Past Questions

According to the Indian Road Congress (IRC) standards for the geometric design of highways, the width of the carriageway is specified based on the number of lanes to ensure safe passage of vehicles. For a single-lane road, the standard width of the carriageway is 3.75 meters. This width is designed to accommodate a single vehicle while providing a small margin of safety. For comparison: - A two-lane road (without a raised median) typically has a width of 7.0 meters (3.5m per lane) or 7.5 meters depending on the road classification. - Option (a) 1.75m and (b) 2.75m are too narrow for standard vehicular traffic. - Option (c) 4.75m does not correspond to a standard IRC single-lane width.

Negative skin friction (also known as down-drag) occurs when the surrounding soil settles more than the pile itself. This relative downward movement of the soil relative to the pile creates a downward shear stress along the pile shaft, which adds to the load on the pile rather than supporting it. This typically happens due to: 1) Consolidation of soft clay layers under a newly placed fill, 2) Lowering of the groundwater table causing soil shrinkage/consolidation, or 3) Settlement of recently deposited alluvial soils. Analysis of other options: - a. Expansive soil: Usually causes 'heave' (upward movement), which can lead to upward forces, though it can be complex; however, it is not the primary cause of negative skin friction. - c. Loose sand: While loose sand can settle, the classic mechanism for negative skin friction is the time-dependent consolidation of cohesive soils (clays) or saturated soils. - d. Mild loose sand: Similar to option c, this is not the definitive cause compared to the consolidation process.

In brick masonry, 'bond' refers to the arrangement of bricks in a way that ensures the load is distributed evenly and the wall remains stable. To prevent continuous vertical joints (which would create a plane of weakness), bricks are overlapped. According to standard engineering practices and building codes (such as IS codes), the minimum overlap or lap required to ensure a satisfactory bond and structural integrity is 1/4 of the length of the brick. - Option (a) 1/4 length: This is the standard minimum requirement for a secure bond. - Option (b) 1/3 length: While providing a better bond than 1/4, it is not the minimum standard requirement. - Option (c) 1/2 length: This is common in English or Flemish bonds (half-lap), but the question asks for the minimum overlap required for a 'better bond' in general terms. - Option (d) Full length: This is physically impossible for an overlap between two adjacent bricks in a single course.

Step-by-step reasoning: 1. Definition: In highway engineering, the transverse slope provided to the road surface is specifically designed to drain rainwater from the center (crown) towards the edges of the pavement to prevent hydroplaning and protect the road structure from water seepage. 2. Terminology: This convexity or transverse slope is technically termed as 'Camber' (also known as cross-slope). 3. Analysis of other options: - (a) Gradient: Refers to the longitudinal slope of the road (rise or fall along the length of the road). - (b) Slope: A general term for any inclined surface; while camber is a type of slope, 'Camber' is the specific technical term for this pavement feature. - (d) Vertical curve: A curve provided to transition between two different longitudinal gradients to ensure comfort and safety.

Footing area = 2 m × 2 m = 4 m². The given bearing capacity is 200 kPa = 200 kN/m². With factor of safety = 3, the safe bearing pressure = 200/3 = 66.67 kN/m². Therefore, maximum safe load = 66.67 × 4 = 266.67 kN. So the correct value is 266.67 kN. Why others are incorrect: - a) 1600 kN uses the full bearing capacity without applying factor of safety. - b) 533.33 kN is twice the correct safe load, so it is too high. - c) 800 kN is also based on an incorrect application of pressure/load conversion and does not account properly for factor of safety. - d) 266.67 kN is the correct safe load, so this option is actually the correct numerical answer; if the option key mapping is followed strictly, the correct key should correspond to 266.67 kN.

For a cohesive soil in an unconfined compression test, the unconfined compressive strength is q_u = 2c_u, where c_u is the undrained shear strength. Given q_u = 120 kPa, the shear strength is c_u = 120/2 = 60 kPa. Hence option (a) is correct. Why others are incorrect: (b) 120 kPa is the unconfined compressive strength itself, not the shear strength; (c) 240 kPa is double the given value and has no basis here; (d) 30 kPa is half of the correct shear strength, so it is too low.

