Gate 2001

GA/QB/PI−III

Elite Academy

1

2001 PI: Production and Industrial Engineering Duration: Three Hours

Maximum Marks: 150 Read the following instructions carefully

1. 2.

This question paper contains TWO SECTIONS: ‘A’ and ‘B’. Section ‘A’ consists of two questions of the multiple-choice type. Question 1 consists of TWENTY FIVE sub-questions of ONE mark each and Question 2 consists of TWENTY FIVE sub-questions of TWO marks each.

3.

Answer Section ‘A’ only on the special machine-gradable OBJECTIVE RESPONSE SHEET (ORS). Questions of Section A will not be graded, if answered anywhere else.

4.

Answer problems of Section ‘B’ in the answer-book.

5.

Write your name, registration number and name of the Centre at the specified location on the right half of the ORS for Section ‘A’.

6.

Using a soft HB Pencil, darken the appropriate bubble under each digit of your registration number.

7. 8.

The objective response sheet will be collected back after 120 minutes have expired from the start of the examination. In case you finish Section ‘A’ before the expiry of 120 minutes, you may start answering Section ‘B’. Questions of Section ‘A’ are to be answered by darkening the appropriate bubble (Marked A, B, C or D) using a soft HB pencil against the question number on the left hand side of the Objective Response Sheet.

9.

In case you wish to change an answer, erase the old answer completely using a good soft eraser.

10.

There is no negative marking.

11.

Section ‘B’ consists of TWENTY questions of FIVE marks each. ANY FIFTEEN out of them have to be answered. If more numbers of questions are attempted, cancel the answers not to be evaluated, else only the first fifteen answers would be considered strictly.

12.

In all 5 marks questions, clearly show the steps. Steps carry partial marks.

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SECTION - A (75 Marks) PI-1. This question consists of TWENTY FIVE sub-questions (1.1-1.25) of ONE mark each. For each of the sub-questions four possible answers (A, B, C and D) are given, out of which only one is correct. Answer each sub-question by darkening the appropriate bubble on the OBJECTIVE RESPONSE SHEET (ORS) by using a soft HB pencil. Do not use the ORS for any rough work. You may like to use the answer book for any rough work, if needed. 1.1

[25 x 1 =25]

The order of matrix to solve the system of linear equations with variable x, y, z and having the form a i x + b i y + c i z + d i = 0 is given by: (A) 4 x 4

1.2

(B) 3 x3

(C) 3 x 4

(D) 4 x 3

5  3 Two vectors called {P} and {Q} are given as  ,   respectively. The product 2  6  {P}T {Q} is equal to: (A) 27

1.3

1.4

(B) 24

(C) -24

(D) 36

A differential equation of the form

∂u ∂ 2u = a 2 2 can be used to represent: ∂t ∂x

(A) Vibration of a string

(B) Propagation of heat

(C) Steady state diffusion of gas

(D) All of these

Mean value of the distribution (A) 1.5

(B) 3

(C) 9

3 n e −3 , for n = 1, 2, 3… ∞ is n! (D) None of these

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Which of the following methods use the augmented matrix to solve simultaneous equations?

1.6

(A) Runge-Kutta method

(B) Euler Method

(C) Milne method

(D) Gauss-Jordon method

The maximum solubility of carbon in austenite in a Fe-C system is close to; (A) 0.02%

1.7

(B) 0.80%

(C) 2.04%

(D) 4.27%

If a body of mass of 10 Kg resting on a horizontal surface is acted upon by a horizontal force of 10 N, the acceleration in m s 2 would be: (A) 9.81 (B) 10.0

1.8

(C) 1.0 (D) 100.0

In Whitworth quick return mechanism (used in shapers), the velocity of the ram is maximum at:

1.9

(A) Middle of the forward stroke

(B) Beginning of return stroke

(C) End of return stroke

(D) None of these

Temperature at two ends of a 3 m long bar are 1000  C and 100  C . Assuming one dimensional heat conduction, the temperature in  C at a section which is 1 m away from the hot end is: (A) 700

1.10

(B) 800

(C) 900

(D) 550

Sprues are tapered to: (A) Reduce air aspiration (B) Improve casting yield (C) Help in faster rate of mould filling (D) Separate slag from molten metal

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GA/QB/PI−III

Minimum required bend radius on the inner side of a bent steel sheet in press bending operation is

1.12

(A) Half of sheet thickness

(B) equal to sheet thickness

(C) 1.5 times sheet thickness

(D) twice the sheet thickness

Which of the following treatments is used for reducing weld cracking in medium carbon low alloy steels?

