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Past Questions — 2010
The language that the computer can understand and execute is called
A)
Machine language
B)
Application software
C)
System program
D)
None of the above
Step-by-step reasoning: 1. Computers are electronic devices that operate using digital logic gates, which recognize only two states: ON (1) and OFF (0). 2. Machine language consists entirely of binary digits (0s and 1s) that correspond directly to the physical operations of the Central Processing Unit (CPU). 3. Because it is the only language the hardware can execute directly without translation, it is the 'native' language of the computer. 4. Evaluating other options: (b) Application software refers to programs designed for end-users (like Word or Chrome), which are written in high-level languages and must be compiled/interpreted into machine code to run. (c) System programs (like Operating Systems) manage hardware but are also written in high-level or assembly languages and eventually converted to machine language for execution. Therefore, only machine language is directly understood and executed by the computer hardware.
Wide area networks (WANs) always require
A)
High bandwidth communication source link
B)
High speed processors
C)
Same type
D)
None of the above
To determine the correct answer, we must analyze the fundamental requirements of a Wide Area Network (WAN). A WAN is a telecommunications network that extends over a large geographical area.
1. Analysis of Option (a): While high bandwidth is desirable for performance, it is not a 'requirement' for a WAN to exist. WANs can operate over low-bandwidth links, such as dial-up connections or low-speed leased lines. Therefore, they do not 'always' require high bandwidth.
2. Analysis of Option (b): High-speed processors are a characteristic of the end-devices (servers/workstations) or high-end routers, but they are not a fundamental requirement for the network architecture itself. A WAN can be established using modest hardware depending on the traffic load.
3. Analysis of Option (c): 'Same type' is ambiguous and technically incorrect. WANs are specifically designed to interconnect different types of networks (LANs, MANs) and can utilize a heterogeneous mix of transmission media (fiber optics, satellite, copper wires) and protocols. They do not require the same type of hardware or media throughout.
Since none of the provided options are absolute requirements for the existence of a Wide Area Network, the correct choice is 'None of the above'.
The intersection of a CFL and a regular language
A)
is always regular
B)
is always context free
C)
both (a) and (b)
D)
need not be regular
This is a fundamental closure property of Context-Free Languages (CFLs).
1. **The Theorem**: The intersection of a Context-Free Language (L1) and a Regular Language (L2) is always a Context-Free Language.
2. **Proof Sketch (Product Construction)**:
- A CFL is accepted by a Pushdown Automaton (PDA).
- A Regular Language is accepted by a Deterministic Finite Automaton (DFA).
- We can construct a new PDA that simulates both the PDA and the DFA simultaneously. The state of the new PDA is a pair (p, q), where 'p' is the state of the original PDA and 'q' is the state of the DFA.
- The new PDA uses the same stack operations as the original PDA, while the state transitions track the progress of both machines.
- Since the resulting machine is still a PDA (it only requires one stack), the resulting language must be context-free.
3. **Evaluating Options**:
- **Option (a)**: Incorrect. The intersection of a CFL and a Regular language is not necessarily regular. For example, if L1 = {a^n b^n | n >= 0} (CFL) and L2 = a*b* (Regular), their intersection is {a^n b^n | n >= 0}, which is context-free but not regular.
- **Option (b)**: Correct. As per the closure property described above.
- **Option (c)**: Incorrect because (a) is incorrect.
- **Option (d)**: While it is true that the result 'need not be regular', in the context of multiple-choice questions regarding closure properties, the most precise and definitive property is that it 'is always context free'. Option (b) defines the class the resulting language belongs to.
Which of the following is a 8 bit Microprocessor?
A)
Intel 4040
B)
Pentium – I
C)
8088
D)
Motorala MC-6801
To determine the correct answer, we must analyze the data bus width (bit-size) of each microprocessor listed:
1. **Intel 4040**: This was a 4-bit microprocessor, an evolution of the Intel 4004.
2. **Pentium-I**: This is a 32-bit microprocessor (with 64-bit external data bus for some versions), designed for high-performance computing.
3. **Intel 8088**: This is a 16-bit microprocessor. Although it has an 8-bit external data bus to reduce system cost, its internal registers and ALU are 16-bit.
