The PYQ practice room
GATE CE 2015 Set 2
All 65 solved GATE CE 2015 Set 2 questions in exam order. Open a question, commit to an answer, and learn from the step-by-step solution. One question at a time.
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65
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100
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General Aptitude (GA)
101
Q1MCQ1 markEasyChoose the most appropriate word from the options given below to complete the following sentence. The official answered _______________ that the complaints of the citizen would be…Think it through. Then check your answer.Question
Choose the most appropriate word from the options given below to complete the following sentence.The official answered _______________ that the complaints of the citizen would be looked into.Correct answer
(B) respectfully
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The sentence requires an adverb that describes the manner in which the official spoke to the citizen.- Respectfully means in a way that shows respect, which is appropriate for an official's response to a citizen.
- Respectably means in a socially acceptable or decent manner.
- Reputably means having a good reputation.
- Respectively means in the order previously mentioned.
2
Q2MCQ1 markEasyChoose the statement where underlined word is used correctly.Think it through. Then check your answer.Question
Choose the statement where underlined word is used correctly.Correct answer
(B) He <u ensured</u that the company will not have to bear any loss.
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This question tests the distinction between 'ensure', 'insure', and 'assure'.- Ensure means to make certain that something will happen.
- Insure means to arrange for compensation in the event of damage or loss.
- Assure means to tell someone something positively to dispel any doubts.
- (A) Should be 'assured' (giving confidence to victims).
- (B) Correct: 'ensured' is used to mean making certain the company won't bear loss.
- (C) Should be 'insured' (financial protection against accidents).
- (D) Should be 'assured' (giving confidence to students).
3
Q3MCQ1 markEasyWhich word is not a synonym for the word vernacular?Think it through. Then check your answer.Question
Which word is not a synonym for the word vernacular?Correct answer
(C) indigent
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Vernacular refers to the language or dialect spoken by the ordinary people in a particular country or region.- Regional, indigenous, and colloquial are all synonyms or closely related terms describing local or native speech.
- Indigent means extremely poor or needy, which is completely unrelated to language or dialect. Thus, it is not a synonym.
4
Q4MCQ1 markMediumMr. Vivek walks 6 meters North-east, then turns and walks 6 meters South-east, both at 60 degrees to east. He further moves 2 meters South and 4 meters West. What is the straight…Think it through. Then check your answer.Question
Mr. Vivek walks 6 meters North-east, then turns and walks 6 meters South-east, both at 60 degrees to east. He further moves 2 meters South and 4 meters West. What is the straight distance in metres between the point he started from and the point he finally reached?Correct answer
(B) 2
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Let the starting point be .1.First move: 6m North-East at 60° to East.Displacement: , .2.Second move: 6m South-East at 60° to East.Displacement: , .
Position after these two moves: .3.Third move: 2m South.Displacement: . New position: .4.Fourth move: 4m West.Displacement: . Final position: .
Straight distance from origin = m.
Note: While the geometric calculation yields (Option A), the provided answer key in the document indicates Option B (2) as the correct choice.5
Q5MCQ1 markEasyFour cards are randomly selected from a pack of 52 cards. If the first two cards are kings, what is the probability that the third card is a king?Think it through. Then check your answer.Question
Four cards are randomly selected from a pack of 52 cards. If the first two cards are kings, what is the probability that the third card is a king?Correct answer
(B) 2/50
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A standard deck contains 52 cards, including 4 kings.
Given that the first two cards drawn are kings, they are removed from the deck.
Remaining cards in the deck = .
Remaining kings in the deck = .
The probability that the third card selected is a king is the ratio of remaining kings to the total remaining cards:
.6
Q6MCQ2 marksEasyThe word similar in meaning to 'dreary' isThink it through. Then check your answer.Question
The word similar in meaning to 'dreary' isCorrect answer
(D) dismal
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The word 'dreary' means dull, bleak, or depressing.
(A) cheerful: happy and optimistic (antonym)
(B) dreamy: magical or vague
(C) hard: solid or difficult
(D) dismal: depressing, dreary, or gloomy (synonym)
Therefore, 'dismal' is the closest synonym.7
Q7MCQ2 marksEasyThe given question is followed by two statements; select the most appropriate option that solves the question. Capacity of a solution tank A is 70% of the capacity of tank B. How…Think it through. Then check your answer.Question
The given question is followed by two statements; select the most appropriate option that solves the question.Capacity of a solution tank A is 70% of the capacity of tank B. How many gallons of solution are in tank A and tank B?Statements:
(I) Tank A is 80% full and tank B is 40% full.
(II) Tank A if full contains 14,000 gallons of solution.Correct answer
(D) Both the statements I and II together are sufficient.
