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A sentence is split into 3 parts in the 3 options given below. If any part of the sentence has an error – grammatical or otherwise, mark that option as your answer. In case there is no error, mark option ‘d’.
In sentences where the structure is "one of + noun + who" is used, the verb after "who" agrees with the noun used after "one of". In the sentence, the noun is "those writers" which is in plural form. Therefore, it must take a plural verb. This makes the use of "has" incorrect in the sentence. The correct plural verb would be "have". The correct sentence would be:
He is one of those writers who have won acclaim the world over.
What is the number of integral solutions for x^{3}x^{2}2x19=0?
Irrespective of the value of ‘x’ (odd or even), x^{3 }x^{2 } 2x will always be even which means the complete expression x^{3 } x^{2 } 2x  19 can NEVER be zero.
All 2 digit numbers where both places occupy the same number, in an ascending order with a vertical line separating their pair of next consecutive numbers. This sequence follows with the same condition of 2 digit numbers. Consider 10 as a first two digit number. The number of these vertical lines separating such pairs is ______.
The pair of such numbers are
11/12; 22/23; 33/34; 44/45; 55/56; 66/67; 77/78; 88/89;
Middle number is the average of the numbers on both sides.
Average of 6 and 4 is 5
Average of (7+4) and (2+1) is 7
Average of (1+9+2) and (1+2+1) is 8
Average of (4+1) and (2+3) is 5
Therefore, Average of (3) and (3) is 3
If x^{2}y^{2}= 702, how many ordered pairs of solutions for x and y exist such that x, y are natural numbers
x^{2}y^{2}= 702 i.e.(xy)(x+y)= 702
Factories 702 as 702= 2×3×3×3×13 Since x and y are both natural numbers, (xy) and (x+y) should be both even or both should be odd. In order to get possible values of x and y we should write 702 as a product of 2 natural numbers which should be a like i.e. both even or both odd. This is not possible since 702 has only 1 even factor leading to (d) as the answer.
What is the respective ratio between the expenditure made by university on research work and purchase of books for library?
Total expenditure is 60 lakhs.
Expenditure made by university on research work
= 8% of 60 lacs = 8 × 60 lacs/100 = 4.8 lacs
Expenditure made by university on purchase of books for library
= 6% of 60 lacs = 6 × 60 lacs/100 = 3.6 lacs
Hence ratio between the expenditure made by university on research work and purchase of books for library
= 4.8 : 3.6 = 4:3
I have been waiting _____ last year September.
Option C is the correct answer.
'Since' is used when a specific time is mentioned.
'For' is used to signify the total time period.
Roll numbers given to a select of students are unique 4 digit numbers. One roll number is picked up at random, what is the probability that the chosen number is divisible by 11 or 13 ?
In the question. We have to count 4 digit numbers which are multiples of 11, 13 and 143 (product of 11 and 13).
The first 4 digit multiple of 11, 13 as well as 143 is 1001 = 11×91= 13×77= 143×7
The last 4 digit multiple of 11 is 9999 =11×909 and that of 13 is 9997 =13×769 and for 143 is 9867 (=143×69)
Number of 4 digit multiple of 11, 13 and 143 are (90991+1), (76977+1) and (697+1)
i.e. 819,693 and 63 respectively.
The number of 4 digit roll numbers which are divisible by 11 and 13 are 819+69363=1449 Number of 4 digit numbers are from 1000 to 9999 i.e. 9000.
Probability of selection the required roll number =
In the solution of the following set of linear equations by Gauss elimination using partial pivoting
5x + y + 2z = 34; 4y – 3z = 12; and
10x  2y + z = – 4
the pivots for elimination of x and y are
5x + y + 2Z = 34
4y  3z = 12
10x  2y + z = 4
5, 0, 10
1st Pivoting
10x  2y + z = 4.......eq1
4y  3z = 12..............eq2
5x + y + 2z = 34.......eq3
eq3  (eq1)/2,
10x  2y + z = 4
4y  3z = 12
2y + 3Z/2 = 36
4, 2
Hence, Pivot elements are x=10 & y=4
Chose the correct expression for f(x) given in the graph.
Substituting the coordinates of the straight line
we get the correct option as C which is
List 1 contains quality parameters of water and List 2 contains reagents required for computation. Choose the correct option.
List 1
P. Bacterial concentration
Q. COD
R. Dissolved Oxygen
S. Chlorine
List 2
1) Silver Nitrate
2) Lactose Broth
3) Sodium Thiosulphate
4) Alkali Iodide Azide
For bacterial concentration in water MPN test is used in which lactose broth is used, In DO test alkali iodide azide is used to get accurate value of dissolved oxygen , in chlorine determination silver nitrate is used as titrating agent and in COD sodium thiosulphate is used as titrating agent.
