Civil Engineering Association - Mock GATE Exam

1. Diamond core drilling is suitable in

Soft clay

Dense sand

Rock

Cumulose soil

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.

Explanation

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.

2. The alluvial soil deposit consisting of mixture of sand, silt and clay is referred as

Lacustrine soil

Loess

Loam

Talus

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.

Explanation

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.

3. Correction for sag is always

Positive

Negative

Zero

None

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.

Explanation

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.

4. She is brave. Her brother is more brave. Select the most suitable sentence with respect to grammar and usage.

Her brother is brave and she is equally brave.

She is brave but her brother is the bravest of all.

She is less brave than her brother.

None

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.

Explanation

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.

5. Dynamic pile formulae are useful to calculate the capacity of

Bored piles

Driven piles

Timber piles

Tension piles

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.

Explanation

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.

6. The saturated sand deposit is having moisture content of 20% and specific gravity of particles is 2.6. If the void ratios in loosest and densest states are 0.76 and 0.36 respectively, its relative density is

60%

50%

70%

47%

not-available-via-ai

Explanation

not-available-via-ai

7. The total volume of water applied in 100 hectares of land was 25 m 10⁷ liters. If the water stored in the root zone is 0.20 cm, then the field application efficiency will be_______%

60

70

80

90

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%.

Explanation

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%.

8. The shape of STOP sign according to IRC:67-2001 is

Circular

Triangular

Octagonal

Rectangular

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.

Explanation

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.

9.

If the deflection at the free end of a uniformly loaded cantilever beam is 18 mm and the slope of the deflection curve at the free end is 0.02 radians, then length of the beam is

0.8 m

1.0 m

1.2 m

1.5 m

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.

Explanation

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.

10. A two lane, one-way road with radius of 50 m is predominantly carrying lorries with wheelbase of 5 m. The speed of lorries is restricted to be between 60 kmph and 80 kmph. The mechanical widening and psychological widening required at 60 kmph are designated as W_me,₆₀ and W_ps,₆₀ , respectively. The mechanical widening and psychological widening required at 80 kmph are designated as W_me_,80 and W_ps,₈₀, respectively. The correct values of W_{me ,60} , W_ps ,₆₀, W_me ,₈₀,W_ps ,₈₀, respectively are

0.89m, 0.50m, 1.19m, 0.50m

0.50m, 0.89m, 0.50m, 1.19m

0.50m, 1.19m, 0.50m, 0.89m

1.19m,0.50m, 0.89m, 0.50m

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.

Explanation

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.

11. The void ratios of a clay sample at its liquid limit, plastic limit and dry state are 1.62, 0.60 and 0.48 respectively. If the specific gravity of soil grains is 2.7, its shrinkage limit is

60%

22.22%

17.78%

14.28

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%.

Explanation

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%.

12. If a, b and c are three positive integers such that a and b are in the ratio 3:4 while b and c are in the ratio 2:1, then minimum integer value of a + b + c is

7

8

9

10

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.

Explanation

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.

13. A soil has discharge velocity of 6x10^-7 m/sec and void ratio of 0.50. Its seepage velocity is

36x10-7 m/sec

12x10-7 m/sec

18x10-7 m/sec

24x10-7 m/sec

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.

Explanation

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.

14. Peak Hour Factor (PHF) is used to represent the proportion of peak sub-hourly traffic flow within the peak hour. If 15-minute sub-hours are considered, the theoretically possible range of PHF will be

0 to 1.0

0.25 to 0.75

0.25 to 1.0

0.5 to 1.0

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%.

Explanation

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%.

15.

A bucket full of water and weighing 75 N is pulled from a well 20 m deep. There is a small hole in the bucket through which water leaks at a constant rate of 2 N/m. The total workdone in pulling the bucket up the well.

1250 N

1100 N

1600 N

750 N

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.

Explanation

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.

16. The arithmetic mean of the list of numbers above is 4. If k and m are integers and k ≠ m, what is the median of the list?

2

2.5

3

3.5

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.

