A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in total from both as interest. The rate of interest per annum is.
Let the rate % = R
According to the question,
100R + 120 R = 2200
Amit rows a boat 9 kilometres in 2 hours down-stream and returns upstream in 6 hours. The speed of the boat (in kmph) is:
Down-stream rate = 9/2 = 4.5 kmph
Upstream rate = 9/6 = 1.5 kmph
The speed of the boat = (4.5 – 1.5) kmph = 3 kmph
The sentences given with blanks are to be filled with an appropriate word(s). Four alternatives are suggested for each question. For each question, choose the correct alternative and click the button corresponding to it.They abandoned their comrades ______the wolves.
When something is left in between more than 2 things or persons, ‘among’ is used. There are a number of wolves in the sentence hence the correct answer is option D.
In the following question, out of the four alternatives, select the word opposite in meaning to the given word.Turgid
Turgid = swollen and distended or congested; pompous
Bloated = inflated
Humble = plain, simple
Puffy = inflated
Tumescent = swollen
Hence, humble is the correct answer.
Direction: Read the information carefully and give the answer of the following questions:
Total number of children = 2000
If two-ninths of the children who play football are females, then the number of male football children is approximately what percent of the total number of children who play cricket?
57.4% (If 2/9 th of children play football are female , then
male children = 1 - 2/9 = 7/9th of children play football;
Let's take z% of male football children equal to cricket children , then
=> z% of cricket children = 7/9 th of football;
=> z% of (23% of 2000) = 7/9 th of (17% of 2000);
=> z% of 23 = 7/9 th of 17
=> z = 7× 17 × 100 = 57.4%;)
Required percentage = (7/9)*17*100/ 23 = 57.4%
Direction: Study the following information carefully and answer the given questions:
The following pie chart shows the percentage distribution of total number of Apples (Dry + Wet) sold in different days of a week :
The pie chart2 shows the percentage distribution of total number of Apples (Wet) sold in different days of a week.
Find the ratio of Apples (Dry + Wet) sold on Tuesday to that of the Apples (Wet) sold on Thursday?
Required ratio = 13% of 7000: 23% of 4500 = 182:207
Directions: In each of the questions below, some statements are given followed by some conclusions. You have to consider the statements to be true even if they seem to be at variance with commonly known facts. You have to decide which of the following conclusions logically follows from the given statements. Give answer.
Some jacket is shirt.
Some shirts are trouser.
No shoes are t-shirt.
All trousers are shoes.
All pants are t-shirt.
I. All shirts being t-shirt is a possibility.
II. Some shoes are trouser.
III. Some jackets are t-shirt.
IV. Some shirts being t-shirt is a possibility.
The question below consists of a set of labelled sentences. These sentences, when properly sequenced form a coherent paragraph. Select the most logical order of sentences from among the options.
P: July 1969 was to see a transformed Indira Gandhi.
Q: Quite a few people contributed with ideas.
R: She sounded the bugle through her historic ‘Note on Economic Policy and Programme’ that was
circulated among delegates at Bangalore on July 9, 1969.
S: But the pivot was P.N.Haksar who gave shape, structure and substance with the help of some of his
colleagues in the Prime Minister’s Secretariat.
P is clearly the first sentence as it introduces the main person of the paragraph, i.e. Indira Gandhi. The pronoun ‘she’ in R refers to Indira Gandhi mention in P. Thus, R forms the second statement. Now, in the sentence R, ‘delegates’ is mentioned which says about the people, who are given in the sentence Q. So, sentence Q must follow R. Thus, the sequence after rearrangement is PRQS and option A is the correct answer.
Direction: Which of the following is the MOST SIMILAR in meaning to the given word?
‘Remnant’ means ‘remainder’ which is same as ‘residue’.
Direction: Study the following information carefully and answer the given questions.
Seven people, P, Q, R, S, T, W and X sitting in a straight line facing north, not necessarily in the same order. R sits at one of the extreme ends of the line. T has as many people sitting on his right, as on his left. S sits third to the left of X. Q sits on the immediate left of W. Q does not sit at any of the extreme ends of the line.
If all the people are made to sit in alphabetical order from right to left, the positions of how many people will remain unchanged?
Hence, only Q's position will remain unchanged.