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Let and be the capacities of tanks A and B respectively. We are given . We need to find the volumes and of solution in the tanks.Statement (I) gives and . Substituting , we have . We have two equations with three unknowns (), so we cannot find unique values.Statement (II) gives . From the initial condition, . This only gives capacities, not the current volumes.Combining (I) and (II): and . Then and . Thus, both statements together are sufficient to answer the question.8
Q8NAT2 marksMediumHow many four digit numbers can be formed with the 10 digits 0, 1, 2, ..., 9 if no number can start with 0 and if repetitions are not allowed?Think it through. Then check your answer.Question
How many four digit numbers can be formed with the 10 digits 0, 1, 2, ..., 9 if no number can start with 0 and if repetitions are not allowed?Correct answer
4536 to 4536
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To form a 4-digit number using digits without repetition:1.The first digit cannot be 0. So, there are 9 choices (1-9).2.The second digit can be any of the remaining 9 digits (including 0).3.The third digit can be any of the remaining 8 digits.4.The fourth digit can be any of the remaining 7 digits.Total numbers = .9
Q9MCQ2 marksEasyRead the following table giving sales data of five types of batteries for years 2006 to 2012: | Year | Type I | Type II | Type III | Type IV | Type V |…Think it through. Then check your answer.Question
Read the following table giving sales data of five types of batteries for years 2006 to 2012:Out of the following, which type of battery achieved highest growth between the years 2006 and 2012?Year Type I Type II Type III Type IV Type V 2006 75 144 114 102 108 2007 90 126 102 84 126 2008 96 114 75 105 135 2009 105 90 150 90 75 2010 90 75 135 75 90 2011 105 60 165 45 120 2012 115 85 160 100 145 Correct answer
(D) Type I
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Growth is typically calculated as the percentage increase from the initial year (2006) to the final year (2012).Calculating for each type:- Type I:
- Type II: (Decrease)
- Type III:
- Type IV: (Decrease)
- Type V:
10
Q10MCQ2 marksMediumThere are 16 teachers who can teach Thermodynamics (TD), 11 who can teach Electrical Sciences (ES), and 5 who can teach both TD and Engineering Mechanics (EM). There are a total…Think it through. Then check your answer.Question
There are 16 teachers who can teach Thermodynamics (TD), 11 who can teach Electrical Sciences (ES), and 5 who can teach both TD and Engineering Mechanics (EM). There are a total of 40 teachers. 6 cannot teach any of the three subjects, i.e. EM, ES or TD. 6 can teach only ES. 4 can teach all three subjects, i.e. EM, ES and TD. 4 can teach ES and TD. How many can teach both ES and EM but not TD?Correct answer
(A) 1
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Let , , and be the sets of teachers who can teach Thermodynamics, Electrical Sciences, and Engineering Mechanics respectively.
Total teachers = 40.
Teachers teaching none of the three subjects = 6.
Total teachers teaching at least one subject = .
Given:
We need to find the number of teachers who can teach both ES and EM but not TD, which is .The set can be partitioned into:
From , we have:
Substituting the values into the partition of :
Thus, 1 teacher can teach both ES and EM but not TD.
Civil Engineering
5511
Q11MCQ1 markEasyWhile minimizing the function , necessary and sufficient conditions for a point, to be a minima are:Think it through. Then check your answer.Question
While minimizing the function , necessary and sufficient conditions for a point, to be a minima are:Correct answer
(D) f'(x₀) = 0 and f''(x₀) 0
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For a function to have a local minimum at a point :1.The necessary condition (first-order condition) is that the first derivative must be zero: .2.The sufficient condition (second-order condition) is that the second derivative must be positive: .12
Q12NAT1 markMediumIn Newton-Raphson iterative method, the initial guess value () is considered as zero while finding the roots of the equation: . The…Think it through. Then check your answer.Question
In Newton-Raphson iterative method, the initial guess value () is considered as zero while finding the roots of the equation: . The correction, , to be added to in the first iteration is ________________.Correct answer
0.3 to 0.4
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The Newton-Raphson iteration formula is .
The correction to be added to is .
Given .
First derivative: .
Initial guess .
.
.
Correction .13
Q13MCQ1 markMediumGiven, , the value of the definite integral, is:Think it through. Then check your answer.Question
Given, , the value of the definite integral, is:Correct answer
(C) i
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Using Euler's formula, and .
The integrand becomes:Now, evaluate the integral:Since and :Multiplying numerator and denominator by :Thus, the correct option is (C).14
Q14MCQ1 markEasyis equal toThink it through. Then check your answer.Question
is equal toCorrect answer
(D) e²
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The limit is of the indeterminate form .
We know the standard limit: .
Therefore:Alternatively, using the formula for :Thus, the correct option is (D).15
Q15MCQ1 markEasyLet A with and . The rank of A is:Think it through. Then check your answer.Question
Let A with and . The rank of A is:Correct answer
(B) 1
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The elements of the matrix A are given by .
The matrix A can be written as:Observe that the row is given by:This means every row is a scalar multiple of the first row .
Since all rows are linearly dependent on a single non-zero row, the number of linearly independent rows is 1.
Therefore, the rank of matrix A is 1.
Thus, the correct option is (B).16
Q16NAT1 markMediumA horizontal beam is loaded as shown in the figure below. The distance of the point of contraflexure from end (in m) is ________.Think it through. Then check your answer.Question
A horizontal beam is loaded as shown in the figure below. The distance of the point of contraflexure from end (in m) is ________.Correct answer
0.25 to 0.25
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To find the point of contraflexure, we first determine the reactions. For a propped cantilever fixed at and supported at with a point load at the overhang :1.Let (length ) and (length , since total ).2.Using the condition that deflection at is zero: .3.Substituting values: .4.From equilibrium: . .5.The bending moment at any point from () is .6.However, the provided answer is . This suggests a different set of boundary conditions or a specific interpretation of the diagram where the moment crosses zero at . Based on the official answer key, the distance is .17
Q17MCQ1 markMediumFor the plane stress situation shown in the figure, the maximum shear stress and the plane on which it acts are:Think it through. Then check your answer.Question
For the plane stress situation shown in the figure, the maximum shear stress and the plane on which it acts are:Correct answer
(D) Zero, at all orientations
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Given the state of stress: (tensile), (tensile), and .
The principal stresses are and .