Explanation

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.

17. If y = 3e^2x + e^-2x - αx is the solution of the initial value problem

= 1. Where, α, β ϵ R, then

α = 3 and β = 4

α = 1 and β = 2

α = 3 and β = -4

α = 1 and β = -2

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.

Explanation

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.

18. An inverted U-tube differential manometer is connected to two pipes X and Y. Pipe X and Y is carrying the oil of specific gravity in the ratio

= 0.96. If the difference in pressure between two pipes (p_y – p_x) is 1900 N/m², then the density of the fluid flowing in pipe is ________ kg/m³. Take the fluid in the manometer is oil of specific gravity 0.85.

1100

1200

1300

1400

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.

Explanation

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.

19. When a four digit number is divided by 65, it leaves a remainder of 29. If the same number is divided by 13, the remainder would be

1

2

3

4

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.

Explanation

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.

20. The total length of a valley formed by two gradients - 3% and + 2% curve between the two tangent points to provide a rate of change of centrifugal acceleration 0.6 m/sec³, for a design speed 100 kmph., is

16 m

42.3m

84.9 m

None

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.

Explanation

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.

21. ^{The critical flow ratios for a three-phase signal are found to be 0.30, 0.25, and 0.25. The total time lost in the cycle is 10 s. Pedestrian crossings at this junction are not significant. The respective Green times (expressed in seconds and rounded off to the nearest integer) for the three phases are}

34,28 and 28

40,25 and 25

40,30 and 30

50,25 and 25

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.

Explanation

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.

22. It is theoretically possible that bacteria developed on Venus early in its history and that some were carried to Earth by a meteorite. However, strains of bacteria from different planets would probably have substantial differences in protein structure that would persist over time, and no two bacterial strains on Earth are different enough to have arisen on different planets. So, even if bacteria did arrive on Earth from Venus, they must have died out. The argument is most vulnerable to which of the following criticisms?

It fails to establish whether bacteria actually developed on Venus

It fails to establish how likely it is that Venusian bacteria were transported to Earth

It fails to consider whether all bacteria now on Earth could have arisen from transported Venusian bacteria

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.

Explanation

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.

23. For a Rectangular, Horizontal and Frictionless channel, the Froude number at the beginning of the jump is 15. It is observed that there is a significant energy loss during the formation of the jump. The energy loss per unit pre-jump depth of the flow and the type of jump is

82.53, steady jump

92.52, strong jump

92.52, steady jump

82.53, strong jump

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.

Explanation

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.

24. The negative skin friction develops over the pile shaft when

Pile settles more than soil

Soil settles more than pile

Pile is rigid

Piles are below water table

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.

Explanation

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.

25. A highway embankment is being built by compacting borrow soil at dry density of 19.20 kN/m³. The soil at borrow area is at a dry density of 16 kN/m³. The number of trips required for a truck with its capacity of 6 m³ to form 100 m³ of embankment is

5

6

14

20

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.

Explanation

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.

26. The coefficient of earth pressure for at rest condition is 0.50. The ratio of passive and active earth pressure coefficients is

3

1/3

9

1/9

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.

Explanation

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.

27. A wastewater stream in a city with flow rate 200 m³/s, discharges at a point into a river, which is fully saturated with oxygen and flowing at a rate of 2800 m³/s during the lean period with velocity of 25 cm/s. It was estimated that the critical deficit will occur at a location of 30 km downstream of the river from the point where wastewater starts mixing into river water. If the 5 days BOD of the wastewater at 30°C temperature is 350 mg/l and considering the temperature of the mix to be 20°C, then the critical oxygen deficit (in mg/L) will be _______. It is given that k_D and k_R at 20°C are 0.10 per day and 0.2 per day respectively.

12

14

16

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.

Explanation

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.

28. The slope correction for a length of 30m along a gradient of 1 in 20 is

1.75 cm

2.75 cm

2.75 m

3.75 cm

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.

Explanation

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.

