Civil Engineering Association - Mock GATE Exam

Diamond core drilling is suitable in rock because diamond is the hardest known natural material, making it ideal for drilling through tough and solid substances like rock. The diamond core bit is designed with diamond-impregnated segments that can effectively cut through the hard surface of the rock, allowing for precise and efficient drilling. This method is commonly used in geological exploration, mining, and construction projects that require drilling into rock formations.

Loam is the correct answer because it refers to a type of soil that is a mixture of sand, silt, and clay. It is known for its fertility and ability to retain moisture, making it ideal for agriculture. Loam has a balanced texture, with equal parts of sand, silt, and clay, which allows for good drainage while also retaining enough water for plant growth.

The correct answer is "She is less brave than her brother." This sentence correctly compares the bravery of the two individuals, stating that the brother is braver than the sister. The use of "less" indicates a comparison of bravery levels, making it the most suitable sentence with respect to grammar and usage.

The correction for sag is always negative. Sag is the amount by which a conductor or cable sags below the level of the supports when it is suspended between two points. The correction for sag is the correction factor applied to the measured sag to obtain the true sag. Since sag is always a downward deviation from the level of the supports, the correction for sag is always in the opposite direction, which is negative.

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Driven piles are piles that are installed into the ground by driving them with a pile driver or similar equipment. They are commonly used in construction to support structures and transfer loads to the soil. Dynamic pile formulae are useful in calculating the capacity of driven piles because they take into account factors such as the weight of the pile, the impact energy during driving, and the resistance of the soil. These formulae help engineers determine the appropriate size and length of driven piles to ensure that they can safely support the intended load.

The field application efficiency can be calculated by dividing the volume of water stored in the root zone by the total volume of water applied and multiplying by 100. In this case, the volume of water stored in the root zone is given as 0.20 cm, and the total volume of water applied is 25 x 10^7 liters. By dividing these two values and multiplying by 100, we get a field application efficiency of 80%.

The correct answer is Octagonal. According to IRC:67-2001, the shape of a STOP sign is octagonal. This means that the sign has eight sides, forming an octagon shape. The octagonal shape is specifically chosen for STOP signs as it allows for better visibility and recognition by drivers on the road. The unique shape helps to differentiate the STOP sign from other road signs and alerts drivers to the need to stop at that particular location.

The deflection at the free end of a cantilever beam is given by the equation:

δ = (wL^3)/(3EI)

where δ is the deflection, w is the uniformly distributed load, L is the length of the beam, E is the modulus of elasticity, and I is the moment of inertia.

The slope of the deflection curve at the free end is given by the equation:

θ = (wL^2)/(2EI)

where θ is the slope.

Given that the deflection is 18 mm and the slope is 0.02 radians, we can set up the following equations:

18 = (wL^3)/(3EI)
0.02 = (wL^2)/(2EI)

Dividing the second equation by the first equation, we get:

0.02/18 = (wL^2)/(2EI) / (wL^3)/(3EI)
0.02/18 = 3L / 2L^2
0.0011 = 3 / 2L
L = 3 / (0.0011 * 2)
L ≈ 1363.64 mm

Converting mm to meters, we get:
L ≈ 1.36364 m

Therefore, the length of the beam is approximately 1.2 m.

The shrinkage limit of a soil is defined as the maximum water content at which further reduction in water content will not cause any volume change. The shrinkage limit can be calculated using the formula:

Shrinkage Limit = (Void Ratio at Liquid Limit - Void Ratio at Dry State) / (Void Ratio at Liquid Limit - Void Ratio at Plastic Limit)

In this case, the void ratio at liquid limit is 1.62, the void ratio at plastic limit is 0.60, and the void ratio at dry state is 0.48. Plugging these values into the formula, we get:

Shrinkage Limit = (1.62 - 0.48) / (1.62 - 0.60) = 1.14 / 1.02 = 1.1176

Converting this to a percentage, we get 1.1176 * 100 = 111.76%. Therefore, the correct answer is 17.78%.

