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His reactions to unpleasant situations tended to _________ everyone’s nerves.
The word that best fills the blank in the above sentence is
The sentence implies that his reactions to unpleasant situations annoyed everyone. The correct word to fill in the blank is aggravate as it means 'to annoy or exasperate'.
The meaning of the other words is:
Deviate: depart from an established course
Revue: a light theatrical entertainment consisting of a series of short sketches, songs, and dances, typically dealing satirically with topical issues
Meander: wander at random
If XY + Z = X(Y + Z) which of the following must be true?
XY + Z = X(Y + Z)
⇒ XY + Z = XY + XZ
⇒ Z = XZ ⇒ Z  XZ = 0 ⇒ Z (1  X) = 0
Either Z = 0 (in this case X can take any value) OR X = 1 (in this case Z can take any value).
X = 1 or Z = 0
What is the number missing from the table?
5 × 3 + 1 = 16
16 × 3 + 1 = 49
9 × 3 + 2 = 29
29 × 3 + 2 = 89
15 × 3 + 3 = 48
48 × 3 + 3 = 147
An ‘archipelago’ is ‘a group of islands’, i.e. it is it is formed of several ‘islands’. So the answer should be the option where the group of the latter word forms the former. Option 1 is incorrect as a group links forms a chain and not the other way round. Similarly, option C and D can be disqualified. An encyclopedia is ‘a book or set of books giving information on many subjects’. The correct answer is option B as many beads form a necklace.
There are three basket of fruits. First basket has twice the number of fruits in the second basket. Third basket has 3/4 th of the fruits in the first. The average of the fruits in all the basket is 30. The number of fruits in the first basket is
Until now only injectable vaccines against Swine Flu have been available. They have been primarily used by older adults who are at risk for complications from Swine Flu. A new vaccine administered in an oral form has proven effective in preventing Swine Flu in children. Since children are significantly more likely than adults to contract and spread Swine Flu, making the new vaccine widely available for children will greatly reduce the spread of Swine Flu across the population.
Which of the following, if true, most strengthens the argument?
The passage points out that the oral form of vaccine will allow Swine Flu to be greatly reduced among the children who are more susceptible to catching the Flu (as compared to adults). Undoubtedly, the non painful method of oral vaccination is more likely to encourage the parents to get their children vaccinated who are skeptical of injections that cause pain. Thus, among all the given options, 4 strengthens the given argument. Options 1, 2 and 3 are completely irrelevant and 5 would weaken the argument as it cites the probable negative effects of the new vaccine which will definitely not encourage parents to get their children vaccinated.
Four friends Rishabh, Keshav, Lavish and Hemang are out for shopping. Rishabh has less money than three times the amount that Keshav has. Lavish has more money than Keshav. Hemang has an amount equal to the difference of amounts with Keshav and Lavish. Rishabh has three times the money with Hemang. Each of them has to buy at least one shorts, or one sleeper, or one sleeveless tshirt, or one goggle that is priced 200, 400, 600, and 1000 a piece, respectively. Lavish borrows 300 from Rishabh and buys a goggle. Keshav buys a Sleeveless tshirt after borrowing 100 from Rishabh and is left with no money. Rishabh buys three shorts. What is the costliest item that Hemang could buy with his own money?
Let, Rishabh has 'P' amount of money, Keshav has 'Q' amount of money, Lavish has 'R' amount of money, and Hemang has 'S' amount of money.
Therefore, according to the question
P < 3 × Q
R > Q,
S = R  Q
P = 3 × S
Also, Lavish borrows 300 from Rishabh and buys a goggle and Keshav buys a Sleeveless tshirt after borrowing 100 from Rishabh and is left with no money. Implies that,
Q = 600  100 (Rishabh) = 500
R ≥ 1000  300 (Rishabh) ≥ 700
Also, Rishabh buys three shorts means he has a minimum of 1000 (100 + 300 + 3 × 200 = 1000) and the maximum is less than 3Q.
Therefore, 1000 ≤ P < 3Q
1000 ≤ P < 3 × 500
1000 ≤ P < 1500
Since Rishabh has three times the money with Hemang
1000 ≤ 3 × S < 1500
333 ≤ S < 500
Hence, Hemang could buy a sleeper with his own money.
I wouldn’t _______ with a soldier who was wearing a metal _______, awarded for a display of _______.
The words that best fill the blanks in the above sentence are:
The sentence implies that I would not poke a soldier who was wearing a metal medal which was awarded for the display of determination.
The meaning of the words is:
Meddle: interfere in something that is not one's concern
Medal: a metal disc typically of the size of a large coin and bearing an inscription or design, made to commemorate an event or awarded as a distinction to someone such as a soldier or athlete
Mettle: a person's ability to cope well with difficulties; spirit and resilience
Therefore, the correct sequence of words to fill in the blanks is option 2.
A paper sheet is in the shape of a right angle triangle and cut along a line parallel to hypotenuse leaving a smaller triangle. There was 25% reduction in the length of the hypotenuse of the triangle. If area of triangle initially was 28 cm^{2} then area of smaller triangle will be ______ cm^{2}.
Let PQR is the initial triangle and SQT is the final triangle.
