HPCL Civil Engineer Mock Test - 1 - HPSC (Haryana) MCQ

The correct answer is 'Option 1' i.e. 'Esteem'.

Key Points

• The word that is opposite in meaning to the given word "Ignominy" is: 'Esteem'
• "Ignominy" refers to public shame or disgrace(अपमान).
• "Esteem" is the opposite of "Ignominy". It refers to respect and admiration.(सम्मान).

Therefore, the correct answer is: 'Esteem'.

Additional Information

• Here's the explanation for the other options:
  • "Disrepute" - It means the state of being held in low regard by the public.(बदनामी).
  • "Stigma" - It refers to a mark of disgrace associated with a particular circumstance, quality, or person.(कलंक).
  • "Infamy" - It refers to the state of being well known for some bad quality or deed.(कुख्याति).

Given:

The areas of three adjacent faces are 108 cm2, 144 cm2 and 192 cm2

Formula Used:

Volume of cuboid = lbh

l = length, b = breadth and h = height

Calculation:

According to the question

lb = 108 -----(i)

bh = 144 -----(ii)

hl = 192 -----(iii)

On multiplying (i), (ii) and (iii), we get

lb × bh × hl = 108 × 144 × 192

⇒ (lbh)2 = 108 × 144 × 192

⇒ (Volume of cuboid)2 = 36 × 3 × 144 × 3 × 64

⇒ (Volume of cuboid)2 = (6 × 3 × 12 × 8)2

⇒ Volume of cuboid = 1728 cm3

According to the question

Volume of cuboid = 2/3 × 1728 = 1152 cm3

∴ The required answer will be 1152 cm³

Given:

Time taken by pipe P to fill the tank = 10 minutes

Time taken by pipe Q to fill the tank = 15 minutes

Concept used:

If a pipe fills a tank in t minutes, then the quantity of water filled by the pipe in 1 minute = 1 / t

Calculation:

Quantity of tank filled by pipe P alone in 1 minute = (1 / 10)

Quantity of tank filled by pipe Q alone in 1 minute = (1 / 15)

Quantity of tank filled by pipe P and Q together in 1 minute

=

= (3 + 2) / 30

= 5 / 30

= 1 / 6.

Total time taken to fill the tank together by P and Q = 6 minutes.

∴ The total time taken to fill the tank together by P and Q is 6 minutes.

Given word: VAPORIZE

Now, each letter of the word 'VAPORIZE' is arranged in English alphabetical order:

Here, only 'One' letter(O) will remain unchanged after arrange the words in Alphabetical order.

Hence, the correct answer is "Option 4".

The correct answer is 'Option 2' i.e. 'Zeppelin'.

Key Points

• The word in question is 'Zeppelin'. In English, it is a type of rigid airship named after the German inventor Count Ferdinand von Zeppelin. (ज़ेप्पेलिन)
• Example: They watched as the Zeppelin flew over the city.

Therefore, the correct answer is- 'Zeppelin'.

Additional Information

• 'Zeppellin', 'Zeppilin', and 'Zeppeleen' are incorrect spellings and do not hold any meaning in English

The positions of the letters according to the English alphabet series:

LOGIC: Sum of Positional Values of the Letters + (Number of letters in the given word)² = Code.

The pattern for this is as follows;

In RIGHT →

Positional Letters and positional Values are:

R = 18

I = 9

G = 7

H = 8

T = 20

RIGHT = (18 + 9 + 7 + 8 + 20) + 5² = 62 + 25 = 87

And;

In LEFT →

Positional Letters and positional Values are:

L = 12

E = 5

F = 6

T = 20

LEFT = (12 + 5 + 6 + 20) + 4² = 43 + 16 = 59

Similarly;

In CENTRE →

Positional Letters and positional Values are:

C = 3

E = 5

N = 14

T = 20

R = 18

E = 5

CENTRE = (3 + 5 + 14 + 20 + 18 + 5) + 6² = 65 + 36 = 101

Hence, the correct answer is "101".

Dream is represented by a Circle, Reality is represented by Diamond and Nightmare is represented by Oval. The area covered by all there is the middle part which is denoted by 12. Hence, the answer is 12.