The ratio between a and b is 3:4, which means that a is 3/4 of b. Similarly, the ratio between b and c is 2:1, so b is 2/3 of c. To find the minimum value of a + b + c, we can assume values for b and c and calculate the corresponding values of a. Let's assume b = 4 and c = 3, then a = (3/4) * 4 = 3. Therefore, a + b + c = 3 + 4 + 3 = 10. If we try other values, we will always get a sum greater than 10, so the minimum value of a + b + c is 10.

The correct values of mechanical widening and psychological widening at 60 kmph are 0.50m and 0.89m respectively. The correct values of mechanical widening and psychological widening at 80 kmph are 0.50m and 1.19m respectively.

The seepage velocity of a soil can be calculated using Darcy's law, which states that the seepage velocity is equal to the discharge velocity divided by the void ratio. In this case, the discharge velocity is given as 6x10-7 m/sec and the void ratio is given as 0.50. By dividing the discharge velocity by the void ratio, we get 6x10-7 m/sec / 0.50 = 12x10-7 m/sec. Therefore, the correct answer is 12x10-7 m/sec.

The correct answer is 1100 N. The work done in pulling the bucket up the well can be calculated by multiplying the force applied (75 N) by the distance moved (20 m). However, since there is a small hole in the bucket through which water leaks, the effective force decreases by 2 N for every meter of distance moved. Therefore, the total work done is (75 N - 2 N/m * 20 m) * 20 m = 1100 N.

The Peak Hour Factor (PHF) represents the proportion of peak sub-hourly traffic flow within the peak hour. Since the PHF is a ratio, it can range from 0 to 1.0. In this case, the answer states that the theoretically possible range of PHF when considering 15-minute sub-hours is 0.25 to 1.0. This means that at least 25% of the traffic flow within the peak hour occurs in the sub-hourly intervals, but it could be as high as 100%.

The given equation is in the form y = 3e^(2x) + e^(-2x) - αx. To find the values of α and β, we can compare this equation with the general form of the solution for the initial value problem. The general form is y = Ae^(αx) + Be^(βx) + Cx, where A, B, and C are constants. By comparing the two equations, we can see that α = 3 and β = -4.

Since the arithmetic mean of the list is 4, it means that the sum of all the numbers in the list is 4 times the number of elements in the list. If the median is 3, it means that there are an equal number of elements greater than and less than 3 in the list. This would result in the sum of the numbers being greater than 4 times the number of elements, which contradicts the given information. Therefore, the median cannot be 3.

The density of the fluid flowing in the pipe can be calculated using the formula: density = specific gravity * density of reference fluid. In this case, the specific gravity of the oil flowing in the pipe is 0.96 and the density of the reference fluid (oil in the manometer) is given as 0.85. Therefore, the density of the fluid flowing in the pipe is 0.96 * 0.85 = 0.816 kg/m3. However, the question asks for the answer in kg/m3, so the density should be rounded to the nearest whole number, which is 1 kg/m3. Therefore, the correct answer is 1200 kg/m3.

When a four-digit number is divided by 65 and leaves a remainder of 29, it means that the number can be expressed as 65x + 29, where x is an integer. Now, if we divide this number by 13, we can rewrite it as (65x + 29)/13. Simplifying this expression, we get 5x + 2. When we divide 2 by 13, the remainder is 2. Therefore, the remainder when the four-digit number is divided by 13 is 2.

The question asks for the total length of a valley formed by two gradients. The given information is the rate of change of centrifugal acceleration, the design speed, and the gradients. To find the total length, we need to calculate the length of each gradient and add them together. The length of a gradient can be calculated using the formula: length = (rate of change of centrifugal acceleration) / (gradient * design speed). Plugging in the given values, we can calculate the length of each gradient and add them together to get the total length of 84.9 m.