4. **Motorola MC-6801**: This is a classic 8-bit microcontroller/microprocessor. It features an 8-bit data bus and is designed for embedded control applications.
Therefore, the Motorola MC-6801 is the only 8-bit processor among the given options.
Which of the following logic expression is incorrect?
A)
1 · 0 = 1
B)
1 · 1 · 0 = 1
C)
1 · 1 · 1 = 1
D)
1 · 1 = 0
1·0=0 in AND, so 1 is incorrect; also 1·1·0=1 and 1·1=0 are incorrect but first listed is A.
Which of the following regular expressions denote zero or more instances of an a and b?
A)
a | b
B)
(a b)*
C)
(a | b)*
D)
a* | b
In Theory of Computation, regular expressions are used to describe sets of strings (languages). Let's analyze the options based on the requirement 'zero or more instances of an a and b':
1. **Option a (a | b):** This denotes a choice between a single 'a' or a single 'b'. It can only generate strings of length 1 ({a, b}).
2. **Option b (a b)*:** This denotes zero or more repetitions of the specific sequence 'ab'. It can generate strings like ε (empty string), 'ab', 'abab', 'ababab', etc. It cannot generate 'aa' or 'bb'.
3. **Option c (a | b)*:** The pipe symbol '|' (or '+') denotes the union (OR), meaning 'either a or b'. The Kleene star '*' denotes zero or more repetitions of the preceding expression. Therefore, (a | b)* generates any combination of a's and b's of any length, including the empty string. This matches the requirement of 'zero or more instances of an a and b'.
4. **Option d (a* | b):** This denotes either zero or more instances of 'a' (e.g., ε, a, aa, aaa...) OR a single 'b'. It cannot generate a string containing both 'a' and 'b' (like 'ab').
Thus, (a | b)* is the correct regular expression for the language consisting of all strings over the alphabet {a, b}.
Which of the following is not accepted by Deterministic Push Down Machine (DPDM) but accepted by Non Deterministic Push Down Machine (NDPDM)?
A)
strings ending with a particular alphabet
B)
all strings in which a given symbol is present at least twice
C)
even palindromes
D)
None of the above
To solve this, we must understand the difference between Deterministic Pushdown Automata (DPDA) and Non-deterministic Pushdown Automata (NPDA).
1. **DPDA vs NPDA**: Unlike Finite Automata, where DFA and NFA are equivalent in power, NPDA is strictly more powerful than DPDA. DPDAs recognize 'Deterministic Context-Free Languages' (DCFLs), while NPDAs recognize all 'Context-Free Languages' (CFLs).
2. **Analyzing Option (a)**: 'Strings ending with a particular alphabet' is a Regular Language. All regular languages are accepted by DFAs, and since every DFA is a DPDA (that doesn't use its stack), this is accepted by DPDA.
3. **Analyzing Option (b)**: 'All strings in which a given symbol is present at least twice' is also a Regular Language. It can be represented by a simple DFA. Therefore, it is accepted by DPDA.
4. **Analyzing Option (c)**: 'Even palindromes' (languages of the form ww^R) are the classic example of a Non-deterministic Context-Free Language. To recognize an even palindrome, the machine must 'guess' where the middle of the string is to start popping from the stack. Since there is no marker (like a center symbol '#') to indicate the middle, a deterministic machine cannot decide when to switch from pushing to popping. Thus, it requires non-determinism.
Therefore, even palindromes are accepted by NPDA but not by DPDA.
The language L = {aibicidi | i >= 1} is
A)
Regular language
B)
Recursively enumerable language
C)
Context free language
D)
None of the above
Step-by-step analysis of the language L = {a^i b^i c^i | i >= 1}:
1. **Analyze the Language Structure**: The language consists of strings where the number of 'a's, 'b's, and 'c's are exactly equal and appear in that specific order. This requires the machine to keep track of the count of 'a's to ensure the same count for 'b's and then again for 'c's.
2. **Check for Regularity (Option a)**: A regular language can be recognized by a Finite Automaton (FA). An FA has finite memory and cannot count an arbitrary number of symbols. Since we need to count 'i' (which can be any integer >= 1), this language is not regular.