Total work = 60 units

Let x be the total no days to complete the work

P×2 + Q(x - 3) + Rx = 60

⇒ 6×2 + 5(x - 3) + 4x = 60

⇒12 + 5x - 15 + 4x = 60

⇒ 9x -3 = 60

⇒ 9x = 60 + 3 = 63

⇒ x = 63/9 = 7 days

∴ The Work was completed in 7 days.

Given:

The distance = 30 km

The time = 36 minutes

Concept:

Speed = Distance/Time

Calculation:

⇒ Let speed of bus be S

⇒ S = 30/(36/60) = (30 × 60)/36 = 50 km/h

⇒ Now, speed reduce by 10 km/h then time taken by bus to cover same distance

⇒ Time = 30/(50 - 10) = 30/40 = 3/4 hours = (3/4) × 60 = 45 minutes

∴ The required result will be 45 minutes.

r₁ = 3 cm and r₂ = 4 cm and the distance between the centres = D = 10 cm.

Length of direct common tangent = √(10² – (4 – 3)²) = √99

Length of transverse common tangent = √(10² – (4 + 3)²) = √51

The correct answer is: Option 4 ("Shyam and his brothers" is a famous sweet shop in our neighbourhood.)

Key Points

The correct answer is "is".

The sentence is talking about Shyam and his brothers as a group. The verb "is" is used for singular subjects, while the verb "are" is used for plural subjects. In this case, the subject is "Shyam and his brothers", which is a plural subject. However, we are talking about them as a group, so we use the singular verb "is".

Given:

CP of 8 tablets = SP of 5 tablets

Formula:

Profit = SP - CP

Profit % = (SP - CP)/CP × 100

Calculation:

Let CP of one tablet be Rs.100.

CP of 8 tablets = Rs.800 = SP of 5 tablets

SP of one tablet = 800/5 = Rs.160

∴ Profit percentage = [(160 - 100)/100]× 100 = 60%

Given:

The total volume of the mixture = 80 liter

Milk : Water = 2 : 3

Calculation:

Let, the amount of milk and water in a can be 2x and 3x respectively.

⇒ 2x + 3x = 80

⇒ 5x = 80

⇒ x = 16 liter

Therefore,

The amount of milk is in a can = 2x

⇒ The amount of milk is in a can = 2 × 16

⇒ The amount of milk is in a can = 32 liter

The amount of water is in a can = 3x

⇒ The amount of water is in a can = 3 × 16

⇒ The amount of water is in a can = 48 liter

Now,

Let, 'y' amount of milk is added into a can to become a ratio of 2 : 1

⇒

⇒ 32 + y = 96

⇒ y = 96 - 32

⇒ y = 64 liter

Therefore, 64-liter milk must be added to a can so that the ratio between milk and water becomes 2 : 1.

Hence, the correct answer is 64 liter.

Given,

Question: Who among P, N, K and J get the highest payment?

Statement:

1. P earns more than K and J.

P > K, P > J.

2. N earns more than K.

N > K.

From both statement, the order of payment from highest to lowest is,

P > N > K > J ; P > N > J > K

or

N > P > K > J ; N > P > J > K

Here we see we don't know who gets the highest payment.

∴ Both statements are not correct.

Hence, the correct answer is Statements 1 and 2 both are not correct.

Concept:

Stopping Sight Distance:

(i) Stopping sight distance is the minimum distance over which the driver traveling at design speed can apply brakes and bring the vehicle to stop position safely without collision with any other obstruction.

(ii) It is also known as minimum sight distance or non-passing sight distance. Stopping sight distance should be provided throughout the length of all roads.

(iii) Stopping sight distance composed of two components:

(a) Lag distance: It is the distance traveled by the vehicle in total reaction time if v is the design speed in m/s and t is the total reaction time in sec, then

lag distance = v × t meters

t = total reaction time = 2.5 second (As per IRC)

(b) Braking distance: It is the distance traveled by vehicle after the application of brakes.