Step-by-step calculation: 1. **Calculate the Actual Bearing Pressure (q):** The actual pressure exerted by the footing on the soil is the total load divided by the area of the footing. Load (P) = 1200 kN Area (A) = 3 m × 2 m = 6 m² Actual Pressure (q) = P / A = 1200 kN / 6 m² = 200 kPa 2. **Understand the relationship between Safe Bearing Capacity (q_safe) and Ultimate Bearing Capacity (q_ult):** The Safe Bearing Capacity is defined as the Ultimate Bearing Capacity divided by the Factor of Safety (FOS), minus the effect of overburden pressure (though in simplified exam contexts, it is often treated as q_safe = q_ult / FOS). Given: Safe Bearing Capacity (q_safe) = 200 kPa 3. **Determine the Factor of Safety (FOS):** In this specific problem, the actual pressure applied (200 kPa) is exactly equal to the safe bearing capacity (200 kPa). However, the question asks for the factor of safety against bearing capacity failure. Usually, the 'Safe Bearing Capacity' is the allowable pressure that ensures the soil does not fail. If the actual pressure (q) = 200 kPa and the safe bearing capacity (q_safe) = 200 kPa, the footing is designed exactly at its safe limit. Wait, looking at the options provided (2.5, 3.0, 3.5, 4.0), there is a standard convention in many textbook problems where the 'Safe Bearing Capacity' is given as a value, and the 'Ultimate Bearing Capacity' is derived. However, if we look at the load and the safe capacity: Actual Pressure = 200 kPa. Safe Bearing Capacity = 200 kPa. This means the design is at the limit of safety. Re-evaluating the problem: In many competitive exams, if the 'Safe Bearing Capacity' is provided as 200 kPa and the actual pressure is 200 kPa, the question is asking for the FOS used to arrive at that safe capacity from the ultimate capacity. Standard engineering practice for footings typically uses a Factor of Safety of 3.0. Given the options and the fact that the applied pressure equals the safe capacity, the intended answer is the standard design FOS for soil foundations, which is 3.0.

In airport engineering, different areas of the airside are designated for specific functions: 1. **Apron (Option b):** This is the defined area of an airport intended to accommodate aircraft for loading or unloading passengers, uploading and downloading mail or cargo, fueling, parking, or staging. It is the correct answer as it matches all the criteria mentioned in the question. 2. **Hangar (Option a):** A hangar is a large enclosed structure used to house aircraft for maintenance, repair, or long-term storage. While parking occurs here, it is a building, not a general area for routine loading and refueling. 3. **Taxiway (Option c):** A taxiway is a paved path used for the movement of aircraft between the runway and the apron. Aircraft are not permitted to park, load, or refuel on taxiways as this would obstruct traffic flow. 4. **Runway (Option d):** The runway is the rectangular area defined in a land aerodrome for the landing and take-off of aircraft. It is strictly for high-speed operations and must be kept clear of stationary aircraft.

Step-by-step solution: 1. **Understand the Relationship**: In soil mechanics, the relationship between porosity (n) and void ratio (e) is given by the formula: e = n / (1 - n) where n is expressed as a decimal. 2. **Calculate Void Ratio for Sample A (e_A)**: Given n_A = 40% = 0.40 e_A = 0.40 / (1 - 0.40) e_A = 0.40 / 0.60 e_A = 2/3 ≈ 0.667 3. **Calculate Void Ratio for Sample B (e_B)**: Given n_B = 60% = 0.60 e_B = 0.60 / (1 - 0.60) e_B = 0.60 / 0.40 e_B = 3/2 = 1.5 4. **Find the Ratio e_A : e_B**: Ratio = e_A / e_B Ratio = (2/3) / (3/2) Ratio = (2/3) * (2/3) Ratio = 4/9 Therefore, the ratio of void ratios e_A : e_B is 4:9. Comparing this to the options: - Option (a) 2:3 is incorrect. - Option (b) 7:3 is incorrect. - Option (c) 4:9 is correct. - Option (d) 9:4 is the inverse ratio (e_B : e_A) and is therefore incorrect.

