Held in 2019
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Daytime temperatures in Delhi can ______ 40°C.
get
stand
reach
peak
Correct answer
reach
Why this answer is correct
Correct answer: 3.
The correct option to be used here is 'reach'. The sentence is talking about a kind of progression and hence the temperature reaches a certain degree.
The growth rate of ABC Motors in 2017 was the same _______ XYZ Motors in 2016.
as off
as those of
as that off
as that of
Correct answer
as that of
Why this answer is correct
Correct answer: Option 4.
The correct phrase to be used in this sentence is 'as that of'.
The sentence compares two values and hence as that of is required to make that happen.
Suresh wanted to lay a new carpet in his new mansion with an area of 70 × 55 sq. mts. However, an area of 550 sq. mts. had to be left out for flowerpots. If the cost of carpet is Rs. 50 per sq. mts., how much money (in Rs.) will be spent by Suresh for the carpet now?
Rs. 1,65,000
Rs. 1,92,500
Rs. 2,75,000
Rs. 1,27,500
Correct answer
Rs. 1,65,000
Why this answer is correct
Carpet area = Total area – Area left out for flowerpots
Carpet area required = [(70 × 55) - 550] = 3300 sq. m
Total cost of carpeting = Carpet area × Cost per square meters
Total cost of carpeting = 3300 × ₹ 50 = ₹ 1, 65,00
A retaining wall with measurements 30 m × 12 m × 6 m was constructed with bricks of dimensions 8 cm × 6 cm × 6 cm. If 60% of the wall consists of bricks, the number of bricks used for the construction is _______ lakhs.
30
40
45
75
Correct answer
45
Why this answer is correct
Volume of the wall = (3000 × 1200 × 600) cu. cm
Volume of all bricks required = 0.6 × (3000 × 1200 × 600) cu.cm
Volume of each brick = (8 × 6 × 6) cu.cm
∴ No of bricks used = \(\frac{{0.6 \times 3000 \times 1200 \times 600}}{{8 \times 6 \times 6}} = 4500000 = 45;lakhs\)
Hima Das was _______ only Indian athlete to win ________ gold for India.
the, many
the, a
an, a
an, the
Correct answer
the, a
Why this answer is correct
Correct option 2.
Hima Das is unique in her achievements therefore the article 'the' is used in the first blank.
Article 'a' is used when we are referring to any member of a group. It is not important which one we are talking about.
Mohan, the manager, wants his four workers to work in pairs. No pair should work for more than 5 hours. Ram and John have worked together for 5 hours. Krishna and Amir have worked as a team for 2 hours. Krishna does not want to work with Ram. Whom should Mohan allot to work with John, if he wants all the workers to continue working?
Amir
Krishna
Ram
None of the three
Correct answer
Krishna
Why this answer is correct
As Krishna does not want to work with Ram, he has to work only with John.
∴ Mohan should allot Krishna to John to work as a pair.
Note: Krishna already worked with Amir as a pair therefore Krishna should work along with John.
Population of state X increased by x% and the population of state Y increased by y% from 2001 to 2011. Assume that x is greater than y. Let P be the ratio of the population of state X to state Y in a given year. The percentage increase in P from 2001 to 2011 is _______.
\(\frac{x}{y}\)
x – y
\(\frac{{100\left( {x - y} \right)}}{{100 + x}}\)
\(\frac{{100\left( {x - y} \right)}}{{100 + y}}\)
Correct answer
\(\frac{{100\left( {x - y} \right)}}{{100 + y}}\)
Why this answer is correct
Let,
Population of state X in 2001 = a
∴ Population of state X in 2011 = a + 0.01 × x × a
Population of state Y in 2001 = b
∴ Population of state Y in 2011 = b + 0.01 × y × b
∵ P is the ratio of the population of state X to state Y in a given year
\({P_{2001}} = \frac{a}{b}\)
\({P_{2011}} = \frac{{\left( {1{\rm{;}} + {\rm{;}}0.01{\rm{;}} \times {\rm{;x;}}} \right) \times {\rm{;a}}}}{{\left( {1 + {\rm{;}}0.01{\rm{;}} \times {\rm{;y}}} \right) \times {\rm{;b}}}}\)
\({\rm{\Delta }}P\left( % \right) = \frac{{{P_2} - {P_1}}}{{{P_1}}} \times 100 = \frac{{\frac{{\left( {1{\rm{;}} + {\rm{;}}0.01{\rm{;}} \times {\rm{;x;}}} \right) \times {\rm{;a}}}}{{\left( {1 + {\rm{;}}0.01{\rm{;}} \times {\rm{;y}}} \right) \times {\rm{;b}}}} - \frac{a}{b}}}{{\frac{a}{b}}} \times 100\)
\({\rm{\Delta }}P\left( % \right) = \left( {\frac{{\left( {1{\rm{;}} + {\rm{;}}0.01{\rm{;}} \times {\rm{;x;}}} \right)}}{{\left( {1 + {\rm{;}}0.01{\rm{;}} \times {\rm{;y}}} \right)}} - 1} \right) \times 100 = \frac{{100\left( {{\rm{x}} - {\rm{y}}} \right)}}{{100 + {\rm{y}}}}\)
The newspaper reports that over 500 hectares of tribal land spread across 28 tribal settlements in Mohinitampuram forest division have already been “alienated”. A top forest official said, “First the tribals are duped out of their land holdings. Second, the families thus rendered landless are often forced to encroach further into the forests”.
On the basis of the information available in the paragraph, ________ is/are responsible for duping the tribals.
forest officials
landless families
The newspaper
it cannot be inferred who
Correct answer
it cannot be inferred who
Why this answer is correct
Correct answer: 4.
The statement talks about the situation of the tribals being forced/duped out of their lands through some nefarious means.
However, the culprit cannot be determined from the sentence.
An oil tank can be filled by pipe X in 5 hours and pipe Y in 4 hours, each pump working on its own. When the oil tank is full and the drainage hole is open, the oil is drained in 20 hours. If initially the tank was empty and someone started the two pumps together but left the drainage hole open, how many hours will it take for the tank to be filled? (Assume that the rate of drainage is independent of the Head)
1.50
2.00
2.50
4.00
Correct answer
2.50
Why this answer is correct
Tank filled by pipe ‘X’ in an hour = 1/5
Tank filled by pipe ‘Y’ in an hour = 1/4
Volume discharged through opening of the tank in an hour = 1/20
If both pipes are open, actual volume filled in an hour = \(\frac{1}{5} + \frac{1}{4} - \frac{1}{{20}} = \frac{2}{5}\)
Total time required to fill tank, \(t = \frac{{{\rm{Volume;of;tank}}}}{{{\rm{Volume;filled;in;an;hour}}}} = \frac{1}{{\frac{2}{5}}} = 2.5;{\rm{hours}}\)
“Popular Hindi fiction, despite – or perhaps because of – its wide reach, often does not appear in our cinema. As ideals that viewers are meant to look up to rather than identify with, Hindi film protagonists usually read books of aspirational value: textbooks, English books, or high value literature.”
Which one of the following CANNOT be inferred from the paragraph above?
Though popular Hindi fiction has wide reach, it often does not appear in the movies
Protagonists in Hindi movies, being ideals for viewers, read only books of aspirational value
Textbooks, English books or high literature have aspirational value, but not popular Hindi fiction
People do not look up to writers of textbooks, English books or high value literature
Correct answer
People do not look up to writers of textbooks, English books or high value literature
Why this answer is correct
The correct answer is option 4.
The paragraph implies that people look up to Hindi film protagonists who usually read books of aspirational value such as textbooks, English books, or high value literature and option 4 clearly contradicts this.
Euclidean norm (length) of the vector [4 – 2 - 6]T is
√48
√24
√12
√56
Correct answer
√56
Why this answer is correct
Let X = \({\left[ {4 - 2 - 6} \right]^T} = \left[ {\begin{array}{*{20}{c}} 4\ { - 2}\ { - 6} \end{array}} \right]\)
∴ Norm X = \(\left| X \right|\) \(= \sqrt {{4^2} + {{\left( { - 2} \right)}^2} + {{\left( { - 6} \right)}^2}} = \sqrt {16 + 4 + 36} = \sqrt {56} \)
The Laplace transform of sin h (at) is
\(\frac{s}{{{s^2} - {a^2}}}\)
\(\frac{s}{{{s^2} + {a^2}}}\)
\(\frac{a}{{{s^2} - {a^2}}}\)
\(\frac{a}{{{s^2} + {a^2}}}\)
Correct answer
\(\frac{a}{{{s^2} - {a^2}}}\)
Why this answer is correct
Concept:
Laplace transform is defined by:
\(F\left( s \right) = ;\mathop \smallint \limits_0^\infty {e^{ - st}}f\left( t \right)dt\)
Where,
f(t) is the function varying with time t
F(s) is the Laplace transform of f(t)
Analysis:
\(sin h(at) = \frac{ e^{at} - e^{-at}}{2}\)
\(F\left( s \right) = ;\mathop \smallint \limits_0^\infty {e^{ - st}} [\frac{e^{at}-e^{-at}}{2}]dt\)
Solving the above, we get:
\(\frac{a}{s^2-a^2}\)
The following inequality is true for all x close to 0.
