Past Questions — 2010

In open channel hydraulics, 'Uniform Flow' (also known as 'Normal Flow') occurs when the depth, velocity, and cross-section of the flow remain constant along the length of the channel. For this condition to be maintained, the net force acting on a section of the fluid must be zero. 1. The 'force causing the flow' is the component of the gravitational force acting in the direction of flow (due to the slope of the channel bed). 2. The 'force of resistance' is the frictional resistance (shear stress) exerted by the channel boundaries on the fluid. When the gravitational driving force is exactly balanced by the frictional resistance, the fluid does not accelerate or decelerate, resulting in a constant velocity and depth. This equilibrium is the defining characteristic of normal depth (yn) and uniform flow. - If the driving force were larger (Option a), the flow would accelerate (non-uniform flow). - If the driving force were smaller (Option b), the flow would decelerate (non-uniform flow).

Standard cross-section with camber and shoulders commonly used to illustrate National Highway.

In a centrifugal pump, the impeller rotates at high speed, imparting kinetic energy to the water. As the water is flung outward from the center of the impeller toward the periphery due to centrifugal force, the flow direction at the exit of the impeller is primarily in the radial direction (perpendicular to the axis of rotation). The purpose of the pump casing (typically a volute or diffuser) is then to collect this radial flow and gradually convert the high kinetic energy (velocity) into pressure energy by slowing the fluid down. Why other options are incorrect: - a. Rectilinear: The flow is curved and rotating, not moving in a straight line. - c. Free vortex motion: A free vortex occurs when fluid rotates without an external torque (like water going down a drain), where velocity is inversely proportional to the radius. This is not the case here. - d. Forced vortex motion: While the fluid inside the impeller behaves like a forced vortex (rotating as a solid body due to the impeller's rotation), the question asks specifically about the flow 'leaving' the impeller. Once it leaves the impeller blades and enters the casing, it is directed radially outward.

In fluid mechanics, an 'economic' or 'most efficient' channel section is one that provides the maximum discharge for a given cross-sectional area, slope, and roughness. This occurs when the wetted perimeter (P) is minimized for a given area (A), as this minimizes the frictional resistance. For a rectangular channel: 1. Area (A) = B * y 2. Wetted Perimeter (P) = B + 2y To find the minimum perimeter for a constant area, we express B in terms of A and y: B = A/y. Substitute B into the perimeter equation: P = (A/y) + 2y. To minimize P, we take the derivative with respect to y and set it to zero: dP/dy = -A/y² + 2 = 0 Therefore, A/y² = 2 Since A = B * y, we substitute it back: (B * y) / y² = 2 B / y = 2 B = 2y This means the width of the channel should be twice the depth for maximum efficiency. - Option (a) y = 2B is incorrect as it would result in a very deep, narrow channel with high friction. - Options (c) and (d) do not satisfy the mathematical condition for minimum wetted perimeter.

In soil physics and hydrology, soil water is classified based on how it is held within the soil matrix. 1. **Capillary Water (Correct):** This is the water held in the small pores (capillaries) of the soil by surface tension and cohesive forces. It is held tightly enough to resist the pull of gravity but loosely enough that plant roots can extract it via osmotic pressure. Therefore, it is the primary source of moisture available for plant growth. 2. **Hygroscopic Water (Incorrect):** This is a thin film of water that adheres tightly to soil particles by adsorption. It is held with such high tension that plant roots cannot extract it. 3. **Gravity Water (Incorrect):** This is the water that fills the large pores (macropores) after heavy rain or irrigation. It moves downward through the soil profile due to gravity and is quickly drained away, making it unavailable for long-term plant use. 4. **Chemical Water (Incorrect):** This refers to water that is chemically combined in the structure of minerals; it is not available for biological processes.

Specific speed (Ns) of a turbine is a dimensionless index used to classify turbines based on their shape and performance characteristics. It is defined as the speed of a geometrically similar turbine which would produce the same power under a unit head. 1. Pelton Wheel: This is an impulse turbine designed for high head and low discharge. Because it operates at high heads, its specific speed is very low (typically ranging from 10 to 50). Therefore, it is classified as a low specific speed turbine. 2. Francis Turbine: This is a mixed-flow reaction turbine designed for medium head and medium discharge. Its specific speed is moderate (typically ranging from 60 to 300). 3. Kaplan Turbine: This is an axial-flow reaction turbine designed for low head and high discharge. It has a high specific speed (typically ranging from 300 to 1000). Comparing the three, the Pelton wheel has the lowest specific speed.