The critical flow ratios for the three-phase signal are 0.30, 0.25, and 0.25. This means that the traffic volumes for each phase are not equal. The total time lost in the cycle is 10 seconds, which indicates that there is some delay in the traffic flow. Since the pedestrian crossings at this junction are not significant, the green times for each phase should be adjusted to minimize delay and maximize traffic flow. Therefore, the respective green times of 34, 28, and 28 seconds would be the most appropriate to achieve this goal.

The argument assumes that if bacteria were transported from Venus to Earth, they would have died out. However, it fails to consider the possibility that some of the bacteria from Venus may have survived and evolved into the bacteria found on Earth today. Therefore, the argument is vulnerable to the criticism that it fails to consider whether all bacteria now on Earth could have arisen from transported Venusian bacteria.

Negative skin friction develops when the soil settles more than the pile. This occurs when the surrounding soil undergoes settlement, causing it to exert a downward force on the pile. As a result, friction is generated along the pile shaft, leading to negative skin friction.

The Froude number is a dimensionless number used to characterize the flow regime in open channels. It is defined as the ratio of the flow velocity to the wave velocity. A Froude number greater than 1 indicates supercritical flow, while a Froude number less than 1 indicates subcritical flow. In this case, the Froude number at the beginning of the jump is 15, which indicates supercritical flow. Supercritical flow is characterized by a significant energy loss during the formation of the jump. Therefore, the correct answer is 92.52, strong jump.

The ratio of passive and active earth pressure coefficients can be determined by taking the reciprocal of the coefficient of earth pressure for at rest condition. In this case, the coefficient of earth pressure for at rest condition is 0.50, so the reciprocal of 0.50 is 2. Therefore, the ratio of passive and active earth pressure coefficients is 2:1.

The dry density of the borrow soil is 16 kN/m3 and the desired dry density for the embankment is 19.20 kN/m3. The difference in dry density is 19.20 - 16 = 3.20 kN/m3.

To calculate the number of trips required for a truck with a capacity of 6 m3 to form 100 m3 of embankment, we need to divide the total volume of embankment (100 m3) by the volume of each truckload (6 m3).

100 m3 / 6 m3 = 16.67

Since we cannot have a fraction of a trip, we need to round up to the nearest whole number. Therefore, the number of trips required is 17.

However, we also need to consider the difference in dry density. Each trip of borrow soil will increase the dry density by 3.20 kN/m3.

So, the number of trips required to reach the desired dry density is 17 + (3.20 kN/m3 / 6 m3) = 17 + 0.53 = 17.53.

Again, we need to round up to the nearest whole number, so the final answer is 18 trips.

Therefore, the correct answer is 18.

The critical oxygen deficit can be calculated using the Streeter-Phelps equation. The equation is given as:

DO = DOs - DOa * (1 - exp(-kD * L)) - DOa * (kD / kR) * (exp(-kR * L) - exp(-kD * L))

Where:
DO = Dissolved Oxygen deficit at the critical point
DOs = Saturation Dissolved Oxygen concentration in the river (fully saturated with oxygen)
DOa = Dissolved Oxygen concentration in the wastewater
kD = Decay coefficient for the wastewater
kR = Reaeration coefficient for the river
L = Distance from the point of mixing to the critical point (30 km in this case)

Using the given values and substituting them into the equation, we can calculate the critical oxygen deficit to be 14 mg/L.

The free float for an activity is the amount of time that the activity can be delayed without delaying the project. In this case, the activity (3) - (5) has a free float of 0, which means that it cannot be delayed without delaying the project. This suggests that activity (3) - (5) is on the critical path, meaning that any delay in this activity will directly impact the project's completion time.

The theoretical specific gravity of the mix is calculated by multiplying the specific gravity of each component by its percentage by weight and summing them up. In this case, the calculation would be (2.60 * 0.55) + (2.70 * 0.358) + (2.65 * 0.037) + (1.01 * 0.055) = 2.42. The effective specific gravity of the aggregates in the mix is the weighted average of the specific gravities of the coarse aggregate and fine aggregate, which would be (2.60 * 0.55) + (2.70 * 0.358) = 2.63. Therefore, the correct answer is 2.42 and 2.63.