3. **Check for Context-Free (Option c)**: A Context-Free Language (CFL) can be recognized by a Pushdown Automaton (PDA), which uses a single stack. A stack can match two dependent counts (e.g., a^i b^i), but it cannot match three dependent counts (a^i b^i c^i) because once the stack is popped to match 'b's against 'a's, the count of 'i' is lost and cannot be used to verify the 'c's. This is a classic example used to demonstrate the Pumping Lemma for CFLs to prove a language is NOT context-free.
4. **Check for Recursively Enumerable (Option b)**: A Recursively Enumerable (RE) language is one that can be recognized by a Turing Machine. A Turing Machine has a read/write tape that allows it to move back and forth, making it capable of counting and comparing multiple groups of symbols. A Turing Machine can easily solve this by marking one 'a', then one 'b', then one 'c', and repeating the process until all symbols are accounted for.
5. **Conclusion**: Since the language is not Regular and not Context-Free, but is decidable by a Turing Machine, it falls into the category of Context-Sensitive Languages (CSL) and, by extension, Recursively Enumerable (RE) languages. Among the given options, 'Recursively enumerable language' is the only correct classification.
The number of paths between any pair of nodes in a tree on n nodes is
A)
0
B)
1
C)
n–1
D)
n
In graph theory, a tree is defined as a connected graph that contains no cycles. One of the fundamental properties of a tree is that for any two distinct vertices (nodes) u and v, there exists exactly one unique simple path connecting them.
Step-by-step reasoning:
1. **Connectivity**: Since a tree is a connected graph, there must be at least one path between any pair of nodes. This eliminates option (a).
2. **Acyclic Nature**: A tree contains no cycles. If there were more than one path between two nodes, the union of those two distinct paths would necessarily form a cycle, which contradicts the definition of a tree. This eliminates options (c) and (d).
3. **Conclusion**: Because the graph is connected (at least one path) and acyclic (at most one path), there is exactly one unique path between any pair of nodes.
Therefore, the number of paths between any pair of nodes in a tree on n nodes is 1.
How many truth tables can be made from one function table?
A)
1
B)
2
C)
3
D)
8
In Boolean algebra and digital logic, a 'function table' (often used interchangeably with a 'truth table') is a mathematical table that defines the output of a Boolean function for every possible combination of its input variables. By definition, a specific Boolean function has one and only one unique truth table. The truth table is the tabular representation of the function's behavior; therefore, for any single given function table, there is exactly one corresponding truth table (as they are essentially the same entity). If the question implies how many truth tables represent a single unique Boolean function, the answer is 1, because the mapping from inputs to outputs for a specific function is deterministic and unique.
Regular sets are not closed under
A)
Union
B)
Intersection
C)
Subset
D)
Complement
In the Theory of Computation, a class of languages is said to be 'closed' under an operation if applying that operation to languages in the class always results in a language that is also in that class.
1. **Union (Option a):** Regular languages are closed under union. If L1 and L2 are regular, L1 ∪ L2 is also regular (can be proven via NFA construction or regular expressions).
2. **Intersection (Option b):** Regular languages are closed under intersection. If L1 and L2 are regular, L1 ∩ L2 is also regular (can be proven via the product construction of DFAs).
3. **Complement (Option d):** Regular languages are closed under complement. If L is regular, its complement Σ* - L is also regular (can be proven by swapping the final and non-final states of a DFA).
4. **Subset (Option c):** Regular languages are NOT closed under the subset operation. If a language L is regular, a subset of L is not necessarily regular. For example, the language L = Σ* (all strings over an alphabet) is regular. However, we can define a subset of L, such as L' = {a^n b^n | n ≥ 0}, which is a well-known non-regular (context-free) language. Since a subset of a regular language can be non-regular, the class of regular sets is not closed under the subset operation.
The Stack pointer holds
A)
16 bit address
B)
16 bit data
C)
8 bit address
D)
8 bit data
The Stack Pointer (SP) is a specialized register in a CPU that stores the memory address of the last program request in a stack. In the context of classic 8-bit architectures (such as the Intel 8085, which is the standard reference for many technical exams of this nature), the CPU processes 8-bit data, but the memory address space is 16 bits wide (allowing it to address up to 64KB of memory). Since the Stack Pointer must be able to point to any location within this memory space, it must be a 16-bit register.
