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Past Questions
Specific heat of air at constant pressure is equal to
A)
0.17
B)
0.21
C)
0.24
D)
1.0
The specific heat of air at constant pressure (Cp) is a standard thermodynamic property. For air at room temperature (approximately 25°C or 298K), the value of Cp is approximately 1.005 kJ/kg·K. Looking at the provided options, 1.0 is the closest and most accurate approximation.
Breakdown of values for air:
- Specific heat at constant pressure (Cp) ≈ 1.005 kJ/kg·K
- Specific heat at constant volume (Cv) ≈ 0.718 kJ/kg·K
- Gas constant (R) = Cp - Cv ≈ 0.287 kJ/kg·K
Comparing this to the options:
- Option a (0.17), b (0.21), and c (0.24) are significantly lower than the known physical constant for air and do not correspond to Cp or Cv.
Thermal power plant works on
A)
Rankine cycle
B)
Otto cycle
C)
Brayton cycle
D)
Carnot cycle
A thermal power plant (steam power plant) operates on the Rankine cycle, which is the idealized thermodynamic cycle for a steam power plant. In this cycle, water is pumped to high pressure, heated in a boiler to become superheated steam, expanded through a turbine to produce work, and finally condensed back into water in a condenser.
Why other options are incorrect:
- b. Otto cycle: This is the ideal cycle for spark-ignition (petrol) internal combustion engines.
- c. Brayton cycle: This is the ideal cycle for gas turbine power plants or jet engines.
- d. Carnot cycle: This is a theoretical, idealized cycle that provides the maximum possible efficiency for any heat engine operating between two temperatures, but it cannot be practically implemented in a power plant due to the difficulty of achieving isothermal heat transfer and isentropic processes in a real-world system.
Iso-octane has octane number of
A)
zero
B)
50
C)
75
D)
100
The Octane Number is a standard measure of the antiknock properties of a fuel, specifically gasoline. It is defined based on a scale using two reference hydrocarbons: n-heptane and iso-octane.
1. Iso-octane (2,2,4-trimethylpentane) is highly resistant to knocking (pre-ignition) under compression, making it the gold standard for high-quality fuel. Therefore, it is assigned an octane number of 100.
2. n-heptane is very prone to knocking and is assigned an octane number of 0.
Any fuel's octane number is the percentage by volume of iso-octane in a mixture of iso-octane and n-heptane that has the same knocking characteristics as the fuel being tested. For example, a fuel with an octane number of 87 behaves like a mixture of 87% iso-octane and 13% n-heptane.
Therefore:
- Option (a) zero refers to n-heptane.
- Option (b) and (c) are arbitrary values for specific fuel blends.
- Option (d) 100 is the defined value for pure iso-octane.
A 75cc engine has the following parameters as 75cc
A)
swept volume
B)
clearance volume
C)
cylinder volume
D)
none of the above
In internal combustion engine terminology, when an engine is described by a specific displacement (e.g., a '75cc engine' or a '1500cc car'), that value refers to the engine's displacement volume. Displacement volume is the volume swept by the piston as it moves from Bottom Dead Center (BDC) to Top Dead Center (TDC), which is formally known as the 'swept volume'.
1. Swept Volume (Vs): The volume displaced by the piston during one stroke. This is the standard measure used to categorize engine size.
2. Clearance Volume (Vc): The small volume remaining in the cylinder when the piston is at TDC. This is much smaller than the swept volume.
3. Cylinder Volume (Total Volume): The sum of the swept volume and the clearance volume (Vt = Vs + Vc).
Since the engine is identified as a '75cc engine', 75cc represents the swept volume.
The process of sublimation is found to occur in
A)
liquid Nitrogen
B)
solid CO2
C)
steel
D)
air
Sublimation is the phase transition where a substance changes directly from a solid state to a gaseous state without passing through the intermediate liquid phase. This occurs when the vapor pressure of the solid exceeds the surrounding atmospheric pressure before the substance reaches its melting point.
1. **Solid CO2 (Dry Ice)**: Carbon dioxide does not exist as a liquid at standard atmospheric pressure (1 atm). It sublimes directly from a solid to a gas at -78.5°C. This is why it is commonly referred to as 'dry ice'.
