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Past Questions
The rotor of a 3 phase induction motor always runs at:
A)
Synchronous speed
B)
Less than synchronous speed
C)
More than synchronous speed
D)
None of these
In a 3-phase induction motor, the rotor must run slightly below synchronous speed so that there is relative motion between the rotating magnetic field and the rotor conductors. This relative motion induces rotor EMF and current, which produce torque. If the rotor reached synchronous speed, no EMF would be induced, so torque would become zero. Hence, the rotor always runs at less than synchronous speed.
The most commonly used semiconductor material is:
A)
Silicon
B)
Germanium
C)
Mixture of Silicon and Germanium
D)
None of the above
Silicon is the most commonly used semiconductor material because it is abundant, inexpensive, thermally stable, and has suitable electrical properties for making diodes, transistors, and integrated circuits. Germanium was used earlier, but silicon replaced it in most applications due to better temperature performance and lower leakage current. A mixture of silicon and germanium is used only in special devices, not as the most common material.
What is the major drawback of wind energy?
A)
High pollution
B)
It is not renewable
C)
Intermittent and variable nature
D)
Low energy efficiency
Wind energy’s main limitation is that it depends on wind availability, which is not constant. Since wind speed changes with weather and time, power generation is intermittent and variable. Option (a) is incorrect because wind energy is clean and has very low pollution during operation. Option (b) is incorrect because wind is a renewable source. Option (d) is not the major drawback; while efficiency and capacity factor matter, the key issue is the unpredictability of supply.
The Solar Cell efficiency is about:
A)
15%
B)
25%
C)
48%
D)
63%
A typical commercial solar cell has an efficiency of about 15% to 20%. Among the given options, 15% is the closest and standard exam answer. 25%, 48%, and 63% are much higher than usual efficiency values for ordinary solar cells.
The value of concentration ratio of flat plate collector is:
A)
1
B)
10
C)
100
D)
1000
A flat plate collector does not use optical concentration; it receives solar radiation directly over its entire absorber area. Therefore, its concentration ratio is 1. Concentrating collectors, such as parabolic troughs or dish collectors, have concentration ratios greater than 1, so options 10, 100, and 1000 are incorrect.
Transformer works on the principal of :
A)
Fleming’s right hand rule
B)
Fleming’s left hand rule
C)
Faraday’s law of Electromagnetic Induction
D)
Ohm’s law
A transformer operates on the principle of mutual induction, which is based on Faraday’s law of electromagnetic induction. When an alternating current flows in the primary winding, it produces a changing magnetic flux in the core, inducing an EMF in the secondary winding. Fleming’s rules apply to motors/generators, not transformers, and Ohm’s law does not describe the transformer’s working principle.
During soil erosion, the deposition occurs when :
A)
Transporting energy is less than resistance energy
B)
Transporting force is dominating the resisting forces
C)
Detachability is high
D)
Detachability is poor
Soil erosion is a three-stage process: detachment, transport, and deposition.
1. Detachment: Occurs when the erosive force (e.g., kinetic energy of raindrops or flowing water) exceeds the soil's cohesive strength (resistance energy).
2. Transport: Occurs when the transporting energy of the medium (water or wind) is sufficient to keep the particles in motion against the forces of gravity and friction.
3. Deposition: Occurs when the transporting energy (velocity/power of the medium) decreases to a point where it is no longer sufficient to overcome the resisting forces (such as gravity, friction, or obstacles).
Therefore, deposition happens specifically when the transporting energy becomes less than the resistance energy, causing the particles to settle out of the medium.
Analysis of other options:
- Option b: If transporting force dominates resisting forces, the soil remains in motion (transport stage) or is further eroded.
- Option c: High detachability leads to more soil being loosened (detachment stage), not deposited.
- Option d: Poor detachability means the soil is resistant to being moved, which prevents erosion from starting but does not trigger the deposition of already transported particles.
In a variable displacement pump, as the pumping head increases:
A)
the rate of pumping decreases
B)
the rate of pumping becomes ideal
C)
the rate of pumping increases
D)
Both (a) & (c)
In fluid mechanics and pump theory, the 'pumping head' refers to the total height or pressure the pump must overcome to move the fluid. For any real-world pump, there is an inverse relationship between the head (pressure) and the flow rate (discharge), as described by the pump's characteristic curve (H-Q curve). As the resistance or the required head increases, the pump's ability to move the same volume of fluid per unit of time decreases. In the specific case of a variable displacement pump, while the displacement can be adjusted to compensate for changes, the inherent physical property remains: for a given power input and setting, an increase in the head (back-pressure) leads to a decrease in the actual flow rate due to increased internal slippage (leakage) and the physics of the system's resistance. Therefore, as the pumping head increases, the rate of pumping (discharge) decreases.