1. Why it is an 'address': The SP does not store the actual data being pushed or popped; it stores the location (address) where that data is found.
2. Why it is '16 bit': In standard 8-bit microprocessor architecture, addresses are 16 bits to accommodate the memory map.
Therefore, options b and d are incorrect because the SP holds an address, not data. Option c is incorrect because an 8-bit address would only allow for 256 bytes of memory, which is insufficient for a system stack.
Precipitation that occurs due to conflict between two air masses is called
A)
frontal precipitation
B)
convective precipitation
C)
orographic precipitation
D)
cyclonic precipitation
Precipitation is classified based on the mechanism that causes the air to rise and cool.
1. **Frontal Precipitation (Correct):** This occurs when two air masses of different temperatures and densities (e.g., a warm moist air mass and a cold dry air mass) meet. The warmer, less dense air is forced to rise over the colder, denser air. As the warm air rises, it cools adiabatically, leading to condensation and precipitation. This 'conflict' zone is known as a front.
2. **Convective Precipitation:** This occurs when the ground is heated intensely by the sun, causing the air above it to become warm, expand, and rise in vertical currents (convection). This typically results in short, intense bursts of rain (thunderstorms).
3. **Orographic Precipitation:** This occurs when moist air is forced to rise over a physical barrier, such as a mountain range. The air cools as it ascends the windward side, leading to precipitation.
4. **Cyclonic Precipitation:** While frontal precipitation is a type of cyclonic precipitation (occurring in extratropical cyclones), 'frontal' specifically describes the conflict between air masses. Cyclonic precipitation is a broader term referring to precipitation caused by low-pressure systems where air converges and rises.
When pressure of fluid is measured with respect to atmospheric pressure as datum, it is called
A)
absolute pressure
B)
vacuum pressure
C)
gauge pressure
D)
all the above
In fluid mechanics, pressure is measured relative to a specific reference point or datum.
1. **Gauge Pressure**: This is the pressure measured relative to the local atmospheric pressure. If the fluid pressure is higher than atmospheric pressure, the gauge pressure is positive. If it is lower, it is often referred to as vacuum pressure. The formula is: Gauge Pressure = Absolute Pressure - Atmospheric Pressure.
2. **Absolute Pressure**: This is the pressure measured relative to a perfect vacuum (zero pressure). It is the sum of the gauge pressure and the atmospheric pressure. The formula is: Absolute Pressure = Gauge Pressure + Atmospheric Pressure.
3. **Vacuum Pressure**: This is a specific type of gauge pressure that occurs when the pressure is below atmospheric pressure. It is measured as the difference between atmospheric pressure and the absolute pressure.
Since the question specifically asks for the term used when atmospheric pressure is used as the datum, the correct term is 'gauge pressure'.
When canal bed level is higher than the natural drainage, then the canal is carried over a natural drainage. Such cross drainage work is called
A)
aqueduct
B)
super passage
C)
siphon
D)
level crossing
In irrigation engineering, cross drainage works are constructed when a canal intersects a natural drainage (like a river or stream). The type of structure depends on the relative levels of the canal bed and the drainage bed:
1. **Aqueduct**: This is used when the canal bed level is higher than the High Flood Level (HFL) of the natural drainage. The canal is carried over the drainage in a trough or trough-like structure, and the drainage flows freely underneath. This matches the condition described in the question.
2. **Super Passage**: This is the opposite of an aqueduct. It is used when the natural drainage is carried over the canal. Here, the canal bed is lower than the drainage bed.
3. **Siphon Aqueduct**: This is used when the canal is carried over the drainage, but the drainage water level is higher than the canal bed, forcing the drainage water to flow under pressure through a siphon.
4. **Level Crossing**: This is used when the canal and the drainage intersect at nearly the same level, and both are allowed to flow under pressure through the structure.
Since the question specifies that the canal bed level is higher and the canal is carried over the natural drainage, the correct structure is an aqueduct.