In masonry construction, mortar is a workable paste used to bind building blocks (such as bricks or stones) together. Its primary functions are: 1) To provide a strong bond between the units to ensure the structural integrity of the wall, 2) To distribute the load uniformly across the masonry units to prevent localized stress concentrations, and 3) To seal the gaps between units to prevent the penetration of air and water. Analysis of options: - (a) Act as a filler: While mortar does fill gaps, its primary engineering purpose is not merely filling space, but creating a structural bond. - (b) Provide bonding: This is the fundamental purpose of mortar; it transforms individual units into a monolithic structural element. - (c) Provide insulation: Mortar generally has higher thermal conductivity than the masonry units themselves (like aerated concrete or certain stones), so it does not serve as the primary insulation layer. - (d) Increase weight: Increasing weight is generally a disadvantage or a neutral byproduct, not a desired function of mortar.

In civil engineering estimation and costing, the unit of measurement for an item depends on the nature of the work. Cement concrete is a three-dimensional material that occupies volume (Length × Width × Depth/Thickness). Therefore, for bulk items like foundations, slabs, and beams, the standard unit of measurement is the cubic metre (m³). Analysis of other options: - Running metre (m): Used for linear items such as pipes, railings, or cornices. - Square metre (m²): Used for surface area items such as plastering, painting, or flooring. - Litre: This is a unit of liquid volume, not used for structural concrete measurement in construction estimation.

When a vehicle traverses a horizontal curve, it is subjected to a centrifugal force acting outwards from the center of the curve. This force tends to push the vehicle away from the curve, which can lead to skidding or overturning. Super-elevation (also known as banking) is the raising of the outer edge of the pavement relative to the inner edge. This tilts the vehicle inward, allowing a component of the vehicle's own weight to act toward the center of the curve, thereby counteracting the centrifugal force and ensuring the vehicle remains stable at design speeds. Analysis of other options: - Option (a) is incorrect because centrifugal force is an inherent physical result of circular motion; super-elevation does not prevent its 'occurrence' but manages its effects. - Option (c) is incorrect because transition curves are introduced to gradually change the centrifugal force from zero to the value required for the circular curve; super-elevation is a cross-sectional feature, not a tool to 'facilitate' the introduction of the curve itself. - Option (d) is incorrect because limiting the rate of change of radial acceleration is the primary purpose of providing a 'Transition Curve', not super-elevation.

In Amplitude Modulation (AM), the bandwidth (BW) is defined as the difference between the highest and lowest frequency components of the modulated signal. An AM signal consists of a carrier frequency (fc) and two sidebands: the Upper Sideband (USB) and the Lower Sideband (LSB). 1. The Upper Sideband frequency is given by: f_USB = fc + fm 2. The Lower Sideband frequency is given by: f_LSB = fc - fm Where 'fm' is the maximum modulating frequency of the message signal. Therefore, the total Bandwidth (BW) is: BW = f_USB - f_LSB BW = (fc + fm) - (fc - fm) BW = 2 * fm Given that the maximum modulating frequency (fm) = 5 kHz: BW = 2 * 5 kHz = 10 kHz. Option (a) is incorrect because it only accounts for one sideband. Options (c) and (d) are incorrect as they do not follow the standard bandwidth formula for conventional AM.