Fuller’s rule is most appropriate for designing mixes where the objective is:
A)
High workability for pumped concrete
B)
Minimum cement content with dense packing
C)
High air entrainment
D)
Rapid setting
Fuller's Rule (also known as the Fuller-Thompson curve) is a mathematical approach used in concrete technology to determine the optimum grading of aggregates. The fundamental principle of Fuller's rule is to achieve the maximum possible density of the aggregate skeleton by ensuring that smaller particles fill the voids between larger particles.
1. Why Option (b) is correct: By achieving 'dense packing' (maximum density), the volume of voids between the aggregates is minimized. Since the cement paste is primarily used to fill these voids and bind the aggregates, a denser packing allows for the use of the minimum amount of cement paste required to fill the gaps, thereby reducing the cement content without compromising the strength or stability of the mix.
2. Why other options are incorrect:
- Option (a): Workability for pumped concrete depends more on the paste volume, use of superplasticizers, and the shape of the aggregates rather than the specific grading curve of Fuller.
- Option (c): Air entrainment is achieved by adding chemical admixtures (air-entraining agents) to create microscopic bubbles; it is not a function of aggregate grading rules.
- Option (d): Rapid setting is controlled by the chemistry of the cement (C3A content) or the addition of accelerators, not by the grading of the aggregates.
A spillway is designed for a peak flood discharge of 2000 cumecs. If the velocity of flow over the spillway is 8 m/s and the depth of flow is 2 metres, what is the required width of the spillway?
A)
100 m
B)
125 m
C)
200 m
D)
250 m
To find the required width of the spillway, we use the fundamental Continuity Equation for fluid flow:
Discharge (Q) = Area (A) × Velocity (V)
Given data:
- Peak flood discharge (Q) = 2000 m³/s (cumecs)
- Velocity of flow (V) = 8 m/s
- Depth of flow (d) = 2 m
Step 1: Define the Area (A) of the flow cross-section.
For a rectangular spillway section, Area = Width (B) × Depth (d).
Step 2: Substitute the Area into the discharge formula:
Q = (B × d) × V
Step 3: Rearrange the formula to solve for Width (B):
B = Q / (d × V)
Step 4: Plug in the values:
B = 2000 / (2 × 8)
B = 2000 / 16
B = 125 m
Therefore, the required width of the spillway is 125 metres.
Comparing with the options:
- Option (a) 100 m is incorrect (would result in Q = 1600 cumecs).
- Option (b) 125 m is correct.
- Option (c) 200 m is incorrect (would result in Q = 3200 cumecs).
- Option (d) 250 m is incorrect (would result in Q = 4000 cumecs).
The discharge per unit drawdown at a well is known as:
A)
specific yield
B)
specific retention
C)
specific capacity
D)
specific storage
In groundwater hydrology, the 'Specific Capacity' of a pumping well is defined as the discharge (Q) obtained per unit drawdown (s). Mathematically, it is expressed as Specific Capacity = Q/s. It represents the productivity of a well and depends on both the aquifer properties (transmissivity) and the well's construction efficiency.
Why other options are incorrect:
- a. Specific Yield: The ratio of the volume of water that a saturated rock will yield by gravity to the total volume of the rock.
- b. Specific Retention: The ratio of the volume of water a saturated rock retains against gravity to the total volume of the rock.
- d. Specific Storage: The volume of water released from storage per unit volume of aquifer per unit decline in hydraulic head.
Where will the centre of gravity of a uniform rod lie?
A)
At its middle point
B)
At its end
C)
At the centre of its cross-sectional area
D)
Depends upon its material
The centre of gravity (CG) is the point through which the entire weight of a body acts, regardless of its orientation. For a body with a uniform distribution of mass (homogeneous material) and a symmetrical shape, the centre of gravity coincides with its geometric centre (centroid). A 'uniform rod' implies that the mass per unit length is constant throughout its length. For a straight rod of length L, the geometric centre is located at L/2, which is the middle point of the rod.
Why other options are incorrect:
- Option b: The end of the rod is not the balance point; the rod would rotate if supported there.
- Option c: The centre of the cross-sectional area describes the axis of the rod, but not the specific point along the length where the total weight is concentrated.
- Option d: Since the rod is specified as 'uniform', the material is consistent throughout; therefore, the CG depends on the geometry, not the specific type of material used.