When the canal and the drainage meet each other at the same level and quality of water of both is almost the same, cross drainage work constructed is
A)
super passage
B)
siphon
C)
level crossing
D)
aqueduct
In irrigation engineering, cross drainage works are constructed when a canal crosses a natural drain. The choice of structure depends on the relative levels of the canal and the drain, and the quality of water.
1. **Level Crossing**: This is used when the canal and the drainage meet at the same level and the water quality of both is similar. In this structure, the canal and the drain are allowed to mix.
2. **Aqueduct**: Used when the canal is carried over the drain via a trough, and the canal bed is higher than the high flood level (HFL) of the drain. There is no mixing of water.
3. **Siphon Aqueduct**: Similar to an aqueduct, but the drain is carried under the canal through a siphon because the HFL of the drain is higher than the canal bed.
4. **Super Passage**: Used when the drain is carried over the canal. The canal is carried under the drain.
5. **Siphon**: Generally refers to a structure where water is carried under pressure through a dipped pipe or conduit.
Since the question specifies that the levels are the same and the water quality is almost the same, a 'level crossing' is the most appropriate and economical choice.
Performance of turbine under unit head is expressed in
A)
unit speed (Nu)
B)
unit power (Pu)
C)
unit discharge (Qu)
D)
all the above
In hydraulic turbine analysis, 'unit quantities' are non-dimensional or normalized parameters used to describe the performance of a turbine regardless of the actual head (H) available. This allows engineers to compare turbines of different sizes.
1. Unit Speed (Nu): Defined as the speed of the turbine divided by the square root of the head (Nu = N / √H). It represents the performance in terms of rotational speed under unit head.
2. Unit Discharge (Qu): Defined as the actual discharge divided by the head raised to the power of 3/2 (Qu = Q / H^(3/2)). It represents the volume of water passing through the turbine under unit head.
3. Unit Power (Pu): Defined as the actual power developed divided by the head raised to the power of 5/2 (Pu = P / H^(5/2)). It represents the power output under unit head.
Since all three parameters (speed, discharge, and power) are expressed as 'unit' values to describe turbine performance under a unit head, the correct answer is 'all the above'.
The road pavement shown in the paper is called
A)
Flexible pavement
B)
Rigid pavement
C)
Semi rigid pavement
D)
Composite pavement
Duplicate of above - typical pavement diagram is flexible pavement.
Remote sensing is the science of making inferences about objects from measurements, made at a distance, without coming into physical contact with the object under study, may be of the following types
A)
active remote sensing
B)
passive remote sensing
C)
both (a) and (b)
D)
none of the above
Remote sensing is broadly classified into two main categories based on the source of the energy used to detect the object:
1. **Passive Remote Sensing**: These sensors detect natural radiation (usually sunlight) that is reflected or emitted by the object. For example, traditional photography and thermal imaging are passive because they rely on existing energy sources.
2. **Active Remote Sensing**: These sensors provide their own source of energy. The sensor emits a signal toward the target and then measures the reflection or backscatter of that signal. Examples include RADAR (Radio Detection and Ranging) and LiDAR (Light Detection and Ranging).
Since both active and passive methods are fundamental types of remote sensing, option (c) is the correct choice.
Two point problem compared to three point problem is
A)
more accurate
B)
quicker
C)
more laborious
D)
all the above
In surveying, specifically in resection problems used to determine the coordinates of an unknown point:
1. The 'Two Point Problem' involves observing angles to two known points and usually requires an additional piece of information (like a known distance or a specific bearing) or a simpler geometric setup to fix the position. Because it involves fewer observations and simpler calculations compared to the three-point problem, it is executed more quickly.
2. The 'Three Point Problem' (also known as the Snellius-Pothenot problem) involves observing angles between three known points. This method is mathematically more complex, requiring more field observations (three angles instead of two) and more rigorous calculations to resolve the position of the point.
Comparing the two:
- Accuracy: The Three Point Problem is generally more accurate because it provides redundancy and doesn't rely on a single distance measurement.
- Labor: The Three Point Problem is more laborious due to the increased number of observations and complex trigonometry.
- Speed: The Two Point Problem is quicker because it requires fewer measurements and simpler processing.
Therefore, option (b) is the correct choice.