ΔPQR is similar to ΔSQT
∵ ST = 0.75 PR
∴ SQ = 0.75 PQ
And QT = 0.75 QR
Initial Area = ×QR×PQ = 28cm^{2 }
Final Area = ×QT×QS
=×0.75QR×0.75PQ
= (0.75)^{2} × 28 = 15.75 cm^{2}
A drinks machine offers three selections  Tea, Coffee or Random but the machine has been wired up wrongly so that each button does not give what it claims. If each drink costs 50p, how much minimum money do you have to put into the machine to work out which button gives which selection?
For the first drink press, the button labelled Random. We know this is NOT the Random button so if it dispenses Tea it is the Tea button likewise if it dispenses Coffee it is the Coffee button. From here it is simple to work out the other buttons knowing that they can't be what they say they are.
There are only two possible combinations each only requires 50p, or one goes to find it.
Let A and B be two 3 × 3 matrices such that A ≠ B, A^{2} = B^{2}, AB = BA and A^{2} + 2A + I = 0 where I is the identity matrix. Let T denote the determinant of any matrix T. Then
(A + B) (A  B) = A^{2}  AB + BA  B^{2}
Given that, AB = BA
⇒ (A + B) (A  B) = A^{2}  B^{2}
Given that A^{2} = B^{2} ⇒ A^{2}  B^{2} = 0
⇒ (A + B) (A  B) = 0
Given that, A ≠ B
⇒ A + B = 0
⇒ A + B = 0
Given that,
A^{2} + 2A + I = 0
Multiply both sides with A^{1}
⇒ (A^{2} + 2A + I) A^{1} = 0
⇒ A + 2I + A^{1} = 0
⇒ A^{1} = A  2I
A^{1} is exists hence A ≠ 0
The positive value of A for which the equation f(x,y) = x^{2} + y^{2 }+ Axy + 5x + 18, does not yields optimum values (i.e. no conclusion can be drawn on the nature of f(x,y) ) and required further investigation ______.
f(x, y) = x^{2} + y^{2} + Axy + 5x + 18
For the case to require further investigation
(r × t) – s^{2} should be equal to zero
∴ (r × t) – s^{2} = 0
⇒ (2 × 2) – A^{2} =0
⇒ A = 2
Customers arrive in a certain store according to a possion process at rate 4 per hour. Given that the store opens at 9.00 am what is the probability that exactly 1 customer has arrived by 9:30
For ½ h
The differential equation 4yy’  12x = 0, satisfying condition y (1) = 3. Then the point (5, 3) will lie:
Since given differential equation is
4yy’  12x = 0
4y dy = + 12x dx
On integrating both the sides
2y^{2} = 6x^{2} + c
2y^{2}  6x^{2} = c
Given y (1) = 3
Thus, 2(3)^{2}  6(1)^{2} = c
18  6 = C
C = 12
Thus, 2y^{2}  6x^{2 }= 12
Now, f (x, y) = 2y^{2}  6x^{2}  12
At point (5, 3)
= 50  54  12
= 16
Thus, the point will lie inside the solution curve.
For a certain distributed data, the middle observation is 10 and mean observation is 8. The maximum occurring data is____
Mode = 3 medium – 2 mean
= 3 (10) – 2(8)
= 30 – 16
=14
As per IS 456:2000, the diagonal tension failure in RCC beams depends upon:
According to Table No. 19 of IS 456:2000,
The allowable shear stress (design shear strength) depends on both percentage of longitudinal steel provided and grade of concrete used in the beam.
The technique utilized for the disposal of biomedical waste, in which it is in contact with steam under controlled pressure and temperature condition with the end goal to complete sterilization is
Autoclaving:  The method used for the disposal of biomedical waste is , in which it is brought in closed contact with steam under controlled pressure and temperature condition in order to carry out disinfection.
Pulverization and Shredding: Basically this is not a method of disposal but in these method heavier solids are broken into the lighter one either by cutting action or by grinding action.
Pulverization refers to the action of crushing and grinding, whereas shredding refers to the action of cutting and tearing.
Pyrolysis: is a method of disposal of the refuse in which burning of refuse is generally done at high temperature in the almost absence of oxygen.
As per Indian Road Congress (IRC) the stripping value of aggregate should not exceed X% for the use in bituminous surface dressing, penetration macadam, bituminous macadam, and carpet construction when aggregate coated with bitumen is immersed in water bath at Y^{o}C for Z hours. The value of X, Y and Z will, respectively, be
The stripping value of aggregates is determined as the ratio of the uncovered area observed visually to the total area of aggregates, expressed as a percentage. If the road aggregates are dry and free from dust then the bitumen adheres well to all normal types of aggregate. In the absence of water there is practically no problem of adhesion with bituminous construction. Adhesion problem occurs when the aggregate is wet and cold. This problem can be dealt with by removing moisture from the aggregate by drying and increasing the mixing temperature. Further, the presence of water causes stripping of binder from the coated aggregates. This problems occur when bitumen mixture is permeable to water.
Static immersion test is one specified by IRC and is quite simple. As per Indian Road Congress (IRC) the stripping value of aggregate should not exceed 5% for the use in bituminous surface dressing, penetration macadam, bituminous macadam, and carpet construction when aggregate coated with bitumen is immersed in water bath at 48^{o}C for 24 hours.
The laboratory test results of the soil sample are given below:
Liquid limit = 37%
Plastic limit = 22%
% passing through 75 μ sieve = 26%
% retained over 4.75 mm IS sieve = 65%
% retained over 0.075 mm but passing through 4.75 mm = 26%
As per IS 1498 – 1970, the soil is classified as:
% passing through 75 μ sieve = 26%
% retained over 75 μ sieve = 100 – 26% = 74 % > 50%
Hence the soil is coarse grained soil.