1.13

1.14

(A) Annealing

(B) tempering

(C) Preheating

(D) post heating

By ultrasonic machining, one can efficiently machine: (A) Low carbon steel

(B) brass

(C) Plastics

(D) glass

The tolerances on 25H6 and 26H8 are 0.013 mm and 0.033 mm respectively. The tolerance on 25.5H7 is: (A) 21µm

1.15

(B) 0.023 mm

(D) None of these

In powder metallurgy, sintering is carried out in: (A) Oxidizing atmosphere (C) Reducing atmosphere

1.16

(C) 30µm

(B) inert atmosphere (D) air

The process capability of a lathe is (A) The maximum diameter of the component that can be accommodated on the lathe (B) The maximum production rate from the lathe (C) The maximum power available for machining (D) The tolerance that is obtained on the parts produced on the lathe

GA/QB/PI−III 1.17

1.18

1.19

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Axes-drives in machining centers use: (A) AC synchronous motor

(B) DC servo motor

(C) Stepper motor

(D) Induction motor

Which of the following is not an assumption in linear programming? (A) Proportionality

(B) Divisibility

(C) Additivity

(D) Uncertainty

Depreciation is meant for (A) Annual maintenance of equipments (B) Tax saving (C) Capital recovery (D) Interest on capital

1.20

1.21

1.22

Which of the following techniques is used in micro-motion analysis? (A) Flow process chart

(B) SIMO chart

(C) Man-Machine chart

(D) String diagram

ABC analysis is done in stores (A) To find EOQ

(B) To make lead time analysis

(C) To make buy / make decisions

(D) To segregate high value items

In Weibull analysis, product reliability is given as (A) (C)

1.23

 t −100  −  e  500   t −100  +  e  500 

0.2

0.2

0.2

(B)

 t +100  −  e  500 

(D)

 t −100  −  e  500 

−0.2

Concurrent engineering applied to project management (A) Performs the stages one after the other

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(B) Minimizes the number of stages by combining wherever possible (C) Performs, the stages simultaneously (D) Minimizes the budgetary allocations 1.24

Identify the pair consisting of two independent variables to be determined in aggregate production planning. (A) Level of inventory and level of production (B) Demand and level of total man power (C) Level of total man power and variable cost of producing one item in overtime (D) Demand and level of inventory

1.25

Load leveling in project management does not reduce (A) Idle labour cost (B) Machinery and equipment cost (C) Cost of resources (D) Hiring and separation cost

PI-2

This question consists of TWENTY FIVE sub-questions (2.1-2.25) of TWO marks each. For each of the sub-questions, four possible answers (A, B, C and D) are given, out of which only one is correct. Answer each sub-question by darkening the appropriate bubble on the OBJECTIVE RESPONSE SHEET (ORS), using a soft HB pencil. Do not use the ORS for any rough work. You may like to use the answer book for any rough work, if needed.

2.1

13 6 Inverse of a matrix [P] =   is given by  4 2 (A)

0.5

2 4  6 13  

(B)

0.5

 2 − 4 − 6 13   

[25 x 2 = 50]

GA/QB/PI−III

(C) 2.2

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2 4  6 13  

 2 − 4 − 6 13   

The magnitude of ∇ 2 φ , when φ is given as, x 2 + y 2 + z 2 is equal to: (A) 3

2.3

(D)

7

(B) 0

(C) 6

(D) –3

 The vector field is defined by the three components of a vector, B given as B x = 4x + 5 y + z By = x + y2 Bz = y 2 + z 2   The magnitude of divergence of B (i.e. div. B ) at a point with coordinates x = 2, y = 5 and z =1, is given a (A) 16

2.4

(C) 20

(D) 0

Six coins are tossed at a time. The probability of obtaining at least five heads is (A) 0.0833

2.5

(B) 5

(B) 0.0937

(C) 0.1093

(D) 0.4166

The formula to determine the square root of a number ‘R’ using Newton Raphson method is given by (where x i +l is the root at i+l the iteration)

2.6

(A) x i +l =

1 R  x i +  2 xi 

(B) x i +1 = x i (2 − x i R )

(C) x i +1 =

1 x i (3 − x i 2 R ) 2

(D) None of these

The diagram in Fig. A1 Shows binary equilibrium diagram of an alloy of X in Y

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Fig. 1 An alloy containing P% of X is allowed to solidify from a temperature Tp. At temperature T the fraction of solid formed is (A)

2.7

AC BC

(B)

BC AB

(C)

AC AB

(D)

BC AC

The stresses on a member are σ x = 0MPa , σ y = 20MPa and τ xy = 44MPa . The principal stresses in MPa are (A) 12 and 22