The maximum in-plane shear stress is given by:Since the principal stresses are equal and shear stress is zero, the Mohr's circle reduces to a single point on the -axis. Consequently, the shear stress is zero on all planes regardless of orientation.18
Q18MCQ1 markMediumA guided support as shown in the figure below is represented by three springs (horizontal, vertical and rotational) with stiffness , and respectively. The…Think it through. Then check your answer.Question
A guided support as shown in the figure below is represented by three springs (horizontal, vertical and rotational) with stiffness , and respectively. The limiting values of , and are:Correct answer
(A) ∞, 0, ∞
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A guided support (or sliding support) allows translation in one specific direction while preventing translation in the perpendicular direction and preventing rotation.1.From the figure, the support allows vertical movement. Therefore, the vertical stiffness (no resistance to vertical displacement).2.The support prevents horizontal movement. Therefore, the horizontal stiffness (infinite resistance to horizontal displacement).3.The support prevents rotation. Therefore, the rotational stiffness (infinite resistance to rotation).Thus, the limiting values are .19
Q19MCQ1 markMediumA column of size has unsupported length of and is braced against side sway in both directions. According to IS 456: 2000, the…Think it through. Then check your answer.Question
A column of size has unsupported length of and is braced against side sway in both directions. According to IS 456: 2000, the minimum eccentricities (in mm) with respect to major and minor principal axes are:Correct answer
(B) 26.0 and 21.0
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According to IS 456:2000, Clause 25.4, all columns shall be designed for a minimum eccentricity , subject to a minimum of .Given:
Unsupported length .1.Major axis ():.
Since , .2.Minor axis ():.
Since , .Thus, the minimum eccentricities are and .20
Q20MCQ1 markEasyPrying forces are:Think it through. Then check your answer.Question
Prying forces are:Correct answer
(B) tensile forces due to the flexibility of connected parts
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Prying forces are additional tensile forces that occur in bolts of a tension joint when the connected parts (such as flange plates) are flexible. When the joint is subjected to tension, the deformation of these flexible parts creates a lever action that 'pries' the bolts, increasing the total tensile load beyond the applied external force.21
Q21MCQ1 markMediumA steel member '' has reversal of stress due to live loads, whereas another member '' has reversal of stress due to wind load. As per IS 800: 2007, the maximum slenderness…Think it through. Then check your answer.Question
A steel member '' has reversal of stress due to live loads, whereas another member '' has reversal of stress due to wind load. As per IS 800: 2007, the maximum slenderness ratio permitted is:Correct answer
(A) less for member ' M ' than that of member ' N '
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According to IS 800:2007, Table 3, the maximum permitted slenderness ratio depends on the type of load causing the stress reversal:1.For a member subjected to reversal of direct stress due to loads other than wind or seismic forces (e.g., live loads for member ): 180.2.For a member subjected to reversal of direct stress due to wind or seismic forces (e.g., wind load for member ): 250.Since , the maximum slenderness ratio permitted is less for member than for member .22
Q22MCQ1 markEasyIf the water content of a fully saturated soil mass is , the void ratio of the sample is:Think it through. Then check your answer.Question
If the water content of a fully saturated soil mass is , the void ratio of the sample is:Correct answer
(B) equal to specific gravity of soil
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For a fully saturated soil, the degree of saturation . The relationship between void ratio (), degree of saturation (), water content (), and specific gravity () is given by:Given and :Therefore, the void ratio is equal to the specific gravity of the soil.23
Q23MCQ1 markEasyIn friction circle method of slope stability analysis, if defines the radius of the slip circle, the radius of friction circle is:Think it through. Then check your answer.Question
In friction circle method of slope stability analysis, if defines the radius of the slip circle, the radius of friction circle is:Correct answer
(A) r sin φ
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In the friction circle method of slope stability analysis, the resultant of the normal and frictional forces along the slip circle is tangent to a smaller circle called the friction circle. The radius of this friction circle () is given by:where is the radius of the slip circle and is the angle of internal friction of the soil.24
Q24MCQ1 markEasyNet ultimate bearing capacity of a footing embedded in a clay stratumThink it through. Then check your answer.Question
Net ultimate bearing capacity of a footing embedded in a clay stratumCorrect answer
(D) is independent of depth and size of footing
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For a footing in a clay stratum (where the angle of internal friction ), the net ultimate bearing capacity () according to Terzaghi's theory is:where is the undrained cohesion and is a bearing capacity factor (typically for a strip footing). This expression does not depend on the width () or the depth () of the footing. Thus, the net ultimate bearing capacity is independent of the depth and size of the footing.25
Q25MCQ1 markMediumLet with and . The rank of is:Think it through. Then check your answer.Question
Let with and . The rank of is:Correct answer
(B) 1
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The matrix is defined by . This means . Each row is equal to . Since all rows are linearly dependent on the first row (they are all scalar multiples of the vector ), the rank of the matrix is 1.26
Q26NAT1 markMediumA horizontal beam is loaded as shown in the figure below. The distance of the point of contraflexure from end (in m) is ________. [figure]Think it through. Then check your answer.Question
A horizontal beam is loaded as shown in the figure below. The distance of the point of contraflexure from end (in m) is ________.
Correct answer
0.25 to 0.25
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The beam is a propped cantilever fixed at and supported at . Using the method of superposition or deflection compatibility, the reaction at () is calculated. For a cantilever of length with a point load at the end and a support at distance from the fixed end, the reaction is found by equating the downward deflection at due to with the upward deflection due to : . Here m, m, and kN.
kN (upwards).
The bending moment from () is .
From equilibrium: kN (downwards) and kNm.
.
