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Which of the following curves represents the Y= In (|e| (sin sin (|x|)|) for |x| < 2π?
= In (|(e| (sin sin(|x|)|)=|sin sin(|x|)|
Now for drawing the graph follow these steps
1) Draw sinx graph on paper
2) Take reflection about x-axis, you will get graph of sin(|x|)
3) Now, take a reflection again but this time
about y-axis, you will get the graph of sin(|x|)
⇒ Option c is correct.
Direction: In the following question, a sentence is given with blanks to be filled in with appropriate word(s). Four alternatives are suggested for the question. Choose the correct alternative out of the four.
The _______ of the moment led Lin to post on Craigslist, describing the guy's _____ shirt and partially dyed blond hair.
Option B: Profanity is socially offensive language, which may also be called bad or offensive language. ‘Profanity of a moment’ makes no sense. Therefore, this option can’t be used.
Option C: Penchant is a strong or habitual liking for something or tendency to do something. ‘Penchant of a moment’ makes no sense. Therefore, this option can’t be used.
Option D: Predilection is a preference or special liking for something; a bias in favour of something. ‘Slurry’ is a thin sloppy mud or cement or, in extended use, any fluid mixture of a pulverized solid with a liquid usually water, often used as a convenient way of handling solids in bulk. Thus, this option makes no sense in the context of the sentence.
Therefore, option a is the apt answer.
Analogy = Situation: Appearance. None of the options accept A showcase any situation. Hence A is the right one. The word moment is present in the question, hence any word which suits situation would be the right option.
A number 18567332145x is divisible by 8. What can be the minimum value of x?
So 45x is divisible by 8
The number is 45 which is near to 48, the number which gets divided by 6 (8 × 6 = 48)
Thus x = 6 is the correct answer
The simple interest on a certain sum for 2 years at 9% per annum is Rs. 225 less than the compound interest on the same sum for 2 years at 10% per annum. The sum is:
% compound interest for 2 year = 10+10+(10×10)/100 = 21%
difference in interest = 225 Rs.
So (21% - 18%) = 3% of sum = 225 Rs.
So Sum = 225×100/3 = 7500 Rs.
Which of the following is the MOST OPPOSITE in meaning to Obstreperous?
Sophisticated= Cultured or one who speaks in a respectable manner
Blusterous = loud and aggressive.
Ribald = indecent or bawdy.
Imperative= essential
Consider a function f(x)=1-|x| on -1≤x≤1.The value of x at which the function attains a maximum, and the maximum value of function are:((mathematics:maxima,minima)
An article is sold with a certain profit percentage such that selling the same at one-third price, there will be a loss of 60%. Find the certain profit percentage.
Let the selling price of the article =Rs.x
If the article is sold in 1/3 price, then selling price =Rs.x/3 and loss =60%
So, the cost price of the article = Rs.( x/3 ) × (100/40) = Rs. (5x/6) Then, profit earned by selling the article at Rs. x = Rs. x – (5x/6) = Rs.x/6
∴ The required profit percentage = [(x/6)/ (5x/6) × 100]% = 20%.
Alternatively:-
Say CP = 100 Rs. say profit X%
SP = 100 + X
New SP = (100 + X)/3 = 100 - 60
X = 20
Alternative Method: (Easier Method)
Let cost price =100 (Because percentages are mentioned here, and the highest number will be 100)
Loss = 60%
Selling price = 40
Actual selling price = 3 × 40 = 120
Profit = 20 %
Rs. 200 are divided among Arun, Bholu and Chetan such that Arun's share is Rs. 30 more than Bholu's and Rs. 20 less than Chetan's. What is Bholu's share?
It is given that Rs. 200 divided among Arun, Bholu and Chetan
Let Bholu’s share = X Rs.
Arun’s Share = X +30 Rs
Chetan’s Share = X+30+20 = X+50
Now, Arun+ Bholu+ Chetan = (X+30) +(X) +(X+50) = 200
3X = 200 – 80
X = 120/3
X = 40
Bholu’s share is 40 Rs.