2.8

(B) 2 and 22

(C) -2 and 22

(D) 2 and 20

In a gear box, the speed reduction of 40 is obtained in single stage. The types of gears used are: (A) Spur gears

(B) Helical gears

(C) Bevel gears

(D) Worm and worm gear

GA/QB/PI−III 2.9

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A rectangular plate of size 0.6 m x 0.2 m is maintained at 600  C . Water is flowing over the plate at 40  C . The convective heat transfer coefficient is 0.002 W / mm 2

2.10



C . The rate of connective heat loss from the plate is

(A) 0.1344 W

(B) 0.1344 k W

(C) 1.344 W

(D) 134.4 k W

The permeability of moulding sand was determined using a standard AFS sample by passing 2000 cc of air at a gauge pressure of 10g / cm 2 . If the time taken for the air to escape was 1 min., the permeability number is: (A) 112.4

2.11

(B) 100.2

(C) 75.3

(D) 50.1

The sequence of operations performed during a closed die forging of a connecting rod is: (A) edging-- fullering-- blocking -- finishing (B) fullering-- edging -- blocking -- finishing (C) edging-- blocking -- fullering -- finishing (D) Fullering -- blocking -- edging --finishing

2.12

Two plates A and B are joined by means of two fillet welds f1andf 2 , having leg length of 8 mm each as shown in Fig. A2. Both these welds have same length, L=150 mm. If the shear strength of the weld metal is 200 MPa and assuming that plate A is strong enough to take such a load, the total load P which the plate A can carry with out failure of the weld is

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GA/QB/PI−III

Fig. A2 (A) 339 k N 2.13

(B) 480 k N

(C) 960 k N

(D) 678 k N

For turning NiCr alloy steel at cutting speeds of 64 m/min and 100 m/min, the respective tool lives are 15 min and 12 min. The tool life for a cutting speed of 144 m /min is: (A) 8 min

2.14

(B) 9 min

(C) 10 min

(D) 11.5 min

Four components P, Q, R and S with the tolerances as shown in Fig. A3 are assembled to make up a dimension F.

Figure: A3 Using statistical tolerance method, the limits on the resultant dimension F are: (A) 140.06 ± 0.062 (B) 140.06 ± 110 (C) 140.00 ± 0.110 (D) 140.00 +−00..170 050 2.15

The following are six materials belonging to thermo-set and thermo-plastic groups:

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(1) Epoxy

(2) Polyethylene

(3) Bakelite

(4) Polystyrene

(5) Polyesters

(6) Polyvinyl chloride

The materials which belong to thermoplastic groups are: (A) 1,2,4,6 2.16

2.17

(B) 3, 4, 5

(C) 2,4,5,6

(D) 2, 4, 6

The best size of wire used to inspect ISO metric thread of 4 mm pitch is (A) 2.000 mm φ

(B) 2.256 mm φ

(C) 2.309 mm φ

(D) 4.000 mm φ

A DC servo motor is directly driving an NC table. The pitch of the lead screw of the table is 5 mm. The motor rotates at 100 rpm for an applied voltage of 10 V. If the voltage-speed characteristic of the motor is linear, the applied voltage for a table speed of 3 m/min is equal to: (A) 30 V

2.18

(B) 60 V

(C) 33 V

(D) 50 V

In the following combinations of needs, which set represents the bottom and top of Maslow’s need hierarchy? (A) Security and Self actualization (B) Physiological needs and Self actualization (C) Physiological needs and Esteem (D) Security and Esteem

2.19

If the fixed cost of production is Rs. 10,000/-, variable cost of production per unit is Rs. 200/- and the unit price is Rs. 300/-, then the break-even volume to be produced will be: (A) 300

2.20

(B) 200

(C) 100

GET in MTM-2 is equivalent to the following in MTM-1: (A) Reach, Move, Release

(B) Reach, Grasp, Position

(C) Reach, Grasp, Move

(D) Reach, Grasp, Release

(D) 150

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GA/QB/PI−III

Annual consumption of a product is 800. The ordering cost is Rs. 50/-. The storage cost is Rs. 2/- per unit per annum. The lowest total incremental cost (TIC) in rupees, in inventory control, is: (A) 400

2.22

(B) 800

(C) 1000

(D) 1200

The end-inventory in the months of May, June and July as indicated in an aggregate plan are 2000, 5000 and 2000 respectively. For a carrying cost of Rs. 100/- per month per item, the total carrying cost in rupees for these three months is: (A) zero