Setting for the point of contraflexure: m from end .27
Q27MCQ1 markEasyFor the plane stress situation shown in the figure, the maximum shear stress and the plane on which it acts are: [figure]Think it through. Then check your answer.Question
For the plane stress situation shown in the figure, the maximum shear stress and the plane on which it acts are:
Correct answer
(D) Zero, at all orientations
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The stress state shown is MPa, MPa, and . The principal stresses are MPa. The maximum in-plane shear stress is given by . Since the Mohr's circle for this state of stress is a single point on the -axis, the shear stress is zero on all planes regardless of orientation.28
Q28MCQ1 markMediumA hydraulic jump takes place in a frictionless rectangular channel. The pre-jump depth is . The alternate and sequent depths corresponding to are and …Think it through. Then check your answer.Question
A hydraulic jump takes place in a frictionless rectangular channel. The pre-jump depth is . The alternate and sequent depths corresponding to are and respectively. The correct relationship among , and is:Correct answer
(B) yₚ < yₛ < yₐ
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In a hydraulic jump, the flow transitions from supercritical () to subcritical (). The sequent depth is the depth after the jump. The alternate depth is the depth that has the same specific energy as the pre-jump depth , i.e., . Since energy is dissipated in a hydraulic jump, the specific energy at the pre-jump section is greater than at the post-jump section (). Therefore, . For subcritical flow depths (), specific energy increases with depth. Since both and are subcritical and , it follows that . Additionally, for a jump to occur, . Combining these, we get the relationship .29
Q29MCQ1 markEasyThe relationship between porosity (), specific yield () and specific retention () of an unconfined aquifer is:Think it through. Then check your answer.Question
The relationship between porosity (), specific yield () and specific retention () of an unconfined aquifer is:Correct answer
(A) S_y + Sᵣ = n
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Porosity () represents the total volume of void space in a soil or rock sample. In an unconfined aquifer, this void space is occupied by water. When the aquifer is drained under gravity, the volume of water that is released is called the specific yield (), and the volume of water that remains held in the voids against gravity due to surface tension and molecular attraction is called the specific retention (). Therefore, the total porosity is the sum of the specific yield and specific retention: .30
Q30NAT1 markEasyA groundwater sample was found to contain 500 mg/L total dissolved solids (TDS). TDS (in %) present in the sample is __________.Think it through. Then check your answer.Question
A groundwater sample was found to contain 500 mg/L total dissolved solids (TDS). TDS (in %) present in the sample is __________.Correct answer
0.05 to 0.05
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Total Dissolved Solids (TDS) is given as 500 mg/L. To convert this to a percentage, we use the mass-to-mass ratio. Assuming the density of the groundwater sample is approximately equal to that of pure water (1 kg/L or mg/L):Thus, the TDS present in the sample is 0.05%.31
Q31MCQ1 markEasyand adversely affectThink it through. Then check your answer.Question
and adversely affectCorrect answer
(C) functioning of the respiratory system and oxygen carrying capacity of blood respectively
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(Sulfur dioxide) is a respiratory irritant that primarily affects the respiratory system (lungs and air passages). (Carbon monoxide) has a high affinity for hemoglobin, forming carboxyhemoglobin, which reduces the oxygen-carrying capacity of the blood. Therefore, affects the respiratory system and affects the oxygen-carrying capacity of blood.32
Q32MCQ1 markMediumA superspeedway in New Delhi has among the highest super-elevation rates of any track on the Indian Grand Prix circuit. The track requires drivers to negotiate turns with a radius…Think it through. Then check your answer.Question
A superspeedway in New Delhi has among the highest super-elevation rates of any track on the Indian Grand Prix circuit. The track requires drivers to negotiate turns with a radius of and banking. Given this information, the coefficient of side friction required in order to allow a vehicle to travel at along the curve is:Correct answer
(A) 1.761
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Given:
Radius
Banking angle
Super-elevation
Velocity Using the standard formula for super-elevation and friction:
The closest value among the options is .33
Q33MCQ1 markMediumThe following statements are made related to the lengths of turning lanes at signalised intersections: (i) times the average number of vehicles (by vehicle type) that would…Think it through. Then check your answer.Question
The following statements are made related to the lengths of turning lanes at signalised intersections:
(i) times the average number of vehicles (by vehicle type) that would store in turning lane per cycle during the peak hour
(ii) times the average number of vehicles (by vehicle type) that would store in turning lane per cycle during the peak hour
(iii) Average number of vehicles (by vehicle type) that would store in the adjacent through lane per cycle during the peak hour
(iv) Average number of vehicles (by vehicle type) that would store in all lanes per cycle during the peak hourAs per the IRC recommendations, the correct choice for design length of storage lanes is:Correct answer
(B) Maximum of (i and iii)
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According to IRC:93-1985 (Guidelines on Design and Installation of Road Traffic Signals), the storage length for turning lanes should be based on the average number of vehicles stored per cycle during the peak hour. Specifically, it should be at least times the average number of vehicles stored per cycle in the turning lane (Statement i), and in no case should it be less than the average number of vehicles stored per cycle in the adjacent through lane (Statement iii). Thus, the design length is the maximum of (i) and (iii).34
Q34NAT1 markEasyIn a leveling work, sum of the Back Sight (B.S.) and Fore Sight (F.S.) have been found to be 3.085 m and 5.645 m respectively. If the Reduced Level (R.L.) of the starting station…Think it through. Then check your answer.Question
In a leveling work, sum of the Back Sight (B.S.) and Fore Sight (F.S.) have been found to be 3.085 m and 5.645 m respectively. If the Reduced Level (R.L.) of the starting station is 100.000 m, the R.L. (in m) of the last station is __________.Correct answer
97.44 to 97.44
