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1. This question paper contains 65 objective & Numerical questions. Bifurcation of the questionare given below:

2. Choose the closest numerical answer among the choices given.

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SECTION-I (General Aptitude)

Multiple Choice Questions : Q. No. 1 to Q. No. 4 carry 1 mark each

Q.1Q.1Q.1Q.1Q.1 Rajat’s house is facing east. One fine morning, after coming out of his house, he walks a distanceof 1 km and then takes a left turn but keeps walking. He takes a left turn after having walked twicethe distance in the previous step. He keeps doing this i.e. taking left turn and walking twice thedistance of the previous step just before he is due to take a left turn a 6th time. Where is he withrespect to the starting point?(a) 13 km E and 26 km N (b) 13 km W, 26 km S(c) 12 km E and 24 km N (d) 26 km E, 13 km N

Q.2Q.2Q.2Q.2Q.2 In the sentence given below, some part of the sentence needs an improvement. Identify theoption which makes a meaningful and grammatically correct sentence:A taller cadet rushed forward than any of his comrades.A taller cadet rushed forward than any of his comrades.A taller cadet rushed forward than any of his comrades.A taller cadet rushed forward than any of his comrades.A taller cadet rushed forward than any of his comrades.(a) A cadet, taller than any of his comrades rushed forward.(b) A cadet rushed forward, taller than any of his comrades.(c) A cadet rushed forward than any of his taller comrades.(d) no need of improvement

Q.3Q.3Q.3Q.3Q.3 In this question, a sentence has been split into 4 parts labeled as ‘a’ to ‘d’. Only one of these partshas an error. Error could be in the form of a missing word, an extra word or in grammar. Identify thepart that has an error.(a) The Indian radio(b) which was previously controlled by the British rulers(c) is free now from the narrow vested interests(d) No error

Q.4Q.4Q.4Q.4Q.4 Choose the most appropriate word to fill in the blanks in the given sentence:

Each occupation has its own _______; bankers, lawyers and IT professionals, for example, all useamong themselves a language which outsiders have difficulty in assimilating.(a) problems (b) jargon(c) rewards (d) compound

Numerical Data Type Question : Q.5 carry 1 mark

Q.5Q.5Q.5Q.5Q.5 In a test given by Prof Virus during the minors, Raju gets 32% marks and fails by 20 markswhereas Chaturbhuj gets 30 marks more than passing marks while scoring 42% marks. Themarks required to get twice the passing marks are ____________.



Multiple Choice Questions : Q. No. 6 to Q. No. 9 carry 2 marks each

Q.6Q.6Q.6Q.6Q.6 Given below are 5 statements out of which 3 are logically related to each other. Out of the 4options having a set of 3 statements each, identify the set where the statements are most closelyrelated to each other. i. All flowers are fragrant.ii. All flowers are majestic.iii. All flowers are plants.iv. All flowers need air.v. All plants need air.(a) iii, v and iv (b) i, iii and ii(c) ii, iv and iii (d) iii, v and ii

Q.7Q.7Q.7Q.7Q.7 There were p pigeons and s sparrows in a cage. One fine morning m birds escaped to freedom.If the caretaker was able to figure out by only looking (not counting) into the cage that at least onepigeon had escaped, based on the knowledge that m = 7, then which of the following cannot bethe value of the number of sparrows and pigeons?(a) 8 and 10 (b) 4 and 12(c) 6 and 25 (d) 2 and 7

Q.8Q.8Q.8Q.8Q.8 Assume that equal number of people are born each day of the year. If the century under considerationis the 20th century, what is the percentage of people who are born on 29th February?[Assume no deaths occur during the period under consideration].

(a) 0.685 (b) 0.0685(c) 0.684 (d) 0.0684

Q.9Q.9Q.9Q.9Q.9 Two trains T1 and T2 are running at speed of 60 km per hour and 45 km per hour respectively. T1leaves from station ‘A’ for station ‘B’ and train T2 leaves from ‘B’ for ‘A’ at the same time – it isgiven that distance between A and B is 1260 km. They cross each other at station ‘C’ between ‘A’and ‘B’. Both strains start their return journey immediately after reaching their respective destinations.After how much time (in hours) of T2 reaching ‘A’, will they cross each other again?(a) 6 (b) 8(c) 10 (d) 12

Numerical Data Type Question : Q.10 carry 2 marks

Q.10Q.10Q.10Q.10Q.10 Rajaram’s monthly income is distributed in 5 expense heads namely house rent, groceries, transportexpenses, electricity and water expenses and finally under the head miscellaneous expenditure.After meeting these expenses, the remaining amount (which is non zero) goes towards savingswhich is a fixed percentage of his income every month. It is known that these amounts can beexpressed as percentage of his income as 6 distinctive integers. The percentage expensesunder 5 heads are 5 integers such that difference between 2 consecutive expenses is a constant.The minimum percentage of his income that he saves per month is ____________.



Section -II (Technical + Engineering Mathematics)

Multiple Choice Questions : Q.1 to Q.12 carry 1 mark each

Q.1Q.1Q.1Q.1Q.1 Match items from groups I, II and III :Group -I Group-II Group-III(Structure) (Effective no. (Co-ordination No.)

of atoms)AAAAA..... Simple cubic PPPPP..... 2 TTTTT..... 8BBBBB..... Body centered QQQQQ..... 1 UUUUU..... 6CCCCC..... Face centered RRRRR..... 4 WWWWW..... 12DDDDD..... Hexagonal closed packed SSSSS..... 6Codes:Codes:Codes:Codes:Codes:

AAAAA BBBBB CCCCC DDDDD(a) (P, T) (Q, U) (S, W) (R, W)(b) (Q, U) (P, T) (R, W) (S, W)(c) (P, U) (Q, T) (R, W) (S, W)(d) (Q, U) (R, T) (S, W) (P, T)

Q.2Q.2Q.2Q.2Q.2 A Carnot heat engine is operating between a source at TH and a sink at TL. If it is desired to doublethe thermal efficiency of this engine, keeping the sink temperature constant. What should be thenew source temperature (TH) ?

(a)��

� �

� �

� �

−(b)

��

� �

� �

� �

−

(c)� ��

� �

� �

� �

−(d)

��

� �

� �

� �−

Q.3Q.3Q.3Q.3Q.3 5 m3 of an ideal gas at 1 bar pressure is compressed isothermally in an adiabatic piston-cylinderdevice to 2 m3. Maximum work input needed by the compressor is

(a) 250 kJ (b) 300 kJ(c) 500 kJ (d) Compression is not possible

Q.4Q.4Q.4Q.4Q.4 The time in which the mass in a damped vibrating system would settle down to 1/50th of its initialdeflection for the given data will be ?

m = 200 kg

ξ = 0.223

k = 40 N/mm(a) 0.221 s (b) 0.455 s(c) 0.622 s (d) 1.260 s

Q.5Q.5Q.5Q.5Q.5 During the process of boiling and condensation, only a phase change take place and one fluidremains at constant temperature throughout the heat exchanger. In terms of number of transferunits (NTU), the effectiveness of such an heat exchanger would be



(a)�

��

��

⎛ ⎞⎜ ⎟⎝ ⎠+ (b) 1– exp (– NTU)

(c)� ��

�

��− −(d) Undefined solution

Q.6Q.6Q.6Q.6Q.6 A beam simply supported at equal distance from the ends carries equal loads at each end. Whichof the following is true ?