2.23

(B) 200,000

Maximise

Z = 3x 1 + 2x 2

Subject to

4 x 1 + x 2 ≤ 60

(C) 500,000

(D) 900,000

8x 1 + x 2 ≤ 90 2 x 1 + 5x 2 ≤ 80 And

x 1 ≥ 0 and x 2 ≥ 0

The number of corner points in feasible solutions of the above LP model is: (A) 3

2.24

(B) 4

(C) 5

(D) 6

The equation for expected length of a queue of a system is given as L q =

λ2 µ( µ − λ )

. The equation is true for: (A) Arrivals follow Poisson distribution and service times are exponential. (B) Arrivals follow Poisson distribution and service times are constant (C) Arrivals follow Earlang distribution and service times are exponential

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(D) Arrivals follow Earlang distribution and service times are constant 2.25

Two motors, one gear box and two pumps are available. One from each category will make a system. The remaining items would be in stand by. The reliabilities of motor, gear box and pump are 0.8607, 0.9231, and 0.7788 respectively. The system reliability is: (A) 0.4148

(B) 0.6187

(C) 0.8895

(D) None of these

SECTION -B (75 Marks) This section consists of TWENTY questions of FIVE marks each. Any FIFTEEN out of these questions have to be answered on the Answer Book provided. If more numbers of questions are attempted, cancel the answers not to be evaluated, else only the first fifteen answers would be considered strictly.

[15 x 5 = 75]

PI-3. The expected relation between two variables is y = a x . Using

least

square

method, determine the value of ‘a’ for the following experimental values. x y

-2 0.22

2 4.50

4 15.00

6 65.00

PI-4. A rectangular block of mass 1 kg is sliding from rest down an inclined plane from a height of 10 m. The angle of inclination is 45  and coefficient of friction is 0.3. Draw the free body diagram. Determine the acceleration and final velocity. PI-5. A square key of plain carbon steel of size 6 mm x 6 mm is fitted in a shaft of 25 mm diameter. The torque of 115 N-m acts on the shaft. Determine the length of

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the key if the yield shear strength for the material of the key is 100 MPa. Also calculate the crushing stress developed in the key. Use factor of safety of 2. Figure, as shown below.

PI-6. Water is flowing through a 600 m long galvanized steel pipe of internal diameter 150 mm with a discharge of 0.05 m 3 / s. Determine the frictional head loss and pumping power to maintain the flow rate. The friction factor is 0.02. PI-7 Ingot castings of 0.4% C steel are to be cast in ingot moulds of 20 cm diameter and 50 cm length. The upper part of the casting in the ingot mould acts as the riser. Show the cross section of the ingot casting with the real shape of the piping defect in the riser. Calculate the shrinkage volume of the liquid metal if the volumetric shrinkage is 3%. Assuming the shrinkage pipe of cylindrical shape and diameter to height ratio of 1:2.5 for the pipe, calculate the height of the pipe. What portion of the casting in the ingot mould will be free from piping defect? PI-8

State the Von Mises yield criterion in terms of deviatoric stresses and explain the meaning of the terms used. State the incompressibility condition in mathematical form. Define deviatoric stress and express it in mathematical form.

GA/QB/PI−III

PI-9

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A 20 mm thick plate with single V-edge preparation, as shown in Fig. B2 is subjected to multi-pass submerged arc welding from one side only. 15% extra weld metal is deposited as weld crown. If the V- groove is filled in 5 equal

Passes, determine the welding speed in mm /min when other welding parameters are: Wire diameter

= 3 mm

Electrode / Wire feed rate

= 2 m/min

Figure: B2 (All dimensions are in mm) PI-10 A cutting tool is designated in ‘Orthogonal Rake System’ as: 0  − 0  − 6  − 6  − 25  − 75  − 0.8mm Determine the main cutting force (Pz ) and cutting power assuming orthogonal machining for the following data: S

= feed

= 0.12 mm /rev

t

= depth of cut

= 2.0 mm

a2

= chip thickness

= 0.22 mm

Vf

= chip velocity

= 52.6 m/min

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τs

= dynamic yield shear strength

= 400 MPa

Pz

= main cutting force

=

S  tτ s (ς sec γ − tan γ + 1)

Where, ς = chip reduction coefficient and γ = orthogonal rake PI-11 A pure metallic block is being machined using Electro Chemical Machining (ECM). The material removal rate (MRR) is 2000 mm 3 / min . The specific Gravity of the metal is 7.8, valency is 2 and atomic weight is 56. If Faraday’s constant is 96500 Coulombs, determine the current requirement. PI-12 Calculate, using Base Tangent Method, the theoretical distance between two parallel measuring faces over 3 teeth on a spur gear of 30 teeth, 3 mm module land 20  pressure angle. PI-13 Design and draw the optimum strip layout for high volume production of sheet metal component shown in Fig. B3. Determine the percentage yield from the sheet.