A rectangular channel had a positive surge of velocity 10 m/s moving down the channel. If depth of flow and velocity after the passing of the surge are 4.0 m and 5.0 m/s respectively, the velocity before the passage of the surge is _____m/s.
Let V_{w} = Velocity of surge wave
Superimpose a velocity V_{w} to the left to simulate steady flow as in figure
Given, y_{2} = 4.0 m, V_{2} = 5 m/s and V_{w} = 10 m/s
By the continuity equation,
⇒ y_{1} (V_{w } V_{1} )= y_{2} (V_{w } V_{2} )
⇒ y_{1} (10  V_{1} )= 4(10  5)
A double cover butt riveted joint is used to connect two flat plates of 250 mm width and 16 mm thickness as shown in figure. There are sixteen power driven rivets of 22 mm diameter at a pitch of 50 mm in both direction on either side of the plate. Two cover plates of 10 mm thickness are used. The capacity of the joint in tension (in kN) with permissible bearing and shearing stresses as 300 MPa and 100 MPa respectively, is
Strength of one rivet in double shear
Strength of the riveted joint in double shear
= 16 x 86.75 = 1388 kN
Strength of one rivet in bearing = d'×t×σb
=(22+1.5)×16×300×10^{3}
=112.8 kN
Strength of rivet joint in bearing
=16 x 112.8 = 1804.8 kN
Strength of plate in tearing = (2504 x 23.5) x 16 x 150 x10^{3}
= 374.4 kN
Thus, the strength of joint will be governed by tearing and it will be equal to 374.4 kN.
A hydraulically efficient trapezoidal channel section has a uniform flow depth of 2m. The bed width (expressed in m) of the channel is
In a floating body I = Moment of inertia of water line area about the longitudinal axis, ⩝= Volume of displaced fluid, B = Centre of pressure, G = Centre of gravity and M = Metacentre. For stable equilibrium of this body
For a beam ABCD having three spans each of lengths 2 m, the rotation θ_{B} and θ_{C} are
Equilibrium of joint C,
Three uniform rods OA, OB and OC having the same length l, cross sectional area A and of the same material remaining in the same vertical plane support a load W at O as shown in figure. The magnitude of forces in the members OA, OB and OC respectively are
Let the compressive force in each of the members OB and OC be P.
Let the tension in OA be R.
Resolving the forces at O vertically, we have
R + 2P cos 60^{0} = W
∴ P = W R
∴ stain energy stored by the frame
R is determined from the condition of minimum strain energy i.e. by the condition,
The cylinder and tube as shown in figure contains oil of specific gravity 0.85. For a gauge reading of 2.5 kg (f)/cm^{2}, the total weight (in kN) of the piston and the slab placed on it is
Let W be the total weight of the piston and the slab.
Since pressure below the piston = Pressure at the same level in the vertical tube
If then the value where V is the closed region bonded by x = 0, y = 0, z = 0 and 2x + 2y + z = 4, is _____.
In the pretensioning method, which of the following statements are correct?
1) The tension in the concrete is induced directly by external force
2) Tension is induced in the tendons before concreting
3) Concrete continues to be in tension after prestressing
In pretension, tension is induced in the tendons before concreting.Once the concrete sets and hardens, tendons are cut and prestress is transferred to concrete. Hence after application of prestressing, concrete is in compression.
A steel bar of 4 m length and uniform crosssectional area of 1200 mm^{2} is suspended vertically and loaded as shown in figure. The elongation of the bar will be [Take, E = 2.05 × 10^{3} N/mm^{2}]
R_{A} = 40+60+80=180 kN
Force P_{1} on portion AB = 180 kN (tensile)
Force P_{2} on portion BC = 18040 = 140 kN (tensile)
Force P_{3} on portion CD = 80 kN ( tensile)
From the particlesize distribution curve, the diameter of particle size corresponding to 10% finer, 30% finer and 60% finer are 0.16mm, 0.18 mm and 0.28mm respectively. Determine the three parameters namely D_{10}, C_{u} & C_{c} required to classify the granular soils
The three soil parameters are the Effective size of the particle, Uniformity Coefficient and Coefficient of Curvature.
The Particle size corresponding to 10% finer is Effective size. i.e. D_{10} = 0.16 mm
Uniformity Coefficient,
Coefficient of Curvature
The flexibility matrix of the following structure is a scalar multiple of
Flexibility Matrix,
Unit force in direction 1,
In a two dimensional stress analysis, the state of stress at a point is as shown below,
The magnitude of ' ' is
Using Mohr’s circle
If the angularity number is 16, find the specific gravity of aggregates if their weight in a cylinder of volume 3000 cc is 4070 gm.