(a) The bending moment is minimum at the mid-span(b) The bending moment is minimum at the support(c) The bending moment varies gradually between the supports(d) The bending moment is uniform between the supports

Q.7Q.7Q.7Q.7Q.7 The diagram below represents the shear force variation over a beam AE, simply supported at Aand D, for a particular set of loading.

B C D E

72 kN

30 kN

A

– 28 kN

– 46 kN

10 m 5 m 5 m 5 m

For the same set of loading, magnitude of bending moment (in kN-m) at mid-span will be(a) 150 (b) 220(c) 225 (d) None of these

Q.8Q.8Q.8Q.8Q.8 The main purpose of chaplets is(a) to ensure directional solidification (b) to provide efficient venting(c) for aligning the mould boxes (d) to support the cores

Q.9Q.9Q.9Q.9Q.9 In MIG welding, the metal is transferred in the form of?(a) A fine spray of metal (b) Molten drops(c) Weld pool (d) Molecules

Q.10Q.10Q.10Q.10Q.10 The cut-off ratio of a diesel cycle is 2. For equal air-standard efficiencies of Otto and Diesel cycle,the ratio of compression ratio of diesel cycle to that of Otto cycle would be(a) 0.5 (b) 1.0(c) 1.5 (d) 2.0

Q.11Q.11Q.11Q.11Q.11 The centre of gravity of the coupler link in a 4-bar mechanism would experience(a) no acceleration (b) only linear acceleration(c) only angular acceleration (d) both linear and angular acceleration



Q.12Q.12Q.12Q.12Q.12 The dia of a hole is given as ��+

+ mm. The upper limit on the dimensions in mm, of the shaft for

achieving maximum interfence of 50 microns is(a) 60.05 mm (b) 60.15 mm(c) 60.25 mm (d) 60.40 mm

Multiple Choice Questions : Q. 13 to Q. 15 carry 1 mark each

Q.13Q.13Q.13Q.13Q.13 The equation uxx + xuxy + yuyy – xyux = 0 is hyperbolic for

(a)�

� >

x(b)

�

� <

x

(c)�

� =

x(d) y = x

Q.14Q.14Q.14Q.14Q.14 The solution of the differential equation

( ) ( )��

�

−+ + − =x

x

(a)� ��

− −= +x (b)

�� − −= +x

(c)� ��

− −= +x (d)

�� −−− =x

Q.15Q.15Q.15Q.15Q.15 The value of �

� ��∫ , where ‘C ’ is the left half of the unit circle, is

(a) 0 (b) 2i(c) –2i (d) None of these

Numerical Data Type Questions : Q.16 to Q.23 carry 1 mark each

Q.16Q.16Q.16Q.16Q.16 A full journal bearing with a journal of 75 mm diameter and a bearing of 100 mm length issubjected to a radial load 3000 N at 500 rpm. If the viscosity of oil is 18 centipoise then thebearing modulus is __________ × 10–7.

Q.17Q.17Q.17Q.17Q.17 Two carnot heat engines are operating in series. The first engine receives heat from the reservoir at1400 K and rejects the waste heat to another reservoir at temperature T. The second enginereceives this energy rejected by the first one, converts some of it to work; and rejects the rest toa reservoir at 300 K. If thermal efficiencies of both engines are the same, then the temperature Tis ____________ K.

Q.18Q.18Q.18Q.18Q.18 Dry saturated steam at a pressure of 10 bar enters a convergent - divergent nozzle and leaves ata pressure of 2 bar. If the flow is adiabatic and frictionless, pressure at the throat of the nozzle is________ bar. (For saturated steam take, n = 1.135)



Q.19Q.19Q.19Q.19Q.19 The circulation around a rectangle defined by

x = 1, y = 1, x = 5 and y = 4

for the velocity field u = 2x + 3y, v = – 2y will be ______ units.

Q.20Q.20Q.20Q.20Q.20 A mild steel plate has to be rolled in one pass such that the final plate thickness is �

�

�⎛ ⎞⎜ ⎟⎝ ⎠ of the

initial thickness, with the entrance speed of 10 m/min and roll diameter of 500 mm. If the platewidens by 2% during rolling, the exit velocity (in m/min) is______ .

Q.21Q.21Q.21Q.21Q.21 In a whitworth quick return motion mechanism, the distance between the fixed centres is 50 mmand the length of the driving crank is 75 mm. The length of slotted lever is 150 mm and lengthof the connecting rod is 135 mm. The ratio of the time of cutting stroke to the time of returnstroke is _________ .

30° C

A

D

P

R

Q.22Q.22Q.22Q.22Q.22 A orange consists of 12 ear-shaped pieces. Two piece diametrically opposite to it are removedand eaten. If initially the mass moment of Inertia about an axis passing through the centre of the

orange, is Ii and after the pieces are eaten is If , then the value of �

i

II is ____________.

Q.23Q.23Q.23Q.23Q.23 A cylindrical pin fin of diameter 5 cm and length 20 cm with negligible heat loss from the tip has anefficiency of 0.5. The effectiveness of this fin is ___________.

Numerical Data Type Questions : Q.24 to Q.25 carry 1 mark each

Q.24Q.24Q.24Q.24Q.24 The probability that A speaks truth is 4/5, while this probability for B is 3/4. The probability thatthey contradict each other when asked to speak on a fact is ___________.

Q.25Q.25Q.25Q.25Q.25

� ��→

−x

xx

is ____________.




Q.26Q.26Q.26Q.26Q.26 Saturated ammonia at 0.2365 MPa enters a single acting compressor and liquid ammonia at 21°Centers the expansion valve of a refrigerating plant.