Show the punch positions with hatching in the strip layout.

Sheet

thickness is 1 mm and bridge allowance (minimum required distance between two edges) is 1.5 t.

Figure: B3 (All dimensions are in mm)

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PI-14 The surface of a metal slab of 278 mm length and 80 mm width is machined by a face milling cutter of 120 mm diameter, having 10 teeth, rotating at 50 rpm. The milling feed is 0.1 mm / tooth of the cutter. The combined time for job loading and unloading is 2 min. The approach and overtravel of the tool are 1 mm each. Calculate the production rate per hour. PI-15 Indian Metals and Alloys Ltd. wishes to invest on a new furnace. The initial required investment is Rs. 400,000/-. The company has to spend another Rs. 400,000/- to renovate the furnace at the end of fifth year. The total life of the furnace is estimated as 10 years with zero salvage value. The annual expected revenue during its life is Rs. 110,000/-. Show the cash flow diagram and determine the internal rate of return of the cash flow. The rate of return factors are given in the following tables. Table for Discrete rate of return factor for 10 year life. MARR 9% 10% 11% 12%

SPCAF 2.3674 2.5937 2.8497 3.1058

SPPWF 0.42241 0.38554 0.35375 0.32197

CRF 0.15582 0.16275 0.16986 0.17698

USPWF 6.4177 6.1446 5.8974 5.6502

SFDF 0.06582 0.06275 0.05986 0.05698

USCAF 15.193 15.937 16.743 17.549

Table for Discrete rate of return factor for 5 year life MARR 9% 10% 11% 12%

SPCAF 1.5386 1.6105 1.6864 1.7623

SPPWF 0.64993 0.62092 0.59417 0.56743

CRF 0.25709 0.26380 0.27060 0.27741

USPWF 3.8897 3.7908 3.6978 3.6048

Where, MARR = Minimum attractive rate of return SPCAF = Single payment capital accumulation factor SPPWF = Single payment present worth factor CRF = Capital recovery factor

SFDF 0.16709 0.16380 0.16060 0.15741

USCAF 5.9847 6.1051 6.2289 6.3528

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USPWF = Uniform series present worth factor SFDF = Sinking fund deposit factor USCAF = Uniform series capital accumulation factor

PI-16 (a)

State the three categories of “ Principles of Motion Economy” and the number of principles in each category.

(b)

Give the classifications of hand and body motions according to the principles of motion economy from simple to complex category.

PI-17 A hypothetical flow shop has only two machines. The processing time matrix is given as: Job A B C D E

Machine 1 2 3 7 9 6

Machine 2 0 8 5 1 4

Determine the job sequence using Johnson’s algorithm. Draw the Gantt chart and determine the minimum makespan. PI-18 A product is available in two sizes: normal size and large size. The customer prefers to buy only the normal size. Large size item is purchased when normal size item is not available. The shopkeeper is forced to buy from the manufacturer one large size item for four of normal size items as a package deal. The unit storage cost is Re. 1/- per week for normal size items and Rs. 2/- per week for large items. Ordering cost is Rs. 900/-. The rates of sale of items are 50 per week for normal size items and 25 per week for large size item. No shortage and no safety stock are permitted. Draw the inventory model showing the combined stock level.

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Derive an equation to find the average incremental cost per week in the inventory control. Find the economic order quantities for both the items. PI-19 The Food World supermarket, in preparation for the upcoming festival season, has just purchased the following: Type of nuts Quantity (kg)

Almond 6000

Cashew 7500

Walnut 7500

Pistachio 8000

It would like to package these nuts in one kg bags and is considering to produce (a) a regular mix containing 15% almond, 215% cashew, 25% walnut and 35% pistachio (b) a deluxe mix containing 30% almond, 30% cashew, 20% walnut and 20% pistachio (c) individual bags of each type of nuts. The expected profits per bag are Rs. 20/- for regular mix, Rs. 25/- for deluxe mix, Rs. 10/- each for cashew and walnut and Rs. 5/- each for almond and pistachio. Sufficient demand exists for each type of bag. The company would like to determine how many bags of each type are to be packaged to maximize the profit. Formulate and LP model to solve this problem. Use the following notations to represent the number of bags of each type: X1 for regular mixture, X 2 for deluxe mix, X 3 for almond, X 4 for cashew, X 5 for walnut and X 6 for pistachio. PI-20 A double sampling inspection plan in operation is specified as: n 1 = n 2 = 50;

c1 = 0, r1 = 3

c 2 = 2, r2 = 3

Determine the average outgoing quality of lots with incoming quality of 2% defective. PI-21 The activities and their estimated durations in a project are:

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Activity Predecessor (s) Optimistic time Pessimistic Time Most likely time

A 8 14 10

B 2 7 5

GA/QB/PI−III

C A 10 15 12

D B, C 6 15 8

Determine the expected time of completion of the project. Determine the least variance path for the project. PI-22 Draw and identify any 10 symbols used in system flow charts.