Angularity number= 67% Solid Volume
Solid Volume= 51%= 0.51 ×3000= 1530 cc
A 2.0 diameter circular culvert is flowing half full and the flow is in critical state. The flow (in m^{3}/s) in the culvert barrel is
For critical flow condition,
For a circular conduit flowing half full,
Water flows at a steady and uniform depth of 2m in an open channel of rectangular cross section having base width of 4m and laid at a slope of 1 in 10000. If it is desired to obtain critical flow in the channel by providing a hump in the bed. Take Manning’s coefficient = 0.01. Calculate the height of hump (in meters) required for the critical flow
Area A = 4 ×2 = 8, Perimeter P = 4 + 4 = 8
R = (A/P) = 1 m
Discharge per unit width q = 8/4 = 2
We know that the minimum specific energy corresponds to the point of critical depth and is given by 1.5 times the critical depth
Critical depth = (q^{2}/g)^{(1/3)} = 0.74 m
Minimum specific energy is 1.5 × 0.74 = 1.11 m
Specific energy should be same at all points (neglecting hydraulic losses)
So, E = E_{minimum} + h (this energy is at position of hump)
E = y + v^{2}/(2*g) = 2.05
h = 2.05 – 1.11 = 0.94 m
An orifice having a diameter of 10cm discharging water and the venacontracta formed has a diameter of 9cm. If the coefficient of velocity is 0.96, then what is the coefficient of discharge?
We know that
C_{d} = C_{c} × C_{v}
C_{d} = coefficient of discharge
C_{c} = coefficient of contraction
C_{v} = coefficient of velocity
Now, C_{c} = area of venacontracta /area of orifice
The direct costs of the three activities along with their duration are given. Activities are completed in the sequence of A then B and then C. Then the minimum possible direct cost for the duration of 21 days will be
Cost slope of A is least, so A is crashed first.
Cost slope of C is less than B so it is crashed before B to get duration 21 days.
For the matrix ONE of the normalized eigen vector is given as
Which gives us the equation
3x_{1} + 3x_{2} = 0
and x_{1} + x_{2} = 0
Which is only one equation,
x_{1} + x_{2} = 0
whose solution is
Which after normalization is
The other eigen vector is obtained by putting the other eigen value λ = 6 I eigen value problem
Since it is a constant multiple of one the normalized vectors which is
A project consists of 7 jobs. Jobs A and F can be started and completed independently. Jobs B and C can start only after job A has been completed. Jobs D, E and G can start only after jobs B, (C and D) and (E and F) are completed, respectively. Time estimates of all the jobs are given in the following table:
The expected completion time of the project is.
The critical path is A – B – D – E – G
Therefore, the maximum completion time is 35 days.
The diagrams below show two similar beams with different loading conditions. The numbers denoted by a, b and q are all positive. How many points of contra flexure are there in each?
Contra flexure point = point with bending moment zero.
One contra flexure point
Two contra flexure points
Which of the following statement is INCORRECT?
Shear parameters ‘c’ and ∅ are dependent of water content of soil
• Vane shear test is a useful method of measuring the shear strength of clay. It is a cheaper and quicker method. The test can also be conducted in the laboratory.
• The laboratory vane shear test for the measurement of shear strength of cohesive soils, is useful for soils of low shear strength (less than 0.3 kg/cm^{2}) for which triaxial or unconfined tests cannot be performed.
• The test gives the undrained strength of the soil. The undisturbed and remoulded strength obtained are useful for evaluating the sensitivity of soil.
NOTE: This test is also useful when the soil is soft and its water content is nearer to the liquid limit.
To determine the relative density of sand sample, the natural moisture content and bulk density were determined in the field and found to be 6% and 1.56 respectively. Sample of this soil was then compacted in proctor mould of 1000 cc and the following data was obtained
Let. of empty mould = 2000g
Let. of mould + soil in loosest state = 3300 g
Let. of mould + soil in densest state = 3850 g
Moisture content used in test = 10%
The relative density of soil is lie % is?