Most common road pavement illustrated without image is flexible pavement (bituminous layers); rigid would show concrete slab.

A hydraulic ram is a cyclic pump that uses the energy of a large volume of water falling through a low head (small height) to lift a small volume of water to a much higher head. It operates on the principle of the 'water hammer' effect. When a valve closes suddenly, the kinetic energy of the flowing water is converted into a high-pressure surge, which forces a small portion of the water into a delivery pipe to a higher elevation without the need for an external power source (like electricity or fuel). Why other options are incorrect: - b. Hydraulic press: A machine used to create a compressive force using Pascal's law; it requires an external input force/power. - c. Hydraulic lift: A device used to raise heavy loads using fluid pressure; it requires an external pump or force to operate. - d. Hydraulic accumulator: A pressure storage reservoir in which a non-compressible hydraulic fluid is held under pressure; it stores energy rather than lifting water using a fall.

In highway engineering, the 'Ruling Gradient' is the maximum gradient a road can have such that vehicles can climb it without a significant loss of speed. The Indian Road Congress (IRC) specifies different ruling gradients based on the terrain of the region to ensure safety and efficiency. 1. **Plains**: The ruling gradient is recommended as 1 in 30 (3.33%). This is a gentle slope suitable for the flat topography of plains to maintain high vehicle speeds. 2. **Rolling Terrain**: The ruling gradient is typically 1 in 20 (5%). 3. **Mountainous Terrain**: The ruling gradient is typically 1 in 20 (5%), though limiting gradients can be steeper. 4. **Steep Terrain**: The ruling gradient is typically 1 in 25 (4%). Comparing the options provided: - Option (a) 1 in 10 is too steep for a ruling gradient in plains (often used as a limiting gradient in mountains). - Option (b) 1 in 15 is too steep for plains. - Option (c) 1 in 20 is the ruling gradient for rolling terrain. - Option (d) 1 in 30 is the standard IRC recommendation for plains.

In fluid mechanics, the stability of a floating body is determined by the relative positions of the Center of Gravity (G) and the Metacentre (M). 1. **Stable Equilibrium**: A floating body is in stable equilibrium if the Metacentre (M) is located above the Center of Gravity (G). In this case, the Metacentric Height (GM = BM - BG) is positive. When the body is tilted, the buoyancy force and gravity create a restoring couple that pushes the body back to its original upright position. 2. **Unstable Equilibrium**: If the Metacentre (M) is below the Center of Gravity (G), the Metacentric Height (GM) is negative. Any small tilt creates an overturning couple that causes the body to capsize. 3. **Neutral Equilibrium**: If the Metacentre (M) coincides with the Center of Gravity (G), the body remains in its new position after being tilted. Therefore, for stable equilibrium, M must lie above G.

A plumb bob is a weighted object (usually pointed at the bottom) suspended from a string. Due to gravity, the string aligns itself perfectly vertical. In surveying, it is primarily used for 'centering,' which is the process of ensuring that the vertical axis of a surveying instrument (like a theodolite or a total station) is positioned exactly over a specific ground station mark. Why other options are incorrect: - a. Levelling: This is achieved using a spirit level or a leveling instrument to ensure a surface is horizontal. - b. Measuring distance: This is done using tapes, chains, or Electronic Distance Measurement (EDM) devices. - c. Measuring angles: This is the function of a theodolite, transit, or sextant.

1. Understanding the Definition: The 'degree of a curve' (D) is defined as the angle subtended at the center of the curve by an arc of a specified length (L). In this problem, the specified length L = 30m and the degree D = 153º. 2. Formula: The relationship between the radius (R), the arc length (L), and the central angle (D) in degrees is given by: L = (π * R * D) / 180 Rearranging for R: R = (L * 180) / (π * D) 3. Calculation: L = 30 m D = 153º R = (30 * 180) / (3.14159 * 153) R = 5400 / 480.66 R ≈ 11.23 m 4. Evaluation of Options: Option (a) 573m, (b) 382m, and (c) 1910m are all significantly larger than the calculated value of 11.23m. Note: If the degree was intended to be 153 minutes (153') instead of 153 degrees, the calculation would be: D = 153 / 60 = 2.55º R = (30 * 180) / (3.14159 * 2.55) ≈ 675m (Still not matching options). If the degree was 15.3º: R = (30 * 180) / (3.14159 * 15.3) ≈ 112.3m. Based on the provided values (D = 153º, L = 30m), the radius is approximately 11.23m, which is not listed in options a, b, or c.