If head pressure is 1.1638 MPa, then the COP of the refrigerating cycle is:

Given: CP (Liquid NH3) = 4.6 kJ/kgK ; CP (Vapour NH3) = 2.75 kJ/kgK

P(MPa) Sat. temp. °C Sp. Vol. of

vapourEnthalpy ofliquid NH3

0.23571

1.1638

– 15°C

+30°C

0.5106

0.11084

168.54

379.3

1481.52

1523.42

5.4387

6.1853

10.526

9.9606

Enthalpy ofvapour NH3

Entropy ofliquid NH3

Entropy ofvapour NH3

m /kg3 kJ/kg kJ/kg kJ/kg kJ/kg

(a) 5.49 (b) 4.94(c) 4.14 (d) 6.14

Q.27Q.27Q.27Q.27Q.27 A mild steel plate 400 mm × 800 mm × 30 mm is to be shaped along its wider face. The ratio ofreturn time to cutting time is 2 : 3 and the feed per cycle is 2 mm. Tool approach and over-travelrespectively are 50 mm latch and cutting speed of tool is 24 m/min. The machining time requiredfor machining the given plate with WSS tool will be

(a) 5 min (b) 20 min(c) 13 min (d) 17 min

Q.28Q.28Q.28Q.28Q.28 A 100-litre electrical radiator containing heating oil is placed in a 100 m3 well sealed room atatmospheric pressure of 1 bar. Initially both room and radiator are at 10°C. The radiator with arating of 1.8 kW is turned on. At the same time, heat is lost from the room at an average rate of0.35 kJ/s. After some time, the average temperature is measured to be 20°C for the air in theroom, and 50°C for the oil in the radiator.

Density and specific heat of the oil are 950 kg/m3 & 2.2 kJ/kg °C respectively. How long the heateris kept on?

(a) 56 min (b) 106.25 min(c) 106.25 sec (d) 5821 sec

Q.29Q.29Q.29Q.29Q.29 A freezing plant requires 50 tons of refrigeration. The freezing temperature is – 33°C and ambienttemperature is 27°C. If the performance of the plant is one-third of the theoretical reversed carnot cycleworking between the same temperature limits, the power input (in kW) required for the plant is

(a) 90.25 (b) 131.25(c) 80.15 (d) 151.15

Q.30Q.30Q.30Q.30Q.30 Two component of a machine are connected by a bolt. The stiffness of bolt is one-third of thecombined stiffness of two components. Pre loading of bolt is 5 kN due to initial tightening of bolt.The bolt material has σyt = 400 MPa and FOS = 2. To sustain the external force of 15 kN, diameterof the bolt should be

(a) 5 mm (b) 8 mm(c) 2 mm (d) 4 mm



Q.31Q.31Q.31Q.31Q.31 For an assignment problem, Initial basic feasible solution is given below.

10

0

10

5

2

10

12

12

6

0

14

25

19

4

0

25

17

0

7

5

0

15

10

0

7

P

Q

R

S

T

I II III IV V

The optimal assignment is(a) P – V , Q – I, R – IV, S - III, T – II (b) P – V, Q – I, R – III, S - IV, T – II(c) P – V, Q – II, R – III, S - IV, T – I (d) P – I, Q – II, R – III, S - IV, T – V

Q.32Q.32Q.32Q.32Q.32 A bush manufacturing company has a contract to supply 5000 bushes to an automobile factory perday. The company has capacity to manufacture 8, 000 bushes and holding cost of 1000 bushesis 8 paise. Set up cost is ` 20. The frequency of production run is approximately

(a) 11 days (b) 16 days(c) 20 days (d) 23 days

Q.33Q.33Q.33Q.33Q.33 The copper thermocouple, spherical in shape, initially at 25°C, which when placed in a gasstream at 200°C measures a temperature of 198°C in 5 seconds. For copper, ρ = 8940 kg/m3,C = 384 J/kg-K, k = 390 W/m-K and the convective heat transfer coefficient = 400 W/m2-K. Thediameter of the thermocouple is

(a) 0.58 mm (b) 0.78 mm(c) 1.23 mm (d) 1.68 mm

Q.34Q.34Q.34Q.34Q.34 For a maximization problem, after some simplex iteration the following table is obtained

� � � �

�

�

��

��

�

� ��

� �

��

��

� ��

�

� �

� � �

� � �

�

�

� �

� �

−

− +

= Σ −

Δ = − −

i

i i

x x i

x

From this, one can conclude that(a) The LPP has a unique optimal solution (b) The LPP is infeasible(c) The LPP is unbounded (d) The LPP has multiple optimal solution



Q.35Q.35Q.35Q.35Q.35 A solid shaft of diameter d and length L is fixed at both the ends. A torque, T0 is applied at adistance, L/4 from the left end as shown in the figure given below.

L/4 3 /4L

To

The difference of magnitude of maximum and minimum shear stress in the shaft is

(a)�

��

�π(b)

�

��

�π

(c)�

��

�π(d)

�

�

�π

Q.36Q.36Q.36Q.36Q.36 The power absorbed by fluid friction in a thrust bearing consisting of a flat disc 10 cm in diameterplaced at the lower end of a vertical shaft is 223.93 W. The oil film is 0.25 mm thick and theviscosity of the oil is 1.3 poise. The angular speed of shaft will be

(a) 1000 rpm (b) 1500 rpm(c) 2000 rpm (d) 2500 rpm

Q.37Q.37Q.37Q.37Q.37 A cylindrical vessel of 1 m diameter has water in it to a height of 2 m. Oil of relative density 0.75is kept over the water column for another 1 m. Above the oil, a dead weight of 5 kN, base diameter1 m is placed. The total hydrostatic force and the ‘centre’ of pressure on a gate 30 cm in diameterplaced along the vertical surface of the vessel such that the lowest point of the gate is on the baseof the vessel will be?

(a) f = 1030 N, zp = 1.25 m (b) f = 2000 N, zp = 2 m(c) f = 2427 N, zp = 3.50 m (d) f = 3000 N, zp = 4 m

Oil

Water

5 kN

Q.38Q.38Q.38Q.38Q.38 A body of mass 1 kg is acted on by a force as shown in the diagram. If the velocity of the bodyis zero at t =0, the velocity and distance traversed to t = 1 min will be?

[The force acts for only 45 sec]



Force

14 N10 N

10s 30s 45sTime

(a) V = 435 m/s, S = 4.3 km (b) V = 235 m/s, S = 8.9 km(c) V = 435 m/s, S = 16.3 km (d) V = 235 m/s, S = 29.8 km

Q.39Q.39Q.39Q.39Q.39 The voltage current characteristics of a dc generator for arc welding is a straight line between anopen-circuit voltage of 80 V and short circuit current of 300 Ampere. The generator settings for themaximum arc power will be

(a) 80 V and 150 A (b) 40 V and 300 A(c) 40 V and 150 A (d) 80 V and 300 A

Q.40Q.40Q.40Q.40Q.40 Match Group-I Group-I Group-I Group-I Group-I with Group-II Group-II Group-II Group-II Group-II and select the correct answer using the codes given below the lists:Group-IGroup-IGroup-IGroup-IGroup-I Group-IIGroup-IIGroup-IIGroup-IIGroup-II

PPPPP..... G09 1.1.1.1.1. Linear InterpolationQ.Q.Q.Q.Q. G41 2.2.2.2.2. RetardationR.R.R.R.R. G01 3.3.3.3.3. Circular interpolationS.S.S.S.S. G03 4.4.4.4.4. Cutter radius compensationCodes:Codes:Codes:Codes:Codes:

PPPPP QQQQQ RRRRR SSSSS(a) 3 1 4 2(b) 4 1 3 2(c) 3 4 1 2(d) 2 4 1 3

Q.41Q.41Q.41Q.41Q.41 A pinion of 20 involute teeth and 125 mm pitch circle diameter drives a rack. The addendum ofboth pinion and rack is 6.25 mm. The length of arc of contact corresponding to the least pressurewhich can be used to avoid interference is

(a) 36.68 cm (b) 36.68 mm(c) 72.44 cm (d) 72.44 mm

Q.42Q.42Q.42Q.42Q.42 Consider the following statements relating to forecasting.