\(2 - \frac{{{x^2}}}{3} < \frac{{x\sin x}}{{1 - \cos x}} < 2\)
What is the value of \(\mathop {\lim }\limits_{x \to 0} \frac{{x\sin x}}{{1 - \cos x}}?\)
2
1/2
0
1
Correct answer
2
Why this answer is correct
\(\mathop {\lim }\limits_{x \to 0} \frac{{x\sin x}}{{1 - \cos x}}\left( {\frac{0}{0}} \right) = \mathop {\lim }\limits_{x \to 0} \frac{{x\cos x + \sin x}}{{ + \sin x}}\left( {\frac{0}{0}} \right);\)
[Using L’ Hospital rule]
\(\Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{x\left( { - \sin n} \right) + \cos x + \cos x}}{{ + \cos x}};\)
[Using L’ Hospital rule]
\(\Rightarrow \frac{{0 + 1 + 1}}{1} = \frac{2}{1} = 2\)
What is the curl of the vector field 2x2yi + 5z2j - 4yzk?
\(- 14zi - 2{x^2}k\)
\(6zi; + ;4xj;-;2{x^2}k\)
\(6zi; + ;8xy; + ;2{x^2}yk\)
\(- 14zi; + ;6yj; + ;2{x^2}k\)
Correct answer
\(- 14zi - 2{x^2}k\)
Why this answer is correct
Concept:
Consider the vector field
\(\vec V = 2{x^2}yi + 5{z^2}j - 4yzk\)
\(curl;\vec V = \nabla \times \vec V = \left| {\begin{array}{*{20}{c}} i&j&k\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\ {2{x^2}y}&{5{z^2}}&{ - 4yz} \end{array}} \right|\)
\(\Rightarrow Curl;\vec V = i\left[ {\frac{\partial }{{\partial y}}\left( { - 4yz} \right) - \frac{\partial }{{\partial z}}\left( {5{z^2}} \right)} \right] - j\left[ {\frac{\partial }{{\partial x}}\left( { - 4yz} \right) - \frac{\partial }{{\partial z}}\left( {2{x^2}y} \right)} \right] + k\left[ {\frac{\partial }{{\partial x}}\left( {5{z^2}} \right) - \frac{\partial }{{\partial y}}\left( {2{x^2}y} \right)} \right]\)
\(\Rightarrow Curl;\vec V = i\left[ { - 4z - 10z} \right] - j\left[ 0 \right] + k\left[ { - 2{x^2}} \right]\)
\(\Rightarrow Curl;\vec V = - 14zi - 2{x^2}k\)
A closed thin walled tube has thickness, t, mean enclosed area within the boundary of the centerline of tube’s thickness, Am, and shear stress t. Torsional moment of resistance, T of the section would be
2τAmt
τAmtok
0.5τAmt
4τAmt
Correct answer
2τAmt
Why this answer is correct
Polar moment of Inertia of Circular tube of diameter ‘D’ and thickness ‘t’ is given by
\({I_p};\left( J \right) = \frac{\pi }{4} \times {D^3} \times {t_t}\)
\(\therefore J = ;\frac{\pi }{4} \times {D^3} \times {t_t} = ;\frac{\pi }{4} \times {D^2} \times D \times {t_t} = {A_m} \times D \times {t_t}\)
Now, we know that
\(\frac{\tau }{r} = \frac{T}{J} = \frac{{G \times \theta }}{L}\)
\(\Rightarrow T = \frac{{\tau \times J}}{r} = \frac{{T \times {A_m} \times D \times {t_t}}}{{\frac{D}{2};}} = 2;{A_m}\tau \times {t_t};\)
A steel column is restrained against both translation and rotation at one end and is restrained only against rotation but free to translate at the other end. Theoretical and design (IS: 800 – 2007) values, respectively, of effective length factor of the column are
1.0 and 1.0
1.2 and 1.0
1.0 and 1.2
1.2 and 1.2
Correct answer
1.0 and 1.2
Why this answer is correct
Explanation:
As per IS: 800 – 2007
For column, restrained against both rotation and translation at one end and restrained only against rotation but free to translate at other end
Theoretical value of effective length factor = 1.0
Design value of effective length factor = 1.2
Design value of effective length factor for various combination are given below:
If the fineness modulus of a sample of the fine aggregates is 4.3, the mean size of the particles in the sample is between
150 μm and 300 μm
2.36 mm and 4.75 mm
300 μm and 600 μm
1.18 mm and 2.36 mm
Correct answer
1.18 mm and 2.36 mm
Why this answer is correct
Fineness Modulus of an aggregate sample represents the average size of the aggregates of the sample. Higher the size of aggregates in the sample, higher is the value of fineness modulus of sample.
| Size of Aggregates | Fineness Modulus |
|---|---|
| 150 μm | 1 |
| 300 μm | 2 |
| 600 μm | 3 |
| 1.18 mm | 4 |
| 2.36 mm | 5 |
| 4.75 mm | 6 |
Since, Fineness modulus of aggregates sample is 4.3, size of the aggregates varies from 1.18 mm to 2.36 mm.
For a channel section subjected to a downward vertical shear force at its centroid, which one of the following represents the correct distribution of shear stress in flange and web?
Correct answer
Why this answer is correct
Explanation:
Shear stress varies linearly along the flange and varies parabolically along the web.
Shear stress for flange is maximum at the intersection of the web and flange.
∴ Correct shear stress distribution for the channel section is given by,
Which one of the options contains ONLY primary air pollutants?
Hydrocarbons and nitrogen oxides
Nitrogen oxides and peroxyacetyl nitrate
Ozone and peroxyacetyl nitrate
Hydrocarbons and ozone
Correct answer
Hydrocarbons and nitrogen oxides
Why this answer is correct
Concept:
Air pollutants:
Pollutants are classified in different ways. According to origin pollutants are two types.
1). Primary pollutants:
2). Secondary Pollutants:
Examples of primary and secondary air pollutants:
| Primary Pollutants | Secondary Pollutants |
|---|---|
| Carbon monoxide (CO) Oxides of nitrogen (NOx, NO) Sulfur oxides (SOx) Volatile organic compounds (VOCs) highly Particulate matter (dust, ash, salt particles) | Ozone PAN Smog Formaldehyde Sulphuric Acid |
Analysis of a water sample revealed that the sample contains the following species.
\(CO_3^{2 - },;N{a^ + },;{H^ + },;PO_4^3,;A{l^{3 + }},;{H_2}C{O_3},;C{l^2},;C{a^{2 + }},;M{g^{2 + }},;HCO_3^ - ,;F{e^{2 + }},;O{H^ - }\)
Concentrations of which of the species will be required to compute alkalinity?
\(CO_3^{2 - },;{H^ + },;HCO_3^ - ,;O{H^ - }\)
\({H^ + },;{H_2}C{O_3},;HCO_3^ - ,;O{H^ - }\)
\(CO_3^{2 - },;{H_2}C{O_3},;HCO_3^ - ,;O{H^ - }\)
\(CO_3^{2 - },;{H^ + },;{H_2}C{O_3},;HCO_3^ -\)
Correct answer
\(CO_3^{2 - },;{H^ + },;HCO_3^ - ,;O{H^ - }\)
Why this answer is correct
Explanation:
Alkalinity is defined as the ability of water to neutralize the acid of a hydronium ion.
Alkalinity (AT) of water = [HCO3-] + [CO32-] + [B (OH)4-] + [H3(SiO4)-] + [HS-] + [organic anions] + [OH-] – [H+]
From the given options of ions in the questions, the Correct answer is \(CO_3^{2 - },;{H^ + },;HCO_3^ - ,;O{H^ - }\)
Structural failures considered in the mechanistic method of bituminous pavement design are
Fatigue and Rutting
Rutting and shear
Shear and slippage
Fatigue and Shear
Correct answer
Fatigue and Rutting
Why this answer is correct
Concept:
As per IRC 37:2012,
Bitumen pavement i.e. flexible pavement is susceptible to failure due to fatigue and rutting.
Fatigue ⇒ Due to the repetition of loads many times, the surface is subjected to cracking which is known as fatigue failure.
Rutting ⇒ It is the longitudinal depression caused due to the movement of most of the vehicles on the same path and also due to a higher percentage of heavy loads. This depression is formed due to excessive shear force at a section.
Note:
Cracking of 20% surface area in flexible pavement and 10% surface area in the rigid pavement will be considered as fatigue failure.
Formation of 20 mm depression will be considered as rutting failure.
A solid sphere of radius, r, and made of material with density ρs, is moving through the atmosphere (constant pressure, p) with a velocity, v. The net force ONLY due to atmospheric pressure (Fp) acting on the sphere at any time, t, is
\(\frac{4}{3}\pi {r^3}{\rho _s}\frac{{dv}}{{dt}}\)
zero
πr2p
4π2p
Correct answer
zero
Why this answer is correct
As body is moving at constant pressure i.e. no additional/disbalancing force is acting upon as body and it is in balanced state.
Note: When object of any shape (not only sphere) is subjected to uniform pressure over entire surface, the net force is zero because pressure force at any point is balanced by an equal and opposite force on opposite end.