2. **Liquid Nitrogen**: This is already in a liquid state. While it evaporates (liquid to gas), it does not undergo sublimation (solid to gas) under standard conditions.
3. **Steel**: Steel is an alloy that melts at very high temperatures to become a liquid; it does not sublime under normal or industrial conditions.
4. **Air**: Air is a mixture of gases. While the components of air can be liquefied or solidified at extremely low temperatures, 'air' as a whole is not a substance that undergoes sublimation in this context.
The air standard efficiency of an Otto cycle compared to diesel cycle for the given compression ratio is
A)
same
B)
more
C)
less
D)
none of the above
To compare the air standard efficiency of the Otto cycle and the Diesel cycle for the same compression ratio (r), we examine their efficiency formulas:
1. Otto Cycle Efficiency: η_otto = 1 - (1 / (r^(γ-1)))
2. Diesel Cycle Efficiency: η_diesel = 1 - [1 / (r^(γ-1))] * [(ρ^γ - 1) / (γ(ρ - 1))]
Where:
- r = Compression ratio
- γ = Adiabatic index (ratio of specific heats)
- ρ = Cut-off ratio (for Diesel cycle)
Comparing the two formulas, the Diesel cycle efficiency is the Otto cycle efficiency multiplied by a factor: [(ρ^γ - 1) / (γ(ρ - 1))].
Since the cut-off ratio (ρ) is always greater than 1 for a functioning Diesel engine, the term [(ρ^γ - 1) / (γ(ρ - 1))] is always greater than 1. This means that for the same compression ratio, the term subtracted from 1 in the Diesel cycle formula is larger than the term subtracted in the Otto cycle formula. Consequently, the resulting efficiency of the Diesel cycle is lower than that of the Otto cycle.
Therefore, for a given compression ratio, the Otto cycle is more efficient than the Diesel cycle.
Triple point of a pure substance on P-V diagram is represented by a
A)
point
B)
curve
C)
line
D)
triangle
In thermodynamics, the triple point of a pure substance is the unique state of temperature and pressure at which three phases (solid, liquid, and gas) coexist in thermodynamic equilibrium. According to Gibbs' Phase Rule (F = C - P + 2), for a pure substance (C=1) with three phases (P=3), the degrees of freedom (F) are: F = 1 - 3 + 2 = 0. This means the triple point is an invariant point; it occurs at a single, specific pressure and temperature. On any thermodynamic coordinate system, such as a P-V (Pressure-Volume) diagram, a P-T (Pressure-Temperature) diagram, or a T-V diagram, a state with zero degrees of freedom is represented as a single, discrete point. Therefore, the triple point is represented by a point, not a line or a curve.
In a diesel engine, the fuel is ignited by
A)
spark
B)
injected fuel
C)
combustion chamber
D)
none of the above
In a diesel engine (Compression Ignition engine), the ignition process occurs as follows: 1. Air is drawn into the cylinder and compressed to a very high ratio. 2. This high compression causes the air temperature to rise significantly, often exceeding the self-ignition temperature of the diesel fuel. 3. When the fuel is injected into this hot compressed air within the combustion chamber, it ignites spontaneously. Therefore, the heat generated within the combustion chamber (due to compression) is what ignites the fuel.
Analysis of options:
- (a) Spark: This is used in Petrol (SI) engines, not Diesel engines.
- (b) Injected fuel: The fuel is the substance being ignited, not the agent that causes the ignition.
- (c) Combustion chamber: This is the correct context, as the heat of compression within the chamber triggers the ignition.
- (d) None of the above: Incorrect, as option (c) describes the mechanism of Compression Ignition.
Critical pressure of a liquid is the pressure
A)
above which liquid will remain liquid
B)
above which liquid becomes gas
C)
above which liquid becomes vapour
D)
none of the above
To solve this, we must define the 'Critical Point' in thermodynamics. The critical point is the end point of the phase equilibrium curve between liquid and vapor. The Critical Pressure (Pc) is the pressure at the critical temperature (Tc) above which a gas cannot be liquefied, regardless of how much it is cooled.
Analyzing the options:
- Option (a): Incorrect. Increasing pressure generally favors the liquid phase, but the critical pressure is not a threshold 'above which' a substance simply 'remains liquid' regardless of temperature.