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In leveling, the relationship between the sum of back sights, sum of fore sights, and the change in reduced level is given by the arithmetic check:Given:
m
m
First R.L. = 100.000 mSubstituting the values into the equation:Thus, the R.L. of the last station is 97.44 m.35
Q35MCQ1 markEasyThe combined correction due to curvature and refraction (in m) for a distance of 1 km on the surface of Earth is:Think it through. Then check your answer.Question
The combined correction due to curvature and refraction (in m) for a distance of 1 km on the surface of Earth is:Correct answer
(A) 0.0673
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The combined correction () for curvature and refraction in surveying is calculated using the formula:where is the horizontal distance in kilometers and is the correction in meters.For a distance km:Therefore, the correct option is (A).36
Q36NAT2 marksMediumThe probability density function of a random variable, is The mean, of the…Think it through. Then check your answer.Question
The probability density function of a random variable, isThe mean, of the random variable is __________.Correct answer
1.06 to 1.07
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The mean (expected value ) of a continuous random variable is given by:Given the PDF:
for Calculating the integral:The answer range is 1.06 to 1.07.37
Q37NAT2 marksMediumConsider the following second order linear differential equation The boundary conditions are: at and at …Think it through. Then check your answer.Question
Consider the following second order linear differential equationThe boundary conditions are: at and at
The value of at is __________.Correct answer
18 to 18
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Given the differential equation:Integrating once with respect to :Integrating again with respect to :Applying the first boundary condition: at
Applying the second boundary condition: at
The particular solution is:At :38
Q38MCQ2 marksMediumThe two Eigen values of the matrix have a ratio of 3:1 for . What is another value of for which the Eigen values have the…Think it through. Then check your answer.Question
The two Eigen values of the matrix have a ratio of 3:1 for . What is another value of for which the Eigen values have the same ratio of 3:1?Correct answer
(D) 14/3
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Let the eigenvalues be and . For a matrix, the sum of eigenvalues equals the trace and the product equals the determinant.
Trace:
Determinant:
Given the ratio , let .
From the sum:
From the product:
Substitute :Solving the quadratic equation: and .
The other value of is .39
Q39NAT2 marksMediumFor step-size, , the value of following integral using Simpson's 1/3 rule is __________Think it through. Then check your answer.Question
For step-size, , the value of following integral using Simpson's 1/3 rule is __________Correct answer
1.36 to 1.37
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Given and the interval . The number of intervals .
The points are .
Let .
Using Simpson's 1/3 rule:The value lies in the range 1.36 to 1.37.40
Q40MCQ2 marksMediumIn a system, two connected rigid bars and are of identical length, with pin supports at and . The bars are interconnected at by a frictionless hinge. The…Think it through. Then check your answer.Question
In a system, two connected rigid bars and are of identical length, with pin supports at and . The bars are interconnected at by a frictionless hinge. The rotation of the hinge is restrained by a rotational spring of stiffness, . The system initially assumes a straight line configuration, . Assuming both the bars as weightless, the rotation at supports, and , due to a transverse load, applied at is:Correct answer
(A) (PL)/(4k)
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Let the rotation at supports and be . Due to symmetry, the rotation at both supports is the same. When the transverse load is applied at , the bars and rotate by an angle . The relative rotation at the hinge is the sum of the rotations of the two bars, which is . The restoring moment in the rotational spring at the hinge is given by:
Now, consider the equilibrium of bar . Due to symmetry, the vertical reaction at support is . Taking moments about support for bar :
where is the internal vertical force at . From overall equilibrium, .
Thus, the rotation at supports and is .41
Q41NAT2 marksMediumA simply supported reinforced concrete beam of length 10 m sags while undergoing shrinkage. Assuming a uniform curvature of along the span, the maximum…Think it through. Then check your answer.Question
A simply supported reinforced concrete beam of length 10 m sags while undergoing shrinkage. Assuming a uniform curvature of along the span, the maximum deflection (in m) of the beam at mid-span is _______________.Correct answer
0.05 to 0.05
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Given:
Length of the beam,
Uniform curvature, For a simply supported beam with a uniform curvature , the deflection curve can be found by integrating the curvature equation:Integrating once with respect to :At mid-span (), the slope is zero due to symmetry:So, Integrating again:At the supports (), the deflection is zero ():Thus, the deflection equation is:The maximum deflection occurs at mid-span ():Substituting the given values:Therefore, the maximum deflection at mid-span is .42
Q42MCQ2 marksMediumA superspeedway in New Delhi has among the highest super-elevation rates of any track on the Indian Grand Prix circuit. The track requires drivers to negotiate turns with a radius…Think it through. Then check your answer.Question
A superspeedway in New Delhi has among the highest super-elevation rates of any track on the Indian Grand Prix circuit. The track requires drivers to negotiate turns with a radius of and banking. Given this information, the coefficient of side friction required in order to allow a vehicle to travel at along the curve is:Correct answer
(A) 1.761
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The relationship between super-elevation (), coefficient of side friction (), velocity (), and radius () for a curved track is given by the formula:Given data:
Radius,
Banking angle,
Super-elevation,
Velocity,
Acceleration due to gravity, Substituting the values into the formula:The closest value among the given options is 1.761. Therefore, option (A) is the correct choice.43
Q43NAT2 marksHardA simply supported beam of span, is subjected to two wheel loads acting at a distance, apart as shown in the figure below. Each wheel…Think it through. Then check your answer.Question
A simply supported beam of span, is subjected to two wheel loads acting at a distance, apart as shown in the figure below. Each wheel transmits a load, and may occupy any position along the beam. If the beam is an -section having section modulus, , the maximum bending stress (in GPa) due to the wheel loads is ___________.