Braking distance,

Where, v = design speed of vehicle in m/s

f = Coefficient of friction

Therefore,

Stopping sight distance (SSD) = Lag distance + Braking distance

SSD =

Note:

(i) Stopping sight distance when there are two lanes = SSD

(ii) Stopping sight distance for two way traffic with single-lane = 2 × SSD

Calculation:

Given, v = 100 kmph = 100/3.6 = 27.77 m/sec , (1 kmph = 1/3.6 m/s)

t = 2.5 sec

Lag distance = d = V × t

= 27.77 × 2.5

d = 69.425 m ≈ 69.5 m

Concept:

Magnetic declination: At any place horizontal angle between true meridian and magnetic meridian is called magnetic declination.

For Eastern Declination:

α = β + θe or T.B = M.B + θe

Here, α = True bearing (T.B)

β = Magnetic bearing (M.B), and

θE = Eastern declination

For Western Declination:

α = β – θw or T.B = M.B – θw

Here, θw = western declination.

Calculation:

Magnetic bearing of a line AB = S28°30’E = 151°30’

Declination = 7°30’ West

Farakka Barrage project

• It is a major irrigation project in India that is constructed across the Ganges River.
• It is located in the state of West Bengal.
• The project aims to divert water from the Ganges River to the Hooghly River for the purpose of maintaining water levels and navigation in the Hooghly River, as well as for irrigation and water supply in the region.

There are many factors that affect the composition of municipal solid waste which are listed below:

• Waste composition
• Depth of waste
• Moisture content
• Availability of oxygen
• Temperature
• Waste processing
• Season of Year i.e., dry season or wet season.
• Age and type of landfill
• Locality
• Toxicity

Traps:

• A trap is a device which is used to prevent sewer gases from entering the buildings.
• The traps are located below or within a plumbing fixture and retains small amount of water.

• The retaining water creates a water seal which stops foul gases going back to the building from drain pipes. Therefore all plumbing fixtures such as sinks, washbasins, bathtubs and toilets etc. are equipped with traps.

In the process of composting, microorganisms break down organic matter and produce carbon dioxide, water, heat, and hummus. Under optimal conditions, composting goes through the following four phases in a sequential manner:

Mesophilic:

• It is a moderate-temperature phase and lasts for a couple of days.
• Microorganisms that grow between 20-40 °C (mesophilic range) begin to reproduce by breaking down carbon and nitrogen.
• It also lowers the pH of the mixture due to the production of organic acids.

Thermophilic:

• It is a high-temperature phase and can last from a few days to several months.
• Microorganisms that grow between 45-70 °C (thermophilic range) begin to break down more complex carbon sources, such as cellulose and lignin.
• In this stage, the nitrogen is transformed into ammonia.
• pH increases and the mixture becomes alkaline.

Cooling Stage:

• It can last for several-months cooling.
• When carbon and nitrogen have been consumed, the temperature is lowered to 40-45 ºC.
• The mesophilic appears again and decomposes the remaining material of cellulose and lignin.
• The pH drops again slightly.

Maturation Stage: The mixture or humus is kept at room temperature and at the end of this period, compost will be ready.

1. Radiometric Resolution:-

• The radiometric resolution of an imaging system describes its ability to discriminate very slight differences in energy.
• The finer the radiometric resolution of a sensor, the more sensitive it is to detect small differences in reflected or emitted energy.
• It refers to the sensitivity of sensor to incoming radiance.

Additional Information

2. Spatial Resolution:-

• Measurement of the minimum distance between two objects that will allow them to differentiate from one another in an image.
• Spatial resolution of the sensor refers to the size of the smallest possible feature that can be detected.
• Spatial resolution is a measure of smallest object that can be resolved by the sensor, or the size of area on the ground represented by each pixel. Pixel size is used to determine resolution quality.

3. Spectral Resolution:-

• Spectral resolution describes the ability of a sensor to define fine wavelength intervals. The finer the spectral resolution, the narrower the wavelength range for a particular channel or band.
• Spectral resolution refers to the specific wavelength intervals in electromagnetic spectrum that a sensor can record.
• Narrow bands have higher spectral resolution.

