JE Departmental • 2026 • Engineering Paper
According to IS:732-1983:
A)
power and heating sub-circuit should be kept Separate and distinct from lighting and fan sub-circuits
B)
power and heating sub-circuit should be in lighting and fan sub-circuits
C)
only power sub-circuit should be common with fan sub-circuits
D)
none of the above
According to IS:732-1983 (the Indian Standard Code of Practice for Electrical Wiring Installations), electrical installations must be divided into sub-circuits to ensure safety and reliability. The standard explicitly mandates that power sub-circuits (which serve high-load appliances like heaters, air conditioners, or power sockets) and heating sub-circuits must be kept separate and distinct from lighting and fan sub-circuits. This is done for several reasons: 1) To prevent a fault in a high-power appliance from causing a blackout of the entire lighting system in a room. 2) To allow for different cable sizes (power circuits require thicker wires to handle higher currents). 3) To ensure that overcurrent protection (MCBs/Fuses) is appropriately rated for the specific load type. Therefore, option (a) is the correct technical requirement, while options (b) and (c) violate safety and design standards.
JE Departmental • 2026 • Engineering Paper
When resistance is added to a circuit having a constant potential, the current is?
A)
reduced in the same proportion
B)
increased in the same proportion
C)
unaltered
D)
made fluctuating
For a circuit with constant potential difference, Ohm’s law gives I = V/R. Since V is constant, increasing resistance R makes the current I decrease inversely. Therefore, when resistance is added, the current is reduced in the same proportion. Option b is incorrect because current does not increase when resistance increases. Option c is incorrect because the current changes when resistance changes. Option d is incorrect because adding resistance does not make the current fluctuate; it simply lowers the current.
JE Departmental • 2026 • Engineering Paper
Electrical energy is measured in:
A)
ohm meter
B)
volt meter
C)
amperemeter
D)
energy meter
Electrical energy is the total power consumed over a specific period of time (Energy = Power × Time). The instrument specifically designed to measure this cumulative consumption is the energy meter (commonly known as a kWh meter).
Analysis of other options:
- a. Ohm meter: Used to measure electrical resistance (measured in Ohms).
- b. Volt meter: Used to measure electrical potential difference or voltage (measured in Volts).
- c. Amperemeter: Used to measure the flow of electrical current (measured in Amperes).
JE Departmental • 2026 • Engineering Paper
According to I.E. rules, the earth resistance for low and medium voltage systems should be:
A)
less than 1V
B)
less than 3V
C)
less than 4V
D)
less than 5V
In the context of Indian Electricity (IE) Rules and standard electrical safety practices for grounding/earthing, the primary objective is to ensure that the touch voltage and step voltage remain within safe limits during a fault. For low and medium voltage systems, the earth resistance is designed such that the voltage rise of the grounding electrode (the potential of the earth electrode relative to remote earth) does not exceed a safe threshold. According to standard safety guidelines derived from IE rules and IS (Indian Standards), the permissible voltage rise for the earthing of low and medium voltage systems is generally kept below 1V to ensure maximum safety against electric shock and to facilitate the rapid operation of protective devices (like MCBs or fuses). Options b, c, and d provide voltage thresholds that are too high, increasing the risk of hazardous touch voltages during a fault condition.
JE Departmental • 2026 • Engineering Paper
The material used for fuse must have:
A)
low melting point and low specific resistance
B)
low melting point and high specific resistance
C)
high melting point and low specific resistance
D)
none of the above
A fuse is a safety device designed to protect an electrical circuit from overcurrent or short circuits by melting and breaking the circuit. To function effectively, the fuse material must satisfy two primary criteria:
1. Low Melting Point: The material must melt quickly when the current exceeds a specific threshold. If the melting point were high, the excessive current would damage the connected appliances or cause wiring fires before the fuse could blow.
2. Low Specific Resistance (Low Resistivity): Under normal operating conditions, the fuse should not cause a significant voltage drop or generate excessive heat (I²R losses) that could lead to premature melting. Therefore, a material with low specific resistance is required to ensure efficiency during normal operation.
Commonly used materials for fuses are alloys of tin and lead, which provide the necessary combination of low melting points and low resistivity.
JE Departmental • 2026 • Engineering Paper
A moving-coil voltmeter has a scale which is:
A)
linear
B)
exponentially increasing
C)
exponentially decreasing
A moving-coil instrument (such as a PMMC - Permanent Magnet Moving Coil meter) operates on the principle that the deflecting torque (Td) is proportional to the current flowing through the coil (Td ∝ I). The restoring torque (Tr) provided by the springs is proportional to the angle of deflection (Tr ∝ θ). At equilibrium, Td = Tr, which means the deflection θ is directly proportional to the current I (θ ∝ I). Since the current in a voltmeter is proportional to the voltage (V = I * R), the deflection is also directly proportional to the voltage. This linear relationship results in a scale with equally spaced markings, known as a linear scale. In contrast, moving-iron instruments have a non-linear (squared) scale because their deflection is proportional to the square of the current.