Hence, (b) is correct option.
Average marks in a test conducted in a class of 24 students gave the average score as 56. It was detected that while computing the average, marks of 3 students were taken as 46, 47 and 43 instead of 64, 74 and 34 respectively. Which of the following is the most appropriate description of the percentage by which the class average goes up now?
The increase in total marks of the class is (64–46)+(74–47)+(34–43) = 36
The average goes up by 36/24 = 1.5 marks
Increase in %age terms = 1.5 x 100/56 = 2.67%.
So, option d is correct.
Present average age of Kanika and Shweta is ‘x’ years. Ratio of the age of Kanika one year ago to the age of Monika four year hence will be 1: 2. Present average age of Kanika, Monika and Shweta is 22 years. Find the value of ‘x’ if the present age of Kanika is 16 years.
Present age of Kanika = 16 years
Age of Kanika one year ago = 16 – 1 = 15 years
Age of Monika after four years = 15 × 2 = 30 years
Present age of Monika = 30 – 4 = 26 years
Sum of the present ages of Kanika, Monika and Shweta = 22 × 3 = 66 years
Present age of Shweta = 66 – 16 – 26 = 24 years
Present average age of Kanika and Shweta = x = (24+16)/2=20 years
So, the value of ‘x’ is 20 years
So option (d) is the correct answer.
The Sludge Volume Index for mixed liquor having suspended solids concentration of 2000 mg/l and showing a settled volume of 200 ml from a one-litre sample would be
S.V.I = 200 ml/2000 mg
= 200 ml/2 gm = 100 ml/gm
The velocity field for flow is given by
and the density varies as p = po exp (-2t). In order that the mass is conserved, the value of λ should be;
By putting the given values in the above equation,
ρoe-2t(-2) + ρ.5+ ρ.5 + ρ. λ = 0
Since ρ = ρo e-2t
Hence
-2ρ + ρ.5+ ρ.5+ ρ. λ = 0
-2+5+5+ λ = 0
Hence λ = -8
In the influence line diagram for mid-span bending moment of a simply supported beam, the ordinate at the quarter span is 0.5 m. If the span of the beam is doubled, the ordinate at the mid-span of the influence line will be;
JLD of bending moment at C
For the above problem [a = b = 1/2]
M = ab/l = 1/4
Since length of the span is doubled
l' = 2l
Hence the ordinate of M1 will now be = 0.5 × 2 = 1 m
M = 2M1
∴ Ordinate of M at mid span = 2 × 1 = 2m
Ca(OH)2 is not a desirable product in the concrete mass because it is soluble in water and gets leached easily, particularly in hydraulic structure, that’s why cement with the small percentage of C3S and more C2S is recommended for use in the hydraulic structure.
The concrete continues to harden over several months. Hardening is not a drying process and can very well take place in water. Heat speeds up the setting and hardening of cement
The stiffening of cement without strength development (flash setting of cement) may cause because of C3A or C4AF.
So option B statement is incorrect.
A portal frame shown in figure (not drawn to scale) has a hinge support at joint P and a roller support at joint R. A point load of 50 kN is acting at joint R in the horizontal direction. The flexural rigidity. EI, of each member is 106 kNm2. Under the applied load, the horizontal displacement (in mm, round off to 1 decimal place) of joint R would be __________
When unit load at R is acting in the direction of 50kN load, then reaction at R = 2 (downward)
Alternatively:
A plane frame is as shown in the figure below. If MBA = 10 kNM (clockwise) and MCD = 21 kNM (counter clockwise) then the value of load P is
Applying ΣMA = 0
–FBA × 5 + 10 = 0
FBA = 2 kN
Similarly, ΣMD = 0
FCD × 7 – 4 = 0
FCD = 3 kN
For horizontal equilibrium of BC
P + FBA – FCD = 0
P = 1 kN
A circular tank of base diameter 12m is subjected to a total load of 12,000 kN. Calculate the vertical stress (in kN/m2) at a point P which is at a depth of 3m and 2m away from the centre of the loaded area.