A trapezoidal section of an open channel has side slope 2H : 1V. If bottom width is 'b’ and depth d’, the relation between b & d for most economical trapezoidal section of the channel is:
For most of the economical open channel Trapezoidal section,
Half of top width should be equal to length of the side slope (n:1=H:V)
The internal and external indeterminacy of the following structure are respectively
No of joints (j) = 6
No of members (m) = 11
External Static Indeterminacy Dse = r – 3 = 4 – 3 = 1
Internal Static Indeterminacy Dsi = m – (2j – 3) = 11 – (12 – 3) = 2
Since (√3 + 1) / (√3  1) = (√3 + 1) (√3 + 1)/ (√3  1) (√3 + 1) (by multiplying numerator and denominator with conjugate of denominator)
Similarly, (√3 1)/(√3 + 1) = (2  √3)
(√2 + 1) / (√2  1) = (√2 + 1)^{2}/ (√2 1) (√2 + 1)
(by multiplying numerator and denominator with conjugate of denominator)
Substituting the above four values in
Two motorcyclist driving at uniform speeds s_{1} and s_{2} respectively in straight line path in opposite directions and come closer 11 m after every second. While running in similar direction, their speeds reduces by 50% and get 0.7 m closer in every second. The speeds s_{1} and s_{2} of the motor cyclist will be:
As the two motor cyclist were running along in opposite directions, so their speed will be:
s_{1}+ s_{2} = 11(1)
As they are running along in the similar direction, so:
s^{1/2} – s^{2/2} = 0.7(2)
= s_{1 }– s_{2} = 1.4 (3)
Also s_{1} = 1.4  s_{2} (4)
Adding 1 and 3, 2s_{1} = 12.4, so s_{1} = 6.2 m
Putting in (1) value of s_{1} for obtaining s_{2} = 4.8
s_{1} = 6.2 m/s and s_{2} = 4.8 m/s
A centrifugal pump needs 1000 W of power when operating at 2000 rpm. If the speed of pump is increased to 4000 rpm, then power requirement is _____ kW.
In a model test of spillway, the discharge and the velocity of flow over the model were 2 m^{3}/s and 1.5 m/s respectively. The discharge over the prototype which is 36 times the model size will be ____ m^{3}/s.
Linear scale ratio 36
A udl of 15 kN/m and length 3m rolls over a simply supported span of 10m length. The maximum bending moment at a section 4m from the right end will be_____ kNm.
Average load on AC = Average load on CB
Maximum BM at C = Load intensity × area of ILD under load
The presence of fats and oils in the waste water are determined with the help of
The presence of fats, oils and greaes in water is measured by making a solution of the evaporated waste water with ether and evaporating the solution so that only fats and oils are left which can be easily weighted.
A trapezoidal section of an open channel has side slope 2H : 1V. If bottom width is 'b’ and depth d’, the relation between b & d for most economical trapezoidal section of the channel is:
For most of the economical open channel Trapezoidal section,
Half of top width should be equal to length of the side slope (n:1=H:V)
In a frame ABCDE, two concentrated loads are applied at B and C. If the length L = 2 m and the plastic moment capacity of the frame is 200kNm. Find the maximum load P that can be applied.
Internal Virtual Work = External Virtual Work
Internal Virtual Work = External Virtual Work
Consider the following equations
x + y + w = 0
y + z = 0
x + y + z + w = 0
x + y + 2z = 0
The system of equations will be
The above system of equations can be written as AX = B
∴ Rank (A) = Rank (A  B ) = number of variables = 4
∴ The system is consistent having unique solution and it can be observed that the solution can only be trivial, i.e., x = y = z = w = 0.
A project has three activities A followed by B which is followed by C. Taking indirect cost of 25 units per week, the cost of completing the project in 15 weeks is given by?
Activity A : 10 weeks400 units, 9 weeks450 units, 8 weeks500 units
Activity B : 5 weeks200 units, 4 weeks225 units
Activity C : 3 weeks700 units, 2 weeks800 units
Activities having the least crash slope are crashed first.
Indirect cost for 15 weeks = 25 × 15 = 375.
Total cost = 1425 + 375 = 1800
A beam ABC shown in given figure is horizontal. The distance to the point of contraflexure from the fixed end A is ________ m. (Take reaction at A is 1 kNm)
Taking moment about A
let point of cantraflexure will be at a distance ‘y’ from free end,
Taking moment about y
So, distance from fixed end = 0.33 m
A high rate trickling filter is used for treating wastewater of a town having population of 40000 persons.