The river which is in the process of building up its own bed and certain slope due to excess sediment is called
A)
degrading river
B)
aggrading river
C)
virgin river
D)
stable river
In fluvial geomorphology and hydrology, the change in the bed level of a river is determined by the balance between sediment transport and sediment supply.
1. **Aggrading River (Option b):** Aggradation occurs when the sediment supply to the river exceeds the river's transport capacity. As a result, the excess sediment is deposited on the river bed, causing the bed level to rise (building up its own bed) and often altering the slope. This is the correct answer.
2. **Degrading River (Option a):** Degradation occurs when the river's transport capacity exceeds the sediment supply. The river erodes its own bed to satisfy this capacity, leading to a lowering of the bed level (scouring).
3. **Stable River (Option d):** A stable river (or a river in equilibrium) is one where the sediment supply is exactly equal to the sediment transport capacity, meaning there is no net gain or loss of bed material over time.
4. **Virgin River (Option c):** This is not a standard technical term for bed-level dynamics; it typically refers to a river that has not been regulated by dams or human intervention.
Source of noise from domestic, loud speaker, construction, transport, crowded market, theaters, religious rituals, projection of satellites, atomic explosion etc. are classified as
A)
industrial
B)
non-industrial
C)
thermal
D)
all the above
The question asks for the classification of various noise sources. Let's analyze the listed sources: domestic (home appliances), loud speakers, construction, transport (traffic), crowded markets, theaters, religious rituals, satellite projection, and atomic explosions.
1. **Industrial Noise**: This refers specifically to noise generated within a factory or industrial plant setting, such as heavy machinery, turbines, compressors, and assembly lines. While 'construction' can sometimes overlap, the majority of the listed items (domestic, markets, theaters, rituals) are community or environmental noise sources.
2. **Non-industrial Noise**: This category encompasses all noise sources that do not originate from industrial manufacturing processes. This includes community noise (transport, markets, theaters), domestic noise, and sporadic high-intensity events (explosions, satellite launches). Since the list is dominated by community and environmental sources, 'non-industrial' is the most accurate classification.
3. **Thermal**: This refers to heat energy and is not a classification for sources of noise pollution.
Therefore, because the sources listed are primarily social, domestic, and environmental rather than factory-based, they are classified as non-industrial noise.
Hardy-Cross method is used to the problems of
A)
boundary layer problem
B)
pipe network
C)
solution of water hammer problem
D)
none of the above
The Hardy-Cross method is an iterative technique used in fluid mechanics and hydraulic engineering to determine the flow rates in a closed-loop pipe network. It is based on two fundamental laws: 1) The Law of Conservation of Mass (the sum of flows entering a junction must equal the sum of flows leaving it), and 2) The Law of Conservation of Energy (the sum of head losses around any closed loop must be zero). By iteratively adjusting the flow in each loop to satisfy these conditions, the method converges on the actual distribution of flow throughout the network.
Why other options are incorrect:
- Option (a) Boundary layer problems are typically solved using the Prandtl boundary layer equations or numerical methods like CFD (Computational Fluid Dynamics).
- Option (c) Water hammer problems involve transient pressure waves and are solved using the Method of Characteristics (MOC) or other wave-propagation equations.
According to Garde and Ranga-Raju incipient motion condition of sediment is based on
A)
competent velocity approach
B)
lift force approach
C)
critical tractive force
D)
all the above
The incipient motion of sediment refers to the condition under which a particle of sediment is first moved by the flow of water. Traditionally, theories like the Shields criterion focus on the 'critical tractive force' (shear stress) acting on the bed. However, Garde and Ranga Raju proposed a different approach to determine the threshold of motion. Their model emphasizes the role of the 'lift force' acting on the sediment particle due to the pressure gradient and turbulence in the flow, rather than relying solely on the drag or tractive force. While competent velocity is a general concept and critical tractive force is the basis for the Shields approach, the specific contribution of Garde and Ranga Raju is associated with the lift force approach to sediment transport initiation.
If D is the duty in ha/cumec, B is the base period of a crop in days, Δ is the depth of water in metre, the relationship among them are
A)
D = 8.64 Δ / B
B)
D = 8.64 B / Δ
C)
Δ = 8.64 B / D
D)
none of the above
Duplicate - Δ=8.64B/D => D=8.64B/Δ.