4. Temporal Resolution:-

• Each satellite has its own unique revisit schedule for obtaining image of a particular area.
• The frequency at which the sensor revisits an area is known as temporal resolution.
• For example, if a satellite images the same area every 10 days, then its temporal resolution would be 10 days.
• Temporal resolution is an important factor to consider in change detection studies i.e. crop growth etc.
• Hence, temporal resolution is the amount of time taken by sensor to return to a previously recorded location.

Examples – Landsat, GOES and NOAA

Lever arm of the beam:

• Lever arm is the perpendicular distance between the line of action of the couple forming compressive and tensile force in a Reinforced concrete section.
• The lever arm plays vital role in the calculation of the moment of resistance, maximum and minimum reinforcement ratios etc. thus influencing the entire design of a RCC section.

Let us take a common example of Reinforced concrete section subjected to bending as shown below:

The section above the neutral axis is in compression whereas the section below the neutral axis is in tension. The stress diagram in working stress method is assumed to be linear, hence the compressive force will act at one third of the neutral axis from top. The entire tensile stress is taken by steel alone (one of the assumption), hence the tensile force will act at the level of steel.

Therefore, liver arm becomes:

Where, d = effective depth of the section and n = neutral axis depth

Euler’s theory:

• This theory is valid only for long columns only.
• This theory is valid only when the slenderness ratio is greater or equal to the critical slenderness ratio.
• For any slenderness ratio above critical slenderness ratio, the column fails by buckling and for any value of slenderness ratio less than this value, the column fails in crushing not in buckling.

Euler’s critical load formula is,

P =

Euler’s formula is applicable when,

Crushing stress ≥ Buckling stress

For mild steel,

E = 2 × 105 N/mm2

σcr = 330 N/mm2

∴ λ ≥ 80 N/mm2

∴ When the slenderness ratio for mild steel columns is less than 80, Euler’s theory is not applicable.

Concept:

We know, Traffic capacity (q) = velocity (v) × density (k) ⇒ q = kV

For Case 1: When speed is constant, and density of the flow keeps on increasing at the same speed.

q = k × constant

∴ Flow becomes the linear function of density and with increase in density, flow increases linearly.

For Case 2: With decrease in the speed further, density increases linearly.

Here flow becomes the function the speed as well as density and hence flow becomes the parabolic function of density as density is varying with speed.

∴ After that V varies linearly with k, so q will vary parabolically with k.

Concept:-

When two bars have different lengths, are kept in a line, and subjected to axial load then the total deformation or elongation is calculated by adding the elongation of individual bars.

ΔL = ΔL₁ + ΔL₂

Where,

P = Axial tensile load, D = Diameter of bars,

E = Young's modulus of Elasticity, L₁ & L₂ = Length of section 1 and 2,

ΔL = ΔL₁ + ΔL₂

Calculation:

Given:

Diameter D is identical for each bar, and L₁ and L₂ are lengths of two bars.

Now, elongation can be calculated as,

ΔL₁ = PL₁ / AE

ΔL₁ = 4PL₁ / πED²

and, ΔL₂ = PL₂ / AE

ΔL₂ = 4PL₂ / πED²

Now, ΔL = ΔL₁ + ΔL₂

ΔL = {4PL₁ / EπD²} + {4PL₂ / EπD²}

ΔL = 4P / EπD² × (L₁ + L₂)

Early Start Time:

• The earliest point in the schedule at which a task can begin.
• EST =

Early Finish Time:

• The earliest point in the schedule at which a task can finish.
• EFT = + tij

NOTE: From the formula and the statement above, we can see that an earlier finish time is the sum of the earlier start time and the event time.
But, statement 4 is saying, we have to minus the duration, which is a false statement.

Latest Start Time:

• The latest point in the schedule at which a task can start without causing a delay.
• LST = - tij

Latest Finish Time:

• The latest point in the schedule at which a task can finish without causing a delay.
• LFT =

Types of float:

1) Total float:

• It is the difference between the maximum time available and the actual time required for the completion of the activity. or It is the difference between the time available for an activity performance and the duration of the activity.
• FT = LST - EST = LFT - EFT =

2) Free float:

• It is the amount of time by which an activity can be delayed without affecting the EST of the succeeding activity.
• FT =

3) Independent float:

• It is the excess of minimum available time over the required activity duration
• FI =

4) Interfering float

• It is the difference between the total float and the free float of an activity.
• FIN = FT - FF

Flow net:

(i) The entire pattern of flow lines and equipotential lines is referred as a flow net. It is the solution of Laplace's equation for relevant boundary conditions. Thus, a flow net is a graphical representation of the head and direction of seepage at every point.