1.1.1.1.1. The time horizon to forecast depends upon where the product currently lies in its life cycle.

2.2.2.2.2. Opinion and Judgemental forecasting method sometimes incorporate statistical analysis.

3.3.3.3.3. In exponential smoothing, low values of smoothing constant, alpha result in more smoothingthan higher value of alpha.



Which of these statements are correct?(a) 1, 2 and 3 (b) 1 and 2 only(c) 1 and 3 only (d) 2 and 3 only


Q.43Q.43Q.43Q.43Q.43 An eigen vector of the matrix, M =

� � �

� � �

� � �

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

, is

(a)

�

�

�

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

(b)

�

�

�

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

(c)

�

�

�

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

(d)

�

�

�

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

Q.44Q.44Q.44Q.44Q.44 The value of � �

�

� � �� +∫ ∫x

x x is

(a) � � (b)� � �

�

−

(c)� � �

�

−(d)

� � �

�

−

Q.45Q.45Q.45Q.45Q.45 Consider the following equations

x + y + w = 0

y + z = 0

x + y + z + w = 0

x + y + 2z = 0The system of equations will be(a) consistent having a unique solution but not trivial(b) consistent having many solutions(c) consistent having a trivial solution(d) inconsistent having no solution

Numerical Data Type Questions : Q.46 to Q.53 carry 2 marks each

Q.46Q.46Q.46Q.46Q.46 Air at 27° and 1 bar is heated at constant volume until the pressure is 5 bar and thencooled at constant pressure back to original temperature. Net decrease in entropy of theprocess is __________ kJ/kgK.



Q.47Q.47Q.47Q.47Q.47 A 20 full depth involute system have a pinion of diameter 300 mm and number of teeth on pinionis 60. If the length of line of action is 20 mm, then the contact ratio is _______.

Q.48Q.48Q.48Q.48Q.48 Air (1 atm; 5° C) with free stream velocity of 2 m/s flowing over a stationary thin 1 m × 1 m flat plate.The flat plate has a uniform surface temperature of 35° C. If the friction force asserted on the flatplate is 0.1 N, the rate of heat transfer from the plate is ________ W.

Properties of air (1 atm; 5° C) is

ρ = 1.269 kg/m3

Cp = 1006 J/kg-K

μ = 1.729 × 10–5 kg/m-s

Pr = 0.7350

Q.49Q.49Q.49Q.49Q.49 A flat steel plate is of trapezoidal form. The thickness of the plate is 15 mm and it tapers uniformlyfrom a width of 60 mm to a width of 10 mm in a length of 300 mm. The elongation of the plate underan axial force of 120 kN is __________ mm.

[Take E = 2.04 × 105 N/mm2]

Q.50Q.50Q.50Q.50Q.50 A symmetric I -section (with width of each flange = 50 mm, thickness of web = 10 mm) of steel issubjected to a shear force of 100 kN. Find the magnitude of the shear stress in the web at itsjunction with the top flange is ___________ N/mm2.

10mm

100 mm

10 mm

10 mm

50 mm

50 mm

Q.51Q.51Q.51Q.51Q.51 A slender rod weighing 140 N is held by a frictionless pin at A and by a spring having a springconstant of 8.80 N/mm at B. The natural frequency of oscillation for small vibrations will be___________ Hz.

3.3 m

1 m

A B



Q.52Q.52Q.52Q.52Q.52 During pure orthogonal turning operation of a hollow cylindrical pipe, it is found that the thicknessof the chip produced is 0.5 m. The feed given to the zero degree rake angle tool is 0.25 mm/rev.The shear strain produced during the operation is________ .

Q.53Q.53Q.53Q.53Q.53 A liquid of specific gravity 1.5 is discharged from a tank through a siphon whose summit point is1.2 m above the liquid level in tank. The siphon pipe has a diameter of 0.1 m and it discharges theliquid into atmosphere at 100 kPa. If liquid vapour pressure is 28 kPa, the safe location of outletis ____________ m below the liquid level in the tank.

1

DictumLevel

1.2 m2

d = 0.1 m

Summit

Numerical Data Type Questions : Q.54 to Q.55 carry 2 marks each

Q.54Q.54Q.54Q.54Q.54 If x is real, the maximum value of �

�

��

�

+ ++ +

x xx x

is ____________.

Q.55Q.55Q.55Q.55Q.55 In an experiment with 15 observations on x, the following results we available� �� ∑ = ∑ =x x

One observation that was 20 was found to be wrong and was replaced by the correct value 30.Then the corrected variance is _____________.



1.1.1.1.1. (b)

2.2.2.2.2. (d)

3.3.3.3.3. (d)

4.4.4.4.4. (d)

5.5.5.5.5. (c)

6.6.6.6.6. (d)

7.7.7.7.7. (a)

8.8.8.8.8. (d)

9.9.9.9.9. (b)

10.10.10.10.10. (c)

11.11.11.11.11. (d)

12.12.12.12.12. (a)

13.13.13.13.13. (b)

14.14.14.14.14. (a)

29.29.29.29.29. (b)

30.30.30.30.30. (b)

31.31.31.31.31. (a)

32.32.32.32.32. (b)

33.33.33.33.33. (b)

34.34.34.34.34. (c)

35.35.35.35.35. (b)

36.36.36.36.36. (c)

37.37.37.37.37. (c)

38.38.38.38.38. (c)

39.39.39.39.39. (c)

40.40.40.40.40. (d)

41.41.41.41.41. (b)

42.42.42.42.42. (c)

15.15.15.15.15. (c)

16.16.16.16.16. 3.75 (3.25-4.25)

17.17.17.17.17. 648 (647-649)

18.18.18.18.18. 5.77 (5.5-6.0)

19.19.19.19.19. –36

20.20.20.20.20. 14.7 (14.6-14.8)

21.21.21.21.21. 2.74 (2.72-2.75)

22.22.22.22.22. 0.83 (0.81-0.85)