JE Departmental • 2026 • Engineering Paper
A lamp has a luminous efficiency of 11 lm/W and gives out a luminous flux of 275 lm. Its wattage is _________ Watt.
A)
20
B)
40
C)
25
D)
60
Option C is correct: 25.
JE Departmental • 2026 • Engineering Paper
Electric shock is:
A)
always fatal
B)
never fatal
C)
sometimes fatal
D)
always disfiguring
Electric shock occurs when an electric current passes through the human body. The severity of the shock depends on several factors: the magnitude of the current (amperage), the path the current takes through the body (e.g., whether it passes through the heart), the duration of exposure, and the resistance of the skin.
1. Why 'a' is incorrect: Not every electric shock is fatal. Many people experience minor shocks (like static electricity or low-voltage leaks) that cause only a tingling sensation or a mild jolt without causing death.
2. Why 'b' is incorrect: High-voltage shocks or currents that cause ventricular fibrillation (heart failure) or respiratory arrest are frequently fatal.
3. Why 'c' is correct: Because the outcome ranges from harmless to lethal depending on the variables mentioned above, it is accurate to say that electric shock is 'sometimes fatal'.
4. Why 'd' is incorrect: While high-voltage burns can cause disfigurement, many shocks leave no visible external marks or disfigurement at all.
JE Departmental • 2026 • Engineering Paper
The formula Power= 3 VI Cos f is applicable to 3-phase circuits having:
A)
star connections
B)
delta connections
C)
both star and delta connections
D)
none of the above connections
In a balanced 3-phase system, the total active power (P) is the sum of the power in all three phases. The general formula for total power is P = 3 × V_phase × I_phase × cos φ.
1. For a Star (Y) connection: The line voltage (V_L) = √3 × V_phase and the line current (I_L) = I_phase. Substituting these into the general formula: P = 3 × (V_L / √3) × I_L × cos φ = √3 × V_L × I_L × cos φ.
2. For a Delta (Δ) connection: The line voltage (V_L) = V_phase and the line current (I_L) = √3 × I_phase. Substituting these into the general formula: P = 3 × V_L × (I_L / √3) × cos φ = √3 × V_L × I_L × cos φ.
Since the derivation leads to the same formula (P = √3 V_L I_L cos φ) regardless of whether the system is connected in star or delta, the formula is universally applicable to both balanced 3-phase configurations.
JE Departmental • 2026 • Engineering Paper
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 the 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 1 short ton (2,000 lbs) of pure ice at 0°C (32°F) in 24 hours.
Step-by-step calculation:
1. The latent heat of fusion for ice is approximately 144 BTU/lb.
2. For 1 short ton (2,000 lbs) of ice: Total heat = 2,000 lbs × 144 BTU/lb = 288,000 BTU.
3. Since this heat is removed over 24 hours, the rate of heat removal is: 288,000 BTU / 24 hours = 12,000 BTU/hr.
Therefore, 1 TR = 12,000 BTU/hr (which is also approximately 3.517 kW). Options a and b are numerically incorrect based on the standard definition of the unit.
JE Departmental • 2026 • Engineering Paper
Ohm’s law states that: (Where ‘I’ is current in ampere, ‘V’ is voltage in volts and ‘R’ is resistance in V)
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 is expressed as V ∝ I. The constant of proportionality is the resistance (R), leading to the formula V = I × R.
Analysis of 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 expression of Ohm's Law.
- Option (c) R=V.I is incorrect; the correct formula for resistance is R = V/I.
Note: The question text contains a typo stating resistance is in 'V' (volts), whereas resistance is actually measured in Ohms (Ω). However, this does not change the mathematical relationship defined by the law.
JE Departmental • 2026 • Engineering Paper
The standard door width for hospital elevators are:
A)
1,100mm
B)
1,300mm
C)
1,200mm
D)
1,400mm
In the design of elevators for healthcare facilities (hospital elevators), the primary requirement is to accommodate stretchers, gurneys, and patients in wheelchairs with accompanying medical staff. Standard architectural and engineering guidelines (such as those found in ASME A17.1 or general building codes for healthcare) specify that the clear door opening must be wide enough to allow a standard hospital bed/stretcher to enter and exit without obstruction. A clear opening width of 1,200mm (approximately 48 inches) is the industry standard for hospital elevators to ensure efficient patient transport and safety. Options a (1,100mm) and b (1,300mm) do not align with the standard metric specifications for these specific installations, and while 1,400mm (option d) would be functional, it exceeds the standard minimum requirement and is not the 'standard' baseline used in most engineering specifications.