Rainfall occurring when moist air rises over a mountain barrier is known as:
A)
Cyclonic rainfall
B)
Convectional rainfall
C)
Orographic rainfall
D)
Frontal rainfall
The question describes a specific meteorological process where moist air is forced to rise over a physical barrier, such as a mountain range.
1. **Orographic Rainfall (Correct):** The term 'oro' comes from the Greek word for mountain. When moist air encounters a mountain, it is forced to ascend (the windward side). As the air rises, it cools adiabatically, leading to condensation and precipitation. This is the exact process described in the question.
2. **Cyclonic/Frontal Rainfall (Incorrect):** These terms are often used interchangeably. This type of rainfall occurs when two air masses of different temperatures and densities meet (a front), forcing the warmer, lighter air to rise over the colder, denser air.
3. **Convectional Rainfall (Incorrect):** This occurs when the sun heats the ground, which in turn heats the air above it. The warm air becomes less dense and rises vertically in currents, cooling and condensing to form clouds (common in tropical regions).
Therefore, the correct term for rainfall caused by a mountain barrier is Orographic rainfall.
Centrifugal pumps transport fluids by converting:
A)
kinetic energy to hydrodynamic energy
B)
hydrodynamic energy to kinetic energy
C)
mechanical energy to kinetic energy
D)
mechanical energy to hydrodynamic energy
A centrifugal pump operates by using an external power source (such as an electric motor or diesel engine) to rotate an impeller. This rotation provides mechanical energy to the fluid. As the impeller spins, it imparts centrifugal force to the liquid, increasing its velocity (kinetic energy) and pressure (potential energy). The combination of pressure and kinetic energy in a flowing fluid is referred to as hydrodynamic energy. Therefore, the pump converts the mechanical energy of the shaft into hydrodynamic energy of the fluid.
Breakdown of options:
- Option (a) and (b) are incorrect because kinetic energy is a component of hydrodynamic energy, not a separate category to be converted into it in this context.
- Option (c) is partially correct in that mechanical energy becomes kinetic energy, but the ultimate goal of a pump is to create 'head' (pressure), making 'hydrodynamic energy' the more comprehensive and accurate term for the resulting energy state of the fluid.
In which of the following type of surveying only linear measurements are made?
A)
Dumpy level
B)
Chain surveying
C)
Theodolite surveying
D)
Contouring
Step-by-step reasoning: 1. The question asks for a type of surveying where ONLY linear measurements are made. 2. Chain Surveying (also known as Chain and Compass surveying when angles are involved, but pure Chain Surveying relies on triangulation) is the simplest method of surveying where only linear measurements are taken in the field using chains or tapes to determine the positions of points. 3. Dumpy Leveling (Option a) is used primarily for determining vertical distances (elevations) and relative heights. 4. Theodolite Surveying (Option c) involves the measurement of both horizontal and vertical angles, as well as linear distances. 5. Contouring (Option d) is the process of mapping lines of equal elevation, which requires both leveling (vertical) and distance (horizontal) measurements. Therefore, only Chain Surveying fits the criteria of relying solely on linear measurements.
The “Area of Influence” of a well is best defined as:
A)
The total surface area of the well pipe.
B)
The area within which the water table or piezometric surface is lowered by pumping.
C)
The depth of the water-bearing stratum.
D)
The distance from the pump to the discharge point
In groundwater hydrology, when a well is pumped, it creates a depression in the water table (for unconfined aquifers) or the piezometric surface (for confined aquifers) known as a 'cone of depression'. The 'Area of Influence' (or Radius of Influence) refers to the spatial extent around the pumping well where the hydraulic head is affected by the pumping action. Beyond this boundary, the water table remains at its original static level.
Analysis of options:
- Option (a) is incorrect because the surface area of the pipe is a physical dimension of the hardware, not a hydrological zone.
- Option (c) is incorrect because the depth of the stratum refers to the thickness of the aquifer, not the lateral area affected by pumping.
- Option (d) is incorrect because the distance to the discharge point is a matter of plumbing/piping layout and has no relation to the aquifer's hydraulic response.
The relation between Transmissibility (T) and Permeability (K) for an aquifer of depth d, is:
A)
K=Td
B)
T=K.d
C)
T=K. log d
D)
T= ln(Kd)
In groundwater hydrology, Transmissibility (T) is defined as the rate at which water is transmitted through a unit width of an aquifer under a unit hydraulic gradient. It is a measure of the aquifer's capacity to transmit water. Mathematically, Transmissibility is the product of the Hydraulic Conductivity (K), also known as permeability in some contexts, and the saturated thickness or depth (d) of the aquifer. The formula is expressed as T = K × d.
- Option (a) K=Td is incorrect as it suggests permeability is the product of transmissibility and depth.