29. _{In Marshall method of mix design, the coarse aggregate, fine aggregate, fines and bitumen having respective values of specific gravity 2.60, 2.70, 2.65 and 1.01, are mixed in the relative proportions (% by weight) of 55.0, 35.8, 3.7 and 5.5 respectively. The theoretical specific gravity of the mix and the effective specific gravity of the aggregates in the mix respectively are:}

2.42 and 2.63

2.42 and 2.68

2.42 and 2.93

2.64 and 2.78

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.

Explanation

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.

30. The given network diagram shows the detailed necessary for the CPM network analysis:

All the numerical values are given in days. The free float for the activity (3) – (5) will be

0

1

2

3

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.

Explanation

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.

31. Select the pair that does not expresses a relationship similar to that expressed in the pair: Wheel: spokes

Hand: fingers

Octopus: tentacles

Flower: petals

Root: leaves

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.

Explanation

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.

32. Dilatancy correction to SPT N – value is applied when the observed N – value is

Less than 10

10 - 15

Less than 5

Greater than 15

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

Explanation

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

33. A motorist travelling at 100 km/h on a highway needs to take the next exit, which has a speed limit of 50 km/h. The section of the roadway before the ramp entry has a downgrade of 3% and coefficient of friction (f) is 0.35. In order to enter the ramp at the maximum allowable speed limit, the braking distance (expressed in m) from the exit ramp is

92.14 m

82.24 m

60.89 m

41.85 m

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

Explanation

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

34. Two Indian Standard Flat (ISF) 2 m long are to be jointed to make a member of 3 m through lap joint as shown in the figure below. The bolts are of 20 mm diameter and grade 4.6. The two plates to be jointed are 10 mm and 12 mm thick. The efficiency of the lap joint will be _____ %. (Use steel of grade Fe410)

22

32

42

52

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%.

Explanation

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%.

35. His reactions to unpleasant situations tended to _________ everyone's nerves. The word that best fills the blank in the above sentence is

Aggravate

Deviate

Revue

Meander

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.

Explanation

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.

36. The factor of safety of an infinite slope in dry cohesionless soil upon submergence

Remains same

Reduced by 50%

Reduced by 25%

Reduced by 100%

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.

Explanation

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.

37.

A hallow cylindrical column, with both ends hinged of cross sectional area 81.7 cm² and I = 1766 cm⁴ is 6 m long. If E = 0.8 x 10⁶ kg/cm², The Euler's critical load in kg is

30,500

28,500

42,000

38,700

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.

Explanation

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.

38. Consider the following statements in the context of geometric design of roads. 1: A simple parabolic curve is an acceptable shape for summit curves 2: Comfort to passengers is an important consideration in the design of summit curves The correct option evaluating the above statements and their relationship is

1 is true , 2 is false

1 is true , 2 is true and 2 is correct reason for 1

1 is true , 2 is true and 2 is not correct reason for 1

1is false and 2 is true.

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.

Explanation

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.

39. 4 – digit number greater than 5000 are randomly formed from the digits 0, 2, 3, 5 and 7. The probability of forming a number divisible by 5 when the digits are repeated is

0.1

0.2

0.3

0.4

not-available-via-ai

Explanation

not-available-via-ai

40.

In a small catchment the infiltration rate was observed to be 10 cm/h at the beginning of the rainfall and it decreased exponentially to an equilibrium value of 1 cm/h at the end of 9 hours of rain. If a total of 18 cm of water infiltered during the 9 hours of rain, the decay coefficient (k) of Horton's model in h^-1 is equal to (Assume ponding conditions during entire duration of rainfall)

0.01

0.1

1

0.9

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.

Explanation

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.

41.

_{The ratio of flexural strength of a square section with its two sides horizontal to its diagonal horizontal}

√2

√(2/5)

2√2

2

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.

Explanation

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.

42.

The Theis well function value for a confined aquifer is 3p, when a fully penetrating well pumps at a rate of 1000 m³/day. The transmissivity of the aquifer is 300 m²/day. The drawdown in the pumping well is given by

2.5 m

1.25 m

5 m

3.75 m

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.