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2001 SOLUTIONS SECTION-A PI-1

Correct options

1.1

(B) 3 × 3 a 1 x + b1 y + c1 z = − d 1

Sol.

a 2x + b2y + c2z = − d2 a 3x + b3 y + c3z = − d 3 a 1 ⇒ a 2 a 3

b1 b2 b3

c1  c 2  c 3 

x   y   z 

=

− d 1  − d   2 − d 3 

which is of the form AX = B where a 1  A = a 2 a  3

b1 b2 b3

1.2

(A) 27

1.3

(B) Propagation of heat

1.4

(B) 3.

1.5

(D) Gauss – Jordon method

1.6

(C) 2.04%

c1   c2 , c 3 

x  X =  y  , z 

− d 1    B = − d 2  − d   3

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1.7

(C) 1.0

1.8

(A) middle of the forward stroke

1.9

(A) 700  C

1.10

(A) reduce air aspiration

1.11

(B) equal to sheet thickness

1.12

(C) preheating

1.13

(D) glass

1.14

(A) 21µm

1.15

(C) reducing atmosphere

1.16

(D) the tolerance that is obtained on the parts produced on the lathe

1.17

(A) AC synchronous motor

1.18

(D) uncertainty

1.19

(C) capital recovery

1.20

(B) SIMO chart

1.21

(D) to segregate high value items

GA/QB/PI−III

GA/QB/PI−III

Elite Academy

 t − 100  −  e  500 

0.2

1.22

(A)

1.23

(C) performs the stages simultaneously

1.24

(A) level of inventory and level of production

1.25

(B) machinery and equipment cost

PI-2

Correct options

2.1

(B)

2.2

(C) 6

2.3

(A) 16

2.4

(C) 0.1093

2.5

(A) x i + 1 =

2.6

(B)

2.7

(C) −2 and 22

2.8

(D) worm and worm gear

− 4 13 

0.5 2

− 6 

1 2

 R  x i +  x i  

BC AB

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Elite Academy

2.9

(D) 134.4 kW

2.10

(D) 50.1

2.11

(B) fullering −edging −blocking −finishing

2.12

(D) 678 kN

2.13

(C) 10 min

2.14

(D) 140.00 +−00..170 050

2.15

(D) 2, 4, 6

2.16

(C) 2.309 mm φ

2.17

(B) 60V

2.18

(B) Physiological needs and Self actualization

2.19

(C) 100

2.20

(D) Reach, Grasp, Release

2.21

(A) 400

2.22

(D) 900,000

2.23

(B) 4

GA/QB/PI−III

GA/QB/PI−III

Elite Academy

2.24

(A) Arrivals follow Poisson distribution and service times are exponential.

2.25

(B) 0.6187

SECTION – B PI-3

y = ax log y = x log a x −2 2 4 6 ∑ x = 10

y 0.22 4.5 15.0 65.0

log e y −1.514 1.504 2.708 4.174 6.872 = ∑ log y

∑ log y = ∑ x log a log a =

∑ log y 6.872 = = 0.6872 ∑x 10 a=2

PI–4

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Elite Academy

Force parallel to inclined surface

= mg sin φ

Force perpendicular to inclined surface = mg cos φ Frictional force (opposes the motion)

= − µ mg cos φ

m / ×ax = m / g (sin φ − µ cos φ) a x = acceleration in x direction a x = 9.81(sin φ4 5  − 0.8 cos 45  ) a x = 4.86 m / sec 2 V12 = V02 + 2a x . s V1 = final velocity m/sec V0 = initial velocity m/ sec = 0 S = Displacement = 10 2 m/s V12 = 0 2 + 2 × 4.86 × 10 2 V1 = 11.72 m/sec Final velocity = 11.72 m/sec

PI-5

Torque acting on the shaft = 115 N.m

GA/QB/PI−III

GA/QB/PI−III

Elite Academy = 115000 Nmm.