The velocity field in a flow system is given by \(v = 2i; + ;\left( {x; + ;y} \right)j; + ;\left( {xyz} \right)k\). The acceleration of the fluid at (1, 1, 2) is
j + k
2i + 10 k
4j + 10 k
4i + 12 k
Correct answer
4j + 10 k
Why this answer is correct
Velocity field \(\left( v \right); = ;2i; + ;\left( {x; + ;y} \right);j; + ;\left( {xyz} \right);k\)
\(Acceleration;\left( {{a_n}} \right) = \left( {\frac{{\delta v}}{{\delta t}}} \right) + \left( {u\frac{{\delta v}}{{\delta x}} + v\frac{{\delta v}}{{\delta y}} + w\frac{{\left( {\delta v} \right)}}{{\delta z}}} \right)\)
Where \(\left( {\frac{{\delta v}}{{\delta t}}} \right) \Rightarrow Local;accelaration\)
\(\left( {u\frac{{\delta v}}{{\delta x}} + v\frac{{\delta v}}{{\delta y}} + w\frac{{\left( {\delta v} \right)}}{{\delta z}}} \right) \Rightarrow Convective;accelaration;\)
Here, \(u; = ;2,;v; = ;x; + ;y,;w; = ;xyz,;t; = ;0,\) n is direction
\({a_x} = 0 + 2 \times \frac{\partial }{{\partial x}}\left( 2 \right) + \left( {x + y} \right) \times \frac{{\partial \left( 2 \right)}}{{\partial y}} + \left( {xyz} \right) \times \frac{{\partial \left( 2 \right)}}{{\partial z}}\)
ax = 0
\({a_y} = 0 + 2 \times \frac{{\partial \left( {x + y} \right)}}{{\partial x}} + \left( {x + y} \right) \times \frac{{\partial \left( {x + y} \right)}}{{\partial y}} + \left( {xyz} \right) \times \frac{{\partial \left( {x + y} \right)}}{{\partial z}}\)
ay = 2 + x + y
\({a_z} = 0 + 2 \times \frac{{\partial \left( {xyz} \right)}}{{\partial x}} + \left( {x + y} \right) \times \frac{{\partial \left( {xyz} \right)}}{{\partial y}} + xyz \times \frac{{\partial \left( {xyz} \right)}}{{\partial z}}\)
= 0 + 2yz + (x + y) × (xz) + xyz × xy
az = 2yz + (x + y) × (xz) + x2y2z
Acceleration = axi + ayj + azk
\(a; = ;0; + ;\left( {2; + ;x; + ;y} \right);j + ;\left( {2yz; + ;\left( {x + y} \right)xz; + ;{x^2}{y^2}z} \right)k\)
\({a_{\left( {1,1,2} \right)}}; = ;0; + ;\left( {2; + ;1; + ;1} \right)j + ;\left( {2; \times ;1; \times ;2; + \left( {1; + ;1} \right) \times 1; \times ;2; + ;{1^2}; \times ;{1^2} \times ;2} \right)k\)
\(\therefore a; = ;4j; + ;\left( {4; + ;4; + ;2} \right)k; = ;4j; + ;10k\)
An inflow hydrograph is routed through a reservoir to produce an outflow hydrograph. The peak flow of the inflow hydrograph is PI and the time of occurrence of the peak is tI. The peak flow of the outflow hydrograph is P0 and the time of occurrence of the peak is t0.
Which one of the following statements is correct?
PI < P0 and tI > t0
PI > P0 and tI > t0
PI < P0 and tI < t0
PI > P0 and tI < t0
Correct answer
PI > P0 and tI < t0
Why this answer is correct
Explanation:
The outflow from the reservoir if uncontrolled will result in the peak of outflow hydrograph at the junction of inflow and outflow hydrograph.
From the given graph, it can be clearly observed that PI > Po and tI < to.
An earthen dam of height H is made of cohesive soil whose cohesion and unit weight are c and γ, respectively. If the factor of safety against cohesion is Fc, the Taylor’s stability number (Sn)
\(\frac{{\gamma H}}{{c{F_c}}}\)
\(\frac{{{F_c}\gamma H}}{c}\)
\(\frac{c}{{{F_c}\gamma H}}\)
\(\frac{{c{F_c}}}{{\gamma H}}\)
Correct answer
\(\frac{c}{{{F_c}\gamma H}}\)
Why this answer is correct
Explanation:
As per Taylors stability number method
The factor of safety with respect to cohesion ⇒ \({F_c} = \frac{C}{{{S_n}\gamma H}}\)
\(\therefore {S_n} = \frac{C}{{\gamma {F_c}H}}\)
Where, C = Cohesion,
γ = Unit weight of soil, and H = safe height
Additional Information
Its maximum value is 0.261 and it occurs for soft clays having an angle of internal friction is zero i.e. for φ = 0 soils.
Important points about Taylors Stability Number:
The notation “SC” as per Indian standard Soil Classification System refers to
Clayey silt
Sandy clay
Clayey sand
Silty clay
Correct answer
Clayey sand
Why this answer is correct
Explanation:
If more than 50% of soil is retained on 4.75 mm sieve ⇒ Gravel Soil
If more than 50% of soil passes through 4.75 mm and more than 50% is retained on 75μ sieve ⇒ Sandy Soil
If sandy soil contains more than 12% of fines with Atterberg Limits above A-Lines & Plasticity Index is greater than 7 ⇒ Clayey Sand (SC)
Additional Information
Well-graded gravel sand mixture with little or no fines = GW (G for gravel and W for well-graded).
Poorly graded sands with little or no fines = SP (S for sand and P for poorly graded).
Inorganic silts and very fine sands or clayey silts with low plasticity = ML (M for silt and L for low plasticity).
Inorganic clays of liquid limit over 50 % = CH (C for clay and H for high plasticity).
Soil classification as per Indian standards:
| Coarse-grained soils (More than half of the material is larger than 75 μ sieve) | ||||||||
|---|---|---|---|---|---|---|---|---|
| Gravels | Sands | |||||||
| More than half of the coarse fraction is larger than the 4.75 mm sieve | More than half of the coarse fraction is smaller than the 4.75 mm sieve | |||||||
| Clean gravel | Gravel with appreciable fines | Clean sands | Sand with appreciable fines | |||||
| GW | GP | GM | GC | SW | SP | SM | SE | |
| Fine-grained soil (More than half of the material is smaller than 75 μ sieve) | ||||||||
| Silts and clay | ||||||||
| Low compressibility (LL < 35) | Medium compressibility (50 < LL > 35) | Higher compressibility (LL > 50) | ||||||
| ML | CL | OL | MI | CI | OI | MH | CH | OH |
| High organic peat soil → (Pt) |
An anisotropic soil deposit has coefficient of permeability in vertical and horizontal directions as kz and kx, respectively. For constructing a flow net, the horizontal dimension of the problem’s geometry is transformed by a multiplying factor of
\(\frac{{{k_z}}}{{{k_x}}}\)
\(\frac{{{k_x}}}{{{k_z}}}\)
\(\sqrt {\frac{{{k_z}}}{{{k_x}}}} \)
\(\sqrt {\frac{{{k_x}}}{{{k_z}}}}\)
Correct answer
\(\sqrt {\frac{{{k_z}}}{{{k_x}}}} \)
Why this answer is correct
Explanation:
The flow net is drawn for a transformed section by shortening the horizontal dimension by \(\sqrt {{{{k_z}}}\over{{{k_x}}}} \)
Where Kz = Co-efficient of permeability in the Z direction
Kx = Co-efficient of permeability in x-direction
.
The value of the function f(x) is given at n distinct values of x and its value is to be interpolated at the point x* using all the n points.
The estimate is obtained first by the Lagrange polynomial, denoted by IL, and then by the
Newton polynomial, denoted by IN.
Which one of the following statements is correct?
IL is always greater than IN
No definite relation exists between IL and IN.
IL is always less than IN
IL and IN are always equal
Correct answer
No definite relation exists between IL and IN.
Why this answer is correct
Explanation:
The speed-density relationship in a mid-block section of a highway follows the Greenshields’s model. If the free flow speed is vf and the jam density is kj , the maximum flow observed on this section is
\(\frac{{{v_f}{k_j}}}{2}\)
\(\frac{{{v_f}{k_j}}}{8}\)
vfkj
\(\frac{{{v_f}{k_j}}}{4}\)
Correct answer
\(\frac{{{v_f}{k_j}}}{4}\)
Why this answer is correct
Explanation:
Flow (q) = Density (k) × Velocity (V)
For Greenshields’s Model
\(V = {V_f} \times \left( {1 - \frac{k}{{{k_j}}}} \right),\)
Where,
V = Velocity at any instant
Vf = Free mean velocity
k = Density of the flow
kj = Jam density of the flow
Maximum capacity (qmax) = Vmax × Kmax
As per Greenshields’s model
\(Vmax; = ;\frac{{{V_f}}}{2};& ;{k_{max}} = \frac{{{k_j}}}{2};\)
Maximum capacity ⇒ \(\left( {{q_{max}}} \right) = \frac{{{V_f}}}{2} \times \frac{{{k_j}}}{2} = \frac{{{V_f}{k_j}}}{4}\)
Additional Information
Free Mean Speed:
From the speed-density diagram, it can be defined as the maximum speed at which the number of vehicles in a unit length is zero i.e. density is zero.
Density:: It can be defined as the number of vehicles per unit length. The unit of measurement is vehicles/km.
Jam Density: From speed-density relation and flow-density curve, it can also be seen that it is the maximum density at which there is no flow on the road.
Traffic Flow: The number of vehicles passing through a particular point in certain time interval is defined as traffic flow. Also, number of vehicles counted in one hour is called traffic flow (q).