As, larger proportion of soil is retained over 4.75 mm IS sieve, so the soil is termed as Gravel.
Now, % fineness = 26% > 12%
So, use I_{p} to further classify the soil
I_{P} = w_{L}  w_{P} = 37 – 22 = 15%
I_{P} = 15% > 7%
So, soil is termed as clayey Gravel
What will be range of plasticity index for the fine grained soil below A  line having intermediate compressibility in the plasticity chart.
As liquid limit (w_{2}) for lined grained soils having intermediate compressibility is:
35% ≤ W_{L} ≤ 50%
I_{p} = 0.73 [W_{L}  20]
For W_{L} = 35%
I_{p} = 10.95
For W_{L} = 50%
I_{p} = 21.9
Range 10.95 ≤ I_{p} ≤ 21.9
A tank full of water is shown in figure below,
Assume width of tank to be 1.5m.
Identify the correct statement made:
Weight of water in the tank = ρ_{w} g × volume of tank
Volume of tank = 3 × 4 × 1.5 + 0.8 × 1.2 × 1.5
Volume of tank = 19.44m^{3}
Weight of water in the tank = 1000 × 9.81 × 19.44
W = 190.706 kN
Hydrostatic Force at the bottom of the tank
P_{bottom }= ρ_{w} gh A
P_{bottom }= 1000 × 9.81 × 5.2 × 0.8 × 1.5
P_{bottom }= 61.21 kN
Hydrostatic Paradox occurs when the total weight of water in the tank is less than the total pressure at the bottom of the tank.
But in the case, total weight of water in the tank is more than the total pressure at bottom of the tank, so there is no Hydrostatic Paradox.
Consider the following Statement:
P) Creep or creep length is the length of the path travelled by the percolating water.
Q) The head loss per unit length of creep is constant throughout the percolating passage.
R) There is no difference between horizontal and vertical creep.
Which of the following is not an assumption of Bligh Creep theory?
Assumptions involved in Bligh Creep Theory:
1) The water percolating through subsurface condition creeps along the base profile of the impervious floor. Creep or creep length is the length of the path travelled by the percolating water.
2) The head loss per unit length of creep is constant throughout the percolating passage. Leading to the conclusion that the loss of head is proportional to the length of creep.
3) There is no difference between horizontal and vertical creep.
The static indeterminacy of the plane frame shown in figure below is?
Concept:
Static indeterminacy of the plane frame.
D_{s} = R  r_{e}  r_{r}
R = No. of unknown reaction components
r_{e} = No. of equilibrium conditions available
r_{r} = Released reaction = m – 1
m = no. of member meeting at Hinge Joint [internal]
R = 13
r_{e} = 3 [∑F_{H} = 0, ∑F_{V} = 0, ∑F_{Z} = 0]
r_{r} = m – 1
r_{r} = 3 – 1 = 2
D_{s} = 13 – 3 – 2
D_{s} = 8
A three hinge parabolic arch is subjected to uniformly distributed load as shown in figure below
If there is an increase in temperature by 10°C, then the final change in rise of crown will be?
Take coefficient of thermal expansion = 10 × 10^{6} per ° C.
Let l = length of the parabolic three hinge arch
Let T be increase by T°C.
Then, rise of crown will increased by
l = 12 m
h = 3 m
∝ = 10 × 10^{6} per °C
T = 10°C
Δh = 1.5 × 10^{3} m
Δh = 1.5 mm
Final rise of crown = 3000 + 1.5
= 3001.5 mm
Initial rise of crown = 3000 mm
Change = 3001.5 – 3000 = 1.5 mm.
Following statements are made with regard to temporary adjustment of a theodolite:
P. Levelling is done before centring of a theodolite.
Q. centring is done before levelling of a theodolite.
R. After centring and levelling, the eye piece is to be focused to make the cross hairs distinct and clear.
S. After focusing the eye piece, the objective is to be focused to bring the image of the object in the plane of crosshairs.
The adjustments to be made at every setting of the instrument are called temporary adjustments.
Following are the five temporary adjustments are required:
1. Setting up: Tripod stand is set on the ground firmly so that its top is at a convenient height.
2. Centring: It is done to place the vertical axis exactly over the station mark
3. Levelling: The levelling is done to make the vertical axis of the instrument truly vertical or to make horizontal plate truly horizontal
4. Focusing the eye piece: The eye piece is focused to make the cross hairs distinct and clear
5. Focusing the objective: This is done to bring the image of the object in plane of the cross hairs
The refraction correction required for taking a staff reading held at 3 km from the instrument is ______ cms.
Refraction correction is given by
C_{R} = +0.0112 D^{2}
Where,
D = distance in (kms)
C_{R} = correction in (metres)
For D = 3 km
C_{R} = 0.1008 m = 10.08 cm
The coefficient of restitution between a snooker ball and side cushion is 0.4. If the ball hits the cushion and then rebounds at an angle of 90° to the original direction then angle made by the ball with side cushion before and after impact will respectively be
The coefficient of restitution,
After impact
From equation (1) and (2)
v = 0.4 μ sin θ and
⇒ tan^{2}θ =2.5
⇒ θ = 57.7°
∴ (90 – θ) = 32.3°
The float that can be used by an activity without affecting successor activities is
Free Float is that portion of total float that can be used by an activity without affecting successor activities i.e. the successor activity can start at their Earliest Start Time (EST). But the preceding activities are affected by free float.