- Option (b) and (c): Incorrect. These options suggest that increasing pressure causes a liquid to become a gas or vapor. In reality, increasing pressure typically compresses a gas into a liquid. The critical pressure is actually the pressure above which the distinction between liquid and gas disappears, resulting in a 'supercritical fluid'.
Since none of the options correctly describe the critical pressure (which is the pressure above which a gas cannot be liquefied by cooling, or the pressure at which the liquid and vapor phases become indistinguishable), the correct choice is 'none of the above'.
Ignition quality of petrol is expressed by
A)
cetane number
B)
octane number
C)
calorific value
D)
none of the above
The ignition quality of a fuel refers to its resistance to premature ignition (knocking).
1. **Octane Number (Option b):** This is the standard measure of the ignition quality of petrol (gasoline). It indicates the fuel's ability to resist 'knocking' or 'pinging' during combustion in a spark-ignition (SI) engine. A higher octane number means the fuel can withstand higher compression before detonating, allowing for higher engine efficiency. It is based on a scale comparing the fuel to iso-octane (100) and n-heptane (0).
2. **Cetane Number (Option a):** This is used to measure the ignition quality of diesel fuel (compression-ignition engines). It indicates the ignition delay period; a higher cetane number means a shorter ignition delay, which is desirable for diesel engines.
3. **Calorific Value (Option c):** This refers to the amount of heat energy released during the complete combustion of a specified amount of fuel. While it determines the energy content, it does not describe the 'ignition quality' or the tendency to knock.
Therefore, for petrol, the correct measure is the octane number.
For reversible adiabatic process, change in entropy is
A)
maximum
B)
minimum
C)
zero
D)
none of the above
Step-by-step reasoning:
1. Definition of Entropy Change: The change in entropy (dS) of a system is defined by the Clausius inequality/equation as dS = dQ/T for a reversible process, where dQ is the heat transferred and T is the absolute temperature.
2. Adiabatic Process: By definition, an adiabatic process is one in which there is no heat transfer between the system and its surroundings (dQ = 0).
3. Reversibility: For a process to be reversible and adiabatic, it must be quasi-static and frictionless.
4. Calculation: Substituting dQ = 0 into the entropy equation: dS = 0 / T = 0.
5. Conclusion: A process that is both reversible and adiabatic is called an 'isentropic process' (iso = same, entropic = entropy). Therefore, the change in entropy is zero.
Why other options are incorrect:
- Option (a) and (b): Entropy change is not maximum or minimum; it is exactly zero for this specific condition.
- Option (d): Incorrect because option (c) is the scientifically accurate answer.
On volume basis, air contains following parts of oxygen
A)
21
B)
23
C)
25
D)
77
Dry air is a mixture of several gases. On a volume basis (which is equivalent to a molar basis according to Avogadro's Law), the composition of Earth's atmosphere is approximately:
1. Nitrogen (N₂): ~78.08%
2. Oxygen (O₂): ~20.95% (commonly rounded to 21%)
3. Argon (Ar): ~0.93%
4. Carbon Dioxide (CO₂): ~0.04%
Analyzing the options:
- Option (a) 21: This is the standard accepted percentage of oxygen in the air by volume.
- Option (b) 23: This is incorrect; it does not correspond to any major atmospheric gas.
- Option (c) 25: This is incorrect.
- Option (d) 77: This is close to the percentage of Nitrogen (~78%), not Oxygen.
Which of the following parameters remains constant during ideal throttling process?
A)
Pressure
B)
Temperature
C)
Enthalpy
D)
Entropy
An ideal throttling process occurs when a fluid flows through a restriction (like a partially closed valve or a porous plug) without any heat transfer (adiabatic) and without any work being done (no moving parts).
1. **Steady Flow Energy Equation (SFEE):** For a steady flow process, the energy equation is:
h1 + PE1 + (V1²/2) = h2 + PE2 + (V2²/2) + q - w
2. **Assumptions for Throttling:**
- Adiabatic process: q = 0
- No work done: w = 0
- Change in kinetic energy is negligible: (V1² - V2²) ≈ 0
- Change in potential energy is negligible: (PE1 - PE2) ≈ 0
3. **Result:** The equation simplifies to h1 = h2. Therefore, the enthalpy remains constant during an ideal throttling process. This is known as an isenthalpic process.
