss). In practice, large power transformers commonly achieve efficiencies around 98% or even slightly higher, so 98% is the correct order of efficiency.
Why the others are incorrect:
- 100% is impossible because some power is always lost as heat in windings and core.
- 75% is far too low for a transformer; that would indicate excessive losses.
- 50% is completely unrealistic for a power transformer and is far below normal operating efficiency.
The power in a 3-phase delta connection is
A)
V I Cos
B)
V I Cos / 2 ph 2
C)
3 V I Cos / ph 2
D)
none of above
In a 3-phase system, the total active power (P) is the sum of the power in all three phases. For a balanced load, the formula is P = 3 × V_ph × I_ph × cosΦ, where V_ph is the phase voltage, I_ph is the phase current, and cosΦ is the power factor.
In a Delta (Δ) connection:
1. Line Voltage (V) is equal to Phase Voltage (V_ph): V = V_ph.
2. Line Current (I) is related to Phase Current (I_ph) by the relation: I = √3 × I_ph, which means I_ph = I / √3.
Substituting these into the power formula:
P = 3 × V × (I / √3) × cosΦ
Since 3 / √3 = √3, the formula simplifies to:
P = √3 × V × I × cosΦ
Comparing this to the options:
- Option (a) 'V I Cos' matches the derived formula (√3VI cosΦ).
- Option (b) and (c) introduce incorrect divisors and subscripts that do not align with standard electrical power formulas for a 3-phase delta system.
Ohm’s law states that (Where ‘I’ is current in ampere, ‘V’ is voltage in volts and ‘R’ is resistance in Ohm)
A)
I=V.R
B)
V=I.R
C)
R=V.I
D)
none of above
Ohm's Law is a fundamental principle in electrical engineering which states that the current (I) flowing through a conductor between two points is directly proportional to the voltage (V) across the two points, provided all physical conditions (like temperature) remain constant. Mathematically, this relationship is expressed as V = I × R, where V is the potential difference in Volts, I is the current in Amperes, and R is the resistance in Ohms.
Evaluating the options:
- Option (a) I=V.R is incorrect; the correct formula for current is I = V/R.
- Option (b) V=I.R is correct as it represents the standard mathematical form of Ohm's Law.
- Option (c) R=V.I is incorrect; the correct formula for resistance is R = V/I.
1(one) Ton of refrigeration is equal to
A)
15,000 BTU/hr
B)
13,000 BTU/hr
C)
12,000 BTU/hr
D)
none of above
A 'Ton of Refrigeration' (TR) is a unit of power used to describe the cooling capacity of air conditioning and refrigeration systems. It is defined as the amount of heat required to melt one short ton (2,000 lbs) of pure ice at 0°C (32°F) in 24 hours.
1. The latent heat of fusion for ice is approximately 144 BTU/lb.
2. For one ton of ice: 2,000 lbs × 144 BTU/lb = 288,000 BTU.
3. To find the hourly rate: 288,000 BTU / 24 hours = 12,000 BTU/hr.
Therefore, 1 TR = 12,000 BTU/hr. Options (a) and (b) are numerically incorrect based on this standard thermodynamic definition.
Transformer tank is not filled with water because
A)
it has bad cooling property but good insulating property
B)
it has bad cooling and bad insulating property
C)
it has good cooling property but bad insulating property
D)
none of above
In a transformer, the medium inside the tank serves two primary purposes: cooling the core and windings, and providing electrical insulation between the high-voltage components and the grounded tank.
1. **Cooling Property**: Water has a very high specific heat capacity and thermal conductivity, making it an excellent cooling medium (better than transformer oil). Therefore, it has 'good cooling properties'.
2. **Insulating Property**: Pure water is a poor conductor, but in practical applications, water easily absorbs impurities and ions, making it highly conductive. More importantly, it cannot provide the high dielectric strength required to prevent electrical arcing/breakdown between the windings and the tank. Therefore, it has 'bad insulating properties'.
Because a transformer requires a medium that can both cool and insulate, transformer oil is used instead of water, as oil provides the necessary dielectric strength (insulation) while still offering adequate cooling. Thus, water is not used because while it cools well, it fails as an insulator.
One-way single pole switch controls
A)
connection of one wire to one position
B)
connection of one wire to two positions
C)
connection of two wire to two positions
D)
connection of two wire to one positions
A one-way single pole switch (also known as a Single Pole Single Throw or SPST switch) is the simplest type of switch used in house wiring. It has two terminals: one for the incoming live wire and one for the outgoing wire to the load. Its primary function is to 'break' or 'make' the connection of a single conductor (the phase/live wire) to a single position (the load).