Free Float = Total Float  Head event Slack
As per Indian road congress, the suggested expression for the design of overlay thickness equivalent to granular material of WBM layer is
Where, h_{0} = thickness of granular or WBM overlay in mm
D_{c} = characteristic deflection.
D_{a} = allowable deflection
The value of D_{a} if the projected design traffic is 450 to 1500 is
D_{a} = Allowable deflection which depends upon the pavement type and the desired design life.
Projected design traffic
Where, P = Present traffic.
R = % annual increase in traffic.
A flow fluid is represented as u = + 2xy, v = x^{2} + 2, w = x^{2} z. What will be the value of ycomponent of vorticity (in radians per unit time) at point (1,  2, 0)_____ (up to two decimal places)
Concept:
Vorticity in ydirection
Ω_{y} = 2 ω_{y}
Calculation:
u = 2xy
at (1, 2, 0)
Ω_{y} = 2 × (0.5) = 1.0
A chlorine dose of 0.8 mg/L is added to treat 12 MLD water and to get 0.4 mg/L of residual chlorine. The bleaching powder requirement per month if it contains 40% of chlorine (in tonnes, up to two decimal place) will be_________.
Bleaching powder dose
= 2mg/L
Bleaching powder requirement (kg/day)
= Q × Bleaching powder dose
= 12× 10^{6 }L/day × 2 × 10^{6} kg/L
= 24 kg/day.
Monthly Requirement = 24 × 30
= 720 kg/month
= 0.720 tonne/month
A beam of triangular crosssection is shown in figure below, ratio of the shear stress at neutral axis to the maximum shear stress in the crosssection will be_____(up to two decimel places).
Maximum shear stress (τ_{max}) in the triangular crosssection is given by:
Where,
τ_{avg} = Average shear stress in the crosssection
Shear stress at neutral axis of the crosssection is given by:
A point load of certain value is applied on the surface of the thick layer of soil. At a 3 m depth at a radial distance of 1.5 m from the point load, what will be the ratio of Boussinesq’s influence factor to the westerguards influence factor.
Concept:
Boussinesq’s influence factor (k_{B}):
Wester guard’s influence factor (k_{W}):
Calculation:
z = 3, r = 1.5 m
A fixed beam of circular crosssection is subjected to a central point load W as shown in figure below. The plastic hinge length (in meters) will be ______.
(∵ shape factor = 2)
x = 0.625 m
∴ L_{P} = 1.25 – x
⇒ L_{P} = 1.25 – 0.625
⇒ L_{P} = 0.625 m
What will be the moment of resistance of the rectangular crosssection of size 250 mm and 550 mm (effective) using limit state method in ____KNm. (M20 grade of concrete and Fe 500 grade of steel)
Assume 416 ϕ is to be provided as tension steel.
Step 1: Calculating the ultimate depth of neutral axis (x_{u})
Total compression = Total tension
0.36 f_{ck} x_{u} B = 0.87 f_{y} A_{st}
0.36 × 20 x_{u} × 250 = 0.87 × 500 × 804.25
x_{u} = 194.36 mm
Step 2: Calculating the limiting depth of neutral axis (x_{u(lim)})
For Fe500,
x_{u(lim)} = 0.46 × 550 = 253 mm.
As x_{u} is less than x_{ulim }, section is under reinforced.
Hence,
Moment of resistance = 0.36 f_{ck} x_{u} B (d – 0.42 x_{u})
Moment of resistance = 0.36 × 20 × 194.36 × 250 × (550 – 0.42 × 194.36)
Moment of resistance = 163.86 kNm.
Let A = (a_{ij}) be a 10 × 10 matrix such that a_{ij} = 1 for i ≠ j and a_{ij} = α + 1, where α > 0. Let λ and μ be the largest and the smallest eigenvalues of A, respectively. If λ + μ = 24, then α equals ______.
A = (a_{ij}) be a 10 × 10 matrix
a_{ij} = 1 for i ≠ j
a_{ij} = α + 1
Let α + 1 = a
Now, the matrix can be formed as
We know that, eigen values are the roots of A  λI = 0
A – λI = 0
⇒ C_{1} → C_{1} + C_{2} + C_{3} + … + C_{10}
R_{2} → R_{2}  R_{1}
R_{3} → R_{3}  R_{1}
⋮
R_{10} → R_{10}  R_{1}
C_{2} → C_{2}  C_{1}
C_{3} → C_{3}  C_{1}
⋮
C_{10} → C_{10}  C_{1}
Where I_{10} is the identify matrix of order 10.
We know that I = 1.
⇒ λ = a  1, λ = a + 9
⇒ λ = α + 1  1, λ = α + 1 + 9
⇒ λ = α, λ = α + 10.
The possible Eigen values are: α, α + 10
Given that, smallest Eigen value = λ = α
Largest Eigen value = μ = ∝ + 10.