- Option (c) T=K. log d is incorrect as the relationship is linear, not logarithmic.
- Option (d) T= ln(Kd) is incorrect as it suggests a natural logarithmic relationship, which does not exist for these parameters.
A centrifugal pump operates at 1800 rpm and discharges 0.72m3/min of water at 6m head. If the pump operates at 2100 rpm, then the discharge will be:
A)
0.61 m3/min
B)
0.77 m3/min
C)
0.84 m3/min
D)
0.98 m3/min
To solve this problem, we use the Affinity Laws for centrifugal pumps. The Affinity Laws describe the relationship between the speed of the pump (N), the discharge flow rate (Q), and the head (H).
1. **The Affinity Law for Discharge (Q):**
The discharge of a centrifugal pump is directly proportional to the rotational speed of the impeller.
Formula: Q1 / Q2 = N1 / N2
2. **Given Data:**
- Initial speed (N1) = 1800 rpm
- Initial discharge (Q1) = 0.72 m³/min
- Final speed (N2) = 2100 rpm
- Final discharge (Q2) = ?
3. **Calculation:**
Rearranging the formula to solve for Q2:
Q2 = Q1 * (N2 / N1)
Q2 = 0.72 * (2100 / 1800)
Q2 = 0.72 * (7 / 6)
Q2 = 0.12 * 7
Q2 = 0.84 m³/min
4. **Analysis of Options:**
- Option (a) 0.61 m³/min: Incorrect. This would occur if the speed decreased.
- Option (b) 0.77 m³/min: Incorrect calculation.
- Option (c) 0.84 m³/min: Correct. Matches the calculated value based on the Affinity Law.
- Option (d) 0.98 m³/min: Incorrect calculation.
Note: The head (6m) provided in the question is extraneous information for calculating the discharge, although it would change proportionally to the square of the speed (H2 = H1 * (N2/N1)²).
A drop spillway is most suitable when the:
A)
slope is very gentle
B)
discharge is very large
C)
vertical drop is significant
D)
soil is highly permeable
A drop spillway is used where the channel or ground has a sudden, appreciable fall in elevation, i.e., a significant vertical drop. It safely conveys water from a higher to a lower level by allowing the flow to drop over a crest and dissipate energy.
Why the others are incorrect:
- b) Very large discharge: Drop spillways are not chosen primarily because of large discharge; they are more suited to small to moderate flows with a distinct drop.
- a) Very gentle slope: A gentle slope does not require a drop structure; it is not the defining condition for a drop spillway.
- d) Highly permeable soil: Soil permeability is not the criterion for selecting a drop spillway; this relates more to seepage control and foundation considerations, not the need for a drop structure.
Revised Universal Soil Loss Equation was developed by:
A)
Wischmeir and Smith
B)
Wischmeir and Kenneth
C)
Kenneth et. al
D)
Williams
To solve this question, it is important to distinguish between the Universal Soil Loss Equation (USLE) and the Revised Universal Soil Loss Equation (RUSLE).
1. **USLE (Universal Soil Loss Equation):** This was developed by **Wischmeir and Smith** in 1978. It provided a mathematical model to predict the average annual soil loss from a given field. Therefore, option (a) is incorrect because it refers to the original USLE, not the Revised version.
2. **RUSLE (Revised Universal Soil Loss Equation):** The original USLE had limitations, particularly in how it handled rainfall erosivity and cover management. To address these, **William H. Williams** developed the Revised Universal Soil Loss Equation (RUSLE) in the 1980s and early 1990s. RUSLE introduced more sophisticated calculations for the R (rainfall erosivity) and C (cover-management) factors, making it more accurate for various climatic conditions and land uses.
3. **Other Options:** Options (b) and (c) are incorrect as they do not correspond to the recognized developers of this specific soil erosion model.
Therefore, the correct developer of the Revised Universal Soil Loss Equation is Williams.
In LUCC, the suitable lands for cultivation are in:
A)
Class I to IV
B)
Class I to V
C)
Class V to VII
D)
Class I to III
In Land Use Capability Classification (LUCC), land is categorized into eight classes (I through VIII) based on its limitations for cultivation.
1. Classes I to IV are considered 'arable' or suitable for cultivation.
- Class I: Very good land with few limitations.
- Class II: Good land with slight limitations.
- Class III: Moderately good land with moderate limitations.
- Class IV: Fairly good land with severe limitations (often used for limited cultivation or pasture).
2. Classes V to VIII are generally considered 'non-arable' or unsuitable for sustainable cultivation due to severe limitations such as steep slopes, poor drainage, or rocky soil.
- Class V to VII are typically used for grazing, forestry, or wildlife habitats.