In antenna theory and electromagnetics, polarization is defined as the orientation of the electric field (E-field) vector of an electromagnetic wave relative to the earth's surface or a specific reference axis as the wave propagates through space. 1. Why option (b) is correct: The electric field is the primary component that interacts with the receiving antenna. Depending on how the E-field oscillates (e.g., vertically, horizontally, or in a circle), the wave is classified as vertically polarized, horizontally polarized, or circularly polarized. 2. Why other options are incorrect: - Option (a): The shape of the antenna (geometry) can influence the polarization, but it is not the definition of polarization itself. - Option (c): The material (e.g., copper, aluminum) affects conductivity and efficiency, not the polarization of the radiated wave. - Option (d): The frequency of operation determines the wavelength and bandwidth, but polarization is independent of the frequency value.

The Central Processing Unit (CPU) is primarily divided into two main components: the Control Unit (CU) and the Arithmetic Logic Unit (ALU). 1. The Control Unit (CU) acts as the 'manager' of the CPU. Its primary functions include: - Fetching instructions from memory (Option a). - Decoding those instructions to determine what operation needs to be performed (Option b). - Generating control signals to coordinate the movement of data between the CPU, memory, and I/O devices (Option d). 2. The Arithmetic Logic Unit (ALU) is the component specifically designed to perform actual data processing. This includes: - Arithmetic operations (Addition, Subtraction, Multiplication, Division). - Logical operations (AND, OR, NOT, XOR). Since arithmetic operations are performed by the ALU and not the Control Unit, option 'c' is the correct answer to the question 'Which of the following is NOT a function of the control unit?'

To convert a decimal number to binary, we use the repeated division-by-2 method. We divide the number by 2 and record the remainder until the quotient becomes 0. The binary equivalent is read from the last remainder (Most Significant Bit) to the first remainder (Least Significant Bit). Step-by-step calculation for (24)10: 1. 24 ÷ 2 = 12, Remainder = 0 2. 12 ÷ 2 = 6, Remainder = 0 3. 6 ÷ 2 = 3, Remainder = 0 4. 3 ÷ 2 = 1, Remainder = 1 5. 1 ÷ 2 = 0, Remainder = 1 Reading the remainders from bottom to top: 11000. Verification using positional weights: (1 × 2^4) + (1 × 2^3) + (0 × 2^2) + (0 × 2^1) + (0 × 2^0) = 16 + 8 + 0 + 0 + 0 = 24. Comparing with options: - Option a: 11000 (Correct) - Option b: 1101111 (Equals 111 in decimal) - Option c: 1111110 (Equals 126 in decimal) - Option d: 11001 (Equals 25 in decimal)

According to the Nyquist-Shannon Sampling Theorem, a continuous-time signal can be perfectly reconstructed from its samples if the sampling frequency (fs) is greater than twice the highest frequency component (fmax) present in the signal. This critical threshold is known as the Nyquist Rate (fs = 2 * fmax). 1. Analysis of the condition: The question states that the sampling frequency is 'slightly more than twice the highest signal frequency' (fs > 2 * fmax). 2. Result: Since the Nyquist criterion is satisfied, the spectral replicas of the signal in the frequency domain do not overlap. This allows a low-pass filter (reconstruction filter) to isolate the original baseband signal perfectly without any distortion. Why other options are incorrect: - Option b: Aliasing occurs only when fs < 2 * fmax (undersampling). Since fs > 2 * fmax, aliasing is avoided. - Option c: Quantization error is a result of the Analog-to-Digital conversion process (bit depth/resolution), not the sampling frequency. - Option d: The bandwidth requirement for transmission depends on the sampling rate, but it does not 'double' simply because the rate is slightly above the Nyquist rate; it scales linearly with the sampling frequency.

Non-uniform quantization is a technique used in digital communication to improve the Signal-to-Quantization Noise Ratio (SQNR) for signals where the amplitude distribution is non-uniform (e.g., speech signals, where low-amplitude signals occur more frequently than high-amplitude ones). 1. **Compression**: To achieve non-uniform quantization, the input signal is first passed through a 'compressor'. This process increases the gain for low-amplitude signals and decreases the gain for high-amplitude signals, effectively 'compressing' the dynamic range. This ensures that small signals are quantized with finer steps, reducing the relative quantization error. 2. **Expansion**: After the signal has been quantized and transmitted/processed, it must be restored to its original dynamic range. This is done using an 'expander', which is the inverse process of compression. Common examples of this process include $\mu$-law and A-law companding (Compression + Expanding). Therefore, non-uniform quantization inherently involves both compression at the transmitter and expansion at the receiver.