Given that, λ + μ = 24
⇒ α + α + 10 = 24
⇒ α = 7
Find the coefficient of x^{3} in the expansion of about x = 1
∴ coefficient of
The numerical value of the integral
= cos 1 + cos 0°
= (1 – cos 1)
The solutions of the differential equation
a. y(x) = 1
b.
c.
d.
y^{1/3} dy = (sin 2x) dx
Integrating on both the sides:
Now, Given y (0) = 0
Or
By using Simpson’s rule find by taking a width of 0.1
Divide the internal (0, 0.6) in to six parts each of width
h = 0.1
By using Simpson’s rule, we have
+4(0.99+0.9139+0.7788)+2(0.9608+0.8521)]
[1.6977+10.7308+3.6258]
[16.0543]=0.5351
In a specific energy curve, discharge is increased from Q_{1} to Q_{2}, keeping specific energy constant. If Y_{A} and Y_{B} are alternate depth corresponding to initial discharge Q_{1}. Y_{C} and Y_{D}, are alternate depth corresponding to discharge Q_{2}. Considering Y_{A} as supercritical depth and Y_{C} is subcritical depth.
Which one of the following is true?
As discharge increases keeping specific energy constant, the specific energy curve will shift right and upward.
Hence, Y_{B} > Y_{C} > Y_{D} > Y_{A}
The different ionic concentration for a water sample is being shown below in a tabulated manner
The total alkalinity and noncarbonate hardness (in mg/l as CaCO_{3}) respectively will be
Total alkalinity (in mg/l as CaCO_{3})
(gm × milliequivalent = 10^{3} mg × 10^{}3 equivalent = mgequivalent)
= 75 + 125
= 200 mg/L as CaCo_{3}
Total hardness (in mg/L as CaCO_{3})
= 100 + 200
= 300 mg/L as CaCo_{3}
Since carbonate hardness is minimum of (total alkalinity or total hardness)
∴ Carbonate hardness = Total Alkalinity = 200 mg/L as CaCo_{3}
Noncarbonate hardness = Total hardness – carbonate hardness = 300 – 200
= 100 mg/l as CaCo_{3}
A prestressed beam of size 250 × 400 mm deep is prestressed by 14 tendons of 6 mm diameter. The cable is located at 150 mm from bottom of the beam. If the initial prestress in the beam is 1250 N/mm^{2}. The loss of prestress due to elastic deformation and creep of concrete respectively will be?
Assume:
M40 Grade concrete
creep coefficient = 1.6
Modulus of elasticity = 2 × 10^{5} mPa.
Neglect the effect due to dead load.
Prestressing force (P) =
Loss due to elastic shortening = mf_{c}
f_{c} = compressive stress at the level of steel
m = modular ration =
At the location of steel tendon, y = e
f_{c} = 4.948 + 0.9277
f_{c} = 5.876 mPa
loss due to elastic shortening = 6.32 × 5.876
= 37.13 mPa.
Loss due to creep of concrete = mϕf_{c}
ϕ = creep coefficient = 1.6
loss due to creep of concrete = 6.32 × 1.6 × 5.876
= 59.42 mPa.
A 30 m chain was found to be 15 cm too short after chaining a distance of 2000 m. It was found to be 30 cm too short at the end of the days work after chaining a distance of 4000 m. Find the true distance if the chain was corrected before the commencement of the day’s work?
For the first 200 m,
Average error (e) =
Incorrect length of chain [L’] = 30 – 0.075
L’ = 29.925 m
Measured length from chain (l’) = 2000 m
True/designated length of chain [L] = 30 m
True length of line (l_{1}):
l_{1} × L = l’ × L’
l_{1} × 30 = 2000 × 29.925
l_{1} = 1995 m
for the next 2000 m,
Average error (e) =
Incorrect length of the chain (L’) = 30 – 0.225
L’ = 29.775 m
Measured length of chain (l’) = 2000 m
True length of the line = l
L’ × l’ = L × l_{2}
29.775 × 2000 = 30 × l
l_{2} = 1985 m
True distance will be = l_{1} + l_{2}
l = 1995 + 1985
l = 3980 m
Consider the following Statement:
P) Tie bars are installed in warping joints in cement concrete pavement.
Q) Radius of the relative stiffness of cement concrete pavement depends on wheel loads.
R) Base course is used in rigid pavements for prevention for pumping
T) Dowels bars are provided at expansion joints and sometimes also at contraction joints in cement concrete pavement.
For the above statement, the correct option is
1) Radius of relative stiffness is given by:
where,
l = radius of relative stiffness (cm)
h = slab thickness (cm)
μ = Poisson’s ratio
k = Modulus of subgrade (kg/cm3)
E = Modulus of elasticity of Cement Concrete (kg/cm2)
Hence, Radius of relative stiffness depends upon
Modulus of elasticity of Cement Concrete
Modulus of subgrade
Poisson’s ratio
2) In Rigid Pavement base course are used for
preventing pumping and
protecting the subgrade from frost action
3) Tie bars are installed in longitudinal joints in cement concrete pavement.
4) Dowels bars are provided at expansion joints and sometimes also at contraction joints in cement concrete pavement.
A 10 m wide rectangular channel has a bed slope of 3.33 × 10^{3} and the Manning's coefficient 0.015. Water in the channel is flowing at a uniform depth of 3 m. At a particular section gradually varied flow (GVF) is observed and the flow depth is measured as 3.1 m. The GVF profile at that section is classified as
Width of the channel, B = 10 m
Slope of the channel, S = 3.33 × 10^{3}
Manning’s coefficient, n = 0.15
Normal depth, y_{n} = 3 m
The discharge of the channel, Q
Where,
R = Hydraulic mean radius = A/P
A = Wetted area of the cross section
S = Bed slope
P = Wetted perimeter of the cross section
∴ A = B × y = 10 × 3 = 30 m^{2}
P = B + 2y = 10 + 2 × 3 = 16 m
Q = 175.58 m^{3}/s
Specific discharge or discharge per unit width,
∴ Critical depth, for the rectangular channel
y_{c} = 3.16 m
Hence, y_{c} > y_{n}
and y_{n} < y < y_{c} leads to the gradually varied profile of S_{2}.