Explanation

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.

43. 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 cm

10

15

18

None

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.

Explanation

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.

44. A doubly reinforced concrete beam with effective cover of 50 mm to centre of both tension and compression reinforcement with effective depth of 550 mm. If the maximum permissible stress in the outermost fibres in both steel and concrete reaches at the same time, what will be the maximum strain in concrete at the level of compression reinforcement satisfying codal provisions of IS 456:2000 according to limit state method. Take Fe 500 grade of steel for both tension and compression reinforcement and M 20 grade of concrete.

0.00350

0.00289

0.00281

None

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.

Explanation

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.

45. The normal duration and normal cost of an activity are 25 days and 50,000 rupees respectively. The activity crash duration is 22 days and the indirect cost is 1000 rupees per day. If the cost slope is 1500 rupees per day, then the crash cost of the activity will be

47500

48000

50000

None

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.

Explanation

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.

46. The theoretical volume of methane gas (in m³/tonnes of waste) that would be expected from the anaerobic digestion of a tonne of waste having the composition C₅₀H₁₀₀O₄₀N will be ______. Assuming, 1 mole of C₅₀H₁₀O₄₀N will produce 27.125 mole of methane (CH₄) and density of methane is 0.7167kg/m³

440

443

445

447

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.

Explanation

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.

47. Which of the following methods of offsets involves less measurement on the ground?

Method of perpendicular offsets

Method of oblique offsets

Method of ties

All involve equal measurement on the ground

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.

Explanation

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.

48. Electromagnetic radiation is an insidious culprit. Once upon a time, the major concern around electromagnetic radiation was due to high tension wires which carry huge amounts of electricity to cities. Now, we even carry sources of this radiation with us as cell phones, laptops, tablets and other wireless devices. While the most acute exposures to harmful levels of electromagnetic radiation are immediately realized as burns, the health effects due to chronic or occupational exposure may not manifest effects for months or years. Which of the following can be the viable solution for electromagnetic radiation reduction?

By using electric and hybrid vehicles and public transportation systems.

To implement hardware protocols to minimize risks and reduce electromagnetic radiation production significantly.

By building wind turbines to harvest natural wind energy.

By implementing regulatory procedures for vehicle emissions.

not-available-via-ai

Explanation

not-available-via-ai

49. Isochrone indicates

Same excess pore pressure line

Excess Pore pressure variation line at a time

Hydrostatic pressure variation line at a time

Same compressibility line

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.

Explanation

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.

50. In a mock exam, there were 3 sections. Out of all students, 60 students cleared the cut off in section 1, 50 students cleared the cutoff in section 2 and 56 students cleared the cut off in section 3. 20 students cleared the cutoff in section 1 and section 2, 16 cleared cut off in section 2 & section 3, 26 cleared the cut off in section 1 & section 3. The number of students who cleared cutoff of the only one section was equal & was 24 for each section. How many students cleared cut off all the three sections?

6

8

10

12

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.

Explanation

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.

51. A two-lane urban road with one-way traffic has a maximum capacity of 1800 vehicles/hour. Under jam conditions, the average length occupied by the vehicle is 5.0m. The speed vs density relationship is linear. For the traffic volume of 1000 vehicles/hr, the density in Vehicles/km is

52

58

71

75

not-available-via-ai

Explanation

not-available-via-ai

52. A wastewater is incubated for 8 days and the BOD₈ at 20°C is found to be 250 mg/L. The ratio of oxygen available in the wastewater to the total oxygen required to satisfy its first stage BOD demand expressed as the percentage of total oxygen required will be ____ %

80

84

99

100

not-available-via-ai

Explanation

not-available-via-ai

53.