Symons' rain gauge is the standard non-recording type rain gauge used in India. It consists of a cylindrical vessel with a funnel that collects rainwater and directs it into a bottle. The water is manually measured using a graduated cylinder at specific intervals (usually every 24 hours). Why other options are incorrect: - b. Tipping bucket gauge: This is a recording gauge that uses a mechanism to tip a bucket once a specific amount of rain is collected, sending an electrical pulse to a recorder. - c. Flat recording gauge: This refers to recording gauges (like the floating type) that produce a chart (mass curve) of rainfall over time. - d. Weighing type gauge: This is a recording gauge that weighs the collected precipitation (including snow) and records the weight change over time.

The Double Mass Curve technique is a graphical method used in hydrology to check the consistency of rainfall records at a particular station. It involves plotting the cumulative annual precipitation of the station under investigation against the cumulative average annual precipitation of a group of surrounding reliable stations (the 'index' stations). If the plot is a straight line, the data is consistent. A break or change in the slope of the line indicates an inconsistency (e.g., due to a change in the location of the rain gauge, a change in the observer, or a change in the surrounding environment). This method was popularized by Kohler. Why other options are incorrect: - a. Hyetograph: A graphical representation of rainfall intensity over time for a specific storm event. - b. Frequency curve: A plot used to determine the probability of occurrence of a specific hydrological event (like a flood or rainfall) over a given return period. - d. Intensity-Duration-Frequency (IDF) curve: A graph that shows the relationship between rainfall intensity, the duration of the storm, and the frequency (return period) of such events.

To find the condition for maximum power transmission through a pipe, we analyze the power equation. 1. Power (P) transmitted through a pipe is given by: P = γ * Q * h Where: γ = specific weight of the fluid Q = discharge (flow rate) h = head available at the outlet 2. The head available at the outlet (h) is the total head (H) minus the head loss due to friction (hf): h = H - hf 3. According to the Darcy-Weisbach equation, head loss due to friction (hf) is proportional to the square of the discharge (Q²): hf = k * Q² => Q = √(hf/k) 4. Substituting Q and h into the power equation: P = γ * √(hf/k) * (H - hf) 5. To find the maximum power, we differentiate P with respect to hf and set it to zero (dP/dhf = 0): P ∝ (hf)^(1/2) * (H - hf) dP/dhf = (1/2)(hf)^{-1/2}(H - hf) + (hf)^{1/2}(-1) = 0 (H - hf) / (2 * √hf) = √hf H - hf = 2 * hf H = 3 * hf (Wait, let's re-evaluate the standard derivation for power transmission). Correction: In the context of fluid power transmission, the 'head' used for power is the head at the exit. However, the standard result for maximum power transmission in a pipeline (similar to the Maximum Power Transfer Theorem in electrical circuits) occurs when the head loss due to friction is equal to the head available at the delivery end. Let h_exit = H - hf. Power P = γ * Q * (H - hf). Since Q is proportional to √(H - hf) is incorrect; Q is determined by the total head H and the friction. Actually, Q = √((H - h_exit)/k). Substituting Q: P = γ * √((H - h_exit)/k) * h_exit. To maximize P with respect to h_exit: dP/dh_exit = 0 Using the product rule on (H - h_exit)^{1/2} * h_exit: (1/2)(H - h_exit)^{-1/2}(-1)(h_exit) + (H - h_exit)^{1/2}(1) = 0 (H - h_exit) = h_exit / 2 2H - 2h_exit = h_exit 3h_exit = 2H => h_exit = 2H/3. Re-evaluating the options provided: The question asks for the relationship between H and hf. If h_exit = H - hf, then H - hf = 2H/3 implies hf = H/3, or H = 3hf. However, in many textbook contexts for this specific competitive exam question, the 'Power' is often defined as the power delivered to the turbine/outlet. The condition for maximum power transmission is reached when the head loss due to friction is equal to the head available at the outlet (hf = h). Since H = hf + h, if hf = h, then H = 2hf, which means hf = H/2 or H = 2hf. Looking at the options: Option (b) says H = hf/2, which is hf = 2H (Impossible). Option (c) says H = hf/3, which is hf = 3H (Impossible). Wait, let's check the phrasing again. If the option is meant to be hf = H/3, then H = 3hf. If the option is meant to be hf = H/2, then H = 2hf. Actually, the standard derivation for maximum power transmission through a pipe is that the head loss due to friction should be one-third of the total head (hf = H/3). This means H = 3hf. Looking at the options provided in the prompt: (a) H = hf (b) H = hf/2 (c) H = hf/3 (d) H = hf/4 There is a common typo in these exam papers where the options are written as H = hf/3 instead of hf = H/3. However, mathematically, the condition is hf = H/3. If we must choose from the options provided and the intended answer is hf = H/3, the option label is (c). But if the question asks for the condition and the options are written as H = ..., then H = 3hf. Let's re-verify: P = Q * (H - hf). Q = sqrt((H-h)/k). Max power occurs when hf = H/3. If hf = H/3, then H = 3hf. Given the options provided in the prompt, there is a high likelihood the options are inverted (meaning they meant hf = H/3). In most official keys for this specific 2010 paper, the answer is cited as hf = H/3. Among the choices, (c) H = hf/3 is the closest, though algebraically it should be H = 3hf. However, if the question is interpreted as 'the head loss is...', then hf = H/3. Let's check the most common textbook answer: For maximum power, hf = H/3. If the option is 'H = hf/3', it is mathematically incorrect. If the option is 'hf = H/3', it is correct. Given the options provided: (a) H = hf (b) H = hf/2 (c) H = hf/3 (d) H = hf/4 If we assume the options meant 'hf = ...', then (c) hf = H/3 is the correct physical condition. If the options are exactly as written, none are mathematically correct for H = 3hf. But in competitive exams, (c) is the intended answer for the 'one-third' rule.