A primary sedimentation tank (PST) is to treat water for 0.1 million population living in a town with Surface Overflow Rate (SOR) of 50000 Litre/m^{2}/d. The diameter (in μm) of the spherical particle which will have 95 percent theoretical removal efficiency in this tank is _______________. Assume the Stokes’s Law is valid for the settling velocity of the particles in water. It is given that:
Given:
Surface Overflow Rate = 50000 L/m^{2}/day
Density of water = 997 kg/m^{3};
The specific gravity of particle = 2.72
Dynamic viscosity of water= 9.97 x 10^{4} m^{2}/s
We know that settling velocity can be written as
Percentage removal =
= 0.95 × Overflow rate
= 2.4254×10^{−5}m
= 24.254 μm
The following observations were taken with a transit theodolite
The reduced level of the staff station P will be ______ m. (up to three decimal places)
In ΔABC
In ΔABD
D = 82.00 m
V = D tan 5°30’
V = 82 tan 5°30’
V = 7.895 m
R.L_{P} = R.L of instrument axis +V S,
R.L_{P} = 286.500 + 7.895 – 0.850
R.L_{P} = 293.545 m
The overtaking sight distance required on a highway is 250 m. The required clearance of obstruction (in meters, up to two decimal place) from centre line of a circular curve of radius 350 m and length 180 m is ________. (Assume two lane highway with d = 1.9 m.)
Concept:
Since length of circular curve (L_{c}) < sight distance (SSD, ISD, OSD)
Therefore, setback distance, for two lane road,
From centre line of the road or centre line of circular curve.
Where R = radius of the curve
S = sight distance
L_{c} = length of the curve
∝ = angle between segment of the curve.
Calculation:
Given:
OSD = 250 m, d = 1.9 m
R = 350 m
L_{c} = 180 m.
Also,
∴ setback distance or the required clearance of obstruction from the centre line of road on circular curve,
M = 350−(350−1.9)×cos(14.814)
M = 350 – 336.53 + 8.95
M = 22.42 m
A driver of a vehicle approaching a signalized intersection at a speed of 45 kmph applied breaks on seeing the signal changing from green to amber and the vehicle was brought to stop on the prescribed stop line during the amber time of 5 seconds. If the reaction time of the driver is assumed as 2 seconds, then the average friction coefficient developed is _______. (up to 3 decimal place)
(Assume, length of vehicle = 6 m and width of road = 7 m)
Concept:
Interval design (most critical time)
There are two intervals, namely the change interval and clearance interval.
The purpose of these intervals is to warn a driver approaching the intersection during the end of a green time about the coming of a red signal. At this moment driver has two choices either to stop before the intersection or to clear the intersection. Amber time is provided such that at least one of these opportunities is available to the driver irrespective of driver position.
If the minimum amber time provided is T_{min}.
Then the distance covered by the driver moving with design speed is
= V × T_{min}
Also,
V × T_{min} = d_{c} + W + L
d_{c} = V × T_{min} – W – L
d_{c} = < x < d_{s} (SSD)
x is the zone of the dilemma
Where V = speed of the vehicle
SSD = stopping sight distance
W = width of the road
L = length of the vehicle
To avoid this dilemma
Calculation:
Given:
V = 45 kmph = 12.5 m/s
Amber time = 5 sec.
Reaction time = 2 sec.
⇒ f = 0.325
The volume of raw sludge fed per day and the volume of digested sludge withdrawn per day of a standard rate sludge digestor is 100 m^{3} and 50 m^{3}. If the mean cell residence time of the digester is one month than the diameter of the tank will be _____ m. (Assuming depth of the tank to be 8 m)
Concept:
Standard rate sludge digester
As per the government of India manual volume of the tank is computed empirically as
Where V_{1} = volume of raw sludge fed per day.
V_{2} = Volume of digested sludge withdrawn per day.
t_{D} = mean cell residence time of the tank.
If monsoon storage is also considered
Where, T = monsoon period.
Calculation:
Given:
volume of raw sludge = 100 m^{3}/day
volume of digested sludge = 50 m^{3}/day
mean cell residence time = 30 days.
The height of the tank = 8m.
Therefore the volume of the tank,
= 2000 m^{3}.
The surface area of the tank =
SA = 250 m^{2}
Now,
⇒ D = 17.84 m
Hence, diameter of the tank = 17.84 m.
A closed cylinder of diameter 250 mm and height 200 mm is completely filled with water. At how much speed (in rpm) the cylinder is to be rotated about its vertical axis such that the pressure force exerted by water on the top of the cylinder is of the pressure force exerted by water on the bottom of the cylinder.
Concept:
Total pressure force on the top of the cylinder =
Total pressure force on the bottom of the cylinder
Where,
ρ = density of water
R = radius of cylinder
Calculation:
P_{1} = 2.1027 × 10^{3} N^{2}
P_{2} = ρgH (π R^{2}) + P_{1}
P_{2} = 1000 × 9.81 × 0.2 × π × (0.125)^{2} + 2.1027 × 10^{3} N^{2}
N = 370.68 rpm
A rough pipe of 0.75 m diameter, 450 m length and average height of roughness 0.35 mm carries water with a velocity of 4.5 m/s. If the coefficient if friction for the pipe is given as:
Where, r = radius of pipe
k = average height of roughness
g = 9.81 m/s^{2}
ϑ = kinematic viscosity of water = 0.95 × 10^{6} m^{2}/s
Then, the power required to maintain this flow will be ____kw.