A wooden block (relative density = 0.65) floats above surface of a liquid of relative density of 1.35, contained in a tank. The volume fraction of metallic piece lies above the liquid surface is
Let V be total volume of wooden block and V’ is the volume that lies immersed in the liquid.
Using principle of floatation,
Weight of wooden block = weight of the liquid displaced by the immersed portion of the wooden block
So, the volume fraction of wooden block below the liquid surface = 1 – 0.4815 = 0.5185
Match the following lists and choose the correct answer using the codes given below
In a 1:36 scale model test of a spillway, discharge of flow over the model is 6 m3/s. If the velocity of flow over model was found to be 5 knots, then the velocity of flow over prototype in m/s will be
Qm = 6 m3/s, Vm = 5 knots = 5 × 0.515 =2.575 m/s Using dimensional analysis
A cylindrical pipe is made up of copper having E=120 Gpa and A=110 mm2. This cylindrical tube is covered with an aluminum tube having E=70 Gpa and A=70 mm2. If the two assembly are in parallel and is loaded with a force of 70 KN. Find the deflection of the whole assembly under this loading. Given length of the tube is 500 mm.
A bar of uniformly tapering section with 75 mm and 150 mm diameter, is held between two parallel end supports 6 m apart. If the length of the bar is 5.998 m, then stress induced in the bar for a temperature change of 30oC will be (Take coefficient of thermal expansion =12×10-6 per oC and E = 2×105 N/mm2)
Expansion of bar due to temperature only,
ΔL = αtL = 12 x 10-6 x 30 x 5.5 = 1.98 x 10-3 m
Since (6 – 5.998 = 0.002 m) of expansion is allowed, but expansion in bar due to temperature change is 0.00198 m, which is less than 0.002 m. Hence there will not be any stress inside the bar.
A spherical body of diameter 750mm of initial temperature 25℃ is immersed in the fluid at temperature 400℃. Thermo physical properties of the body are k=22W/m-K, c=400 J/kg and ρ=9000kg/m3 (Regular notations used). Determine the time (in seconds) when the temperature will become 390℃, given the convective heat transfer coefficient between body and fluid is 300W/m2.
Which of the following statement is correct regarding transportation
• North-West corner method gives initial feasible solution
• Hungarian method is used to solve transportation problem
Consider a project given below
The variance for the project will be
From the network, we can see that critical path is A – C – E (1 – 2 – 4 – 6).
Now variance of the project is
Guest’s theory of failure is best suited for
Maximum shear stress theory is given by Guest. According to this theory,
Since the shear stress at yield point in a simple tension test is equal to one-half the yield stress in tension, therefore the equation may be written as
This theory is mostly used for designing members of ductile materials.
In automobiles, the brakes used in hand braking systems are usually
In automobiles, the brakes used in hand braking systems are usually externally contracting brakes.
In AJM process nozzle tip distance changes with MRR
A 50mm thick strip rolled to 30mm thickness in 3 passes with equal reduction in each pass by using rolls of 600mm diameter, what is the minimum coefficient of friction required for rolling operation________.
A rectangular hole of size 50mm x 60mm is to be made on a 8 mm thick sheet having ultimate tensile strength and sheer strength of 500MPa & 300MPa respectively the punching force required (in KN) is__________.
Which of the following is not related to soldering operation:
Soldering operation is a solid-liquid (heterogeneous) welding process.
A metric thread of pitch 3mm and diameter 30mm, inspected for its pitch diameter using 2- wire method. The diameter of best size wire in mm is
For metric thread included angle is taken 600
Hardenability of steel can be improved by adding which of the following element
Molybdenum, Chromium, Nickel, Boron are used to increase hardenability.
The value of sensitivity for an isochronous governor is
Isochronous governor is that governor which has same speed for all position of sleeves and sensitivity equal to infinity. Therefore right option is (c).