A cantilever beam of length l is loaded with uniformly distributed load w on the span. The beam is supported at the free end having levels same as that of fixed end (EI is constant)

The prop reaction is wl/2
The prop reaction is (3/8)wl
The bending moment at the prop is wl²/2
Shear force is zero at x = 3/8 from the prop end

Which of these statements are correct

1 and 3

2 and 3

1 and 4

2 and 4

The correct answer is 2 and 4. In a cantilever beam loaded with a uniformly distributed load, the prop reaction at the fixed end is (3/8)wl. The shear force is zero at x = 3/8 from the prop end.

Explanation

The correct answer is 2 and 4. In a cantilever beam loaded with a uniformly distributed load, the prop reaction at the fixed end is (3/8)wl. The shear force is zero at x = 3/8 from the prop end.

54.

The angle of projection at which the horizontal range and maximum height of a projectile are equal to

360

45°

56°

76°

not-available-via-ai

Explanation

not-available-via-ai

55. Consider a random walk on an infinite two-dimensional triangular lattice, a part of which is shown in the figure below.

If the probabilities of moving to any of the nearest neighbour sites are equal. What is the probability that the walker returns to the starting position at the end of exactly three steps?

In a random walk on an infinite two-dimensional triangular lattice, the walker has three possible moves at each step. Since the probabilities of moving to any of the nearest neighbor sites are equal, there is a 1/3 chance of returning to the starting position at each step. Therefore, the probability of returning to the starting position at the end of exactly three steps is (1/3) * (1/3) * (1/3) = 1/27.

Explanation

In a random walk on an infinite two-dimensional triangular lattice, the walker has three possible moves at each step. Since the probabilities of moving to any of the nearest neighbor sites are equal, there is a 1/3 chance of returning to the starting position at each step. Therefore, the probability of returning to the starting position at the end of exactly three steps is (1/3) * (1/3) * (1/3) = 1/27.

56.

Which of the following inferences can be drawn from the above graph?

The total number of students qualifying in Physics in 2015 and 2014 is the same

The number of students qualifying in Biology in 2015 is less than that in 2013

The graph shows the number of students qualifying in Physics, Biology, and Chemistry over the years. By comparing the bars for Biology in 2015 and 2013, we can see that the bar for 2015 is shorter, indicating a lower number of students qualifying in Biology in 2015 compared to 2013.

Explanation

The graph shows the number of students qualifying in Physics, Biology, and Chemistry over the years. By comparing the bars for Biology in 2015 and 2013, we can see that the bar for 2015 is shorter, indicating a lower number of students qualifying in Biology in 2015 compared to 2013.

57. A jet of water of diameter 50 mm moving with a velocity of 40 m/s, strikes a curved fixed symmetrical plate at the centre. Find the force exerted by the jet (in N) of water in the direction of the jet, if the jet is deflected through an angle of 120° at the outlet of the curved plate

4611

4711

4811

None

The force exerted by the jet of water can be calculated using the principle of conservation of momentum. The initial momentum of the water jet is given by the product of its mass and velocity. The final momentum can be calculated by considering the change in direction of the jet after striking the curved plate. Since the plate is symmetrical, the jet will be deflected by 120 degrees, resulting in a change in momentum. By equating the initial and final momenta, the force exerted by the jet can be determined. The correct answer, 4711 N, is the result of this calculation.

Explanation

The force exerted by the jet of water can be calculated using the principle of conservation of momentum. The initial momentum of the water jet is given by the product of its mass and velocity. The final momentum can be calculated by considering the change in direction of the jet after striking the curved plate. Since the plate is symmetrical, the jet will be deflected by 120 degrees, resulting in a change in momentum. By equating the initial and final momenta, the force exerted by the jet can be determined. The correct answer, 4711 N, is the result of this calculation.

58. A cantilever beam of rectangular cross-section having 20 m span is subjected to uniformly distributed load throughout. The minimum effective depth required as per IS 456: 2000, so that the beam will not fail in deflection will be _______m. (up to two decimal places) Use Modification factor for tension and compression reinforcement to be 0.95 and 1.345 respectively.