**Why other options are incorrect:**
- **Pressure (a):** Pressure always drops significantly across a throttling device; this is the primary purpose of throttling.
- **Temperature (b):** Temperature usually changes. For an ideal gas, since enthalpy depends only on temperature, it would remain constant, but for real gases or liquids (like in a Joule-Thomson expansion), the temperature changes.
- **Entropy (d):** Throttling is an irreversible process due to friction and turbulence. According to the Second Law of Thermodynamics, entropy must increase (s2 > s1).
One ton of refrigeration is equal to the refrigeration effect corresponding to melting of 1000 Kg of ice in
A)
24 hours
B)
12 hours
C)
8 hours
D)
10 hours
The term 'Ton of Refrigeration' (TR) is a unit of power used in refrigeration and air conditioning. It is historically defined as the rate of heat removal required to freeze one short ton (2000 lbs) of water at 0°C into ice at 0°C within a period of 24 hours. In metric terms, this corresponds to the refrigeration effect produced by melting 1000 kg (1 metric tonne) of ice at 0°C into water at 0°C over a period of 24 hours.
Calculation:
1. Latent heat of fusion of ice = 334 kJ/kg (approximately).
2. Total heat for 1000 kg = 1000 kg * 334 kJ/kg = 334,000 kJ.
3. Time period = 24 hours = 24 * 3600 seconds = 86,400 seconds.
4. Refrigeration effect (Power) = 334,000 kJ / 86,400 s ≈ 3.86 kW (Standard TR is often approximated as 3.5 kW or 12,000 BTU/hr depending on the ton definition used).
Therefore, the standard definition specifically requires the time interval to be 24 hours. Options b, c, and d are incorrect as they do not align with the standard definition of a 'Ton' of refrigeration.
Which of the following quantities is not the property of the system?
A)
Pressure
B)
Temperature
C)
Density
D)
Heat
In thermodynamics, a 'property' of a system is a characteristic that depends only on the current state of the system and not on the path taken to reach that state (state functions).
1. Pressure (a), Temperature (b), and Density (c) are all state properties. If you know the state of a substance, these values are fixed and uniquely defined regardless of how the substance reached that state.
2. Heat (d) and Work are not properties of a system; they are 'path functions'. Heat is energy in transit across the boundary of a system due to a temperature difference. A system does not 'contain' heat; it contains internal energy. Heat only exists during a process (the transfer of energy). Therefore, heat cannot be a property of a system because it depends on the process/path taken to change the system's state.
Since the question asks which quantity is NOT a property, the correct answer is Heat.
Which of the following is not an extensive property?
A)
Entropy
B)
Enthalpy
C)
Internal energy
D)
Density
In thermodynamics, properties are classified into two categories: Extensive and Intensive properties.
1. Extensive Properties: These are properties whose values depend on the amount of matter (mass) present in the system. If the system is divided into two equal parts, the value of an extensive property is halved. Examples include Mass, Volume, Entropy, Enthalpy, and Internal Energy.
2. Intensive Properties: These are properties that are independent of the amount of matter present in the system. They remain constant regardless of the size of the system. Examples include Temperature, Pressure, and Density.
Analyzing the options:
- a. Entropy: Depends on the mass of the system (Extensive).
- b. Enthalpy: Depends on the mass of the system (Extensive).
- c. Internal energy: Depends on the mass of the system (Extensive).
- d. Density: Defined as mass per unit volume (ρ = m/V). Since it is the ratio of two extensive properties, the mass cancels out, making density independent of the amount of matter (Intensive).
Therefore, Density is not an extensive property.
Vapour compression refrigeration is somewhat like
A)
Carnot cycle
B)
Rankine cycle
C)
Reversed Rankine cycle
D)
none of the above
The Vapour Compression Refrigeration Cycle (VCRC) consists of four main processes: isentropic compression, constant pressure heat rejection (condensation), isenthalpic expansion, and constant pressure heat absorption (evaporation). This sequence of processes is fundamentally the reverse of the Rankine cycle, which is used in steam power plants to produce work. In the Rankine cycle, water is pumped (compression), heated to steam (evaporation), expanded in a turbine to produce work, and then condensed. In VCRC, the cycle is reversed to move heat from a low-temperature region to a high-temperature region by consuming work. While the Carnot cycle is the theoretical ideal for any refrigeration cycle, the actual mechanical implementation of VCRC (using a compressor and expansion valve) closely mirrors the reversed Rankine cycle. Specifically, the use of a liquid-suction expansion valve instead of a turbine (which would be used in a true reversed Carnot cycle) makes it a practical application of the reversed Rankine principle.