Concept:
Power required to maintain the above flow is:
Q = Rate of flow (m^{3}/s)
ρ = density of flowing fluid (kg/m^{3})
h_{f} = head loss due to friction (m)
D = Diameter of pipe
V = velocity of the flowing fluid
f = friction factor = 4f^{’}
Calculation:
Using the given relation,
f^{’} = 0.07631
f = 0.3053
Check for the type of flow
Flow is turbulent
h_{f} = 189.06 m
Q = A × V
Q = 1.989 m^{3}/sec
P = 3688.95 kW
A project is composed of eight activities, the time estimate for which are given below.
If a 30week deadline is imposed, what is the probability that the project will be finished within the time limit?
Concept:
Mean expected completion time of an activity
And variance is
Where,
t_{0} = optimistic time
t_{m} = most likely time
t_{p} = pessimistic time
The standard deviation of the project will be
along the critical path
along critical path
Earliest expected completion time of the project as a whole
T_{E} = t_{1} + t_{2} + … + t_{n} along the critical path
The probability factor,
Where, T_{S} = Scheduled completion time of the project
T_{E} = Earliest expected completion time of project
σ = Standard deviation of project
Calculation:
Network diagram
The critical path is A – D – G – H
σ_{critical path }=
= 2.45 weeks
Earliest expected completion time of the project is 29 weeks
If the scheduled completion time is 30 weeks then the probability of completion of the project
Z = 0.408
∴ Probability of completion (interpolation)
= 0.6591
The infiltration capacity of a soil follows the Horton’s exponential model, f = 13 + 29 e^{(3/2)t }where f is in mm/hr and t is the time in hour. The total depth of infiltration and average infiltration rate during the first 40 min of a storm respectively is
Concept:
Horton’s equations:
Horton expressed the decay of infiltration capacity with time as an exponential decay
f_{t} = f_{c} + (f_{0}  f_{c}) e^{kt} for 0 < t < t_{c}
where,
f_{t} = Infiltration capacity at any time t from the start of rainfall
f_{0} = Initial infiltration capacity at t = 0
f_{c} = Final infiltration capacity occurring at t = t_{c}
k = Horton’s decay constant which depends upon soil characteristic and vegetation cover
The above Horton’s equation is applicable only when the rainfall intensity is greater than or equal to f_{t}.
Calculation:
Infiltration ratio f_{t} = 13 + 29e^{(3/2)t}
The total depth of infiltration for the first 40 minutes
F_{40} = 20.888 mm
The average infiltration rate for the first 40 min will be
f_{avg} = 31.33 mm/hr
The following composition of a solid waste is given below:
The difference in the energy content on ashfree dry bases and on a dry basis (in kJ/kg) is ______.
(Consider moisture content of waste as 25% and ash content as 6%.)
The energy content of the waste (as a discarded basis) is given by:
The energy content on dry waste basis is given by:
= 15166.67 kJ/kg
The energy content on ashfree dry basis is given by
Where,
mc = moisture content
ac = ash content
The difference between energy content on the ashfree dry basis and on a dry waste basis
= (16485.5 – 15166.67) kJ/kg
= 1318.83 kJ/kg
Consider a simply supported beam as shown in figure below. A moment of 20 kNm applied at C will produce 15 mm deflection at point D, then at how much load (kN) should be applied at D, So that the rotation at C will be limited to 0.00345 radian.
M_{C} = 20 kNm
δ_{D} = 15 mm = 0.015 m
θ_{C} = 0.0345 radians
P_{D} = unknown
Using Maxwell’s law of reciprocal deflection
M_{C} × θ_{C} = P_{D} × δ_{D}
20 × 0.0345 = P_{D} × 0.015
P_{D} = 46 kN
A 200 mm wide plate is to be jointed to a column flange as shown in the figure. If the ultimate strength for the weld metal is 400 MPa, the size of the weld (in mm, up to two decimal place) is _________. (It is given that moment of inertia about the perpendicular axis to the plane around C.G is 10^{7} times the throat thickness of the weld)
Concept:
The effect of eccentricity load at the center of gravity of weld group will be due to direct load and due to a twisting moment (P × e)
Due to direct load P, direct shear stress f_{1} is developed at point A and due to twisting, torsional shear stress f_{2} is developed at A.
At point A
Where t = throat thickness of the weld
Also,
∴ f_{rA} = Resultant shear stress
f_{rA} =
⇒ t ≥ 3.899 mm
∴ size of weld =
= 5.57 mm
= 6 mm
A solid circular shaft is subjected to Bending moment and Torsional moment of 100 kNm and 50 kNm respectively as shown in figure below.
The maximum shear stress developed at point ‘P’ of the shaft will be _________ MPa.
Maximum shear stress in given by:
At Point P:
σ_{1} = 17.81 MPa
σ_{2} = 319.61 MPa
For the subsoil conditions shown below, determine the change in effective stress at 7 m below the ground level when the water level is permanently lowered for long term conditions _____ kN/m^{2}.