- Option (a) is correct because the switch controls the flow of one wire (the live wire) to one specific destination/position (the light or appliance).
- Option (b) is incorrect because controlling one wire to two positions would describe a selector switch or a change-over switch.
- Option (c) and (d) are incorrect because a single pole switch does not control two wires simultaneously; controlling two wires (Live and Neutral) would require a double-pole switch.
Causeway is one type of-
A)
Bridge
B)
Subsurface drainage
C)
Cross drainage
D)
Culvert
A causeway is a raised road across a stream or marshy area, designed to allow water to flow over the road surface during high-water periods (floods). In civil and highway engineering, any structure used to allow water to pass across a road or railway embankment is categorized as a 'Cross Drainage (CD) work'.
Reasoning for options:
- (a) Bridge: While a causeway is a crossing, it differs from a bridge because a bridge is designed to keep the traffic above the high-flood level (HFL), whereas a causeway is intentionally submerged during floods.
- (b) Subsurface drainage: This refers to systems (like perforated pipes) used to remove excess water from the soil beneath the road surface, not structures that cross a stream.
- (c) Cross drainage: This is the correct umbrella term. CD works include culverts, bridges, causeways, siphons, and aqueducts.
- (d) Culvert: A culvert is a specific type of CD work that allows water to pass *under* the road through a pipe or box. A causeway allows water to pass *over* the road.
The function of base in paint is-
A)
To provide film on surface
B)
To hide imperfections of surface
C)
To reduce shrinkage cracks in paint film
D)
To bring down the overall cost
In the composition of paint, the 'base' refers to the extender or filler (such as whiting, barytes, or talc). The primary functions of the base/extender are: 1) To reduce the cost of the paint by replacing a portion of the expensive pigment with cheaper materials, and 2) To control the gloss and improve the mechanical properties of the film.
Analysis of other options:
- Option (a) 'To provide film on surface' is the function of the Binder/Vehicle (e.g., linseed oil, acrylic resin).
- Option (b) 'To hide imperfections of surface' is primarily the function of the Pigment (for opacity) or the Primer/Putty layer applied before painting.
- Option (c) 'To reduce shrinkage cracks' is managed by the formulation of the binder and the addition of specific additives, not the base/extender.
The main three constituents of cement are-
A)
Aluminium, lime and Sulphur
B)
Lime, alumina and magnesium
C)
Lime, magnesium and manganese
D)
Lime, silica and alumina
Portland cement is manufactured by burning a mixture of calcareous (lime-bearing) and argillaceous (clay-bearing) materials. The primary chemical constituents that make up the bulk of the cement clinker are:
1. Lime (CaO): The most significant constituent, providing strength and stability.
2. Silica (SiO2): Provides strength through the formation of calcium silicates.
3. Alumina (Al2O3): Acts as a flux to lower the clinkering temperature and contributes to the setting properties.
While other elements like Iron Oxide, Magnesium Oxide, and Sulphur (added as Gypsum) are present, they are considered minor constituents.
Analysis of options:
- Option (a) is incorrect because Sulphur is a minor additive (gypsum) and Aluminium is part of Alumina, but Lime/Silica/Alumina are the primary trio.
- Option (b) is incorrect because Magnesium is a minor constituent (usually kept low to prevent unsoundness).
- Option (c) is incorrect because Manganese is only present in trace amounts.
- Option (d) is correct as it lists the three dominant oxides that form the structural basis of cement.
The aggregate impact value of aggregate to be used in wearing course of pavement is-
A)
less than or equal to 30%
B)
less than or equal to 40%
C)
less than 35%
D)
more than 25%
The Aggregate Impact Value (AIV) test is used to determine the resistance of aggregates to sudden shock or impact. In highway engineering, the wearing course is the topmost layer of the pavement and is subjected to the highest impact loads from vehicle tires. According to standard specifications (such as IRC - Indian Roads Congress), the maximum permissible Aggregate Impact Value for aggregates used in the wearing course of a pavement is 30%. A lower AIV indicates a tougher aggregate that can withstand impact without crushing.
- Option (a) is correct as it aligns with the standard limit of ≤ 30% for wearing courses.
- Option (b) is incorrect because 40% is generally the limit for lower layers (like the base course) but too high for the wearing course.
- Option (c) is incorrect as the standard limit is specifically 30%, not 35%.
- Option (d) is incorrect because a value 'more than 25%' would allow for aggregates that are too weak (e.g., 45%), which would lead to premature pavement failure.
Which has the highest Coefficent of steel for estimation, Beam, Column, slab or footing?
A)
Slab
B)
Column
C)
Beam
D)
Footing
In the context of estimation and costing for reinforced concrete structures, the 'coefficient of steel' refers to the quantity of steel required per unit volume of concrete (kg/m³).