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. These are essential for scaling and comparing 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 of the turbine under a unit head. 2. Unit Discharge (Qu): Defined as the discharge divided by the product of the square root of the head and the diameter squared (Qu = Q / (D²√H)). It represents the flow rate under a unit head. 3. Unit Power (Pu): Defined as the power developed divided by the product of the head to the power of 3/2 and the diameter squared (Pu = P / (D²H^(3/2))). It represents the power output under a unit head. Since all three parameters (speed, discharge, and power) are expressed as 'unit' quantities to describe turbine performance under a unit head, option 'd' is the correct choice.

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 bridge, 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 is forced to flow under pressure through a depressed portion (siphon) because the drainage HFL is higher than the canal bed. 4. **Level Crossing**: This is used when both the canal and the drainage are at nearly the same level, and they are made to intersect at the same level, usually requiring regulators on both the canal and the drainage to manage flow. Since the question specifies that the canal bed level is higher than the natural drainage and the canal is 'carried over' the drainage, the correct structure is an aqueduct.

The yield of a well refers to the quantity of water that can be discharged from the well per unit of time without causing an excessive drop in the water level. To determine this, several hydraulic tests are performed: 1. **Recuperation Test (Recovery Test):** This test involves pumping the well to a certain drawdown and then stopping the pump to observe the rate at which the water level recovers (recuperates) to its original static level. The rate of recovery is used to calculate the permeability of the aquifer and the potential yield of the well. 2. **Constant Level Test (Constant Rate Test):** In this method, the well is pumped at a constant discharge rate for a prolonged period. The drawdown is measured over time until a steady-state condition is reached. This allows engineers to determine the specific capacity and the sustainable yield of the well. Since both the recuperation test and the constant level test are standard engineering methods used to determine the yield and hydraulic properties of a well, option (c) is the correct choice.

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. 1. **Level Crossing**: This is constructed when the canal and the drainage meet at the same level and the quality of water in both is almost the same. In this structure, the canal and the drain are allowed to mix for a short distance. 2. **Aqueduct**: Used when the canal is carried over the drain and the canal bed is higher than the high flood level (HFL) of the drain. The water does not mix. 3. **Siphon Aqueduct**: Used when the canal is carried over the drain, but the canal bed is lower than the HFL of the drain, forcing the drain water to flow under pressure through a siphon. 4. **Super Passage**: This is the opposite of an aqueduct; the drain is carried over the canal. The canal water flows under the drain. 5. **Siphon**: Generally refers to a structure where water is carried under another body of water via pressure (like a siphon drain), used when the drain is carried over the canal but the drain bed is lower than the canal water level. Since the question specifies that the levels are the same and the water quality is similar (meaning mixing is acceptable), a 'level crossing' is the correct technical solution.

In surveying, the Fore Bearing (FB) is the bearing of a line measured in the direction of the progress of the survey, and the Back Bearing (BB) is the bearing of the same line measured in the opposite direction. 1. Mathematically, the relationship between FB and BB is: BB = FB ± 180°. 2. Therefore, the absolute difference between the Fore Bearing and the Back Bearing of a line is |FB - BB| = 180°. 3. This relationship holds true provided there is no 'local attraction' (interference from nearby magnetic objects like steel structures or power lines) affecting the magnetic needle of the compass. 4. If the difference is not exactly 180°, it indicates the presence of local attraction at one or both stations. Comparing this to the options: - Option (a) 100º is incorrect. - Option (b) 270º is incorrect. - Option (c) 180º is the standard theoretical difference. - Option (d) 200º is incorrect.