JE Departmental • 2026 • Engineering Paper
1(one) horse-power is equal to:
A)
546 W
B)
656 W
C)
746 W
D)
none of the above
Horsepower (hp) is a unit of power. In the electrical and mechanical engineering context (specifically the Imperial or Mechanical horsepower), 1 hp is defined as the power needed to lift 550 pounds by 1 foot in 1 second. When converted to the International System of Units (SI), 1 mechanical horsepower is approximately equal to 745.7 watts, which is standardly rounded to 746 W for most engineering examinations.
Analysis of options:
- Option (a) 546 W: Incorrect value.
- Option (b) 656 W: Incorrect value.
- Option (c) 746 W: Correct standard conversion for 1 hp.
- Option (d) none of the above: Incorrect, as option (c) is the standard value.
JE Departmental • 2026 • Engineering Paper
Standard frequency of supply in India is:
A)
25Hz
B)
100 Hz
C)
50 Hz
D)
none of the above
In India, the standard frequency for the alternating current (AC) power supply is 50 Hz (Hertz). This means the current changes direction 100 times per second (50 full cycles). This standard is followed across the national grid to ensure compatibility between power generation, transmission, and consumer electrical appliances.
Analysis of options:
- Option (a) 25Hz: This was used in some early railway systems or specific industrial applications in the past, 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 standard for India, the UK, and most of Europe and Asia.
- Option (d) none of the above: Incorrect, as option (c) is correct.
JE Departmental • 2026 • Engineering Paper
Which of the following is the poorest conductor?
A)
aluminium
B)
copper
C)
iron
D)
carbon
Electrical conductivity is the measure of a material's ability to allow the flow of electric current. To determine the poorest conductor, we compare the electrical resistivity (or conductivity) of the given materials:
1. **Copper (b):** One of the best electrical conductors known, used extensively in wiring due to its very low resistivity.
2. **Aluminium (a):** An excellent conductor, though slightly less conductive than copper. It is widely used in overhead power lines due to its light weight.
3. **Iron (c):** A conductor, but significantly poorer than copper and aluminium. However, it is still a metallic conductor with relatively high free-electron mobility.
4. **Carbon (d):** Carbon (specifically in its graphite form) is a semi-metal/non-metal. While it can conduct electricity, its resistivity is orders of magnitude higher than that of metals like copper, aluminium, and iron.
Since carbon has the highest resistivity (and thus the lowest conductivity) among the listed options, it is the poorest conductor.
JE Departmental • 2026 • Engineering Paper
Which of the following processes is used in summer air conditioning?
A)
heating and humidification
B)
cooling and dehumidification
C)
humidification
D)
dehumidification
In summer air conditioning, the main objective is to remove heat and excess moisture from the air to make the indoor environment comfortable. Therefore, the air is typically subjected to cooling and dehumidification. Option (b) is correct.
Why the others are incorrect:
- (a) Heating and humidification are used in winter air conditioning, not summer.
- (c) Humidification alone adds moisture and does not address the high temperature of summer air.
- (d) Dehumidification alone removes moisture but does not cool the air, so it is incomplete for summer air conditioning.
JE Departmental • 2026 • Engineering Paper
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 the above
Transformer tanks are not filled with water because water is a good coolant but a very poor insulating medium. In a transformer, the insulating liquid must both remove heat and provide electrical insulation between live parts and the tank; water would cause short-circuiting and insulation failure. Hence option (c) is correct: it has good cooling property but bad insulating property. Option (a) is incorrect because water does not have good insulating property. Option (b) is incorrect because water does have good cooling property. Option (d) is incorrect because one of the given statements is correct.
JE Departmental • 2026 • Engineering Paper
If voltage supply to the energy meter is more than the rated value, energy meter will run:
A)
slow
B)
fast
C)
neither slow or fast
D)
none of the above
An induction-type energy meter operates based on the interaction of two magnetic fluxes: the driving flux (proportional to the load current) and the braking flux (proportional to the supply voltage). The speed of the aluminum disc is proportional to the power (P = VI cosφ). However, the braking torque is provided by the voltage coil. If the supply voltage increases beyond the rated value, the magnetic flux in the voltage coil increases. In a typical energy meter, the torque equation is such that the driving torque is proportional to the current and the voltage. If the voltage increases while the load remains constant, the energy meter tends to register more energy than is actually consumed because the increased voltage increases the flux and the resulting torque on the disc, causing it to rotate faster than it would at the rated voltage. Therefore, the meter runs 'fast'.