The degree of static indeterminacy of the plane frame is shown in the figure is _______
Correct answer
Why this answer is correct
Explanation:
Number of support reactions, (r) = 3 + 2 + 2 = 7
Number of members (m) = 21
Number of closed loops (C) = 4
Degree of static indeterminacy (Ds) = Dse + Dsi = 3C + (r - 3)
Dse = 7 – 3 = 4
Dsi = 3 × 4 = 12
Number of Force Release = (m’ – 1) = 2 – 1 = 1
(where m’ = number of members connected to the internal hinge)
Ds = 4 + 12 – 1 = 15
The characteristic compressive strength of concrete required is a project is 25 MPa and standard deviation in the observed compressive strength expected at site is 4 MPa. The average compressive strength of cubes tested at different water-cement (w/c) ratios using the same material as is used for the project is given in the table.
| Water content (%) | 45 | 50 | 55 | 60 |
|---|---|---|---|---|
| Average compressive strength of cubes (MPa) | 35 | 25 | 20 | 15 |
The water-cement ratio (in percent, round off to the lower integer) to be used in the mix is
Correct answer
Why this answer is correct
Explanation:
Given, Grade of concrete M25
fck = 25 MPa
We know, Target mean strength is
fm = fck + 1.65σ
= 25 + 1.65 × 4 = 31.6 MPa
From the given table
| fm | w/c |
|---|---|
| 35 | 45 |
| 31.6 | ? |
| 25 | 50 |
From linear interpolation
\(\frac{{35 - 25}}{{35 - 31.6}} = \frac{{45 - 50}}{{45 - x}}\)
\(45 - x = \left( {45 - 50} \right) \times \left( {\frac{{35 - 31.6}}{{10}}} \right)\)
45 – x = -1.7
⇒ x = 45 + 1.7 = 46.7 ≈ 46 (as rounded up to lowest integer has been asked)
The data from a closed traverse survey PQRS (run in the clockwise direction) are given in the table
| Line | Included angle (in degree) |
|---|---|
| PQ | 88 |
| QR | 92 |
| RS | 94 |
| SP | 89 |
The closing error for the traverse PQRS (in degrees) is ________.
Correct answer
Why this answer is correct
Explanation:
In a closed traverse
Sum of interior angles = (2n - 4) × 90
⇒ (2 × 4 - 4) × 90 = 360°
The sum of angles given = 88 + 92 + 94 + 89 = 363°
Error (e) = Measured Value – Actual Value
⇒ Error (e) = 363 – 360 , e = 3°
A vehicle is moving on a road of grade +4% at a speed of 20 m/s. Consider the coefficient of rolling friction as 0.46 and acceleration due to gravity as 10 m/s2. On applying brakes to reach a speed of 10 m/s, the required braking distance (in m, round off to nearest integer) along the horizontal, is _________.
Correct answer
Why this answer is correct
Explanation:
Given,
Initial speed (Vi) = 20 m/sec
Coefficient of friction (μ) = 0.46
Final speed (Vf) = 10 m/sec
Gradient (n%) = + 4% (upward)
Braking distance is
\( \Rightarrow ;{D_{braking}} = \frac{{V_i^2}}{{2g\left( {\mu + 0.01n} \right)}} - \frac{{V_f^2}}{{2g\left( {\mu + 0.01n} \right)}}\)
\({D_{braking}} = \frac{{V_i^2 - V_f^2}}{{2g\left( {\mu + 0.01n} \right)}} = \frac{{{{20}^2} - {{10}^2}}}{{2 \times 10\left( {0.46\ +\ 0.01 \times 4} \right)}} = 30:m\)
The command area of a canal grows only one crop, i.e., wheat. The base period of wheat is 120 days and its total water requirement, Δ, is 40 cm. If the canal discharge is 2 m3/s, the area, in hectares, rounded off to the nearest integer, which could be irrigated (neglecting all losses), is
Correct answer
Why this answer is correct
Concept:
Duty: It is the number of hectares of land irrigated for full growth of a given crop by a supply of 1 cumec of water continuously during the entire base period of that crop.
Delta: The total water depth required by a crop to attain its full maturity in its base period.
Base Period: The time period that elapsed from the instant of the showing of the crop to the instant of its harvesting is called base period or crop period.
The relation between Duty (D), Delta (Δ), and Base period (B) is given as:
\({\bf{Δ }}\left( {{\bf{metres}}} \right) = \frac{{8.64B}}{D}\)
Calculation:
Given,
Base period (B) = 120 days
Delta (Δ) = 40 cm = 0.40 m
Discharge (Q) = 2 m3/sec
\(Duty\left( D \right) = \frac{{8.64B}}{{\rm{\Delta }}} = \frac{{8.64 \times 120}}{{0.40}} = 2592\)
Discharge (Q) = Area/Duty
\(\Rightarrow A = Q \times D = 2 \times 2592 = 5184;hectares\)
Construction of a new building founded on a clayey soil was completed in January 2010. In January 2014, the average consolidation settlement of the foundation in clay was recorded as 10mm. The ultimate consolidation settlement was estimated in design as 40 mm. Considering double drainage to occur at the clayey soil site, the expected consolidation settlement in January 2019 (in mm, round off to the nearest integer) will be_________.
Correct answer
Why this answer is correct
January - 2014
Consolidation settlement (S) = 10 mm
Ultimate settlement (Sf) = 40 mm
Drainage Condition → Single Drainage
Degree of consolidation ⇒ \(\left( {{U_1}} \right) = \frac{S}{{{S_f}}} \times 100 = \frac{{10}}{{40}} \times 100 = 25% \)
January - 2019
Time = 9 years
Drainage Condition → Double Drainage
\(\because U;\left( % \right) \le 60\)
\(\Rightarrow {\left( {{T_v}} \right)_1} = \frac{\pi }{4}{\left( {{U_1}} \right)^2} \Rightarrow {\left( {{T_v}} \right)_1} = \frac{\pi }{4}{\left( {0.25} \right)^2} = 0.0491\)
\({T_v} = \frac{{{C_v}.t}}{{{d^2}}} \Rightarrow ;\frac{\pi }{4}{\left( U \right)^2} = ;\frac{{{C_v} \times t}}{{{d^2}}} \Rightarrow {U^2} \propto t\)
\(\frac{{U_1^2}}{{U_2^2}} = \frac{{{t_1}}}{{{t_2}}} = \frac{{{{\left( {0.25} \right)}^2}}}{{U_2^2}} = \frac{4}{9};\)
\(\Rightarrow U_2^2 = \frac{9}{4} \times {0.25^2}\)
\({U_2} = \sqrt {\frac{9}{4} \times {{0.25}^2}} = 0.375\)
Consolidation in 2019 (S2) = 40 × 0.375 = 15 mm
The probability density function of a continuous random variable distributed uniformly between x and y (for y > x) is
\(\frac{1}{{x - y}}\)
x - y
y - x
\(\frac{1}{{y - x}}\)
Correct answer
\(\frac{1}{{y - x}}\)
Why this answer is correct
Concept:
A probability density function is a function of continuous random variable, whose integral across an interval gives the probability that the value of the variable lies within the same interval.
The probability density function for uniformly distributed random variable
\(f\left( x \right) = \frac{1}{{\left( {y - x} \right)}}for;y > x\)
Consider the hemi-spherical tank of radius 13 m as shown in the figure (not drawn to scale). What is the volume of water (in m3) when the depth of water at the centre of the tank is 6 m?
78 π
468 π
156 π
396 π
Correct answer
396 π
Why this answer is correct
Volume \(\Rightarrow V = \mathop \smallint \limits_7^{13} \pi \times {r^2}dh = ;\mathop \smallint \limits_7^{13} \left( {{R^2} - {h^2}} \right)dh = \pi \left[ {{R^2}h - \frac{{{h^3}}}{3}} \right]_7^{13};\)
\(\Rightarrow V = \left{ {\left[ {{{13}^2} \times 13 - \frac{{{{13}^3}}}{3}} \right] - \left[ {{{13}^2} \times 7 - \frac{{{7^3}}}{3}} \right]} \right} \times \pi \)
\(\therefore V = 396;\pi \)
An ordinary differential equation is given below:
\(\left( {\frac{{dy}}{{dx}}} \right)\left( {xlnx} \right) = y\)
The solution for the above equation is (Note: K denotes a constant in the options)
y = Klnx
y = Kxlnx
y = Kxex
y = Kxe-x
Correct answer
y = Klnx
Why this answer is correct
Given D.E
\(\left( {\frac{{dy}}{{dx}}} \right)\left( {xlnx} \right) = y\)
\(\Rightarrow \frac{{dy}}{y} = \frac{{dx}}{{xlnx}} \to {\rm{variable}} - {\rm{separable;Differential;Equation}}\)
Integrating on both sides; then
\(\Rightarrow \smallint \frac{1}{y}dy = \smallint \frac{{1/x}}{{lnx}}dx\)
\(\Rightarrow lny = \ln \left[ {lnx} \right] + lnk;;\left[ {\because;\smallint \frac{{{f^1}\left( x \right)}}{{f\left( x \right)}}dx = lnf\left( x \right)} \right]\)
⇒ \(lny;-;lnK; = ;ln\left[ {lnx} \right]\)
\(\Rightarrow \ln \left[ {\frac{y}{k}} \right] = \ln \left[ {\ln x} \right] \Rightarrow \frac{y}{k} = \ln x\)
⇒ \(y = k;ln;x\)
For a plane stress problem, the state of stress at a point P is represented by the stress element as shown in the figure
By how much angle (q) in degrees the stress element should be rotated in order to get the planes of maximum shear stress?