23.23.23.23.23. 8 (7.9-8.1)

24.24.24.24.24. 0.35

25.25.25.25.25. 1.41 (1.40-1.42)

26.26.26.26.26. (b)

27.27.27.27.27. (c)

28.28.28.28.28. (b)

43.43.43.43.43. (a)

44.44.44.44.44. (d)

45.45.45.45.45. (c)

46.46.46.46.46. 0.46 (0.45-0.47)

47.47.47.47.47. 1.35 (1.3-1.4)

48.48.48.48.48. 1852.76 (1850-1855)

49.49.49.49.49. 0.42 (0.40-0.44)

50.50.50.50.50. 71.12 (70-73)

51.51.51.51.51. 7.88 (7.86-7.90)

52.52.52.52.52. 2.5 (2.4-2.6)

53.53.53.53.53. 3.69 (3.67-3.71)

54.54.54.54.54. 41

55.55.55.55.55. 78

Section-II (Technical + Engineering Mathematics)

Section-I (General Aptitude)1.1.1.1.1. (a)

2.2.2.2.2. (a)

3.3.3.3.3. (c)

4.4.4.4.4. (b)

5.5.5.5.5. 360

6.6.6.6.6. (a)

7.7.7.7.7. (a)

8.8.8.8.8. (d)

9.9.9.9.9. (b)

10.10.10.10.10. 5

ANSWERSANSWERSANSWERSANSWERSANSWERS

MOCK TEST-1Mechanical Engineering



EXPLANATIONS

Section -I (General Aptitude)

1.1.1.1.1. (a)(a)(a)(a)(a)

L6

L5L4

L1

L3 L2

S

In the above diagram, ‘S’ denotes the starting point and L1 to L6 denote the left turns taken byRajat. He stops just when he is ready to take the 6th left turn. This point is 26 units above thestarting line and 13 units to the right of the starting vertical line i.e. 13 east and 26 north.

5.5.5.5.5. (360)(360)(360)(360)(360)

The difference in percentage marks obtained by Raju and Chaturbhuj is 42 – 32 = 10% whereas thedifference in their marks is 20 + 30 = 50 (Raju gets 20 marks less while Cahturbhuj gets 30 marksmore than pass marks).

This gives us 1% = 5 marks and the passing marks as 20 + 32% = 180.

The required marks = 360 (twice the passing marks)



6.6.6.6.6. (a)(a)(a)(a)(a)

Because all flowers are plants and all plants need air, it follows that flowers need air.

7.7.7.7.7. (a)(a)(a)(a)(a)

The maximum number of sparrows flew away are 6 because at least 1 pigeon has escaped as perto the caretaker’s observation.

The caretaker was able to make the judgement without counting that at least 1 pigeon has escapedbecause all the sparrow were gone.

This means number of sparrow can have a maximum value of 6.

8.8.8.8.8. (d)(d)(d)(d)(d)

In 20th century i.e. 1901 to 2000, there are 25 leap years.

The total number of people born on 29th February = 25p and the total number of people born inentire 20th century = 36500p + 25p = 36525p; giving the required percentage of people born on29th February = (25p/36525p)*100 = 0.0684

Caution:Caution:Caution:Caution:Caution: A common error likely to be committed is to divide 25p by 365p which will give therequired percentage as 0.0685 which is also one of the choices OR forget to multiply 25p/36525pby 100 getting a value = 0.000684 which also happens to be one of the options.

9.9.9.9.9. (b)(b)(b)(b)(b)

T1 takes 1260/60 = 21 hours to reach ‘B’ and train T2 takes 1260/45 = 28 hours to reach ‘A’.

Before train ‘T2 starts its return journey, T1 has already covered 60*(28 – 21) =420 Km.

At this instant, distance between T1 & T2 is 1260 – 420 = 840 Km.

The time taken now for the trains to cross each other is 840/(45 + 60) = 8 hours.

10.10.10.10.10. (5)(5)(5)(5)(5)

A careful reading of the question reveals that the expenses under 5 given heads forms an arithmeticprogression.

Let ‘S’ denote the savings percentage and ‘a’ is the percentage expenditure under the head withminimum expenditure.

Percentage expense under the 5 heads can be taken as a, a + d, a + 2d, a + 3d and a + 4drespectively.

Sum of these expenses is 5a + 10d and we can say that

5a + 10d + S = 100 or 5a + 10d = 100 – S

Since each expense head is a distinct integer, it means that 100 – S has to be a multiple of 5 andif we have to get minimum value for ‘S’, it is 5.



Section -II (Technical + Engineering Mathematics)

2.2.2.2.2. (d)(d)(d)(d)(d)

Denoting the new source temperature by TH

ηth-I = � �

�

�

�−

and ηth-II =�

�

�

��

�− = 2ηth-I

⇒�

�

�

��

�− = � � �

�

�

�

⎛ ⎞−⎜ ⎟⎝ ⎠

Solving for TH*, TH* =�

� �

� �

� �

� �−

3.3.3.3.3. (d)(d)(d)(d)(d)Isothermal adiabatic compression process is not possible because:ΔU = 0 for isothermal processand Q = 0 for adiabatic process.So, the first law relation, Q – W = ΔU. Reduces to W = 0.Therefore, this adiabatic system cannot-receive any net work at constant temperature.

4.4.4.4.4. (d)(d)(d)(d)(d)

�

�

�

�= � ��ξω

ωn =� � ��

� ��

�

�

×= =

50 = e0.22 × 14.14 × (NTd)

⇒ NTd = 1.26 s

Total time = 1.26 s

7.7.7.7.7. (a)(a)(a)(a)(a)

B C D E

72 kN

30 kN

A

– 28 kN

– 46 kN

10 m 5 m 5 m 5 m



(M)x = 12.5 m = Area of shear force diagram= 2.5 × 28 + 5 × 46 – 30 × 5= 150 kN-m

Alternate:Alternate:Alternate:Alternate:Alternate:Shear force at A is 72 kN⇒ Support reaction at A = 72 kNThere is a linear variation of shear force from A to B.

⇒ UDL between A and B = ��

��

− − = 10 kN/m. There is a sudden decrease in shear force at C.

⇒ Downward load at C = 46 – 28 = 18 kN⇒ Change in shear force at D gives support reaction⇒ Support reaction at D = 30 – (–46) = 76 kN. Shear force suddenly dropped at E.⇒ Downward load at E = 30 kN.Loading diagram

10 kN/m 18 kN 30 kN

76 kN72 kN

10 m 5 m 5 m 5 mMid span is 12.5 m from both ends A and E

10 kN/m

M

72 kN

10 m 2.5 m

Bending moment at mid span, M = 72 (12.5) – 10 × 10 (2.5 + 5)= 150 kN-m

8.8.8.8.8. (d)(d)(d)(d)(d)

Chaplets are used to support core inside the mould cavity to take care of its own weight andovercome the static forces.