Given load, Q = 12000 kN
Depth, z = 3m
Radial distance, r = 2m
vertical load, q = Q/A
Area of tank, A = πr2 = 113.04 m2
q = 12000/113.04 = 106.16 kN/m2
σz =18.05 kN/m2The results of two plate load tests performed on a given location with two circular plates are given below:
1. Diameter = 750 mm, S = 15 mm, Q = 150 kN
2. Diameter = 300 mm, S = 15 mm, Q = 50 kN
Use Housel’s equation i.e. Q = Aq + Ps
A = contact area
q = bearing pressure beneath area A (constant)
P = perimeter of footing
s = Perimeter shear (constant)
Determine the load (in kN) on circular footing 1.2 m diameter that will cause a settlement of 15 mm.
⇒ 150 = 0.44q + 2.36s …(i)
Similarly
Q2 = 50 kN
⇒ 50 = 0.071q + 0.94s ...(ii)
Solving (i) and (ii)
S = 46.13 kN/m
q = 93.48 kN/m2
Now for footing
Q = 279.6 kN
The net safe bearing capacity of a purely cohesive soil
Thus according to Terzaghi’s equation the net safe bearing capacity will be independent of both width and depth of the footing.
A scale representing either three units or only one unit and its fractions up to second place of decimal point is _______.
If f(t) = (eat - cos bt) 1/t , then the Laplace transform of f(t)
We know that
A concrete beam pre-stressed with a parabolic tendon is shown in the fig. The pre-stressing force applied is 1620 kN. The ULD includes self-weight of the beam. The stress in the top fibre of the middle section is (take compressive stress as positive)
Bending Moment at min-span, M = (45 × 7.32)/8
= 299.7 kNm
Stress At top fibre fct
= 5.7 N/mm2
Consider a signal x (t) given by x(t) L.T. ↔ x(s) = log(s+5/s+6) , then x(t) is given by
Then
Given F(S) = x(S) = log log (S+5) -log log (5+6)
Differentiate both sides
Multiply by negative sign on both
By property (1)
Consider the following simultaneous equations (with c1, and c2 being constants):
3x1 + 2x2 = c1
4x1 + x2 = c2
The characteristic equation for these simultaneous equations is;
3x1 + 2x2 = c1
4x1 + x2 = c2 Matrix From is (3 2 4 1 )(x1 x2) = [c1 c2]
AX = B
Characteristic equations of above systems is
|A - λI| = 0
|3 - λ 2 4 1-λ | = 0
By expanding λ2 - 4λ - 5 = 0
The degree of static indeterminacy of the truss shown below is.
= 14 + 3 - 2 × 8
= 17 - 16
= 1
A pressure gauge reads 62.3 kPa and 90 kPa respectively at heights of 7 m and 4 m fitted on the side of a tank filled with liquid. What is the approximate density of the liquid on kg/m3?
Which one of the statement is INCORRECT in design of hourly traffic volume?
A lift irrigation scheme using a discharge of 81m3/hr is planned to raise a crop whose delta is 45cm. Intensity of irrigation is 55%. Assuming 4500 hours of working of water supply for a year, area(hectares) required for irrigation would be;
Now the area to be irrigated = 364500/0.45 = 810000 = 81hectare
Required area = 81 /0.55 = 147.2 hectares
An 8 m long simply-supported elastic beam of rectangular cross-section 100mm × 200 mm is subjected to a uniformly distributed load of 10 kN/m over its entire span. The maximum principal stress (in MPa, up to two decimal places) at a point located at the extreme compression edge of a cross-section and at 2 m from the support is ____________.