1. **Columns**: Columns are primary vertical load-bearing members that must resist high axial compressive loads and potential buckling. To ensure structural stability and ductility (especially in seismic zones), they require a higher percentage of longitudinal reinforcement and closely spaced lateral ties. Typically, the steel percentage in columns ranges from 0.8% to 6% (though usually 1-4% in standard practice), leading to the highest kg/m³ value among the listed options.
2. **Beams**: Beams primarily resist bending and shear. While they require significant steel, the overall volume of concrete is usually larger relative to the steel compared to a column. Steel percentages typically range from 0.5% to 2%.
3. **Slabs**: Slabs are thin elements that distribute loads. They generally have the lowest steel density, often ranging from 0.3% to 1% of the concrete volume.
4. **Footings**: Footings distribute the load to the soil. While they contain reinforcement to resist bending, the volume of concrete is very large, which significantly lowers the overall coefficient of steel (kg/m³).
Therefore, among the given options, the Column has the highest coefficient of steel.
Which of the following is not the desirable property of soil as highway material?
A)
Stability
B)
Good drainage
C)
Ease of compaction
D)
Compressibility
In highway engineering, the subgrade and base layers must be able to support heavy traffic loads without significant deformation.
1. Stability (Option a): This is a highly desirable property. The soil must be stable to prevent the pavement from shifting or failing under load.
2. Good drainage (Option b): This is desirable because water accumulation in the soil layers leads to a loss of shear strength, causing potholes and structural failure (rutting).
3. Ease of compaction (Option c): This is desirable because soil that can be easily compacted to its maximum dry density ensures a dense, strong foundation with minimal voids.
4. Compressibility (Option d): This is NOT a desirable property. High compressibility means the soil will undergo significant volume reduction (settlement) under load. This leads to uneven pavement surfaces, cracking, and overall structural failure of the highway. Therefore, low compressibility is desired, making 'Compressibility' the correct answer as the non-desirable property.
Joints in cement concrete pavement are provided.
A)
to take care of wheel load.
B)
to reduce temperature stresses
C)
to allow drainage
D)
for ease of construction
Cement concrete is a rigid material with a low coefficient of thermal expansion, but it still expands and contracts with changes in temperature. In a continuous slab, this expansion (during summer) and contraction (during winter) would create massive internal stresses, leading to uncontrolled cracking and warping. Joints (such as expansion joints and contraction joints) are strategically provided to create 'controlled' gaps or breaks, allowing the concrete slabs to expand and contract freely without causing structural failure.
Why other options are incorrect:
- Option (a): Wheel loads are managed by the thickness of the slab and the quality of the base/sub-base, not by the joints. In fact, joints are often the weakest points under wheel loads.
- Option (c): Drainage is managed by the camber (cross-slope) of the pavement and side drains, not by the joints.
- Option (d): While joints are part of the construction process, their primary engineering purpose is structural integrity against thermal stress, not the 'ease' of building the road.
What is the number of bricks taken in analysis of cost for 1sqm half brick wall?
A)
49.5 nos
B)
52 nos
C)
52.5 nos
D)
56.5nos
To calculate the number of bricks for 1 sqm of a half-brick wall, we use the standard modular brick size and mortar joints.
1. **Standard Modular Brick Size**: The standard size of a modular brick is 190 mm x 90 mm x 90 mm.
2. **Size with Mortar**: Adding a standard mortar joint of 10 mm to the length and height, the effective size of one brick in the wall becomes 200 mm x 100 mm (0.2 m x 0.1 m).
3. **Calculation for 1 sqm**:
- Area of one brick with mortar = 0.2 m × 0.1 m = 0.02 sqm.
- Number of bricks = Total Area / Area of one brick = 1 sqm / 0.02 sqm = 50 bricks.
However, in Indian Standard (IS) codes and PWD practice for estimation, the 'standard' brick size often used for calculations is the non-modular size (approx. 230mm x 115mm x 75mm) or specific regional standards. For a half-brick wall (thickness of 115mm or 100mm), the standard analysis of cost for 1 sqm typically accounts for wastage and specific joint thicknesses.
According to standard PWD analysis of rates for a half-brick wall (using modular bricks with a slight allowance for wastage or specific jointing), the value is taken as 52.5 bricks per square meter.
- Option (a) 49.5 is slightly under the theoretical 50.
- Option (b) 52 is close but not the standard analysis figure.
- Option (c) 52.5 is the standard figure used in many PWD cost analysis tables to account for the modular size plus a small percentage of wastage.
- Option (d) 56.5 is too high for a standard half-brick wall.