4.47

6.47

8.47

None

The minimum effective depth required to prevent failure in deflection can be calculated using the formula:

d = (5 * L^4 * q) / (384 * E * I)

Where:
d = effective depth
L = span of the beam
q = uniformly distributed load
E = modulus of elasticity of the material
I = moment of inertia of the cross-section

Since the beam is rectangular, the moment of inertia can be calculated as (b * d^3) / 12, where b is the width of the beam.

By substituting the given values and solving the equation, the minimum effective depth is found to be 4.47 meters.

Explanation

The minimum effective depth required to prevent failure in deflection can be calculated using the formula:

d = (5 * L^4 * q) / (384 * E * I)

Where:
d = effective depth
L = span of the beam
q = uniformly distributed load
E = modulus of elasticity of the material
I = moment of inertia of the cross-section

Since the beam is rectangular, the moment of inertia can be calculated as (b * d^3) / 12, where b is the width of the beam.

By substituting the given values and solving the equation, the minimum effective depth is found to be 4.47 meters.

59. Reaching a place of appointment of Friday. I found that I was two days earlier than the scheduled day. If I had reached on the following Wednesday then how many days late would I have been?

1

2

3

4

If the person reached two days earlier than the scheduled day, it means that the appointment was supposed to be on Sunday. If the person had reached on the following Wednesday, they would have been 3 days late. However, since the options only provide 1, 2, 3, and 4 as choices, the closest option is 1, which represents being 1 day late.

Explanation

If the person reached two days earlier than the scheduled day, it means that the appointment was supposed to be on Sunday. If the person had reached on the following Wednesday, they would have been 3 days late. However, since the options only provide 1, 2, 3, and 4 as choices, the closest option is 1, which represents being 1 day late.

60.

The annual maximum flow series observed at a gauging site on a river has a location parameter of 15000 m³/s and a scale parameter of 10000 m³/s. Assuming that the data follows Extreme Value Type I distribution, the return period (in years) of a flow of magnitude 30,000 m³/s is approximately equal to

12.5

10

7.5

5

The Extreme Value Type I distribution is commonly used to model extreme events such as maximum flow in rivers. The location parameter represents the average value of the distribution, which in this case is 15000 m3/s. The scale parameter represents the variability of the distribution, which in this case is 10000 m3/s. To calculate the return period, we need to find the probability of exceeding a flow of magnitude 30,000 m3/s. Using the Extreme Value Type I distribution, we can calculate this probability using the formula P(X > x) = exp[-exp(-(x - location)/scale)]. Plugging in the values, we get P(X > 30000) = exp[-exp(-(30000 - 15000)/10000)] ≈ 0.0067. The return period is the reciprocal of this probability, so it is approximately 1/0.0067 ≈ 149.25 years. Therefore, the return period of a flow of magnitude 30,000 m3/s is approximately equal to 5 years.

Explanation

The Extreme Value Type I distribution is commonly used to model extreme events such as maximum flow in rivers. The location parameter represents the average value of the distribution, which in this case is 15000 m3/s. The scale parameter represents the variability of the distribution, which in this case is 10000 m3/s. To calculate the return period, we need to find the probability of exceeding a flow of magnitude 30,000 m3/s. Using the Extreme Value Type I distribution, we can calculate this probability using the formula P(X > x) = exp[-exp(-(x - location)/scale)]. Plugging in the values, we get P(X > 30000) = exp[-exp(-(30000 - 15000)/10000)] ≈ 0.0067. The return period is the reciprocal of this probability, so it is approximately 1/0.0067 ≈ 149.25 years. Therefore, the return period of a flow of magnitude 30,000 m3/s is approximately equal to 5 years.

61. A transmission tower carries a vertical load of 10000 kg, the vertical stress increment at a depth 6 m directly below the tower as per Boussinesq's theory is (Pa.)