- Class VIII is reserved for recreation or wildlife.
Therefore, the range of lands suitable for cultivation is Class I to IV.
The length of a survey line measured with a 30 cm chain was found to be 350 m. Calculate the true length of the line if the chain was 12 cm too long.
A)
349.92 m
B)
350.8 m
C)
351.4 m
D)
352.2 m
To find the true length of a line when the measuring chain is incorrect, we use the standard formula:
True Length = (L' / L) × Measured Length
1. **Identify the given values:**
- Nominal length of the chain (L) = 30 m (Note: The question states '30 cm chain', but in the context of a 350 m survey line, this is a typographical error for a 30 m chain. A 30 cm chain would be impractical for such a distance).
- Error in chain = 12 cm = 0.12 m (too long).
- Actual length of the chain (L') = L + error = 30 m + 0.12 m = 30.12 m.
- Measured length of the line = 350 m.
2. **Apply the formula:**
True Length = (30.12 / 30) × 350
True Length = 1.004 × 350
True Length = 351.4 m
3. **Conclusion:**
Since the chain was longer than the standard length, each 'unit' measured was actually longer than 30 m, meaning the actual distance covered is greater than the recorded measurement. Therefore, 351.4 m is the correct true length. Option (c) matches this calculation.
The drilling fluid commonly used in hydraulic rotary drilling is:
A)
Cement slurry
B)
Bentonite mud
C)
Diesel
D)
Lime water
In hydraulic rotary drilling, a drilling fluid (often called 'drilling mud') is circulated through the drill pipe and returned up the annulus. Bentonite, a type of swelling clay, is the most commonly used material for this purpose.
1. Why Bentonite Mud is correct: Bentonite creates a viscous fluid that serves several critical functions: it carries drill cuttings (rock fragments) from the bottom of the hole to the surface; it forms a 'filter cake' on the borehole walls to prevent the fluid from leaking into the surrounding formation and to prevent the walls from collapsing; and it cools and lubricates the drill bit.
2. Why other options are incorrect:
- a. Cement slurry: This is used for 'grouting' or sealing the annulus after the drilling is complete to prevent surface contamination of the aquifer, not as a drilling fluid during the drilling process.
- c. Diesel: While used in some specialized oil drilling operations (oil-based muds), it is not the common fluid for standard groundwater hydrology/water well drilling due to environmental contamination risks.
- d. Lime water: This does not possess the necessary viscosity or thixotropic properties required to transport cuttings or stabilize the borehole walls.
Revised Universal Soil Loss Equation estimates:
A)
Average sediment yield
B)
Average annual sediment yield
C)
Average annual soil erosion
D)
All (a), (b) & (c)
The Revised Universal Soil Loss Equation (RUSLE) is a mathematical model used to predict the long-term average annual rate of soil erosion caused by rainfall and overland flow. The formula is expressed as A = R × K × LS × C × P, where 'A' represents the average annual soil loss per unit area.
1. Why 'c' is correct: RUSLE specifically calculates 'gross erosion' (the total amount of soil detached and transported from a slope), which is defined as the average annual soil erosion.
2. Why 'a' and 'b' are incorrect: There is a critical technical distinction between 'soil erosion' and 'sediment yield'. Soil erosion refers to the detachment and transport of soil from a site, whereas sediment yield refers to the actual amount of sediment that reaches a specific point (like a stream or a reservoir) after accounting for deposition during transport. RUSLE estimates the erosion at the source, not the yield at the outlet. Therefore, it does not estimate sediment yield.
Detachment of soil particles is maximum, when the depth of overland flow is between:
A)
1/5 and 1/3 of drop diameter
B)
1/5 and 1/3 of soil’s particle’s diameter
C)
Both (a) & (b)
D)
0.15 to 0.20mm
In the study of soil erosion, specifically rainsplash erosion, the detachment of soil particles is influenced by the interaction between falling raindrops and the existing layer of overland flow (sheet flow). When there is no water on the surface, the raindrop's energy is absorbed directly by the soil. When the water depth is very high, the water acts as a cushion, absorbing the impact energy. Research in soil mechanics and hydrology indicates that the maximum detachment of soil particles occurs when the depth of the overland flow is between 1/5 and 1/3 of the diameter of the falling raindrop. At this specific depth, the impact energy is optimally transmitted to the soil particles to dislodge them, while the flow is sufficient to transport them away, but not deep enough to fully cushion the impact. Option (b) is incorrect because soil particle diameter is far too small to be the governing factor for raindrop impact energy. Option (d) is a specific measurement that does not account for the variability in raindrop sizes.