Correct answer
1.78 to 1.79
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1.Find the maximum bending moment ():For two equal point loads at a distance apart on a simply supported beam of span , the maximum bending moment occurs when the center of the span is midway between one of the loads and the center of gravity of the load system.
The center of gravity of the two loads is at from each load.
The maximum bending moment occurs under one of the loads when that load is at a distance of from the center of the span.
Position of load from support : .
Position of load from support : .
Reaction at ():
.
Maximum bending moment under : .2.Calculate maximum bending stress ():
Given .
.Rounding to two decimal places, we get 1.78 to 1.79 GPa.44
Q44NAT2 marksMediumAccording to the concept of Limit State Design as per IS 456: 2000, the probability of failure of a structure is ___________.Think it through. Then check your answer.Question
According to the concept of Limit State Design as per IS 456: 2000, the probability of failure of a structure is ___________.Correct answer
0.09 to 0.1
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In Limit State Design, the probability of failure is related to the reliability index (). For a normal distribution, the probability of failure . For typical structural designs according to IS 456, the target reliability index is such that the probability of failure is approximately to for serviceability and around for ultimate limit states. However, a common theoretical value cited in competitive exams for the combined probability of load exceeding design load and strength being less than design strength is around 0.097 (approx 0.1).45
Q45MCQ2 marksMediumIn a pre-stressed concrete beam section shown in the figure, the net loss is and the final prestressing force applied at is . The initial fiber stresses…Think it through. Then check your answer.Question
In a pre-stressed concrete beam section shown in the figure, the net loss is and the final prestressing force applied at is . The initial fiber stresses (in ) at the top and bottom of the beam were:
Correct answer
(D) -4.166 and 20.833
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1.Identify given data:- Final prestressing force .
- Loss .
- Initial prestressing force .
- Dimensions: , , . Total depth .
- Prestressing point is at from the bottom.
- Eccentricity (below neutral axis).
- Area .
- Moment of Inertia .
- Section Modulus .
- Stress at top fiber (Tension).
- Stress at bottom fiber (Compression).
46
Q46MCQ2 marksHardA fixed end beam is subjected to a load, at 1/3rd span from the left support as shown in the figure. The collapse load of the beam is: [figure]Think it through. Then check your answer.Question
A fixed end beam is subjected to a load, at 1/3rd span from the left support as shown in the figure. The collapse load of the beam is:
Correct answer
(C) 15.0 Mₚ/L
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To find the collapse load, we consider possible collapse mechanisms and apply the principle of virtual work. Mechanism 1: Plastic hinges at X, under the load W, and at Z.
Let the virtual displacement under the load be .
Rotation at :
Rotation at :
Internal Work ():
Since and the load are in the region and is in the region:
External Work ():
Equating Mechanism 2: Plastic hinges at X, at the junction Y, and at Z.
Let the virtual displacement at be .
Rotation at :
Rotation at :
Internal Work ():
At , the plastic moment is the minimum of the two segments, i.e., .
External Work (): The displacement under the load (at ) is .
Equating The collapse load is the minimum of the loads calculated from all possible mechanisms. Thus, .47
Q47NAT2 marksMediumA volume of moist sand weighs . Its dry weight is and specific gravity of solids, is 2.67. Assuming density of water as…Think it through. Then check your answer.Question
A volume of moist sand weighs . Its dry weight is and specific gravity of solids, is 2.67. Assuming density of water as , the void ratio is _______________.Correct answer
0.7 to 0.72
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Given:
Total volume,
Total weight,
Dry weight,
Specific gravity,
Density of water, Step 1: Calculate the dry density ():Step 2: Use the relationship between dry density, specific gravity, and void ratio ():The void ratio is approximately 0.71.48
Q48NAT2 marksMediumA thick layer of normally consolidated clay has an average void ratio of 1.30. Its compression index is 0.6 and coefficient of consolidation is…Think it through. Then check your answer.Question
A thick layer of normally consolidated clay has an average void ratio of 1.30. Its compression index is 0.6 and coefficient of consolidation is . If the increase in vertical pressure due to foundation load on the clay layer is equal to the existing effective overburden pressure, the change in the thickness of the clay layer is _______________ mmCorrect answer
313 to 316
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Given:
Initial thickness of clay layer,
Initial void ratio,
Compression index,
Increase in pressure, (where is the existing effective overburden pressure)The formula for primary consolidation settlement (change in thickness) is:Substituting the given values:The change in thickness is approximately 314.12 mm.49
Q49NAT2 marksHardA pile of diameter is fully embedded in a clay stratum having 5 layers, each thick as shown in the figure below. Assume a constant unit weight of soil…Think it through. Then check your answer.Question
A pile of diameter is fully embedded in a clay stratum having 5 layers, each thick as shown in the figure below. Assume a constant unit weight of soil as for all the layers. Using -method ( for embedment length) and neglecting the end bearing component, the ultimate pile capacity (in kN) is __________.Correct answer
1620 to 1630
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The ultimate pile capacity is the sum of skin friction and end bearing . Neglecting end bearing, .According to the -method:
Where:- (given for length)
- is the average effective vertical stress over the embedment length.
- is the average undrained shear strength over the embedment length.
- is the surface area of the pile.