26.6
48.3
31.7
13.3
Correct answer
31.7
Why this answer is correct
Concept:
Angle of Major principle plane (θp):
\(\rm \tan 2\theta_p = \frac{{2;{τ _{xy}}}}{{{σ _x}; - {σ _y}}}\)
τxy - Shear stress
σx, σy - Principal stress acting in x, y directions
Calculation:
Given,
σx = 80 MPa, σy = 20 MPa, τxy = 25 MPa
Angle of Major principle plane (θp):
\(\tan 2;{\theta _p} = \frac{{2;{τ _{xy}}}}{{{σ _x}; - {σ _y}}} = \frac{{2\left( { - 25} \right)}}{{80 - \left( { - 20} \right)}}\)
\(\Rightarrow {\theta _p} = ; - 13.28^\circ \)
Angle of maximum shear stress plane = θp + 45° = -13.28 + 45 = 31.7°
The critical bending compressive stress in the extreme fiber of a structural steel section is 1000 MPa. It is given that the yield strength of the steel is 250 MPa, width of flange is 250 mm and thickness of flange is 15 mm. As per the provisions of IS: 800-2007, the non dimensional slenderness ratio of the steel cross-section is
0.50
2.00
0.25
0.75
Correct answer
0.50
Why this answer is correct
Explanation:
Given,
Yield stress of steel (fy) = 250 N/mm2
Critical bending compressive stress in the extreme fiber (fcrb) = 1000 N/mm2
The non-dimensional effective slenderness ratio of the steel cross-section as per IS 800:2007 is given by
\(\lambda = \sqrt {\frac{{{f_y}}}{{{f_{crb}}}}} \)
\(\lambda = \sqrt {\frac{{{f_y}}}{{{f_{crb}}}}} = \sqrt {\frac{{250}}{{1000}}} = 0.5\)
In the context of provisions relating to durability of concrete, consider the following assertion:
Assertion (1): As per IS 456-2000, air entrainment to the extent of 3% to 6% is required for
concrete exposed to marine environment.
Assertion (2): The equivalent alkali content (in terms of Na2O equivalent) for a cement
containing 1% and 0.6% of Na2O and K2O respectively, is approximately 1.4% (rounded to 1
decimal place).
Which one of the following statements is correct?
Assertion (1) is TRUE and Assertion (2) is FALSE
Both Assertion (1) and Assertion (2) are TRUE
Both Assertion (1) and Assertion (2) are FALSE
Assertion (1) is FALSE and Assertion (2) is TRUE
Correct answer
Assertion (1) is FALSE and Assertion (2) is TRUE
Why this answer is correct
As per clause 8.2.2.3 of IS 456-2000, entrained air percentage of 3 to 6% is required to resist freezing and thawing i.e. not meant for marine environment.
∴ Assertion 1 is wrong.
Equivalent alkali content in terms of Na2O = [Na2O] + 0.685 [K2O]
= 1 + 0.685 × 0.6 = 1.41 %
Hence, Assertion 2 is correct.
Chlorine is used as the disinfectant in a municipal water treatment plant. It achieves 50 percent of disinfection efficiency measured in terms of killing the indicator microorganisms (E-Coli) in 3 minutes. The minimum time required to achieve 99 percent disinfection efficiency would be
9.93 minutes
11.93 minutes
21.93 minutes
19.93 minutes
Correct answer
19.93 minutes
Why this answer is correct
Concept:
As per Watson and Chicks law
\({N_t} = {N_o}{e^{ - K{C^n}t}}\)
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:
Disinfection efficiency η1 = 50%, t1 = 3 minutes
\({\eta _1} = \left[ {;1 - {e^{ - k \times ;{t_1}}}} \right] \times 100\)
\(50 = \left[ {;1 - {e^{ - k \times ;3}}} \right] \times 100\)
K = 0.231 min-1
η2 = 99%, t2 = ?
\({\eta _2} = \left[ {;1 - {e^{ - k \times ;{t_2}}}} \right] \times 100\)
\(99 = \left[ {;1 - {e^{ - 0.23 \times ;{t_2}}}} \right] \times 100\)
t2 = 19.93 minutes
A camera with a focal length of 20 cm fitted in an aircraft is used for taking vertical aerial photographs of a terrain. The average elevation of the terrain is 1200 m above mean sea level (MSL). What is the height above MSL at which an aircraft must fly in order to get the aerial photographs at a scale of 1:8000?
3200 m
2800 m
2600 m
3000 m
Correct answer
2800 m
Why this answer is correct
Given that
f = 0.2 m, h = 1200 m, \({\rm{s}} = \frac{1}{{8000}}\)
Scale of photograph \(\Rightarrow S = \frac{f}{{H - h}}\)
\(\Rightarrow \frac{1}{{8000}} = \frac{{0.2}}{{H - 1200}}\)
H – 1200 = 8000 × 0.2
H – 1200 = 1600 = 2800 m
A flexible pavement has the following class of loads during a particular hour of the day.
i. 80 buses with 2-axles (each axle load of 40 kN);
ii. 160 trucks with 2-axles (front and rear axle loads of 40 kN and 80 kN, respectively)
The equivalent standard axle load repetitions for this vehicle combination as per IRC:37-2012 would be
320
250
240
180
Correct answer
180
Why this answer is correct
eXPLANTION:
For single axle and single wheel on either side
\(({{Axle\ load\ in\ kN} \over 65})^4\)
For single axle and double wheel on either side
\(({{Axle\ load\ in\ kN} \over 80})^4\)
For tandem axle and double wheel on either side
\(({{Axle\ load\ in\ kN} \over 148})^4\)
For tridem axle and single wheel on either side
\(({{Axle\ load\ in\ kN} \over 224})^4\)
Standard axle load = 80kN
80 buses with 2 axles, each axles load is 40 kN
∴ Number of repetitions of 40 kN, N1 = 80 × 2 = 160
160 trucks with front axle load = 40 kN
∴ Number of repetitions of 40 kN, N2 = 160
160 trucks with rear axle load = 40 kN
∴ Number of repetitions of 40 kN, N2 = 160
Equivalent standard axle load \(= 160 \times {\left( {\frac{{40}}{{80}}} \right)^4} + 160 \times {\left( {\frac{{40}}{{80}}} \right)^4} + 160 \times {\left( {\frac{{80}}{{80}}} \right)^4}\)
\(= 160 \times {\left( {\frac{1}{2}} \right)^4} + 160 \times {\left( {\frac{1}{2}} \right)^4} + 160 \times {1^4}\)
= 20 + 160 = 180
The inverse of the matrix \(\left[ {\begin{array}{*{20}{c}} 2&3&4\ 4&3&1\ 1&2&4 \end{array}} \right]\) is
\(\left[ {\begin{array}{*{20}{c}} 2&{ - \frac{4}{5}}&{ - \frac{9}{5}}\ { - 3}&{\frac{4}{5}}&{\frac{{14}}{5}}\ 1&{ - \frac{1}{5}}&{ - \frac{6}{5}} \end{array}} \right]\)
\(\left[ {\begin{array}{*{20}{c}} { - 2}&{\frac{4}{5}}&{\frac{9}{5}}\ 3&{ - \frac{4}{5}}&{ - \frac{{14}}{5}}\ { - 1}&{\frac{1}{5}}&{\frac{6}{5}} \end{array}} \right]\)
\(\left[ {\begin{array}{*{20}{c}} {10}&4&{ - 9}\ { - 15}&4&{14}\ 5&{ - 1}&6 \end{array}} \right]\)
\(\left[ {\begin{array}{*{20}{c}} {10}&4&9\ {15}&{ - 4}&{ - 14}\ { - 5}&1&6 \end{array}} \right]\)
Correct answer
\(\left[ {\begin{array}{*{20}{c}} { - 2}&{\frac{4}{5}}&{\frac{9}{5}}\ 3&{ - \frac{4}{5}}&{ - \frac{{14}}{5}}\ { - 1}&{\frac{1}{5}}&{\frac{6}{5}} \end{array}} \right]\)
Why this answer is correct
Let,
\(A = \left[ {\begin{array}{*{20}{c}} 2&3&4\ 4&3&1\ 1&2&4 \end{array}} \right]\)
\({A^{ - 1}} = \frac{1}{{\left| A \right|}} \times \left( {adjoint;A} \right)\)
\({\rm{Co}} - {\rm{factor;of;Matrix}} = {\rm{;}}\left[ {\begin{array}{*{20}{c}} {10}&{ - 15}&5\ { - 4}&4&{ - 1}\ { - 9}&{14}&{ - 6} \end{array}} \right]{\rm{;}}\)
\(\therefore Adj\left( A \right) = ;\left[ {\begin{array}{*{20}{c}} {10}&{ - 4}&{ - 9}\ { - 15}&4&{14}\ 5&{ - 1}&{ - 6} \end{array}} \right]\)
\(\left| A \right| = \left| {\begin{array}{*{20}{c}} 2&3&4\ 4&3&1\ 1&2&4 \end{array}} \right| = 2\left[ {12 - 2} \right] - 3\left[ {16 - 1} \right] + 4\left[ {8 - 3} \right]\)
⇒ |A| = 20 – 45 + 20 = -5
\(\therefore {A^{ - 1}} = \frac{{Adj\left( A \right)}}{{\left| A \right|}} = - \frac{1}{5}\left[ {\begin{array}{*{20}{c}} {10}&{ - 4}&{ - 9}\ { - 15}&4&{14}\ 5&{ - 1}&{ - 6} \end{array}} \right]\)
\(\Rightarrow {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} { - 2}&{\frac{4}{5}}&{\frac{9}{5}}\ 3&{\frac{{ - 4}}{5}}&{\frac{{14}}{5}}\ { - 1}&{\frac{1}{5}}&{\frac{6}{5}} \end{array}} \right]\)
A long uniformly distributed load of 10 kN/m and a concentrated load of 60 kN are moving together on the beam ABCD shown in the figure (not drawn to scale). The relative positions of the two loads are not fixed. The maximum shear force (in kN, round off to the nearest integer) caused at the internal hinge B due to the two loads is _______
Correct answer
Why this answer is correct
Influence line for shear force at B.