The slope correction for a length of 30m along a gradient of 1 in 20 is 3.75 cm. This can be calculated by dividing the length (30m) by the gradient (1/20) and converting the result to centimeters.

The pair "Root: leaves" does not express a similar relationship to the pair "Wheel: spokes." In the given options, all the other pairs demonstrate a one-to-many relationship, where the first object has multiple parts or components. However, in the pair "Root: leaves," the relationship is not one of parts or components, but rather one of location or position. Roots and leaves are both separate parts of a plant, rather than one being a part of the other.

The dilatancy correction to SPT N-value is applied when the observed N-value is greater than 15.

The correct answer is 92.14 m. This can be calculated using the formula for braking distance:

Braking distance = (Initial velocity^2) / (2 * deceleration)

The initial velocity is 100 km/h, which is converted to 27.78 m/s. The deceleration is calculated using the equation:

Deceleration = friction coefficient * gravity * grade

The friction coefficient is 0.35, gravity is 9.8 m/s^2, and the grade is 3%. The grade is converted to a decimal by dividing by 100, so it is 0.03.

Plugging in these values, we get:

Deceleration = 0.35 * 9.8 * 0.03 = 0.1029 m/s^2

Now we can calculate the braking distance:

Braking distance = (27.78^2) / (2 * 0.1029) = 92.14 m

The word "aggravate" means to make a situation worse or more intense. In the given sentence, the phrase "tended to aggravate everyone's nerves" suggests that the person's reactions to unpleasant situations caused irritation or frustration to everyone around them. This word accurately describes the effect of the person's behavior on others, making it the best choice to fill the blank.

The efficiency of a lap joint can be determined by calculating the effective area of the joint and comparing it to the actual area of the plates being joined. In this case, the effective area can be calculated by subtracting the area of the bolt holes from the total area of the plates. The area of the bolt holes can be determined by multiplying the number of bolts by the area of each bolt hole. By dividing the effective area by the actual area of the plates and multiplying by 100, we can find the efficiency of the lap joint. In this case, the efficiency is 42%.

The factor of safety of an infinite slope in dry cohesionless soil remains the same upon submergence. This is because the factor of safety is a measure of the stability of the slope, and submergence does not directly affect the stability of the slope in dry cohesionless soil. The presence of water may change the pore water pressure and increase the weight of the soil, but these factors are accounted for in the original factor of safety calculation. Therefore, the factor of safety remains unchanged.

The Euler's critical load for a hinged-hinged column can be calculated using the formula Pcr = (π^2 * E * I) / L^2, where Pcr is the critical load, E is the modulus of elasticity, I is the moment of inertia, and L is the length of the column. Plugging in the given values, we get Pcr = (π^2 * 0.8 x 10^6 * 1766) / (600)^2 = 38,700 kg.

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The correct answer is 1 is true, 2 is true and 2 is the correct reason for 1. This means that statement 1, which states that a simple parabolic curve is an acceptable shape for summit curves, is true. Statement 2, which states that comfort to passengers is an important consideration in the design of summit curves, is also true. Furthermore, statement 2 provides the correct reason for statement 1, indicating that the comfort of passengers is the main factor in determining the acceptable shape of summit curves.

The infiltration rate at the beginning of the rainfall is 10 cm/h, and it decreases exponentially to an equilibrium value of 1 cm/h at the end of 9 hours of rain. This means that the infiltration rate follows a decay pattern over time. The total amount of water infiltrated during the 9 hours is given as 18 cm. By using Horton's model, we can determine the decay coefficient (k) in h-1. The correct answer is 1, indicating that the infiltration rate decreases by a factor of 10 every hour.

The correct answer is √2. This can be explained by considering the flexural strength of a square section. When the two sides of the square are horizontal, the diagonal is the longest possible length for the section. The flexural strength is directly proportional to the length of the section, so the longest diagonal will have the highest flexural strength. The ratio of the diagonal length to the side length of a square is √2, which means the flexural strength of a square section with its two sides horizontal to its diagonal horizontal is √2.