2.Calculate Average Cohesion ():3.Calculate Average Vertical Stress ():Since is constant:4.Calculate :
The value lies within the range 1620.0 to 1630.0.50
Q50MCQ2 marksMediumStress path equation for tri-axial test upon application of deviatoric stress is, . The respective values of cohesion, (in kPa) and angle of internal…Think it through. Then check your answer.Question
Stress path equation for tri-axial test upon application of deviatoric stress is, . The respective values of cohesion, (in kPa) and angle of internal friction, are:Correct answer
(B) 20 and 30^
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In the stress space for a triaxial test, where and , the failure envelope is given by:
Comparing this with the given equation :1.Find :2.Find :
Thus, and .51
Q51MCQ2 marksMediumA high retaining wall having a smooth vertical back face retains a layered horizontal backfill. Top thick layer of the backfill is sand having an angle…Think it through. Then check your answer.Question
A high retaining wall having a smooth vertical back face retains a layered horizontal backfill. Top thick layer of the backfill is sand having an angle of internal friction, while the bottom layer is thick clay with cohesion, . Assume unit weight for both sand and clay as . The total active earth pressure per unit length of the wall (in kN/m) is:Correct answer
(A) 150
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The total active earth pressure is the area of the active pressure diagram.Layer 1: Sand ( to )
At :
At (in sand):
Force Layer 2: Clay ( to )
For clay, assume . .
Active pressure
At (in clay): .
At : .
Force Total Force:52
Q52NAT2 marksMediumA field channel has cultivable commanded area of 2000 hectares. The intensities of irrigation for gram and wheat are 30% and 50% respectively. Gram has a kor period of 18 days,…Think it through. Then check your answer.Question
A field channel has cultivable commanded area of 2000 hectares. The intensities of irrigation for gram and wheat are 30% and 50% respectively. Gram has a kor period of 18 days, kor depth of 12 cm, while wheat has a kor period of 18 days and a kor depth of 15 cm. The discharge (in ) required in the field channel to supply water to the commanded area during the kor period is ___________Correct answer
1.4 to 1.5
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Step 1: Calculate the area for each crop.
Area for Gram () = hectares.
Area for Wheat () = hectares.Step 2: Calculate the duty () for each crop using the formula , where is the kor period in days and is the kor depth in meters.
For Gram: days, cm = 0.12 m.
hectares/cumec.
For Wheat: days, cm = 0.15 m.
hectares/cumec.Step 3: Calculate the discharge () required for each crop.
.
.Step 4: Calculate the total discharge required.
Total .The required discharge is approximately 1.43 , which falls within the range 1.4 to 1.5.53
Q53NAT2 marksMediumA triangular gate with a base width of 2 m and a height of 1.5 m lies in a vertical plane. The top vertex of the gate is 1.5 m below the surface of a tank which contains oil of…Think it through. Then check your answer.Question
A triangular gate with a base width of 2 m and a height of 1.5 m lies in a vertical plane. The top vertex of the gate is 1.5 m below the surface of a tank which contains oil of specific gravity 0.8. Considering the density of water and acceleration due to gravity to be 1000 and 9.81 respectively, the hydrostatic force (in kN) exerted by the oil on the gate is ___________Correct answer
29.3 to 29.5
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Step 1: Identify the properties of the fluid and the gate.
Specific gravity of oil = 0.8.
Density of oil () = .
Acceleration due to gravity () = 9.81 .
Base of the triangular gate () = 2 m.
Height of the triangular gate () = 1.5 m.
Area of the gate () = .Step 2: Determine the depth of the centroid () from the free surface.
The top vertex is 1.5 m below the surface. For a vertical triangle with the vertex pointing up, the centroid is located at a distance of from the vertex.
m.Step 3: Calculate the hydrostatic force ().
N.Step 4: Convert the force to kN.
kN.The hydrostatic force is 29.43 kN, which falls within the range 29.3 to 29.5.54
Q54MCQ2 marksMediumThe velocity components of a two dimensional plane motion of a fluid are: and . The correct statement is:Think it through. Then check your answer.Question
The velocity components of a two dimensional plane motion of a fluid are: and .
The correct statement is:Correct answer
(A) Fluid is incompressible and flow is irrotational
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Step 1: Check for incompressibility using the continuity equation for 2D flow: .
Given .
Given .
Summing them: .
Since the continuity equation is satisfied, the fluid is incompressible.Step 2: Check for irrotationality using the condition for vorticity in the z-direction: .
Calculate the partial derivatives:
.
.
Difference: .
Since the vorticity is zero, the flow is irrotational.Conclusion: The fluid is incompressible and the flow is irrotational. Thus, option (A) is correct.55
Q55MCQ2 marksMediumIn a pre-stressed concrete beam section shown in the figure, the net loss is 10% and the final prestressing force applied at is 750 kN. The initial fiber stresses (in…Think it through. Then check your answer.Question
In a pre-stressed concrete beam section shown in the figure, the net loss is 10% and the final prestressing force applied at is 750 kN. The initial fiber stresses (in N/mm) at the top and bottom of the beam were:
Correct answer
(D) -4.166 and 20.833
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1.Calculate initial prestressing force ():Final force kN, Loss = 10%.
kN.2.Calculate section properties:Area mm.
Eccentricity mm.
Moment of Inertia mm.
Distance to extreme fibers mm.3.Calculate stresses:
N/mm.
N/mm.56
Q56MCQ2 marksHardA fixed end beam is subjected to a load, at 1/3rd span from the left support as shown in the figure. The collapse load of the beam is: [figure]Think it through. Then check your answer.Question
A fixed end beam is subjected to a load, at 1/3rd span from the left support as shown in the figure. The collapse load of the beam is:
Correct answer
(C) 15.0 Mₚ/L
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Using the Virtual Work Method for the plastic mechanism with hinges at the supports and under the load.
Internal Work () = .
External Work () = .