From Muller’s Breslau principle
ILD for obtaining maximum shear force at B
Maximum shear force \(V = 60 \times 1 + \left( {\frac{1}{2} \times 1 \times 2} \right) \times 10\)
∴ V = 60 + 10 = 70kN
A plane frame shown in the figure (not to scale) has linear elastic springs at node H. The spring constants are kx = ky = 5 × 105 kN/m3 and kθ = 3 × 105 kNm/rad.
For the externally applied moment of 30 kNm at node F, the rotation (in degrees, round off to 3 decimals) observed in the rotational spring at node H is ___________
Correct answer
Why this answer is correct
Explanation:
Since horizontal reaction at E = Hc = 0
Bending moment at F for member FE = MFE = 0
Member FH behaves like propped cantilever beam
\({M_\theta } = {k_\theta } \times \theta\)
30 kNm = 3 × 105 kNm/rad × θ
⇒ θ = 1 × 10-4 radians = 0.0057°
A rolled I-section beam is supported on a 75 mm wide bearing plate as shown in the figure. Thickness of flange and web of the I-section are 20 mm and 8 mm, respectively. Root radius of the I-section is 10 mm. Assuming: material yield stress, fy = 250 MPa and partial safety factor for material, γmo = 1.10
As per IS: 800-2007, the web bearing strength (in kN, round off to 2 decimal places) of the beam is ________
Correct answer
Why this answer is correct
Explanation:
Given,
Thickness of web (tw) = 8 mm
Thickness of flange (tf) = 20 mm
Radius of roof (r) = 10 m
Width of support (b1) (or support reaction) = 75 mm
As per IS 800:2007
Load distribution angle (α) = 21.8° [1V: 2.5H]
Design bearing strength of steel = \(\frac{{{f_y}}}{{{\gamma _o}}} = \frac{{250}}{{1.1}};MPa\)
⇒ Width of critical section@ root resisting web crippling due to high bearing stress (w1) = b1 + 2.5 (tf + r)
= 75 + 2.5 (20 + 10)
= 150 mm
⇒ Total load carrying capacity of web with respect to web crippling
⇒ \({w_1} \times {t_w} \times \frac{{{f_y}}}{{{\gamma _{mo}}}} = 272.72;kN\)
When a specimen of M25 concrete is loaded to a stress level of 12.5 MPa, a strain of 500 × 10–6 is recorded. If this load is allowed to stand for a long time, the strain increases to 1000 × 10–6. In accordance with provisions of IS: 456-2000, considering the long-term effects, the effective modulus of elasticity of the concrete (in MPa) is ________
Correct answer
Why this answer is correct
Concept:
Modulus of Elasticity of Concrete: It can be defined as the ratio of applied stress to corresponding strain within elastic limits.
It may be given as:
i) Short Term Modulus of elasticity = 5000 √fck
ii) Long Term Modulus of elasticity = \(\left( {{E_\theta }} \right) = \frac{{{E_C}}}{{1 + \theta }}\) \(=\frac{5000√{{{f}_{ck}}}}{1+\theta }\)
where θ is the creep coefficient
Calculation:
Given,
Grade of concrete M25
Initial Strain = 500 × 10–6 and strain increases to 1000 × 10–6
Ec = 5000 √fck = 5000 √25 = 25000 MPa
Creep coefficient \(\left( \theta \right) = \frac{{creep;strain}}{{elastic;strain}}\)
\(\theta = \frac{{longterm;strain - elastic;strain}}{{elastic;strain}} = ;\frac{{1000 \times {{10}^{ - 6}} - 500 \times {{10}^{ - 6}}}}{{500 \times {{10}^{ - 6}}}} = 1\)
\({E_\theta } = \frac{{250000}}{{1 + 1}} = \frac{{25000}}{2} = 12500;MPa.\)
A water treatment plant treats 6000 m3 of water per day. As a part of the treatment process, discrete particles are required to be settled in a clarifier. A column test indicates that an overflow rate of 1.5 m per hour would produce the desired removal of particles through settling in the clarifier having a depth of 3.0 m. The volume of the required clarifier, (in m3, round off to 1 decimal place) would be
Correct answer
Why this answer is correct
\(Flow\left( Q \right) = 6000{m^3}/day = \frac{{6000}}{{24}}{m^3}/hr = 250{m^3}/h\)
Overflow rate (Vo) = 1.5 m3/h
\(Area = \frac{Q}{{{V_0}}} = \frac{{250}}{{1.5}} = 166.67{m^2}\)
\(Volume = Area \times depth = \frac{{250}}{{1.5}} \times 3 = 500{m^3}\)
Raw municipal solid waste (MSW) collected from a city contains 70% decomposable material that can be converted to methane. The water content of the decomposable material is 35%. An elemental analysis of the decomposable material yields the following mass percent.
C: H : O : N : other = 44 : 6 : 43 : 0.8 : 6.2
The methane production of the decomposable material is governed by the following stoichiometric relation
\({C_a}{H_b}{O_c}{N_d}; + ;nH2O; \to ;mCH4; + ;sCO2; + ;dNH3\)
Given atomic weight: C = 12, H = 1, O = 16, N = 14. The mass of methane produced (in grams, round off to 1 decimal place) per kg of raw MSW will be _______.
Correct answer
Why this answer is correct
Let 1 kg of raw municipal solid waste (MSW)
Decomposable material = 70% of 1 kg = 0.7 kg
Water content = 35%
Weight of solids in \(MSW = \left( {\frac{{100 - 35}}{{100}}} \right) \times 0.7 = 0.455;kg = 455;gms\)
Given: C: H: O: N: other = 44 : 6 : 43 : 0.8 : 6.2
\({\rm{Weight;of;C}} = \frac{{\left( {44} \right)}}{{\left( {44 + 6 + 43 + 0.8 + 6.2} \right)}} \times 455 = 200.2;gms\)
\({\rm{Weight;of;H}} = \frac{{\left( 6 \right)}}{{\left( {44 + 6 + 43 + 0.8 + 6.2} \right)}} \times 455 = 27.3;gms\)
\({\rm{Weight;of;O}} = \frac{{\left( {43} \right)}}{{\left( {44 + 6 + 43 + 0.8 + 6.2} \right)}} \times 455 = 195.6;gms\)
\({\rm{Weight;of;N}} = \frac{{\left( {0.8} \right)}}{{\left( {44 + 6 + 43 + 0.8 + 6.2} \right)}} \times 455 = 3.64;gms\)
⇒ 12 a = 200.2 ⇒ a = 16.68
⇒ 1 × b = 27.3 ⇒ b = 27.3
⇒ 16 × c = 195.6 ⇒ c = 12.225
⇒ 14 × d = 3.64 ⇒ d = 0.26
Now, balancing the given stoichiometric r × n
\({C_a}{H_b}{O_c}{N_d}; + ;nH2O; \to ;mCH4; + ;sCO2; + ;dNH3\)
12 a = 12 m + 12 s
⇒ 12 m + 12 s = 200.2 (∵ 12 a = 200.2)
b + 2n = 4 m + 3d
⇒ 4m – 2n = 26.52 (∵ 6 = 27.3)
16 c + 16 n = 325
⇒ n = 25 – 12.225 (∵ 16 c = 195.6)
∴ 4 m – 2 (2s – 12.225) = 26.52
⇒ 4 m – 4s = 2.07
⇒ 12 m – 12s = 6.21
Also, 12 m + 12 s = 200.2
⇒ m = 8.6
∴ Mass of methane produced = 16 × m = 16 × 8.6 = 137.6 gms
Consider the reactor shown in the figure. The flow rate through the reactor is Q m3/h .The concentration (in mg/L) of a compound in the influent and effluents are C0 and C, respectively. The compound is degraded in the reactor following the first order reactions. The mixing condition of the reactor can be varied such that the reactor becomes either a completely mixed flow reactor (CMFR) or a plug-flow reactor (PFR). The length of the reactor can be adjusted in these two mixing conditions to LCMFR and LPER while keeping the cross-section of the reactor constant. Assuming steady state and for C/C0 = 0.8, the value of LCMFR/LPER (round off to 2 decimal places) is _______
Correct answer
Why this answer is correct
\(\frac{c}{{{C_o}}} = 0.8\)
For CMFR (Completely mixed flow reactor)
\(c = \frac{{{C_o}}}{{1 + kt}} \Rightarrow 0.8 = \frac{1}{{1 + kt}} \Rightarrow {t_{CMFR}} = \frac{{0.25}}{k}\)
For PMFR (Plug Flow Reactor)
\(c = {C_o} \times {e^{ - kt}} \Rightarrow 0.8 = {e^{ - kt}} \Rightarrow {t_{PFR}} = \frac{{0.223}}{k}\)
For steady state, v = constant
\(\because L = V \times t \Rightarrow \frac{{{L_{CMFR}}}}{{{L_{PFR}}}} = \frac{{V \times {t_{CMFR}}}}{{V \times {t_{PFR}}}} = \frac{{0.25}}{{0.223}} = 1.12\)
A series of perpendicular offsets taken from a curved boundary wall to a straight survey line at an interval of 6 m are 1.22, 1.67, 2.04, 2.34, 2.14, 1.87, and 1.15 m. The area (in m2 , round off to 2 decimal places) bounded by the survey line, curved boundary wall, the first and the last offsets, determined using Simpson’s rule, is _______.