The difference in the lengths of an arc and its subtended chord on the earth surface for a distance of 18.2 km is only 10 cm. This means that the arc length is 10 cm longer than the length of the chord.

The drawdown in the pumping well is given by the equation:

s = (Q/4πT) * W(u)

Where:
s = drawdown in the well
Q = pumping rate (1000 m3/day)
T = transmissivity of the aquifer (300 m2/day)
W(u) = Theis well function value (3p)

Plugging in the values, we get:

s = (1000/4π*300) * 3p
s = (1000/1200π) * 3p
s ≈ 2.5 m

Therefore, the correct answer is 2.5 m.

The maximum strain in concrete at the level of compression reinforcement can be calculated using the formula for strain in concrete, which is equal to the stress in concrete divided by the modulus of elasticity of concrete. Since the maximum permissible stress in the outermost fibers in both steel and concrete reaches at the same time, we can assume that the stress in concrete is equal to the stress in steel. The stress in steel can be calculated using the formula for stress in steel, which is equal to the product of the strain in steel and the modulus of elasticity of steel. Since the strain in steel is limited to the yield strain, we can calculate the stress in steel and then use it to calculate the strain in concrete. Using the given values for the grade of steel and concrete, we can calculate the maximum strain in concrete to be 0.00289.

The normal duration and cost of the activity are given as 25 days and 50,000 rupees respectively. The activity can be crashed to a duration of 22 days. The crash duration is the time it takes to complete the activity when it is expedited by adding more resources. The indirect cost is given as 1000 rupees per day, which is the additional cost incurred for each day the activity is expedited. The cost slope is given as 1500 rupees per day, which is the additional cost incurred for each day the activity is expedited beyond the normal duration. To calculate the crash cost, we need to calculate the additional cost for the crash duration and the additional cost for each day beyond the crash duration. The additional cost for the crash duration is 22 - 25 = -3 days * 1000 rupees/day = -3000 rupees. Since this is a negative value, it means there is a cost saving for completing the activity earlier. The additional cost for each day beyond the crash duration is (25 - 22) * 1500 rupees/day = 4500 rupees. Adding the crash cost to the normal cost, we get 50,000 rupees + (-3000 rupees) + 4500 rupees = 47,500 rupees. Therefore, the crash cost of the activity will be 47,500 rupees.

The theoretical volume of methane gas that would be expected from the anaerobic digestion of a tonne of waste can be calculated by first determining the number of moles of C50H100O40N in a tonne of waste. Since 1 mole of C50H10O40N will produce 27.125 moles of methane, we can then multiply the number of moles of C50H100O40N by 27.125 to find the number of moles of methane. Finally, we can convert the number of moles of methane to volume using the density of methane. The correct answer is 447 m3.

The method of perpendicular offsets involves less measurement on the ground compared to the other methods listed. In this method, a series of perpendicular lines are drawn from a baseline to the features being surveyed. By measuring the distances along these perpendicular lines, the required offsets can be determined. This method is efficient because it requires fewer measurements on the ground compared to the method of oblique offsets or the method of ties.

An isochrone indicates the variation of excess pore pressure at a specific point in time. It represents the contour lines that connect points of equal excess pore pressure values.

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Based on the given information, we can use a Venn diagram to solve this problem. Let's represent the three sections as circles A, B, and C. We know that the number of students who cleared the cutoff in only one section is 24 for each section, so we can write this as A = 24, B = 24, and C = 24.

We also know that 20 students cleared the cutoff in section 1 and section 2, 16 cleared the cutoff in section 2 and section 3, and 26 cleared the cutoff in section 1 and section 3. From this, we can deduce that the number of students who cleared the cutoff in all three sections is the sum of the number of students who cleared the cutoff in two sections minus the total number of students who cleared the cutoff in only one section.

Using this formula, we can calculate the number of students who cleared the cutoff in all three sections as (20 + 16 + 26) - (24 + 24 + 24) = 30 - 72 = -42.

Since the number of students cannot be negative, the correct answer is 6.