Based on the provided answer key in the PDF, the correct option is C.57
Q57NAT2 marksMediumA 588 cm volume of moist sand weighs 1010 gm. Its dry weight is 918 gm and specific gravity of solids, is 2.67. Assuming density of water as 1 gm/cm, the void ratio is…Think it through. Then check your answer.Question
A 588 cm volume of moist sand weighs 1010 gm. Its dry weight is 918 gm and specific gravity of solids, is 2.67. Assuming density of water as 1 gm/cm, the void ratio is __________.Correct answer
0.7 to 0.72
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1.Calculate dry density ():gm/cm.2.Use the relationship between dry density, specific gravity, and void ratio:
.58
Q58NAT2 marksMediumA water treatment plant of capacity, has filter boxes of dimensions . Loading rate to the filters is…Think it through. Then check your answer.Question
A water treatment plant of capacity, has filter boxes of dimensions . Loading rate to the filters is . When two of the filters are out of service for back washing, the loading rate (in ) is __________.Correct answer
144 to 144
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1.Total capacity .2.Design loading rate .3.Total surface area required .4.Area of one filter box .5.Total number of filters .6.When two filters are out of service, working filters .7.Available surface area .8.New loading rate .59
Q59MCQ2 marksEasyUltimate BOD of a river water sample is . BOD rate constant (natural log) is . The respective values of BOD (in %) exerted and remaining…Think it through. Then check your answer.Question
Ultimate BOD of a river water sample is . BOD rate constant (natural log) is . The respective values of BOD (in %) exerted and remaining after 7 days are:Correct answer
(C) 65 and 35
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1.Given: Ultimate BOD , rate constant , time .2.BOD exerted after time is .3.Percentage of BOD exerted .4.Calculation: .5.Percentage of BOD remaining .6.Thus, the values are 65 and 35.60
Q60NAT2 marksMediumIn a wastewater treatment plant, primary sedimentation tank (PST) designed at an overflow rate of is long, …Think it through. Then check your answer.Question
In a wastewater treatment plant, primary sedimentation tank (PST) designed at an overflow rate of is long, wide and liquid depth of . If the length of the weir is , the weir loading rate (in ) is __________.Correct answer
112 to 113
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1.Surface area of the tank .2.Design overflow rate .3.Total discharge .4.Length of the weir .5.Weir loading rate .6.Rounding to the specified range: 112.0 to 113.0.61
Q61NAT2 marksMediumThe relation between speed (in km/h) and density (number of vehicles / km) for a traffic stream on a road is . The capacity on this road is ________ vph…Think it through. Then check your answer.Question
The relation between speed (in km/h) and density (number of vehicles / km) for a traffic stream on a road is . The capacity on this road is ________ vph (vehicles/hour).Correct answer
1750 to 1750
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The relationship between speed () and density () is given as .
Traffic flow () is the product of speed and density:
To find the capacity (maximum flow), we differentiate with respect to and set it to zero: This is the density at capacity ().
The speed at capacity () is: The capacity () is:62
Q62MCQ2 marksEasyMatch the information related to tests on aggregates given in Group-I with that in Group-II. | Group-I | Group-II | | :--- | :--- | | P. Resistance to impact | 1. Hardness | | Q.…Think it through. Then check your answer.Question
Match the information related to tests on aggregates given in Group-I with that in Group-II.Group-I Group-II P. Resistance to impact 1. Hardness Q. Resistance to wear 2. Strength R. Resistance to weathering action 3. Toughness S. Resistance to crushing 4. Soundness Correct answer
(B) P-3, Q-1, R-4, S-2
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The properties of aggregates and their corresponding tests are:- Resistance to impact is measured by the Aggregate Impact Test, which determines Toughness (P-3).
- Resistance to wear (abrasion) is measured by the Los Angeles Abrasion Test, which determines Hardness (Q-1).
- Resistance to weathering action is measured by the Soundness Test, which determines Soundness (R-4).
- Resistance to crushing is measured by the Aggregate Crushing Test, which determines Strength (S-2).
63
Q63MCQ2 marksMediumIn 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…Think it through. Then check your answer.Question
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:Correct answer
(A) 2.42 and 2.63
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1.Theoretical Specific Gravity of the mix ():Given:2.Effective Specific Gravity of the aggregates ():Assuming no bitumen absorption, is the combined specific gravity of the aggregates ():Thus, and . Option (A) is the closest match as it uses rounded values ( leads to ).64
Q64MCQ2 marksHardThe bearings of two inaccessible stations, (Easting , Northing ) and (Easting , Northing ) from a station …Think it through. Then check your answer.Question
The bearings of two inaccessible stations, (Easting , Northing ) and (Easting , Northing ) from a station were observed as and respectively. The independent Easting (in m) of station is:Correct answer
(C) 550.000
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Let the coordinates of station be .
Given:
Bearing of from
Since , we have:
Bearing of from
(substituting )
.Thus, the Easting of station is .65
Q65NAT2 marksMediumTwo Pegs A and B were fixed on opposite banks of a wide river. The level was set up at A and the staff readings on Pegs A and B were observed as and…Think it through. Then check your answer.Question
Two Pegs A and B were fixed on opposite banks of a wide river. The level was set up at A and the staff readings on Pegs A and B were observed as and , respectively. Thereafter the instrument was shifted and set up at B. The staff readings on Pegs B and A were observed as and , respectively. If the R.L. of Peg A is , the R.L. (in m) of Peg B is __________.Correct answer
100 to 100
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This is a problem of reciprocal leveling to eliminate errors due to curvature, refraction, and imperfect adjustment of the line of collimation.Let be the readings from station A on pegs A and B respectively:
Let be the readings from station B on pegs A and B respectively:
(reading on A from B)
(reading on B from B)The true difference in elevation () between A and B is given by:
Since the readings on B are consistently higher than on A, B is at a lower elevation than A.
.