Which of the following cycles uses air as the refrigerant?
A)
Ericsson
B)
Bell-Coleman
C)
Carnot
D)
Stirling
The Bell-Coleman cycle, also known as the Reversed Brayton cycle or the Air Refrigeration cycle, specifically uses air as the working fluid (refrigerant). It operates on the principle of compressing air, cooling it at constant pressure, expanding it to lower the temperature, and then absorbing heat from the space to be cooled.
Analysis of other options:
- a. Ericsson Cycle: This is a theoretical thermodynamic cycle that uses regeneration; while it can use gas, it is not the standard 'air refrigeration cycle' used in practical applications like aircraft cooling.
- c. Carnot Cycle: This is an ideal, theoretical cycle that defines the maximum possible efficiency for any heat engine or refrigerator regardless of the working fluid used.
- d. Stirling Cycle: This is a theoretical cycle used primarily for heat engines (not refrigeration) utilizing a regenerator and a gaseous working fluid, but it is not the standard cycle associated with air refrigeration systems.
One ton refrigeration corresponds to
A)
50 KCal/hr
B)
50 KCal/min
C)
50 KCal/sec
D)
100 KCal/min
1. Definition: One Ton of Refrigeration (TR) is defined as the amount of heat removal required to freeze one short ton (2000 lbs) of pure water at 0°C into ice at 0°C in 24 hours.
2. Calculation in SI/Imperial units: The latent heat of fusion of ice is approximately 144 BTU/lb. Therefore, 1 TR = (2000 lbs × 144 BTU/lb) / 24 hours = 12,000 BTU/hr.
3. Conversion to Calories: 1 BTU is approximately equal to 252 calories. Thus, 1 TR = 12,000 × 252 = 3,024,000 kcal/hr.
4. Converting to per minute: To find the value per minute, divide by 60: 3,024,000 kcal / 60 min = 50,400 kcal/min.
5. Approximation: In engineering textbooks and competitive exams, this is commonly approximated to 50 kcal/min (or more precisely 50.4 kcal/min) for simplicity.
6. Evaluating options:
- Option (a) 50 KCal/hr is far too low.
- Option (b) 50 KCal/min matches the calculated approximation (50.4 kcal/min).
- Option (c) 50 KCal/sec would be 3000 kcal/min, which is too high.
- Option (d) 100 KCal/min is double the required value.
The relative coefficient of performance is
A)
Actual COP/Theoretical COP
B)
Theoretical COP/Actual COP
C)
Actual COP × Theoretical COP
D)
none of the above
In thermodynamics and refrigeration, the Coefficient of Performance (COP) is a measure of the efficiency of a refrigeration system. The 'Theoretical COP' (often referred to as the Carnot COP) represents the maximum possible efficiency that any refrigeration system can achieve operating between two specific temperature limits. The 'Actual COP' is the efficiency achieved by a real-world system, which is always lower than the theoretical COP due to irreversibilities such as friction, pressure drops, and heat losses. The 'Relative COP' (also known as the relative efficiency) is defined as the ratio of the actual performance to the maximum theoretical performance. Mathematically: Relative COP = (Actual COP) / (Theoretical COP). This value is always less than or equal to 1 (or 100%).
Specific heat of air at constant volume Cv is equal to
A)
0.17
B)
0.21
C)
0.24
D)
1.0
Cv of air ~0.171 kcal/kgK =0.17; Cp~0.24.