10.10.10.10.10. (c) (c) (c) (c) (c)ρ = 2

ηOtto = ηDiesel

⇒�

��

��

γ−⎛ ⎞− ⎜ ⎟

⎝ ⎠=

��

� ��

γ− γ⎛ ⎞ ρ −− ×⎜ ⎟ γ ρ −⎝ ⎠



⇒�

�

��

γ−⎛ ⎞⎜ ⎟⎝ ⎠

=� ��

��

γ −⎛ ⎞ −×⎜ ⎟ −⎝ ⎠

⇒ �

⎛ ⎞⎜ ⎟⎝ ⎠

=

��

��

−⎡ ⎤− =⎢ ⎥⎣ ⎦

12.12.12.12.12. (a)(a)(a)(a)(a)Interference = Maximum shaft diameter – Minimum hole diameter

0.05 = Shaft maximum– 60⇒ Shaftmaximum = 60.05 mm.

13.13.13.13.13. (b)(b)(b)(b)(b)

uxx + xuxy + yuyy – xyux = 0

Given that the P.D.E. is hyperbolic

x 2 – 4y > 0

⇒ x 2 > 4y

⇒ 4y < x 2

∴ y <�

x

14.14.14.14.14. (a)(a)(a)(a)(a)

From the given equation

( )��

��

+ +x x = e tan–1y

⇒ �

�

�

�

� +

+x x =

��

��

�

�

−

+

⇒ x · IF = !� �� ∫

where, IF =��

�

��

�� −+

∫=

⇒��

−x =

�� −

+

15.15.15.15.15. (c)(c)(c)(c)(c)

Given C: � = 1

⇒ z = e i θ

⇒ dz = e iθ d θ



and�

" ��##� �

π πθ

I =

� ��

��

��

ππ θθ

θ=π π

⎛ ⎞θ = ⎜ ⎟⎝ ⎠∫

iii i

i

⇒ I = ( ) ( )� ��

θ π π−i i

∴ I = – 2 i

16.16.16.16.16. (3.75)(3.75)(3.75)(3.75)(3.75)

Bearing load, p =��

��

�

� �=

× × = 4 × 105 N/m2

Z = 18 CP = 0.018 N s/m2

Bearing modulus =�

��

� � �

�

×=× ×

= 3.75 × 10–7

17.17.17.17.17. (648)(648)(648)(648)(648)

ηth, I = ��

�

�−

ηth, II = � ��

�−

Equating both:

TH

T

HE – I

HE – II

TL

W1

W2�

�

�

�− = � ��

�−

⇒ T = � �� = ×

= 648.07 K

18.18.18.18.18. (5.77)(5.77)(5.77)(5.77)(5.77)

�

�

�

�

⎛ ⎞⎜ ⎟⎝ ⎠

=��

�

�

�

�

−⎛ ⎞⎜ ⎟+⎝ ⎠

⇒ P2 =

��

��

��

⎛ ⎞ ×⎜ ⎟⎝ ⎠+ = 5.77 bar



19.19.19.19.19. (–36)(–36)(–36)(–36)(–36)

0

Y

y = 4

x = 5x = 1

y = 1X

ωz =�

�

�

⎡ ⎤∂ ∂−⎢ ⎥∂ ∂⎣ ⎦x

ωz =� �

��

− = −

Vorticity = 2ωz = –3

Area of rectangle = (5 – 1) × (4 – 1) = 12 units

Circulation = Vorticity × Area = – 3 × 12 = – 36 units

20.20.20.20.20. (14.7)(14.7)(14.7)(14.7)(14.7)There is no change in metal volume at a given point per unit time throughout the process.∴ bi hi vi = bf hf vf

Given, �!

!i=

�

�

��

�i= 1.02

vf =� �

� !�

� !× ×i i

i

vf =� �

��

× ×

vf = 14.7 m/min.

21.21.21.21.21. (2.74)(2.74)(2.74)(2.74)(2.74)

�

β =�

��

�

��

�"= =

�

β= 48.2°



β = 96.4°

∴ $�� %#�&��'#��(�

$��#�%#��&��#��(�=====

��

��

− β −= =β

22.22.22.22.22. (0.83)(0.83)(0.83)(0.83)(0.83)Let k be the moment of Inertia contributed by a angle piece.∴ Ii = 12 k

If = 10 k

⇒ �

i

II

=��

��

�

�=

23.23.23.23.23. (8)(8)(8)(8)(8)

ηfin =��

�!��

��

⎛ ⎞=⎜ ⎟⎝ ⎠

εfin =��

�� !��

!��!" � "

= ×

� �

� �

εη

=�

��

��

� �

"

× × π ×= π × = 8

24.24.24.24.24. (0.35)(0.35)(0.35)(0.35)(0.35)

Probability of A speaking a truth, P (A) =

∴ Probability of A speaking a lie, ( )� " =�

Probability of B speaking a truth,P (B) =�

∴ Probability of B speaking a lie, ( ) # =�

Now, we needed P(A) ( )� � + P (B) ( )� " = � � � �

��

× + × = =

25.25.25.25.25. (1.41)(1.41)(1.41)(1.41)(1.41)

� ��→

−x

xx

=�

�

��→x

xx

=

� �� →x

xx

Using L′ Hospital’s rule

⇒

� �� →x

xx

=

� ��

�→= =

x

x



26.26.26.26.26. (b)(b)(b)(b)(b)

1′ 1–15°C

35°C

3′3 21°C

2′ 2

4′ 4

h

P

21°C 30°C

30°C

S1 = S2

= S2′ + CP ln

�

�

�

� ′

⇒ 10.526 = 9.9606 + 2.75 ln �

��

�

⇒ T2 = 372 K

h2 = h ′2 + CP (T2 – T ′2)

= 1523.42 + 2.75 (372 – 303)

= 1713.17 kJ/kg

Again, h3 = h ′3 – CP (T ′3 – T3)

= 379.3 – 4.6 (303 – 294)

= 337.9 kJ/kg

h3 = 337.9 kJ/kg

COP = � �

� �

��

��

! !

! !