Correct answer
Why this answer is correct
Using Simpson’s rule, we have
\(Area;\left( A \right) \approx \frac{d}{3} \times \left[ {\left( {{y_0} + {y_n}} \right) + 2\left( {{y_2} + {y_4} + \ldots + {y_{n - 2}}} \right) + 4\left( {{y_1} + {y_3} + \ldots + {y_{n - 1}}} \right)} \right]\)
\(A = \frac{6}{3} \times {\rm{[}}\left( {1.22 + 1.15} \right) + 2\left( {2.04 + 2.14} \right) + 4\left( {1.67 + 2.34 + 1.87} \right){\rm{} }};\)
\(A = 68.5;{m^2}\)
The uniform arrival and uniform service rates observed on an approach road to a signalized intersection are 20 and 50 vehicles/minutes, respectively. For this signal, the red time is 30 s, the effective green time is 30 s, and the cycle length is 60s. Assuming that initially there are no vehicles in the queue, the average delay per vehicle using the approach road during a cycle length (in seconds, round off to 2 decimal places) is _______ .
Correct answer
Why this answer is correct
Explanation:
Webster’s average delay is given by,
\(\left( {{d_i}} \right) = \frac{{\frac{{{C_o}}}{2}{{\left( {1 - \frac{{{g_i}}}{{{C_o}}}} \right)}^2}}}{{1 - \frac{{{q_i}}}{{{S_i}}}}}\)
Where,
di is an average delay, Co is Cycle length, gi is effective green time
qi is normal flow.
Given,
Cycle length (C0) = 60 seconds
Normal flow (qi) = 20 veh/min
Saturation flow (Si) = 50 veh/min
Effective green time (gi) = 30 seconds
\({d_i} = \frac{{\frac{{60}}{2}{{\left[ {1 - \frac{{30}}{{60}}} \right]}^2}}}{{1 - \frac{{20}}{{50}};}} = 12.5;seconds\)
A broad gauge railway line passes through a horizontal curved section (radius = 875 m) of length 200 m. The allowable speed on this portion is 100 km/h. For calculating the cant consider the gauge as centre-to-centre distance between the rail heads, equal to 1750 mm, The maximum permissible cant (in mm, round of to 1 decimal place) with respect to the centre-to-centre distance between the rail heads is ______.
Correct answer
Why this answer is correct
Explanation:
For a railway track
Allowable cant
, where V is in km/hr = 100 kmph
Additional Information
Equilibrium superelevation/cant:
It is the difference between the height of the outer rail and the inner rail on a curve.
Inner rail is taken as the reference rail and with respect to inner rail, outer rail is raised.
Objective of providing a superelevation:
Superelevation or cant C = GV2/127R
Where
R = radius of rail in meter
V = Speed of train in km/hr
G = gauge of track
Important Points
The maximum permissible limit for actual Cant or superelevation
| Track | Speed < 120 Kmph | Speed > 120 Kmph |
|---|---|---|
| Broad Gauge | 16.5 cm | 18.5 cm |
| Meter Gauge | 10 cm | |
| Narrow Gauge | 7.6 cm |
The speed-density relationship of a highway is given as u = 100 – 0.5 k where, u = speed in km per hour, k = density in vehicles per km. The maximum flow (in vehicles per hour, round off to the nearest integer) is ______.
Correct answer
Why this answer is correct
Method-I
Given speed density relationship as u = 100 – 0.5 K
Capacity (q) = K × u = K × (100 – 0.5 K) = 100 K – 0.5 K2
For capacity to be maximum, \(\frac{{dq}}{{dk}} = 0\)
q = 100 K – 0.5 K2
\(\frac{{dq}}{{dk}} = 100 - 0.5\left( {2K} \right) = 0 \Rightarrow K = 100;veh/km\)
Maximum capacity = 100 k – 0.5 k2 = 100 × 100 – 0.5 × 1002 = 5000 veh/hr
Method – 2:
Given u = 100 – 0.5 k
At the speeds of vehicle zero then, the density will be jam density
u = 0 ⇒ K = KJ
\(100 - 0.5;K = 0 \Rightarrow K = \frac{{100}}{{0.5}} = 200;veh/hr\)
Similarly, Vf = 100 – 0.5 × 0 = 100 km/hr
\(Maximum;capacity\left( {{q_{max}}} \right) = \frac{{{V_f} \times {K_J}}}{4} = \frac{{100 \times 200}}{4} = 5000;veh/hr\)
A confined aquifer of 15 m constant thickness is sandwiched between two aquicludes as shown in the figure (not drawn to scale)
The heads indicated by two piezometers P and Q are 55.2 m and 34.1 m, respectively. The aquifer has a hydraulic conductivity of 80 m/day and its effective porosity is 0.25. If the distance between the piezometers is 2500 m, the time taken by the water to travel through the aquifer from piezometer location P to Q (in days, round off to 1 decimal place) is ______.
Correct answer
Why this answer is correct
Discharge Velocity, V = k × i
\(V = k \times \left( {\frac{{{h_1} - {h_2}}}{L}} \right) = 80 \times \left( {\frac{{55.2 - 34.1}}{{2500}}} \right) = 0.6752;m/day\)
Porosity, n = 0.25
∴ Seepage Velocity, \({V_s} = \frac{V}{n} = \frac{{0.6752}}{{0.25}} = 2.70;m/day\)
Time taken \(t = \frac{{Length}}{{{V_s}}} = \frac{{2500}}{{2.7008}} = 925.65;days\)
At the foot of a spillway, water flows at a depth of 23 cm with a velocity of 8.1 m/s, as shown in the figure.
The flow enters as an M-3 profile in the long wide rectangular channel with bed slope 1 : 1800 and Manning’s n = 0.015. A hydraulic jump is formed at a certain distance from the foot of the spillway. Assume the acceleration due to gravity, g = 9.81 m/s2. Just before the hydraulic jump, the depth of flow y1 (in m, round off to 2 decimal places) is ________.
Correct answer
Why this answer is correct
Explanation:
Given,
At a given section, y = 0.23 m and velocity (v) = 8.1 m/s
q (discharge per unit width) = 8.1 × 0.23 = 1.863 m2/s
Now,
For a wide rectangular channel (B ≫≫ y)
A = B × y
P = B + 2y
Hydraulic Mean Radius \(;\left( R \right) = \frac{A}{P} = \frac{{By}}{{B + 2y}} = \frac{{By}}{B} = y\)
For Mild slope ⇒ NDL (Normal Depth Line) > CDL (Critical Depth Line)
By manning’s equation
\(Q = \frac{1}{n} \times A \times {R^{\frac{2}{3}}} \times {S^{\frac{1}{2}}};\)
\(Q = \frac{1}{n} \times \left( {By} \right) \times {y^{\frac{2}{3}}} \times {S^{\frac{1}{2}}}\)
\(\frac{Q}{B} = \frac{1}{n} \times {y^{\frac{5}{3}}} \times {S^{\frac{1}{2}}}\)
\(q = \frac{1}{n}.{y^{5/3}}{S^{1/2}}\)
\(\Rightarrow 1.863; = \frac{1}{{0.015}} \times y_n^{5/3}{\left( {\frac{1}{{1800}}} \right)^{1/2}}\)
yn = 1.107 m
Before the hydraulic jump, depth → y1 = 1.107 m
yn is the conjugate depth of y1
\(\frac{{{y_1}}}{{{y_n}}} = \frac{{ - 1}}{2}\left[ {1 - \sqrt {1 + 8 \times Fr_n^2} } \right]\)
\({y_1} = \frac{{ - {y_n}}}{2}\left[ {1 - \sqrt {1 + 8 \times \frac{{{q^2}}}{{g.{y^3}}}} } \right]\)
\({y_1} = \frac{{ - 1.107}}{2}\left[ {1 - \sqrt {1 + 8 \times \frac{{{{1.863}^2}}}{{9.81 \times {{1.107}^3}}}} } \right] = 0.42:m\)
Two identical pipes (i.e. having the same length, same diameter, and same roughness) are used to withdraw water from a reservoir. In the first case, they are attached in series and also discharge freely into the atmosphere. In the second case, they are attached in parallel and friction factor is same in both the cases, the ratio of the discharge in the parallel arrangement to that in the series arrangement (round off to 2 decimal places) is ________.