Match list I (terms) with list II (definition) and select the correct answer using the codes given below the lists:
A) Base circle
B) Trace point
C) Pitch point
D) Prime circle
I. Smallest circle that can be draw from the centre of cam and tangent to pitch curve
II. Smallest circle that can be draw from the centre of cam through the pitch points
III. Point on pitch curve having the maximum pressure angle
IV. Point on pitch curve having the minimum pressure angle
V. Smallest circle that can be draw to the cam profile
VI. Reference point on follower and used to generate pitch curve
A B C D
Base circle → smallest circle that can be draw to the cam profile
Trace point → Reference point on follower and used to generate pitch curve
Pitch point → Point on pitch curve having the maximum pressure angle
Prime circle → smallest circle that can be draw from the centre of cam and tangent to the pitch curve
A milk chilling unit can remove heat from the milk at the rate of 42 MJ/h. The rate of heat leakage into the milk from the surrounding is 4.2 MJ/h. The specific heat of milk above the freezing point is 4.2 kJ/kg K and below the freezing point is 2.1 kJ/kg K. The freezing point of milk is -2 oc and the latent heat is 100 kJ/kg. The time required (in min) for cooling a batch of 100 kg milk from 47 oc to -5 oc is
Suppose the time required is t hrs then
The amount of heat removed
= heat removed to decrease the temperature from 47 oc to -2 oc + latent heat + heat removed to decrease the temperature from -2 oc to -5 oc
(42000 – 4200)t = 100 ×[4.2 × (47+2) + 100 + 2.1× (−2+5)]
t = 0.825 hr = 49.5 min
A power plant operating on a closed Brayton cycle operates between pressure of 1 bar and 8 bar. Compression and Expansion both are isentropic. The heat absorbed by the coolant per kg kW output of air which is used to cool the air after expansion in turbine is_____
Heat supplied per kg-kW output,
Heat supplied=2.23 kW
Hence, heat rejected = heat absorbed by the coolant= 1.23
The matrix 3×3 matrix A is transformed to the diagonal matrix D = P-1AP, where, P The possible Eigen vector of A is
The model matrix P, which diagonalizes A is obtained by grouping the Eigen vectors of A into square matrix
So, the Eigen vectors of A are (-1,-2,1), (2,-1,0), (3,0,1).
Out of these options only option 2 is correct which can be the Eigen vector of A.
The value of the integral evaluated around the closed curve |z-3| = 1 is
The poles are determined as
z2 - 4z + 3 = 0
(z−3)(z−1) = 0
z = 3, 1
There is only one pole z =3 lies inside the given circle |z-3| =1
So, the value of integral
If 5 unbiased dice are thrown simultaneously, the probability of obtaining exactly two sixes is
The velocity vector is given as The divergence of this velocity vector at (1,1,1)
The volume of solid bounded by the surface x = 0, y = 0 and x + y + z =4 and z = 0 is
z varies as 0 to 4 – x – y
y varies as 0 to 4 – x
x varies as 0 to 4
An airplane, travels through still air at 300 km/hr, is having two propellers of 2.5 m diameter. Take density of still air as 1.22 kg/m3. The difference of pressure (in kPa) across a propeller, if discharge through each propeller is 480 m3/s, will be
Let V1 and V2 be velocities at upstream and downstream section.
So, theoretical efficiency,
Water of mass density 1000 kg/m3 is transported using a pipe of diameter 40 cm. Flow in the pipe is turbulent. If the flow velocity at the centre and that at a 5 cm distance from centre of pipe are 5 m/s and 4 m/s respectively, then the value of wall shear stress (in Pa) will be
As we know, in pipe flow, maximum velocity occurs at the pipe centre.