− −= =− −

27.27.27.27.27. (c)(c)(c)(c)(c)Length of the stroke, L = 800 + 50 + 50 = 900 mm = 0.9 m

Cutting stroke time = ��

� = 0.0375 min

)�� #��

*�� #��= �

�

Return time = �

� × Cutting time =

�

� × 0.0375 = 0.025

Total time per cycle = 0.0375 + 0.025 = 0.0625Shaping width, B = 400 + (2 × 5) = 410 mm



Number of cycle required = ��

� = 205 cycle

Total machining time, 205 × 0.0625 = 13 min (say)

28.28.28.28.28. (b)(b)(b)(b)(b)

Mass of air, ma =�

�

� � �

�� $ �%

&�

× ×=×

= = = = = 123.12 kg

Mass of oil, mo = ρoil × Voil = 950 × 0.1 = 95 kg

We consider the room and the oil in the radiator as a closed system since no mass is crossing theboundary. The energy balance for this stationary constant volume system can be expressed as:

Ein – Eout = ΔEsystem

⇒ (Win – Qout).Δt = ΔUair + ΔUoil

= [m CV (T2 – T1)]air + [m C (T2 – T1)]oil

⇒ (1.8 – 0.35).Δt = [123.12 × 0.718 (20 – 10)] + [95 × 2.2 × (50 – 10)]

⇒ Δt = 6375.17 sec

= 106.25 min

29.29.29.29.29. (b)(b)(b)(b)(b)

Cooling required = 50 × 3.5 = 175 kJ/sec

T1 = 27 + 273 = 300 K

T2 = – 33 + 273 = 240 K

(COP)ref. = �

� �

� �

��

�

� �= =

− −

Actual COP of plant =

�

(COP)ref. =�'

�

�=

��

�

⇒ W =��

×kJ/s = 131.25 kW

30.30.30.30.30. (b)(b)(b)(b)(b)

Given, kc = 3 kb

P = 15000 N, P1 = 5000 N

Load shared by bolt, ΔP = ��

� �

� � � �

� ��

� � � �

⎛ ⎞ ⎡ ⎤=⎜ ⎟ ⎢ ⎥+ +⎝ ⎠ ⎣ ⎦

= 3750 N



Net load on the bolt, Pb = P1 + ΔP

= 5000 + 3750 = 8750 N

Stress in the bolt, σ = ��

"

⇒ �(

��)

σ=

�

��

�π

⇒ ��

�=

�

��

�

×π ×

d = 7.46 mm

31.31.31.31.31. (a)(a)(a)(a)(a)

By applying the procedure of optimality, we get the following table

6

0

10

1

2

6

12

12

2

0

10

25

19

0

0

21

17

0

3

5

0

19

14

0

11

P

Q

R

S

T

I II III IV V

Optimal assignment is

P – V, Q – I, R – IV, S - III, T – II

32.32.32.32.32. (b)(b)(b)(b)(b)

We are given,

D = 5000 bushes per day

P = 8000 bushes per day

Co = ` 20

Ch = ` 0.08 per 1000 bushes

= ` 0.00008 per item

EOQ = ��

��

��

��

�

⎛ ⎞−⎜ ⎟⎝ ⎠

⇒ EOQ = ��

��

��

× × = ×⎛ ⎞−⎜ ⎟⎝ ⎠

Hence, Production run =��

�

× = 16.3 days



33.33.33.33.33. (b)(b)(b)(b)(b)

Using lump capacitance method

� �

� �θ

θ−−i

=��

(��

−ρ

��

� ��

−−

= �

��

��

�

�

× ×−× × −=

−

��

�

�−

= 0.0114286

∴ ��

�

�+

= 87.5

��

"

= 4.47164

∴

"= 1.30285 × 10–4 =

�

�

⇒ r = 3.909 × 10–4 m

⇒ d = 7.817 × 10–4 m = 0.7817 mm

34.34.34.34.34. (c)(c)(c)(c)(c)As the current problem is for maximization we select the highest positive value in Δ j row andselected column is called key column and the variable in the column head is termed as incomingvariable. Now divide the bi values from the corresponding elements of key column to get replacementratio column.All the values in replacement ratio column are negative and it indicate that the problem hasunbounded solution.

35.35.35.35.35. (b)(b)(b)(b)(b)

T1 =

��

��

�

×=

T2 =

��

�

×=

L/4 3 /4L

To

d(1) (2)

Maximum shear stress =

�

�

� �

��

��

��

� �

×=

π π

At, r = 0Shear stress = 0

Difference = � �

��

� �

� �− =

π π



36.36.36.36.36. (c)(c)(c)(c)(c)

We have

∴ T = f × r

∴ dT = d f × r

Considering a strip of thickness dr at distance ‘r’ from central axis.

h

ω

∴ dT = +��

��

μ × ×

dT = ��

� �� !

ωμ × π ×

∴ Total torque = ��

* *

��

μω π=∫ ∫

T =�

&

�

μω π × =

�

&

�

μπ ω

∴ P = Tω

⇒

�

&

�

μπ ω × ω= P

⇒� �

�

��

� �� −× π × ω

× ×= 223.93

⇒ ω = 209.437

�

��

�π= 209.437

N = 1999.98 rpm

37.37.37.37.37. (c)(c)(c)(c)(c)

Let us consider an equivalent column of water H corresponding to water, oil and dead weight.

H = 2 + 1 × 0.75 + �

��

� �

��

π ××



H = 3.65 m of water

Depth of centroid C, � =��

��

− = 3.5 m

Total force on gate, F = +"!ρ

= 1000 × 9.81 × π × 0.32 × 3.5

F = 2427 N

Zp =�

�

� �

�

,�

�"�

�

π

= = + π ×

I

=�

��

�+

Zp = 3.502 m, which is 0.148 m from base of vessel.

38.38.38.38.38. (c)(c)(c)(c)(c)∴ Impulse = Change in momentum

Area of graph = m (Vf – Vi)

�

� × 10 × 10 + (20 × 14) + (

�

� × 15 × 14) = m (V – O)

But, m = 1⇒ V = 435 m/s

Average force =��#�%#'��,

$��#�� =

��-

=

Acceleration = 9.666 m/s

S = ��

�× × + ×

(∵ Body will travel with constant velocity of 435 m/s for the next 15 seconds after the removal of force)

S = 16312.5 m = 16.312 km

39.39.39.39.39. (c)(c)(c)(c)(c)The voltage length characteristics of DC current is given by

V = �� .��

��

⎧ ⎫−⎨ ⎬⎩ ⎭

I

P = V I = ��

��

⎛ ⎞−⎜ ⎟⎝ ⎠I I

For maximum pores;��

� I= 0



⇒ ��

��− × I = 0

∴ I = 150 Aand V = 40 V

41.41.41.41.41. (b)(b)(b)(b)(b)For no interference, rack addendum,

sin2 φ = �"

�

sin2 φ =��

��

⇒ sin φ = 0.3162φ = 18.435°

Length of path of contact = � �� + − ° = 34.8 mm

Length of arc of contact =� � � �

�� =

φ ° = 36.68 mm

42.42.42.42.42. (c)(c)(c)(c)(c)

Opinion and Judgement forecasting is done where their are no available previous data of demandand forecast, so in absence of previous data, opinion and judgement forecasting is done.