MA = (−10 × 2 × 1) + 40 × 2 = 60 k Nm
= 90/mm3
τ = 0N/mm2 {point is at top}
So principal stress = 90N/mm2 = 90MPa
The anchorage value of a standard bend, of a reinforcement bar of diameter φ
Anchorage value for 90° bend = 8φ
What is the anchorage length of a tie bar, which is bent through 135° round a bar of diameter 12mm?
So, 6φ = 6 x 12 = 72 mm
If two centrifugal pumps (I and II) with heads H1, HII and discharge Q1, QII are connected in series, then the correct option is:
A laboratory model of a river is built to a geometric scale of 1:200. The fluid used in the model is oil of mass density 850 kg/m3. The highest flood in the river is 12000 m3/s. the corresponding discharge in the model shall be
Flow in river is based on gravitational force. So, model will be based on Froude number.
A stone weights 450 N in air and 250 N in water. The volume of the stone is:
250 = 450 - 9810 × V
V = 0.0204 m3
A beam is built using M25 concrete and Fe-250 bars of 16 mm dia bars. In limit state design, the development length, Ld, in terms of diameter of reinforcement, for compression zone is _______ mm.
[τbd = 1.4 N/mm2]
Factor 1.25 is used for compression.
The system of equations x - 4y + 7z = 12 , 3x + 8y - 2z = 10, 26 z - 8y = 6
Here, Rank [A : B] = Rank[A] = 3 = n
And n = 3
Therefore, the system has unique solution.
A steel bar of 20 mm Fe415 grade is embedded in a concrete block of M20 grade. The bond stress τbd = 1.2 MPa as per IS : 456-2000, for mild steel in tension. The length of embedment is 1m. The maximum pulling force that can be applied on the bar is ___________ kN
P1 = τbd × 1.6 × π × d × l = 1.2 × 1.6 × π × 20 × 1000 × 10-3
P1 = 120.63 kN
Force required to break the bar
P2 = 113.43 kN
Hence maximum pulling force = Min. of P1 & P2
= 113.43 kN
A single-stage impulse turbine with a diameter of 120 cm runs at 3000 rpm. If the blade speed ratio is 0.42, then, the inlet velocity of steam will be
D = 120 cm = 0.120 m , N = 3000 rpm
Blade speed ratio: 0.42
= 60π
Actual Inlet Velocity
= 60π/0.42 = 448.79m/s
A penstock is 3000 meters long. Pressure wave travels in it with a velocity of 1500 m/s. If the turbine gates are closed uniformly and completely in a period of 4.5 seconds, then it is called
L = 3000 m
C = 1500 m/s
t = 4.5 sec > tc hence slow closure
Choose the correct statement(s) with respect to the Lag distance in computing the absolute minimum sight distance of the vehicle.
i. The reaction time is taken as 0.15sec
ii. The distance is dependent only on the coefficient of friction, reaction time and gradient
iii. It is depended on reaction time and velocity.
iv. The reaction time is the time required to observe the obstruction.
The lag distance is given by the simple equation.
Lag distance = Velocity x Reaction Time
Reaction time as per IRC recommendation is 2.5 sec
Lag distance is dependent mainly on Velocity of the moving vehicle and the reaction time and independent on coefficient of longitudinal friction, gradient and efficient of brake.
The reaction time of the driver is the time required to apply the brake after observing the obstruction
Considering two vehicles moving at a speed of 85kmph and 60 kmph on a road surface with skid resistance of 0.72 and braking efficiency of 55%, what will be the non-passing sight distance? Assume necessary data.