A single acting reciprocating pump has a piston of 15 cm diameter and 22.5 cm stroke. The pump is required to deliver 0.29 m3 of water per minute at 75 rpm. If the suction and delivery heads are 3 m and 10 m respectively, then the percentage of slip of pump is
For a single acting pump,
The tensile stress at the bottom edge of a cast iron beam shown in figure is 50 MPa when it is subjected to a bending moment. The value of stress (in MPa) at the top will be
Consider the beam in figure given below. The reaction at point C and rotation at point B will be (flexural rigidity of beam = EI)
The moment at point B = 4Pa
In the cantilever beam ABC, the deflection at C due to moment 4Pa will be given as
The reaction at C will be upwards
Rotation at B due to moment
Rotation at B due to reaction at C
Rotation at B = rotation due to moment + rotation due to reaction
A thin cylindrical vessel of 3.2 m diameter and 5 m long is subjected to a circumferential tensile stress of 140 N/mm2 and axial tensile stress of 70 N/mm2. If the volumetric strain of the vessel is 14×10-4, then the Poisson’s ratio of the material will be (Neglect the compressive stress on the walls and take E = 2×105 N/mm2)
Now, the volumetric strain for cylindrical vessel
A 5 mm diameter and 4 m length steel wire is submerged in a fluid at 60oC. The conductivity and resistivity of wire is 20 W/m-deg and 70 micro-ohm-cm. If 600 amperes electric current passes through it and conductance at wire surface is 4 kW/m2-deg, then maximum temperature of the wire is
Given that, L = 4 m =400 cm, d = 5 mm = 0.5 cm
Heat generated per unit volume,
Maximum temperature occurs at geometric centre line of the wire, and can be computed from the relation,
In a certain heat exchanger, cold fluid enters at 40oC and rate of the flow is 1650 kg/hr. The hot fluid is at a temperature of 130oC and cold fluid leaves the heat exchanger at 90oC. Specific heat of the cold fluid is 3.5 kJ/kg-K. If heat transfer area is 0.707 m2, then the overall heat transfer coefficient is
Find the net value of heat flux between two parallel, infinite plates having emissivity of 0.75 and 0.7. The planes are maintained at 450 K and 500 K.
The grey body factor is given by,
Since infinite long plates can see each other. So, F12 = 1 and A1 = A2 So, the grey body factor will become
Now, the rate of heat interchange between the plates
And net heat flux,
A parallelepiped enclosure measures 3 m × 3.5 m with 4 m height. The bottom floor having 0.75 emissivity is maintained at 380 K and all other surfaces (emissivity = 0.8) are maintained at 505 K. The net radiation (in kW) to the floor will be
As per the given data,
A1 = area of four walls and ceiling
A hot plate of 40 cm × 40 cm maintained at 90oC is exposed to ambient air at 30oC. The appropriate correlation for the convective coefficient is
The properties of ambient air at 60oC are given as
thermal conductivity = 0.028 W/m-deg, cp = 1.008 kJ/kg-K,
mass density = 1.06 kg/m3, kinematic viscosity = 18.97×10-6 m2/s.
If the plate is kept vertical, the value of convective heat transfer coefficient will be
All the properties are measured at mean temperature.
A ball pen manufacturer produces refill and barrel. The work department is capable of producing 20000 refills or 10000 barrels or any other combinations of these per week. The demand for refills and barrels is limited to 15000 and 8000 units per week respectively. The profit per unit from refill is 1 rs. And that from barrel is 3 rs. The maximum profit to the manufacturer in rupees will be
Let x1 and x2 be number of refills and barrels respectively and Z denotes total profit earned.
Formulation of problem:
From the above tableau, we can conclude that x2 is the entering variable because it is having most negative Zj – Cj value. And S2 is exit variable because it has least positive ratio.
Revised simplex tableau:
Since there is no negative Zj – Cj value, so this is the final simplex table. The optimum mix of products for maximizing profit is
Refills = 4000, and barrels = 8000
The maximum profit is given by
Exponential smoothening method with smoothening index 0.4 is used to determine the forecast of certain product. The demand for August, September, October and November were 210, 220, 200 and 230 units respectively. Find the forecast for the month of December if the forecast of August was 250 units.