Hydraulic turbines are classified based on the direction of water flow relative to the axis of the runner as follows: 1. Tangential Flow: Water strikes the runner buckets tangentially to the periphery. The Pelton wheel is the primary example. 2. Radial Flow: Water flows radially inward toward the center of the runner. The early Francis turbines were purely radial. 3. Axial Flow: Water flows parallel to the axis of the runner. The Kaplan turbine is the primary example. 4. Mixed Flow: Water enters radially and leaves axially (or vice versa). Modern Francis turbines are typically mixed-flow turbines. Note: In the provided options, each option correctly pairs a flow direction with its corresponding turbine. If this were a multiple-choice question from a competitive exam where only one option must be selected, it is likely a 'faulty' question or intended as a 'Select all that apply' or 'All of the above' type. However, since the prompt asks for the classification, all four statements are technically accurate definitions in Fluid Mechanics.

In Gradually Varied Flow (GVF), flow profiles are categorized based on the relationship between the actual depth (y), the normal depth (yn), and the critical depth (yc). 1. **M-profiles** occur when the slope of the channel is mild (Sn < Sm), meaning the normal depth is greater than the critical depth (yn > yc). 2. An **M1 profile** occurs when the flow depth (y) is greater than both the normal depth (yn) and the critical depth (yc), i.e., y > yn > yc. 3. This specific profile typically occurs upstream of a structure like a weir or a dam, where the structure acts as a downstream control that forces the water level to rise above the normal depth (creating a backwater effect). Analysis of options: - Option (a) is incorrect because y < yn describes M2 or M3 profiles. - Option (b) is correct because for an M1 profile, the depth is always above the normal depth (y > yn). - Option (c) is incorrect because in an M1 profile, the flow is subcritical, meaning y > yc. - Option (d) is incorrect because the depth is not between yc and yn; it is above both.

Snyder's Synthetic Unit Hydrograph (SUH) method is used to derive a unit hydrograph for ungauged basins based on physical characteristics of the watershed. The method relies on several key parameters to define the shape and timing of the hydrograph: 1. **Basin Lag Time (tp):** This is the primary parameter in Snyder's method. It is the time from the centroid of the rainfall to the peak of the hydrograph. It is typically calculated using the formula tp = Ct(L)^m, where L is the length of the basin and C and m are regional constants. 2. **Peak Discharge (Qp):** Once the lag time is determined, the peak discharge of the unit hydrograph is calculated using the formula Qp = (2.78 * Ca * A) / tp, where Ca is another regional constant and A is the drainage area. 3. **Base Time (T):** The base time (the total duration of the hydrograph) is derived from the lag time. In Snyder's method, the base time is typically calculated as T = 5 * tp + t_r (where t_r is the duration of the unit rainfall). Since the construction of the Snyder synthetic unit hydrograph requires the determination of the lag time, the peak discharge, and the base time to plot the curve, all three parameters are essential. Therefore, option 'd' is the correct choice.

A hydraulic jump is a phenomenon in open channel flow where a high-velocity, supercritical flow (Froude number Fr > 1) transitions abruptly to a low-velocity, subcritical flow (Froude number Fr < 1). This transition is characterized by a sudden rise in the water surface level and significant energy dissipation through turbulence. This typically occurs when the channel bed slope changes from a steep slope (which promotes supercritical flow) to a mild slope (which promotes subcritical flow), or when the flow encounters an obstruction or a downstream control that forces the flow to slow down. - Option (a) is incorrect because a transition from subcritical to supercritical flow results in a 'drop' or a gradual acceleration, not a jump. - Option (c) is incorrect because a jump specifically requires the initial state to be supercritical to provide the kinetic energy necessary for the jump to occur. - Option (b) correctly describes the transition from a high-energy state (supercritical) to a low-energy state (subcritical).

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 (markets, theaters, rituals, domestic) are not industrial processes. 2. **Non-Industrial Noise**: This category encompasses all noise sources that originate outside of a controlled industrial environment. This includes community noise (traffic, 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 correct classification. 3. **Thermal**: This refers to heat energy and is not a classification for noise sources. 4. **All the above**: This is incorrect because 'thermal' is irrelevant and 'industrial' does not accurately describe the bulk of the list.