Tag v1 = 85 kmph
V2 = 60 kmph
Skid resistance = 0.72
Brake efficiency, ŋ = 55%
Coefficient of friction, f = Skid resistance x Brake efficiency
f = 0.72 x 55/100
f = 0.396 = 0.40
Non- passing sight distance = SSD1 + SSD2
Assume reaction time , t = 2.50 secc
length of runway under standard condition is 1800m. the airport site has an elevation of 300m. it has reference temperature of 33°C. if the runway is to be constructed with effective gradient of 0.6%. then determine the corrected runway length
Corrected length = 1926 m
ii) Standard atmospheric temp of the given elevation
= 15° = 0.0065 × 300 = 13.05° C
Rise in temp = 33°C – 13.05° C = 19.95° C
Correction = 1926 / 100 × 19.95 = 384.237 m
Corrected length = 1926 + 384.237 = 2310.23 m
Check for combine correction = Total correction in %
= 2310.23 – 1800 / 1800 = 28.34%
28.34% < 35%
Hence, OK
Correction for gradient = 20/100 × 2310.23 × 0.6 = 277.22 m
Corrected length = 2310.23 + 277.22 = 2587.45 m
A vehicle moving at 40 kmph speed was stopped by applying the brake and the length of skidmark was 18 m. If the average skid resistance of the pavement is known to be 0.70, the brake efficiency of the the test vehicle (in %) is
Average skid resistance developed,
∴ Brake efficiency
A rectangular cross-section as shown below, experiences a maximum bending stress of 30 N/mm2. The magnitude of force experienced by the shaded region is _____ kN.
Force experienced by infinity small area at distance x = dF
Flexibility matrix of the beam shown below is If the support B settles by Δ/EI units, what is the reaction at B?
From relation Δ = f.P
In matrix form
Δ1 = f11P1 + f12P2 .................i
Δ2 = f21P1 + f22P2...............ii
Using given data,
Solving eqn (iii) and (iv)
∴ Reaction at B P1 = 3.75 Δ
The stiffness matrix of a beam element is given as . Then the flexibility matrix is
is stiffness matrix, then
The matrix of co-factor
Transpose matrix,
Inverse of Matrix
What are the critical activities in the project network shown below. (Use PERT analysis)
First find the expected time of the activities. Then found slack of the activities. Activities having zero slack are the critical activities.
The critical activity is 1-3-5-6.
A rectangular concrete beam of width 150 mm and depth 250 mm is prestressed by pretensioning to a force of 200 kN at an eccentricity of 30 mm. the cross-sectional area of the prestressing steel is 200 mm2. Take modulus of elasticity of steel and concrete as 2.1 x 105 MPa and 3 x 104 MPa respectively. The percentage loss of stress in the prestressing steel due to elastic deformation of concrete is ________.
Loss of prestress due to elastic deformation,
Total stress in steel,
Percentage loss of stress
A concrete beam prestressed with a parabolic tendon is shown in the sketch. The eccentricity of the tendon is measured from the centroid of the cross-section. The applied prestressing force at service is 1620KN. The uniformly distributed load of 45 KN/m includes self-weight.
The stress (in N/mm2) in the bottom fibre at Mid Span is
Stress due to load
= -6.3948N/mm2
Stress due to pre-stres
= 9.3312N/mm2
Total stress = 9.3312 – 6.3948 = 2.9364 N/mm2 = 2.94 Compression
A thick cylinder of internal diameter 10 cm is subjected to an internal pressure of 50 N/mm2, then the thickness of the cylinder according to the maximum shear stress theory, if yield stress is 270 N/mm2 and factor of safety is 2, is _______mm
Inner radius, R1 = 50 mm
Let external radius is R2 mm
Radial pressure, p = 50 N/mm2
Principal stresses at inner surface of thick cylinder are
Maximum shear stress, τmax
Allowable normal stress, σa
Allowable shear stress.
Undefined control sequence \therefore
Thickness of cylinder,
t = R2 − R1 = 98.2 − 50 = 48.2mm
Calculate the standard error of the volume of a cuboid whose sides are A, B, and C have values given below and standard error as +/-___________ in cm3
A = 50.0 cm +/-.04 cm
B = 40.0 cm +/-.03 cm
C = 30.0 cm + /-.02 cm
A water mains 140 cm inner diameter contains water at a pressure head of 200 m. The minimum thickness of the metal shell required for the water mains, given that maximum permissible stress in the metal is 400 kg/cm2, will be ___ mm.