A steel shaft of 75 mm diameter and 630 MPa ultimate tensile strength is subjected to a torque which varies from 1500 N-m to -950 N-m. Assume the yield stress for steel in reversed bending as 510 N/mm2, surface finish factor as 0.87, size factor as 0.85 and fatigue stress concentration factor as 1. The factor of safety using Soderberg criteria will be
Consider the following figure of welded joint. If the allowable shear stress is 105 MPa for static loading and value of static load P is 350 kN, then the length (in mm) of the bottom weld will be
Given : x + y = 200 mm ; P = 200 kN = 200 × 103N ; ï = 75 MPa = 75 N/mm2
Let L1 = Length of weld at the top,
L2= Length of weld at the bottom, and
L = Total length of the weld = L1 + L2
Since the thickness of the angle is 20 mm, therefore size of weld, s = 20 mm
We know that for a single parallel fillet weld, the maximum load (P),
A shaft of 300 mm diameter is carrying a load of 20 kN at 2000 rpm. A bearing of 450 mm length supporting the shaft. The absolute viscosity of lubricating oil is 0.011 kg/m-s at the operating temperature and diametral clearance of the bearing is 0.30 mm. The power loss in bearing due to friction will be
Given : d = 300 mm = 0.3 m ; W = 20 kN = 20000 N ; N = 2000 r.p.m. ;
l = 450 mm; c = 0.30 mm ; Z = 0.011 kg/m-s
We know that, bearing pressur
A 150 mm diameter blind hole is to be drilled upto 60mm depth at a speed of 500 rpm and feed 0.2 mm per revolution, point angle of drill is 1100 . Machining time per hole (minute) __
A cast steel slab of dimension 40 x 20 x 10 cm, is poured horizontally using side riser. The riser is cylindrical in shape with diameter and height both equal to 20 cm.Then the freezing ratio is___.
The epicyclic gear arrangement with the compound planets 3-4 is shown in figure. The number of teeth on gears 2, 3, 4 and 5 are 60, 20, 30 and 20 respectively. The annular gear 2 is keyed to the propeller shaft X which rotates at 650 rad/s (CCW). The shaft Y is free to rotate with respect to gear 5 which is fixed and makes the same no. of revolution as the arm. If 100 kW power is supplied to gear 2 then the holding torque (in N-m) on gear 5 is (assume 100% efficiency throughout)
Given, T2 = 60, T3 = 30, T4 = 30, T5 = 20,ω2 = 650 rad/s,
ω5 = 0 rad/s, P =100 kW
First of all we need to find the angular velocity of arm 1.
Using algebraic method,
ω1 = 520 rad/s (CCW)
Since the shaft Y will make same no. of revolution as arms, so
Speed of shaft X = 520 rad/s (CCW)
Torque on driving shaft X or gear 2
Since 100 % efficiency throughout, so power available at gear 5 of shaft Y = 100 kW
Torque on driving shaft Y or gear 5
So, the holding torque on gear 5 = 192.3−153.84 = 38.46 N -m
A homogeneous solid disc of mass M and radius r is connected to a spring (spring constant =K) at a point a above the centre of the disc on an inclined plane as shown in figure. The other end of the spring is fixed to the vertical wall. If the disc rolls without slipping then the natural frequency of the spring in rad/s is
A two stage vapour compression plant with a direct contact heat exchanger uses R-12 as refrigerant. The evaporator and condenser temperature are -300C and 400C respectively. The condenser and evaporator pressure are 15.5 bars and 1.2 bars respectively. If the capacity of the plant is 25 tonnes of refrigeration then the COP of the plant is
Compression work = m1 (h2−h1) + m2 (h4−h3)
Capacity of plant = 25 tonnes = (25×14000)/ (3600) = 97.22 kJ/s
m1 (h1−h8) = 97.22 (∵ h7 = h8)
m1 (1404.6−181.5) = 97.22
m1 = 0.08 kg/s
m1 (h2−h7) = m2 (h3−h6)
0.08 × (1574.3−181.5) = m2 (1443.5−371.7) (∵ h5 = h6)
m2 = 0.104 kg/s
Compression work, W = 0.08 × (1574.3−1404.6) + 0.104 × (1628.1−1443.5)
W = 32.77 kW
So, COP = 97.22/32.77 = 2.97
A quantity of air initially at 1.5 bar, 270C brought to a final temperature of 127 0C by heat transfer from a thermal reservoir at 227 0C. The irreversibility is (take atmospheric temperature = 27 0C, atmospheric pressure = 1 bar, Cp = 1.005 kJ/kgK, Cv = 0.718 kJ/kgK)