The brake mean effective pressure of an I.C. engine with increase in speed will
A)
increase
B)
decrease
C)
remain inaffected
D)
none of the above
Brake Mean Effective Pressure (BMEP) is a theoretical constant pressure that, if acting on the piston for the entire power stroke, would produce the same brake power as the actual engine. The formula for Brake Power (BP) is: BP = (BMEP × L × A × n) / (k), where L is stroke length, A is area of piston, n is number of power strokes per second, and k is a constant. While Brake Power generally increases with speed (RPM) up to a certain point, BMEP behaves differently. As engine speed increases, several losses increase: 1. Volumetric Efficiency: At higher speeds, the time available to fill the cylinder with air-fuel mixture decreases, leading to a drop in volumetric efficiency and a reduction in the mass of charge per cycle. 2. Frictional Losses: Mechanical friction (piston rings, bearings) increases significantly with speed. 3. Pumping Losses: The energy required to push exhaust gases out and pull fresh air in increases. Because BMEP represents the 'net' pressure available after these losses, the combined effect of decreased volumetric efficiency and increased frictional/pumping losses causes the BMEP to decrease as speed increases beyond the peak torque point. Therefore, for a given engine configuration, BMEP typically decreases with an increase in speed.
A full binary tree with n non-leaf nodes contains
A)
log2 n nodes
B)
n+1 nodes
C)
2n nodes
D)
2n+1 nodes
Step-by-step reasoning:
1. Definition of a Full Binary Tree: In a full binary tree (sometimes called a proper or plane binary tree), every node has either 0 or 2 children. This means no node has only one child.
2. Let 'n' be the number of non-leaf nodes (internal nodes) and 'L' be the number of leaf nodes.
3. Total nodes (N) = n + L.
4. In any binary tree where every internal node has exactly 2 children, the number of leaf nodes is always one more than the number of internal nodes. Mathematically: L = n + 1.
5. Substitute the value of L into the total nodes equation:
N = n + (n + 1)
N = 2n + 1.
Example:
- If n = 1 (root only), L = 2. Total nodes = 1 + 2 = 3. (Formula: 2(1)+1 = 3).
- If n = 3, L = 4. Total nodes = 3 + 4 = 7. (Formula: 2(3)+1 = 7).
Analysis of options:
- a. log2 n: Incorrect; this relates to the height of a balanced tree, not total nodes.
- b. n+1: Incorrect; this is the number of leaf nodes, not total nodes.
- c. 2n: Incorrect; this does not account for the root/leaf relationship in a full binary tree.
- d. 2n+1: Correct based on the derivation above.
Which data structure is needed to convert infix notations to postfix notations?
A)
Linear list
B)
Queue
C)
Tree
D)
Stack
To convert an infix expression (e.g., A + B) to a postfix expression (e.g., AB+), the Shunting-yard algorithm is typically used. This algorithm requires a Stack to temporarily hold operators and parentheses. The stack is used to manage the precedence and associativity of operators: when an operator is encountered, it is pushed onto the stack if it has higher precedence than the operator currently on top; otherwise, the top operator is popped and added to the output string.
Why other options are incorrect:
- a. Linear list: While a stack is a type of linear list, 'Linear list' is too generic. A simple list does not inherently provide the Last-In-First-Out (LIFO) mechanism required for operator precedence management.
- b. Queue: A queue follows First-In-First-Out (FIFO), which would process operators in the order they appear, ignoring mathematical precedence rules.
- c. Tree: While an expression can be represented as an Expression Tree, the process of converting infix to postfix is fundamentally handled by a stack. A tree is a result or a different representation, not the primary data structure used for the conversion algorithm itself.
A complete binary tree with the property that the value at each node is at least as large as the values at its children in known as
A)
binary search tree
B)
AVL-tree
C)
completely balanced tree
D)
heap
The question describes the fundamental properties of a Max-Heap.
1. **Structural Property**: The question specifies a 'complete binary tree'. A complete binary tree is one where every level, except possibly the last, is completely filled, and all nodes are as far left as possible. This is a requirement for a binary heap.
2. **Heap Property**: The condition that 'the value at each node is at least as large as the values at its children' is the definition of the Max-Heap property. In a Max-Heap, for any given node I, the value of I is greater than or equal to the values of its children.
Why other options are incorrect:
- **a. binary search tree**: In a BST, the left child must be smaller than the parent, and the right child must be larger. It does not require the tree to be complete, nor does it require the parent to be larger than both children.
- **b. AVL-tree**: An AVL tree is a self-balancing binary search tree. While it maintains a height balance, it follows BST ordering rules, not the heap property.
- **c. completely balanced tree**: This refers to the structural balance of a tree (where the heights of subtrees differ by at most 1), but it does not define a specific relationship between the values of parents and children.