Torque = Force × F=

D 25 = F× = 115000 Nmm 2 2

115000 × 2 = 9200 N 25

K = yield strength in shear of key material = 100 N/mm 2 . Allowable shear stress = =

K factor of safety 100 = 50 N / mm 2 2

Length of the key in mm = L,

Width of key = 6mm

∴ L × 6 × 50 = 9200 N,

L=

9200 6 × 50

Length of the key = 30.7 mm Crushing stress developed =

F L×t/2

=

9200 30.7 × 3

= 99.9 N/mm 2

PI-6 Diameter of the pipe = 150mm = 0.15m = d Length of the pipe = L = 600m Cross sectional area of pipe (A) =

π × (0.15) 2 4

= 0.01767 m 2

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Elite Academy

Velocity of water in pipe =

Discharge area of cross section

V= Frictional head loss =

0.05 = 2.829m/sec. 0.01767

f × L × V2 = hf 2gd

f = friction factor = 0.02

hf =

0.02 × 600 × (2.829) 2 2 × 9.81 × 0.15

h f = 32.63 m

Power required to maintain the flow = P P = W × Q × h f watts. W = Wt density of water = 9.81 × 1000 kg / m 3 = 9810 kg / m 3 Q = flow rate = 0.05 m 3 / sec ∴ P = 9810 × 0.05 × 32.63 = 16000 W. P = 16 kW.

PI-7 Volume of the ingot = V=

πd 2 ×h 4

π × 20 2 × 50 = 15707.96 cm 3 4

GA/QB/PI−III

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29

Volume of shrinkage = 3% i.e. 0.03 × 15707.96 = 471.23 cm 3 Volume of Shrinkage pipe =

πd 2p 4 =

× hp

π 2 d p × 2.5d = 471.23 4

d p = 6.21 cm = diameter of pipe h p = 2.5 d p = 15.5 cm length of the pipe

Portion of the ingot which is free from piping = 50 – 15.5 = 34.5 cm 34.5 cm from bottom of the ingot is free from piping defect

PI-8 Deviatoric stress: Total stress tensor can be divided into: i) Hydrostatic stress tensor and ii) Deviator stress tensor Hydrostatic stress involves only pure tension or compression and produces only elastic volume changes and not plastic deformation. Yield stress of the metal is independent of hydrostatic component of stress although fracture strain is influenced by it. A deviator stress tensor represent shear stresses and important in causing plastic deformation. Hydrostatic stress =

σx + σy + σz 3

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GA/QB/PI−III

σ x τxy τxz  σ m 0 0  σ x − σ m     σm 0  + τ yx τ yx σ y τ yz  = 0   0 σm  τzx τzx τzy σz  0 

τxy σ y − σm τzy

  τ yz   σz − σm  τxz

Stress deviator Tensor τ ij =  2σ x − σ y − σ z  3   τ yx   τ zx 

τ xy 2σ y − σ x − σ z 3 τ zy

Expressing in terms of principal stresses 2σ1 − σ 2 − σ 3 2  σ − σ 2 σ1 − σ 3  =  1 +  3 3 2 2  =

2 [ τ3 + τ 2 ] 3

Where τ 3 & τ 2 are principal shear stresses. σ 3 − J 1σ 2 + J 2 σ − J 3 = 0 Where J 1 , J 2 and J 3 are the invariants of stress tensor. J 1 = (σ x − σ m ) + (σ y − σ m ) + (σ z − σ m ) = 0 J 2 = τ xy 2 + τ yz 2 + τ zx 2 − σ x σ y − σ y σ z − σ z σ x In term of principal stresses. J2 =

[

    τ yz   2σ z − σ y − σ x   3  τ xz

1 ( σ1 − σ 2 ) 2 + ( σ 2 − σ 3 ) 2 + ( σ 3 − σ1 ) 2 6

]

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Elite Academy

31

Vow Mises proposed that yielding will occur when second invariant J 2 of the stress deviator exceeds some critical value. J 2 = C 2 = Constant to evaluate the constant, and to relate it to yielding in simple tension, σ1 = σ 0 , σ 2 = σ 3 = 0 1 C 2 = σ 02 3

Then

∴ σ0 =

[(σ 2

1

1

− σ 2 ) 2 + (σ 2 − σ 3 ) 2 + ( σ 3 − σ 1 ) 2

In compressibility condition It means total volumetric strain is zero. ε1 + ε 2 + ε 3 = 0 Or εx + εy + εz = 0 PI-9

Cross section of the weld

]