Domestic sewage = 150 lpcd having 200 mg/l of BOD
Industrial sewage = 0.25 mlD having 600 mg/l of BOD
BOD removal in primary Clarifier = 35%
Permissible BOD loading = 8000 kg/hec m/day( including recirculated sewage)
Take recirculation ratio = 1
Use formula for efficiency = (100)/(1 + 0.44(√(W/V*F)))
V is volume, F is recirculation factor and W is BOD applied to filter
Find the efficiency of filter
Domestic sewage = 150 × 40000/1000 = 6000 m^{3}/day
Industrial sewage = 0.25 × 1000000/1000 = 250 m^{3}/day
Total BOD of domestic = 6000 × 200/1000 = 1200 kg/day
Total BOD industrial = 250 × 600/1000 = 150kg/day
Total BOD = 1200 + 150 = 1350 kg/day
BOD applied to filter = 1350 × (1 – 0.35) = 877.5 kg/day
Filter volume = BOD applied to filter / permissible BOD loading
= 877.5 × 10000/8000 = 1096.9 m^{3}
Recirculation factor (F) = (1 +R)/(1 + 0.1R)^{2} = 1.653 ( R = 1)
Efficiency using above formula = 76.56 %
Consider a spatial curve in threedimensional space given in parametric form by
The length of the curve is ________
A car starts from rest and accelerates at a speed of 35 m/s in time 18 seconds. Find the average acceleration of the car?
From the question:
Acceleration of car can be obtained using:
Now:
=(35 m / s)/18s
= 1.944 m/s^{2}
The force in member DC in the given truss, if support A settles vertically by 25mm is ____________ kN. (take tension as positive)
Select RA as redundant, the direction of which will be taken as vertically downward. Apply unit load at A vertically downward.
For the given project network, where numbers along various activities represent normal time.
Find the project completion expected duration (Te)?
Critical path is 123455
And expected duration is 32 days.
The degree of freedom of the beam shown is?
kinematic determinacy D_{K}= 3jr+m
j=no of joint =4
r= no of reaction = 7
m= no of hinge = 1
D_{K} = 3×4 7+1
D_{K} = 6
A water course commands an irrigated area of 800 hectares. The intensity of irrigation of rice this area is 55%. The transplantation of tice takes 12 days and total depth of water required by the crop is 45 cm on the field during the transplantation period. During the transplantation period, the useful rain falling on the field is 10 cm, then the discharge required in the water course, if losses of water is to be 20% in the water course, is _______cumec.
As we know the relation
Here, B = 12 days
∆ = 45 – 10 = 35 cm or 0.35 m
D = Duty of irrigation water in hectares/cumec
=296.23 hactares/ cumec
Since, the losses in the canal are 20%
Hence, the duty of water at the head of the water course = 296.23 x 0.80 = 236.98 hectares/cumec
Now, total area under rice plantation = 800 x 0.55 = 440 hectares
So, discharge at the head of water course
The prejump Froude Number for a particular flow in a horizontal rectangular channel is 10. The ratio of sequent depths (i.e., postjump depth to prejump depth) is ___________
Consider
Prejump depth =y_{1}
Postjump depth =y_{2}
Irrigation water is to be provided to a crop in a field to bring the moisture content of the soil from the existing 18% to the field capacity of the soil at 28%. The effective root zone of the crop is 70 cm. If the densities of the soil and water are 1.3 g/cm^{3} and 1.0 g/cm^{3} respectively, the depth of irrigation water (in mm) required for irrigating the crop is ________
Depth of irrigation water
Following data refers to the activities of a project, where, node 1 refers to the start and node 5 refers to the end of the project.
The critical path (CP) in the network is
The critical path is the sequence of activities with the longest duration. A delay in any of these activities will result in a delay for the whole project.
Critical path = 1‒4‒3‒5 = 3+3+2 = 8
Note: Critical path is the longest path through a network of activities, which inturn, determines the shortest time possible to complete all tasks in the network (i.e. the shortest time to complete the project).
In a tidal model, the horizontal scale ratio is 1/750 and the vertical scale is 1/75.The model period (in minutes), corresponding to a prototype period of 18 hours, would be
For any discrete random variable X, with probability mass function
define the polynomial function
For a certain discrete random variable Y, there exists a scalar β ∈ [0,1] such that g_{Y} (z)= (1  β + βz)^{N}. The expectation of Y is
Derivative of g_{x}(z) evaluated at z=1 gives expectation E(X) of X.
Therefore, take derivative of g_{Y}(z) with respect to z, and plug in z=1
Derivative is N.β.(1  β + βz)^{(N1)}, plug in z = 1, gives Nβ.
The maximum BOD (in mg/l) present in a solution of glucose of concentration 330 mg/l.
The reaction of Glucose with oxygen,
C_{6}H_{12}O_{6} + 6O_{2}→ 6CO_{2} + 6H_{2}O + energy
180 gms 192 gms
BOD per 320 mg of glucose,
At joint B
The shear force diagram of a loaded simply supported beam is shown in the following figure:
The maximum bending moment (in KN m) of the beam is
M_{B} – M_{A} = ( BM)_{B}
= Area of shear force diagram from A to B since M_{A} = 0
Due to simply supported beam
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