Given, T1 = 270C =300 K, T2 = 1270C = 400 K, T = 2270C = 500 K, To = 300 K
Entropy change of universe,
An unsaturated air enters in an insulated chamber at 320C where it flows over a long sheet of water and becomes cooled to 260C, which is adiabatic saturation temperature. The total pressure of the mixture remains constant at 100 kPa. The corresponding pressure to saturation temperature is 3.363 kPa. The humidity ratio at the entry of chamber is (in kg vap/kg dry air)
Given data: Cp of air = 1.005 kJ/kgK
At 320C, hv1 = 2559.9 kJ/kg,
At 260C, hfg2 = 2439.9 kJ/kg, hf2 = 109.1 kJ/kg
Given, p =100 kPa, ps = 3.363 kPa
Since the system is insulted and no work interaction, so using conservation of energy
ma1ha1 + mv1hv1 + (mv2−mv1) hf2 = ma2ha2 + m2hv2 (i)
ma = mass of dry air (kg)
mv = mass of water vapour (kg)
ha = specific enthalpy of dry air (kJ/kg)
hf = specific enthalpy of liquid water (kJ/kg)
hv = specific enthalpy of water vapour in air (kJ/kg)
Divide the eqn (i) by ma1, we get
ha1 + W1 hw1 + (W2− W1) hf2 = ha2 + W2hv2 (∵ ma1 = ma2) (ii)
Where, W = specific humidity
Humidity ratio at exit,
(ha1 − ha2) + W2 (hf2 − hv2) = W1 (hf2−hv1)
Cp (T2 – T1) + W2 hfg2 = W1 (hv1−hf2) (∵ hv2 = hg2 and hfg2 = hg2 − hf2)
1.005 × (26−32) + 0.0216×2439.9 = W1 × (2559.9−109.1)
W1 = 0.019 kg vap/kg dry air
A four stroke double acting petrol engine is used 90 kW brake power on a job. The engine used 40 kg/h of fuel under working condition and its mechanical efficiency is 75%. If the engine friction power reduces by 10 kW then the saving of fuel (in kg/h) is (assume, indicated thermal efficiency remain constant)
Given, BP = 90 kW, m1= 40 kg/h, ηmech= 0.75
Initial indicated power,
(IP)1 = 90/0.75 = 120 kW
Frictional power, FP =IP−BP = 30 kW
New frictional power, FP = 30−10 =20 kW
(IP)2 = BP + IP =90 + 20 =110 kW
According to question,
The force in the member BD and CD respectively are
Suppose R1, R2 and R3 are the reactions at A and F as shown in figure.
R3 = 4 kN
R1 + R2 = 12
R1 × 8 = 12 × 4 + 4 × 6
R1 = 9 kN
R2 = 3 kN
cos θ = 6/7.21
sin θ = 4/7.21
At joint A
TAB = −9 kN (∵ considered tensile as positive)
At joint B
TBC cos θ + TAB = 0
TBC sin θ + TBD = 0
TBD = −10.815 × 4/7.21 = −6 kN = 6 kN (compressive)
A ball of mass M strikes the floor with a speed of V m/s at an angle of incidence θ with the normal as shown in figure. The coefficient of restitution is e. The speed of the reflected ball and the angle of reflection of ball respectively are
let the angle of reflection is α and the speed after collision is V2 m/s.
As there is no force parallel to the surface so,
V sin θ = V2 sin α (i)
In the vertical direction, the floor exerts a force on the ball.
As we know that
Velocity of separation = e (velocity of approach)
V2 cos α= e V cos θ (ii)
From (i) and (ii)
The function, f (x) =2 Inx, satisfies the Lagrange’s Mean Value Theorem in the interval x∈ [1, 5] then the value of x is
According to Lagrange’s Mean Value Theorem, if the function f (x) is continuous in the interval x∈ [a, b] and f’(x) exist in the interval x∈ (a, b) then
A function follows the following equation
The Laplace transform of f(t) is
Taking Laplace transform on both the sides
Consider the following differential equation
If y (2) = 1 then the value of y (2.2) using Runge- Kutta third order method is
An ideal gas of mass 0.5 kg has a pressure of 600 kPa, a temperature of 770C and a volume of 0.08 m3. This gas undergoes an irreversible process to a final pressure of 600 kPa and final volume of 0.12 m3, during which the heat addition and work done on gas are 2 kJ and 30 kJ respectively. The value of Cv of gas is
Given, m =0.5 kg, p1 = p2 =600 kPa, T1 = 770C = 350 K, V1 = 0.08 m3, V2 = 0.12 m3, Q =2 kJ, W = −30 kJ
From ideal gas eqn,
p1V1 = mRT1