JE Departmental • 2026 • Engineering Paper
Continuity of an electrical circuit is usually checked by:
A)
ammeter
B)
voltmeter
C)
ohm meter
D)
all of above
Continuity testing is the process of verifying if an electrical path is complete (closed) or broken (open).
1. **Ohm meter (Correct):** An ohmmeter measures electrical resistance. In a continuity test, the meter sends a small current through the circuit. If the circuit is continuous, the resistance will be very low (close to 0 ohms), indicating a closed path. If the circuit is broken, the resistance will be infinite (OL), indicating an open circuit. This is the standard tool for checking continuity.
2. **Ammeter (Incorrect):** An ammeter measures the flow of current in a live circuit. Using an ammeter to check continuity is impractical and dangerous, as it requires the circuit to be powered, and a short circuit could blow the ammeter's fuse or damage the device.
3. **Voltmeter (Incorrect):** A voltmeter measures the potential difference between two points. While a voltmeter can sometimes be used to infer a break in a circuit (by detecting a voltage drop across a break), it does not directly measure the physical continuity of the conductor in the way an ohmmeter does.
Therefore, the ohmmeter is the specific instrument designed for this purpose.
JE Departmental • 2026 • Engineering Paper
The minimum width of single lane village road is:
A)
3.5m
B)
4.25m
C)
3.0m
D)
3.75m
In rural road standards, the minimum width of a single-lane village road is generally taken as 3.0 m. This is the standard carriageway width for a single-lane road in village/rural road practice. 3.5 m is typically used for a single-lane road carriageway in some highway standards, while 4.25 m and 3.75 m are larger widths and do not represent the minimum for a village road.
JE Departmental • 2026 • Engineering Paper
As construction material, plywood is preferred to thin planks of timber because of:
A)
Good strength and dimensional stability in both lateral and longitudinal directions
B)
Good dimensional stability in both longitudinal and lateral directions
C)
Good strength in both longitudinal and lateral directions
D)
Savings in cost and environmental considerations
Plywood is made by gluing thin layers of wood with adjacent grains at right angles. This cross-laminated structure reduces shrinkage and warping, giving plywood good strength and dimensional stability in both directions. Thin planks of timber are more prone to swelling, shrinking, splitting, and warping along the grain and across the grain. Option (b) is incomplete because it mentions only dimensional stability, while the key advantage also includes strength. Option (c) is less precise because plywood is valued not just for strength but especially for stability in both directions. Option (d) is not the main technical reason.
JE Departmental • 2026 • Engineering Paper
Complete hydration of cement the water cement ratio required is about:
A)
0.25
B)
0.37
C)
0.15
D)
0.40
For complete hydration of cement, the theoretical water-cement ratio required is about 0.38 by weight; in many exam questions it is taken approximately as 0.37. This amount is enough for both chemical combination and gel pore filling. Option (a) 0.25 is too low, (c) 0.15 is far below the requirement, and (d) 0.40 is slightly higher than the standard approximate value.
JE Departmental • 2026 • Engineering Paper
Roadway width (Formation width) for Major District Road (Single lane road) in mountainous and steep terrain is:
A)
12m
B)
4.75m
C)
6.25m
D)
4.0m
As per IRC standards for hill roads, the formation width for a single-lane Major District Road in mountainous and steep terrain is 4.0 m. The wider values given in the options (4.75 m and 6.25 m) correspond to larger or different roadway configurations/terrain classes, not this specific case.
JE Departmental • 2026 • Engineering Paper
The bond strength of steel and concrete is enhanced by various measures. Which of the following do not enhance bond strength?
A)
Deformed bars are used instead of Plain bars
B)
Bigger bar diameters are used
C)
Higher grade of concrete is used
D)
Increased cover is provided.
Bond strength between steel and concrete is improved by increasing mechanical interlock and adhesion. Deformed bars develop better bond than plain bars, higher-grade concrete generally provides better grip/anchorage, and smaller bar diameters are preferable to larger diameters because they usually give better bond. However, increased concrete cover does not enhance bond strength; it mainly improves protection against corrosion and fire. Therefore, the option that does not enhance bond strength is (d).
JE Departmental • 2026 • Engineering Paper
The standard weight of 1m length 20mm dia reinforcing bar is:
A)
1.59 kg
B)
1.58kg
C)
2.47 kgs
D)
2.8 kgs
The standard unit weight of a steel reinforcing bar is calculated by the formula d²/162, where d is the diameter in mm. For a 20 mm bar: 20²/162 = 400/162 ≈ 2.47 kg per metre. Hence, the correct answer is 2.47 kgs. The other options do not match the standard weight formula.