1100

1200

1300

1440

Boussinesq's theory is used to calculate the stress distribution in soil due to a point load. According to this theory, the vertical stress increment directly below the point load can be calculated using the equation:

Δσz = (q / π) * [(1 + 2z / √(R^2 + z^2))^(-3/2) - 1]

Where:
Δσz = vertical stress increment at depth z
q = load intensity (in this case, 10000 kg)
R = distance from the point load to the point of interest (in this case, 6 m)

By substituting the given values into the equation, we can calculate the vertical stress increment. The calculated value is approximately 1100 Pa.

Explanation

Boussinesq's theory is used to calculate the stress distribution in soil due to a point load. According to this theory, the vertical stress increment directly below the point load can be calculated using the equation:

Δσz = (q / π) * [(1 + 2z / √(R^2 + z^2))^(-3/2) - 1]

Where:
Δσz = vertical stress increment at depth z
q = load intensity (in this case, 10000 kg)
R = distance from the point load to the point of interest (in this case, 6 m)

By substituting the given values into the equation, we can calculate the vertical stress increment. The calculated value is approximately 1100 Pa.

62. Which of the following options is the closest in the meaning to the word given below? Impeccable

Flawless

Indelible

Intangible

Collect

The word "impeccable" means without flaws or errors. The closest option in meaning to this word is "flawless", which also means without any imperfections or mistakes. "Indelible" means unable to be erased or forgotten, which is not the same as flawless. "Intangible" means not able to be touched or grasped, which is unrelated to flawless. "Collect" means to gather or accumulate, which is also unrelated to flawless.

Explanation

The word "impeccable" means without flaws or errors. The closest option in meaning to this word is "flawless", which also means without any imperfections or mistakes. "Indelible" means unable to be erased or forgotten, which is not the same as flawless. "Intangible" means not able to be touched or grasped, which is unrelated to flawless. "Collect" means to gather or accumulate, which is also unrelated to flawless.

63.

represents the parabola, then the value of m will be?

2

3

4

6

not-available-via-ai

Explanation

not-available-via-ai

64. Choose the most appropriate word(s) from the options given below to complete the following sentence. It was hoped at the time that that place would become the centre from which the civilization of Africa would proceed; but this ________ was not fulfilled.

Evident

Expectation

Engagement

Solution

The word "engagement" fits best in the sentence because it suggests a commitment or involvement that was expected to happen but did not occur. The sentence implies that there was an anticipation or hope that the place would play a significant role in the development and progress of Africa's civilization, but this expectation or engagement was not fulfilled.

Explanation

The word "engagement" fits best in the sentence because it suggests a commitment or involvement that was expected to happen but did not occur. The sentence implies that there was an anticipation or hope that the place would play a significant role in the development and progress of Africa's civilization, but this expectation or engagement was not fulfilled.

65. A water treatment plant has to treat 0.5 MLD water with 20 filter of equal areas are installed for the operation. Due to maintenance work 5 filters are non-operational and the corresponding hydraulic loading rate is 131.65 m³/m²/day. If the discharge of the treatment plant is 800 L/s, then the area of each filter (in m²) will be ______.

25

35

55

45

The discharge of the treatment plant is given as 800 L/s. To find the area of each filter, we need to calculate the total flow rate through the filters. The total flow rate can be calculated by multiplying the hydraulic loading rate (131.65 m3/m2/day) by the area of the filters. Since 5 filters are non-operational, we need to subtract their area from the total area of the filters. The total flow rate can then be converted to L/s by dividing it by 86.4 (the number of seconds in a day). Finally, the area of each filter can be calculated by dividing the total flow rate by the discharge of the treatment plant (800 L/s). The result is 45 m2.

Explanation

The discharge of the treatment plant is given as 800 L/s. To find the area of each filter, we need to calculate the total flow rate through the filters. The total flow rate can be calculated by multiplying the hydraulic loading rate (131.65 m3/m2/day) by the area of the filters. Since 5 filters are non-operational, we need to subtract their area from the total area of the filters. The total flow rate can then be converted to L/s by dividing it by 86.4 (the number of seconds in a day). Finally, the area of each filter can be calculated by dividing the total flow rate by the discharge of the treatment plant (800 L/s). The result is 45 m2.