Use γ_{W} = 10 kN/m^{3}
When water table is not lowered:
σ̅ at 7 m below ground level
σ̅ = 18 × 4 + 17.5 × 2 + 17 × 1 + 50  10 × 1
σ̅ = 164 kN/m^{2}
when water table is permanently lowered:
clay will become in bulk state
γ = 17.5 kN/m^{3}
σ̅ = 18 × 4 + 17.5 × 3 + 50
σ̅ = 174.5 kN/m^{2}
change = 174.5  164 = 10.5 kN/m
A shaft of 350 mm diameter is subjected to the angle of twist of 1.8° in a length of 9 m, if the permissible shearing stress is not to exceed 50 MPa, the maximum torque that can be applied safely to the shaft is _____kNm Take G = 90 GPa
Concept:
Using, Torsion formula
Where, T = Applied Torsion (kNm)
I_{P} = Polar moment of inertia
R = Radius of the shaft
G = Shear modulus
θ = permissible angle of twist
L = Length of the shaft
ττ = permissible shear stress in the shaft
Calculation:
θ = 1.8° = 0.03141 radian
L = 9 m
τ_{developed} _{ }= 54.9675 N/mm^{2} > 50 N/mm^{2}
Hence, limit the τmaxτmax to 50 N/mm^{2}
So,
T = 420.857 kNm
A simply supported beam of length 6m is loaded with uniformly distributed load of 15 KN/m. The Beam is made of two materials which rests on one over ther such that ratio of modulus of Elasticity of top beam is twice of modulus of elasticity of bottom beam as shown in figure below. The difference between the maximum compressive stress in the bottom beam and top beam will be ________ MPa.
Let M_{1} = Maximum Bending moment resisted by top section
M_{2} = Maximum Bending moment resisted by bottom section.
M = M_{1} + M_{2}
= 67.5Kn−m
As the curvature of both the beam is same
M_{1} = 18.34 kNm
M_{2} = M – M_{1} = 67.5 – 18.34 = 49.16 kNm
z_{2} = 510.4 × 10^{4} mm^{3}
(σ_{2}) – (σ_{1}) = 9.63 – 3.59
= 6.04 MPa
A 12 long reinforced concrete pile of diameter 450 mm was driven by drop hammer having total mass of 24000 kg and having an effective fall of 0.9 cm. The average penetration for the last five blows was recorded as 0.6 cm per blow. If the total elastic compression is 2.3 cm, then the safe load on the pile will be _____ kN.
Assume coefficient of restitution and factor of safety to be 0.30 and 2 respectively. Use Hiley’s formula.
Take γ_{concrete} = 25 kN/m^{3}. Efficiency of hammer (η_{n}) = 1.0.
Concept:
By Hiley’s formula
Ultimate load carrying capacity of the pile is given by:
Where,
η_{n }= efficiency of hammer = 1.0 (drop hammer)
η_{b} = efficiency of hammer blow =
e = coefficient of restitution
p = weight of pile
w = weight of hammer
c = constant (elastic compression between pile and soil)
s = set (penetration of pile per blow of hammer)
h = height of fall
calculation:
W = 24000 × 9.81 = 235.44 kN
ep = 0.30 × 47.71 = 14.31 kN
w > ep
η_{h} = 1.0
s = 0.6 cm blow
c = 2.3 cm
Q_{up} = 102.44 kN
A soil sample was consolidated under a cell pressure of 8 kg/cm^{2}, pore water pressure of 4.5 kg/cm^{2}. With no drainage permitted the cell pressure was raised to 10 kg/cm^{2} results in increased pore water pressure to 5.75 kg/cm^{2}. The axial load was then increased to give a deviator stress of 6.25 kg/cm^{2} (with the cell pressure at 10 kg/cm^{2}), pore pressure is now noted to be 7.15 kg/cm^{2}. The ratio of pore pressure coefficients B to A is ______.
Concept:
Pure pressure parameter (B)
σ_{3 }= cell pressure
U_{C }= pore pressure
Δσ_{3} = change in cell pressure in cell pressure stage
ΔU_{C} = change in pore pressure in cell pressure stage
Pore pressure parameter (A)
ΔU_{d} = change in pore in deviator state
Δσ_{d} = change in deviator stress
Calculation:
ΔU_{c} = 5.75  4.5 = 1.25 kg/cm^{2}
Δσ_{3} = 10  8 = 2 kg/cm^{2}
ΔU_{d} = 7.15  5.75 = 1.4 kg/cm^{2}
Δσ_{d} = 6.25 kg/cm^{2}
It is given that:
Modulus of elasticity of cement concrete = 210 × 10^{3} kg/cm^{2}
Poisson’s ratio for concrete = 0.20
Modulus of subgrade reaction, K = 5 kg/m^{3}
Radius of wheel load distribution = 15 cm.
Slab thickness = 20 cm
The distance from the apex of slab corner to section of maximum stress along the corner bisector is,
Concept:
Wastergaard defines this term as the radius of relative stiffness as
Where l = radius of relative stiffness, cm
E = modulus of elasticity of cement concrete, kg/cm^{2}
μ = Poisson’s ratio
h = slab thickness
k = subgrade modulus or modulus of subgrade reaction, kg/cm^{3}.
The radius of relative stiffness is used to calculate the distance from the apex of slab corner to the section of maximum stress along the corner bisector (X).
Where a = radius of wheel load distribution
L = radius of relative stiffness.
Calculation:
= 73.5 cm
Now, the distance, X
X = 85.66 cm
X = 85.66 cm
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