In the context of microprocessor architecture (specifically referring to the 8085 microprocessor, which is the standard for these types of engineering exams), the 'sequence of operations' refers to the priority or the standard order of machine cycles used to execute an instruction. 1. **Opcode Fetch**: This is always the first operation of every instruction cycle. The microprocessor must first fetch the operation code (opcode) from memory to determine what action to perform. 2. **Memory Read/Write**: After the opcode is decoded, the processor may need to read data from memory (Memory Read) or store a result back into memory (Memory Write) to complete the instruction. 3. **I/O Read/Write**: Finally, the processor may interact with external peripherals via input/output ports (I/O Read to get data from a sensor/switch, or I/O Write to send data to a display/actuator). Option (a) correctly lists these in the logical order of execution: first fetching the instruction, then handling internal memory operations, and finally handling external I/O operations. Options (b), (c), and (d) are incorrect because they either place I/O operations before the opcode fetch or swap the logical flow of memory access.

In microprocessor architecture, an 'Interrupt' is a signal sent by a hardware device or a software program to the CPU, requesting that the CPU temporarily suspend its current execution flow to handle a specific event or task. Once the Interrupt Service Routine (ISR) is completed, the CPU returns to the original task it was performing. Analysis of other options: - a. Disturbance: This is a general term and not a technical term used in microprocessor architecture to describe CPU requests. - b. Attenuation: This refers to the gradual loss in intensity of any kind of flux (like signal strength in a cable) as it passes through a medium; it is a telecommunications/electronics term, not a CPU request mechanism. - d. Noise: This refers to unwanted electrical signals that interfere with the transmission of data; it is a problem to be filtered out, not a request for CPU time.

In optical fiber communication, the light source must meet several critical criteria: it must be compact, efficient, capable of high-speed modulation (switching on and off rapidly), and emit light at wavelengths where optical fibers have low attenuation (typically in the near-infrared region, such as 1310 nm or 1550 nm). 1. Semiconductor Lasers (e.g., Laser Diodes): These are the most common sources because they are extremely small, can be directly modulated at GHz speeds, and their emission wavelengths can be precisely engineered to match the low-loss windows of silica fibers. 2. CO2 Lasers: These emit in the far-infrared region (~10.6 μm), where silica fibers are opaque (extremely high absorption), making them unsuitable for fiber communication. 3. Nd:YAG Lasers: While they emit in the infrared, they are bulky, expensive, and cannot be modulated at the high speeds required for data transmission compared to semiconductor lasers. 4. Argon Lasers: These emit primarily in the visible spectrum (blue/green), which suffers from high attenuation in standard optical fibers, and they are too large for practical communication hardware.

The 8086 microprocessor has a pin named MN/MX' (Minimum/Maximum). This pin is used to determine the operating mode of the processor based on the logic level applied to it: 1. **Minimum Mode**: When the MN/MX' pin is tied to Vcc (Logic 1), the processor operates in Minimum Mode. In this mode, the 8086 generates all the necessary control signals (like ALE, RD, WR) itself, and it is typically used in single-processor systems. 2. **Maximum Mode**: When the MN/MX' pin is tied to Ground (Logic 0), the processor operates in Maximum Mode. The term 'active low' (indicated by the prime symbol ' or the bar over the signal) means the function is activated when the signal is at a low voltage level (0). In Maximum Mode, the 8086 requires an external bus controller (like the 8288) to generate the control signals, allowing for multi-processor configurations or co-processor integration. Therefore, applying an active low signal to the MN/MX' pin puts the 8086 into Maximum Mode.