Correct answer
Why this answer is correct
1st case: series connection
Total head loss due to friction
\({h_f} = {h_{{f_1}}} + {h_{{f_2}}}\)
\({h_f} = \frac{{{f_1}{L_1}Q_S^2}}{{12.1d_1^5}} + \frac{{{f_2}{L_2}Q_S^2}}{{12.1d_2^5}}\)
\(\Rightarrow \left( {{h_{{f_S}}}} \right) = \frac{{2fLQ_S^2}}{{12.1{d^5}}}\)
2nd case: Parallel connection
Total head loss due to friction is same in both pipes \({Q_1} = {Q_2} = \frac{{{Q_P}}}{2}\)
\({h_{fP}} = \frac{{fL{{\left( {\frac{{{Q_P}}}{2}} \right)}^2}}}{{12.1{d^5}}} = \frac{{fLQ_P^2}}{{4 \times 12.1 \times {d^5}}}\)
As given
\(\left( {{h_{fS}}} \right) = \left( {{h_{fP}}} \right) \Rightarrow \frac{{2fLQ_S^2}}{{12.1{d^5}}} = ;\frac{{fLQ_P^2}}{{4 \times 12.1 \times {d^5}}} \Rightarrow \frac{{{Q_P}}}{{{Q_S}}} = \sqrt 8 = 2.83\)
The ordinates of a 2-hour unit hydrograph (i.e. for 1 cm of effective rain), for a catchment, are shown in the table.
| Time | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Ordinates | 0 | 2 | 8 | 18 | 32 | 45 | 30 | 19 | 12 | 7 | 3 | 1 | 0 |
A 6-hour storm occurs over the catchment such that the effective rainfall intensity is 1 cm/hour for the first two hours, zero for the next two hours, and 0.5 cm/hour for the last two hours. If the base flow is constant at 5 m3/s, the peak flow due to this storm (in m3/s, round off to 1 decimal place) will be ________.
Correct answer
Why this answer is correct
Explanation:
Rainfall excess in 1st two hours for rainfall intensity of 1 cm/hr ⇒ 2 cm
Rainfall excess in 2nd two hours for rainfall intensity of 0 cm/hr ⇒ 0 cm
Rainfall excess in 3rd two hours for rainfall intensity of 0.5 cm/hr ⇒ 1 cm
| Time | 2-hr Unit Hydrograph Ordinates | 1st DRH R1 = 2 cm | 1st DRH R1 = 2 cm | 1st DRH R1 = 2 cm | Total DRH |
|---|---|---|---|---|---|
| A | B = 2×A | C = 0×A with 2-hour lag | D = 0.5×A with 4-hour lag | E = B+C+D | |
| 0 | 0 | 0 | - | - | 0 |
| 1 | 2 | 4 | - | - | 4 |
| 2 | 8 | 16 | 0 | - | 16 |
| 3 | 18 | 36 | 0 | - | 36 |
| 4 | 32 | 64 | 0 | 0 | 64 |
| 5 | 45 | 90 | 0 | 2 | 92 = Peak |
| 6 | 30 | 60 | 0 | 8 | 68 |
| 7 | 19 | 38 | 0 | 18 | 56 |
| 8 | 12 | 24 | 0 | 32 | 56 |
| 9 | 7 | 14 | 0 | 45 | 59 |
| 10 | 3 | 6 | 0 | 30 | 36 |
| 11 | 1 | 2 | 0 | 19 | 21 |
| 12 | 0 | 0 | 0 | 12 | 12 |
Peak of flood ⇒ 92 + base flow = 92 + 5 = 97 m3/s
The dimensions of a soil sampler are given in the table.
| Parameter | Cutting edge | Sampling tube |
|---|---|---|
| Inside diameter (mm) | 80 | 86 |
| Outside diameter (mm) | 100 | 90 |
For this sampler, the outside clearance ratio (in percent, round off to 2 decimal places) is __________.
Correct answer
Why this answer is correct
D1 = 80 mm D3 = 86 mm
D2 = 100 mm D4 = 90 mm
\(Outside;clearance;ratio = \frac{{{D_2} - {D_4}}}{{{D_4}}} \times 100 = \frac{{100 - 90}}{{90}} \times 100 = 11.11% \)
A 2 m × 4 m rectangular footing has to carry a uniformly distributed load of 120 kPa. As per the 2 : 1 dispersion method of stress distribution, the increment in vertical stress (in kPa) at a depth of 2 m below the footing is _______.
Correct answer
Why this answer is correct
\({\sigma _z} = \frac{Q}{{\left( {B + Z} \right)\left( {L + Z} \right)}};{\rm{for;}}2{\rm{V;to;}}1{\rm{H;dispersion}}\)
\({\sigma _z} = \frac{{2 \times 4 \times 120}}{{\left( {2 + 2} \right)\left( {4 + 2} \right)}} = \frac{{960}}{{24}} = 40;kPa\)
Constant head permeability tests were performed on two soil specimens, S1 and S2. The ratio of height of the two specimens (LS1: LS2) is 1.5, the ratio of the diameter of specimens (DS1:DS2) is 0.5, and the ratio of the constant head (hs1:hs2) applied on the specimens is 2.0. If the discharge from both the specimens is equal, the ratio of the permeability of the soil specimens (ks1 : ks2) is ________
Correct answer
Why this answer is correct
\(\frac{{{{\left( {{L_s}} \right)}_1}}}{{{{\left( {{L_s}} \right)}_2}}} = 1.5;;\frac{{{{\left( {{D_s}} \right)}_1}}}{{{{\left( {{D_s}} \right)}_2}}} = 0.5,;\frac{{{{\left( {{h_s}} \right)}_1}}}{{{{\left( {{h_s}} \right)}_2}}} = 2.0\)
By constant head permeability test
Q = k × i × A
\(Q = k \times \frac{h}{L} \times A = k \times \frac{h}{L} \times \frac{\pi }{4} \times {d^2}; \Rightarrow Q \propto \frac{{kh}}{L} \times {d^2}\)
\(\frac{{{Q_1}}}{{{Q_2}}} = \frac{{\frac{{{k_1}{h_1}}}{{{L_1}}} \times d_1^2}}{{\frac{{{k_2}{h_2}}}{{{L_2}}} \times d_2^2}} = \frac{{{k_1}}}{{{k_2}}} \times \frac{{{h_1}}}{{{h_2}}} \times \frac{{{L_2}}}{{{L_1}}} \times {\left( {\frac{{{d_1}}}{{{d_2}}}} \right)^2}\)
\(\because;{Q_1} = {Q_2}\)
\( \Rightarrow 1 = \frac{{{k_1}}}{{{k_2}}} \times 2 \times \frac{1}{{1.5}} \times {\left( {0.5} \right)^2}\)
\(\frac{{{k_1}}}{{{k_2}}} = \frac{{1 \times 1 \times 5}}{{2 \times {{0.5}^2}}} = 3\)
A timber pile of length 8m and diameter of 0.2m is driven with a 20 kN drop hammer, falling freely from a height of 1.5 m. The total penetration of the pile in the last 5 blows is 40mm. Use the Engineering News Record expression. Assume a factor of safety of 6 and empirical factor (allowing reducing in the theoretical set, due to energy losses) of 2.5 cm. The safe load carrying capacity of the pile (in kN, round off to 2 decimal places) is ______.
Correct answer
Why this answer is correct
Using Engineering News Formula
Safe load carrying capacity \(\left( {{Q_s}} \right) = \frac{{Wh;{\eta _h}}}{{F\left( {S + C} \right)}}\)
Where,
W = weight of hammer = 20 kN
h = Height of fall = 1.5 m
\(S = set;value = \frac{{40}}{5}mm = 8;mm = 0.8;cm\)
C = 2.5 cm
F = Factory of safety = minimum of 6 to be taken
Efficiency (η) = 100%
\({Q_{safe}} = \frac{{Wh}}{{F\left( {S + C} \right)}} = \frac{{20 \times 1.5}}{{6\left( {0.008 + 0.025} \right)}} = 151.51;kN\)
A square footing of 2m sides rests on the surface of a homogeneous soil bed having the properties: cohesion c = 24 kPa, angle of internal friction Φ = 25°, and unit weight 3 γ =18kN/m3. Terzaghi’s bearing capacity factor Φ = 25° are Nc = 25.1, Nq = 12.7, Nγ = 9.7, N’c = 14.8, N’q = 5.6 and N’q = 5.6 and N’γ = 3.2. The ultimate bearing capacity of the foundation (in kPa, round off to 2 decimal places) is ________.
Correct answer
Why this answer is correct
Φ = 25° < 29° ⇒ Local shear failure
\({C_m} = \frac{2}{3}C = \frac{2}{3} \times 24 = 16kPa\)
As per Terzaghi’s theory
For square footing
Ultimate bearing capacity \( \Rightarrow {q_u}; = ;1.3;{C_m}N{'_c}; + ;\gamma \times {D_f} \times N{'_q} + ;0.4 \times \gamma \times B \times ;N{'_\gamma }\)
∴ qu = 1.3 × 16 × 14.8 + 0 + 0.4 × 18 × 2 × 3.2
∴ qu = 307.84 + 46.08 = 353.92 kPa
65 questions in the full paper