Compatibility equation θA = 0
Hence, the correct option is (A).
A 50 cm diameter pipe carries water (v = 0.294 × 10–6 m2/s) at an average velocity of 10 m/s . The roughness of pipe surface is 0.8 mm. For fully rough turbulent flow, What is the boundary shear stress.
F = 0.022
Shear velocity
Determine the distance from the pipe wall at which local velocity is equal to the average velocity for turbulent flow in pipe
A water supply system is designed for a town with population of 2,50,000. Determine the fire demand (in ml/min) using empirical relationship and the Match List I and List II
Kulching Formula
Q = 3182 √P
P = population in thousand; Q = demand in lit/min
Q = 50311.8 lit/min
Freeman Formula:
Q = 1136[(P/10 )+ 10]
Q = 39760 lit/min
USA- Underwriters formula
Q = 4637 √P [1- 0.01√P ]
Q = 61725 lit/min
Buston Formula:
Q = 5663 √P = 89540 lit/min
Determine the critical deficit (in ppm) of dissolved oxygen when the stream flowing at a rate of 210 cumec with saturated dissolved oxygen limit of 10.6ppm is diluted with a sewage flowing at the rate of 1.35 cumec. Assume the Dissolved oxygen limit of waste water as 4ppm.
D.O. of mix = 10.55 ppm
Critical deficit of D.O. = Saturated D.O. of stream – D.O. of mix = 0.05 ppm
Which of the following are correctly matched:
1. Turbulent flow = Moment Integral equation
2. Laminar flow = Hagen Poiseuille equation
P = Pressure difference across tube
Μ = viscosity of liquid.
In turbulent flow, prantis mixing length theory is used to calculate shear stress.
12m3/s of water is applied to a field having area of 30 hectares for 4 hours. About 0.3m of water stored in the root zone. Water application efficiency(%)in this case would be;
The volume of water in root zone Vf = 0.3 * 30 * 104 = 90000m3
The volume of water supplied Vs = 12 * 4 * 3600 = 172800 m3
Water application efficiency = Vf/ Vs *100 = 90000/172800 = 52%
A body was subjected to two mutually perpendicular stresses of -6MPa and 24MPa respectively. What is the shear stress on the plane of pure shear?
Shear stress on plane of pure shear
The figure shows reducing area conduit carrying water. The pressure p and velocity V are uniform across sections 1 and The density of water is 1000 kg/m3. If the total loss of head due to friction is just equal to the loss of potential head between the inlet and the outlet, then V2 in m/s will be _____
A job having a processing time of 11 days is to be processed in scheduling operation. The job is to be delivered on 18th july 2018 but the job is received on 7th july. Find the critical ratio of the job to be processed.
= 18-7/11
= 11/11
= 1
A rectangular beam of cross-section B × D has moment of resistance equal to M. A smaller square section is removed from the centre such that the moment of resistance of the remaining area is equal to half the initial MOR. The length of side of section removed is
For solid section
Moment of resistance of the solid section
For Hollow Section
Moment of inertia
Section modulus
Moment of Resistance
Given
Mh = M/2
During the construction of a highway bridge, the average permanent load on the clay layer is expected to increase by about 95 kN/m3. The average effective overburden pressure at the middle of the clay layer is 190 kN/m3. Here, H is 4.5 m, Compression index is 0.26, e0 = 0.78, and coefficient of consolidation is 0.23 m2/month. The clay is normally consolidated. Determine the total primary consolidation settlement (in mm) and the ratio of pre-compression pressure to the effective stress of the bridge.
The total primary consolidation settlement will be calculated as
On substituting the values in above equation Sp = 0.115m = 115mm
Ratio of pre-compression pressure to the effective stress is
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