43.43.43.43.43. (a)(a)(a)(a)(a)

Eigen values are 2, 2, 2

Consider (A – λ I )X = 0

(A – 2 I )X = 0

�

�

�

�

�

⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

xxx

=

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

x 2 = 0 and x 3 = 0

∴ x 1 = k arbitrary

∴ Eigen vector is

�⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

k = 1, X =

�

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦



44.44.44.44.44. (d)(d)(d)(d)(d)

By changing the order of integration

=�

�

��

� �� +∫ ∫ · x x

=� �

�

� �

��

�

� ��⎛ ⎞

+ ⎜ ⎟⎝ ⎠∫x

=� �

�

� ��

�� +∫

Let 1 + y 3 = t⇒ 3y 2 dy = dt

⇒ y 2 dy =�

��

at y = 0, t = 1at y = 1, t = 2

⇒ Integral =�

�

��

��∫

= ( )�

��

�

� � � � � ��

� ��

�⎛ ⎞ −= − =⎜ ⎟⎝ ⎠

45.45.45.45.45. (c)(c)(c)(c)(c)

The above system of equations can be written as AX = B

where, A =

� � � � �

� � � � �

� � � � �

� � � � �

� �

�

-

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

x

⇒ [A | B ] =

� � � � �

� � � � �

� � � � �

� � � � �

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

Applying, R3 → R3 – R4 and R4 → R4 – R1

⇒ [A | B ] =

� � � � �

� � � � �

� � � � �

� � � � �

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥−⎣ ⎦



Applying, R4 → R4 + 2R3

⇒ [A | 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 canonly be trivial, i.e., x = y = z = w = 0.

46.46.46.46.46. (0.46)(0.46)(0.46)(0.46)(0.46)

T k( )

300 k

V = constantP = constant

31

S3 S1 S2

2

S (kJ / kgK)

�

�

�

�= �

�

�

� �

�

� �� /

��

�⇒ = ⇒ =

Net entropy change = (S1 – S3) = (S1 – S2) + (S2 – S3) ... (i)

(S1 – S2)V= C = CV ln �

�

��

��

�

�= × l

= – 1.15557 kJ/kgK

(S2 – S3)P=C = CP ln �

�

�

�

= 1.005 ln ��

��

= 1.61748 kJ/kgK

Net change (decrease) in entropy,

(S1 – S3) = – 1.15557 + 1.61748 = 0.462 kJ/kgK



Alternate:Alternate:Alternate:Alternate:Alternate:

Since entropy is a property, the decrease in entropy is given by (S1 – S3) is independent of theprocess undergone between states 1 & 3. Applying isothermal process:

(S1 – S3) = – R ln �

�

�

� = – 0.287 × ln

�

= 0.462 kJ/kgK

47.47.47.47.47. (1.35)(1.35)(1.35)(1.35)(1.35)

Circular pitch, PC = π m = ��

��

�

�

�

�π × = π × = π

Length of arc of contact =0��'�, �% �� % ��

�� =

��

�� = 21.28 mm

Contact ratio = !� ��" ��

*�� 1��" =

��

=

π

48.48.48.48.48. (1852.76)(1852.76)(1852.76)(1852.76)(1852.76)

Ff =�

�� %

� ρ

Cf =�

� � ��

� � ��

�

�

"

×= =× × ×ρ

St × Pr 2/3 =��

⇒ ��

�

!��

�×

ρ=

��

⇒ ��

! ×× ×

=��

�

h = 61.7586 W/m2-K

Q = h × As × ΔT = 61.7586 × 1 × 1 × 30 = 1852.76 W

49.49.49.49.49. (0.42)(0.42)(0.42)(0.42)(0.42)

δ l ===== �

� � �

��'� �

�

��

'� � � �

⎛ ⎞⎜ ⎟− ⎝ ⎠

=�

�

��

��

× ×× × × −

= 0.422 mm



50.50.50.50.50. (71.12)(71.12)(71.12)(71.12)(71.12)

10mm

100 mm

10 mm

10 mm

50 mm

50 mm

60 mm y1 = 55 mm

60 mm

From symmetry, � = 60 mm from top

INA =� ��

��

× ×−

= 3.866 × 106 mm4

τ =�

�

� � � � ��

��

��

�

× × × ×=× ×I

= 71.12 N/mm2.

51.51.51.51.51. (7.88)(7.88)(7.88)(7.88)(7.88)

1 m

θθ

x

2.3 m

∴ (kx) × 2.3 = – I α

k (2.3) θ × 2.3 = – θI��

I =�

��

��+ ×l

m =� �

� ��

=

I =�

��

��

× + × = 18.97

⇒ 8.80 × 103 × 2.32 θ = – 18.97 θ��

θ�� + 2453.97 θ�� = 0



On comparing, θ�� + ωn2 θ�� = 0

⇒ ωn = � �� = 49.537 rad/s

⇒ 2π f = 49.37

⇒ f = 7.88 cycles/s

52.52.52.52.52. (2.5)(2.5)(2.5)(2.5)(2.5)γ = cot φ + tan (φ – α)α = 0t = 0.25 mm

tc = 0.5 mm

r =��

��

�

�= =

tan φ =��

� ��

α− α

⇒ r = 0.5

⇒ r = 0.5γ = cot φ + tan (φ – α)γ = cot φ + tan φ

γ = ��

�+ =

53.53.53.53.53. (3.69)(3.69)(3.69)(3.69)(3.69)

1

DatumLevel

1.2 m2

d = 0.1 m

Summit

3

h

Using Bernaulli’s equation at point 1, 2, 3.

��

�

�!

+ ++ + +

ρ =�

� �

�

�

+ ++

ρ

( )��

��

�×

+ +× ×

=��

��

×+

× ×[∵ V2 = V3]

h = 3.69 m



54.54.54.54.54. (41)(41)(41)(41)(41)

�

�

� � ��

� � �

+ ++ +

x xx x

=�

�

� � � ��

� � �

+ + ++ +

x xx x

= �

��

� � �+

+ +x x

For computing the maximum value, we need to find minimum value of 3x2 + 9x + 7

f (x) = 3x 2 + 9x + 7

⇒ f ′(x) = 6x + 9 = 0

⇒ x = –1.5

and f ′′(x) = 6 > 0

∴ At x = –1.5, f (x) is minimum

∴ f (–1.5) = 3 (–1.5)2 + 9 (– 1.5) + 7

= 0.25

∴ Maximum value of �

�

� � ��

� � �

+ ++ +

x xx x

=��

� ��

+ =

55.55.55.55.55. (78)(78)(78)(78)(78)

∑ x = 170 and ∑ x2 = 2830

Increase in ∑ x = 10 and ∑ x ′ = 170 + 10 = 180

Increase in ∑ x2 = 900 – 400 = 500 then

∑ x′2 = 2830 + 500 = 3330

∴ Variance =�

��

��

⎛ ⎞× − × = − =⎜ ⎟⎝ ⎠

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