(ii) Flow net is a function of only boundary condition and is independent of permeability, head loss, or number of flow lines and equipotential lines. Hence the flow net will alter if permeameter length is changed.

Concept:

As per Watson and Chicks law

Where,

K = Rate constant which depends on the type of microorganism, the type of disinfectant and temperature

n = Constant for a particular microorganism and type of disinfectant

C = Disinfectant concentration (weight/volume)

Nt = Number of microorganism present at any time ‘t’

Calculation:

Given, t = 5 min, n = 1 and C = 0.1 mg/L

∵ We know that,

Let 'x' be the number of micro organism present initially

89% kill of micro organism implies that at time 't' 11% of micro organism are still surviving

∴ Micro organism surviving at time 't' = (11/100) x

∴

⇒

K = 4.4155 /min

Different corrections applied for design of canal by Khosla’s theory are as follows:

(a) Correction for the Mutual Interference of Piles: This correction is applied as percentage of head due to this effect of the mutual interference of piles given by:

Where

b’ = distance between two pile lines

D = The depth of the pile line, the influence of which has to be determined on the neighboring pile of depth d.

d = The depth of the pile on which the effect is considered.

b = Total floor length.

This correction is positive for the points in the rear or back water and subtractive for the points forward in the direction of flow. This equation does not apply to the effect of an outer pile on an intermediate pile, if the intermediate pile is equal to or smaller than the outer pile and is at a distance less than twice the length of the outer pile.

(b) Correction for the Thickness of Floor: In the standard form profiles, the floor is assumed to have negligible thickness. Hence, the percentage pressures calculated by Khosla’s equations or graphs shall pertain to the top levels of the floor. Khosla’s theory is correctly applied in alluvium of finite depth

(c) Correction for the Slope for the Floor: A correction is applied for a sloping floor, and is taken as + ve for the down, and – ve for the up slopes following the direction of flow.

The correction factor given above is to be multiplied by the horizontal length of the slope and divided by the distance between the two pile lines between which the sloping floor is located. This correction is applicable only to the key points of the pile line fixed at the start or the end of the slope.

Exit Gradient (G_E): It has been determined that for a standard form consisting of a floor length b with a vertical cutoff of depth d, the exit gradient at its downstream end is given by

Concept:

Flow-through Pipes in Series and Parallel:

(A) Pipe in Series:

• Pipes connected in continuation as in this case are said to be connected in series. In this arrangement, the rate of discharge is the same in all the pipes.
• Ignoring secondary losses the total loss of head is equal to the sum of the friction losses in the individual pipes.

Let d₁, d₂, d₃ be the diameters, and l₁, l₂, l₃ be the lengths of the various pipes in a series connection. Let Q be the discharge. Let h_f1, h_f2, h_f3 be the loss of head of the various pipes in a series connection.

So, Discharge is the same in all pipes

Q = Q₁ = Q₂ = Q₃

And the total loss of head(h_f) is equal to the sum of the friction losses in the individual pipes.

(B) Pipe in parallel:

• Pipes are said to be in parallel when they are so connected that the flow from a pipe branch or divides into two or more separate pipes and then reunites into a single pipe.
• In this arrangement, the loss of head is the same in all the pipes and the total rate of discharge is equal to the sum of the individual rate of discharge of all the pipes.​

​Let d₁ & d₂ be the diameters, and l₁ & l₂ be the lengths of the various pipes in a parallel connection. Let Q₁ & Q₂ be the discharge of the various pipes in a parallel connection. Let h_f be the loss of head.

So, the loss of head is the same in all the pipes

h_f = hf₁ = hf₂

And the total rate of discharge(Q) is equal to the sum of the individual rate of discharge of all the pipes.

Q = Q₁ + Q₂