32

Elite Academy x = h tan 30  = 18.4 tan30  = 10.62  1  2 × (106 × 18.4) + (1.85 × 20)1.15  2  = 267.26 mm 2 Assuming one meter of weld length, Volume of metal required = 267.26 × 1000 mm 3 Groove is filled in 5 equal passes. ∴ Metal deposited in each pass = 267.26 × 1000 5 = 267.26 × 200 mm 3 Wire diameter : 3mm, wire feed rate : 2m/min. Rate of deposition of metal =

π × 32 × 2000 = 14137 mm 3 / min 4

Hence time required for one pass =

Welding speed =

267.26 × 200 = 3.78 min 14137

1000 = 264.5 mm / min 3.78

PI-10 In Orthogonal Rake System, 0  − 0  − 6  − 6  − 25  − 75  − 0.8mm Side rake angle or orthogonal rake angle. = 0  ∴ γ = orthogonal rake angle = 0  t = depth of cut = 2mm,

S 0 = feed rate = 0.12mm/rev

S = Chip reduction coefficient =

Chip thickness after cut Uncut chip thickness

GA/QB/PI−III

GA/QB/PI−III ξ γ=

Elite Academy

t1 0.22 = = 1.83 t0 0.12

τ s = dynamic yield shear strength = 400 mPa = 400 N / mm 2 Pz = main cutting force = S 0 t τ s (ξ sec r − tan γ + 1) = 0.12 × 2 × 400 (1.83 × sec 0 − tan 0 + 1) Pz = 272 N Chip reduction coefficient =

Cutting velocity V = 1.83 = Chip velocity Vc

V = 1.83 × Vc = 1.83 × 52.6 = 96.26 m / min = 1.6 m/sec Cutting power =

Pz × V 272 × 1.6 kW = = 0.437 kW 1000 1000

Cutting power = 0.437 kW

PI-11 Q=

AI cm 3 / sec ρzF

Q = metal removal rate cm 3 / sec A = Atomic weight = 56 I = current applied in amps ρ = density or specific gravity = 7.8 gm/cm 3 z = valency = 2 F = Faraday’s constant in coulombs Q = 2000 mm 3 / min

33

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Elite Academy

=

GA/QB/PI−III

2000 1 = cm 3 / sec 1000 × 60 30

∴

1 56 × I = 30 7.8 × 2 × 96500

I=

7.8 × 2 × 96500 = 896 Amp 30 × 56

Current required = 896 Amps

PI-12 By base tangent method, BD = Theoretical distance between two parallel measuring surface π πs   +  = N.m. cos φtan φ − φ − 2N N   N = No. of teeth = 30 m = module = 3 φ = Pressure angle = 20  =

20 π = 0.349 radian 180

S = No. of teeth between parallel faces = 3 π 3π    +  BD = 30 × 3 × cos 20 tan 20 − 0.349 − 60 30   BD = 23.4 mm

PI-13 For a rectangular shaped component, single row, single pass strip layout gives maximum material utilisation

GA/QB/PI−III

Elite Academy

Pitch = 35 + 1.5 = 36.5 mm Strip width = 40 + 2 × 1.5 = 43 mm. Yield =

area of the part × 100 pitch × width

Area of the part = 40 × 35 − 10 × 10 − 10 × 15 = 1150 mm 2 Yield =

1150 × 100 = 73.27 43 × 36.5

yield = 73.27 %

PI-14

D = dia of cutter = 120 mm z = No. of teeth on cutter = 10 S z = feed/tooth = 0.1mm N = RPM of cutter = 50 A = approach = =

1 D − D 2 − B2   2  1 120 − 120 2 − 80 2  = 15.28 mm   2

Total travel of cutter, assuming 1mm each on initial approach and overtravel

35

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Elite Academy

GA/QB/PI−III

= 1 + 15.28 + 278 + 1 = 295.28 mm

Tm = machining time = =

Cutter travel Sz × Z × N 295.28 = 5.91 min 0.1 × 10 × 50

Total time required to complete the job = time for loading and unloading + Tm = 2 + 5.91 = 7.91 min Hence production rate/hour =

60 = 7.58 7.91

Production rate = 7.58 jobs/hour

PI-15 1. Annual expected revenue during 10 years of life is Rs 1,10,000. It forms uniform series

2. Initial investment is Rs 4,00,000 and investment at the end of five years Rs 400,000 constitutes single payment. 3. Converting investment and revenues showing in Present worth term

GA/QB/PI−III

Elite Academy Investment = 4,00,000 +

4,00,000 (1 + i) 5

 1 = 4 1 + (1 + i) 5 

   

= 4[1 + SPPWF] for five years. Expected revenue stream (uniform series)  (1 + i) n − 1 P=  × A n  i (1 + i) 

Annual revenue = 1.1 lakhs

Uniform series present worth for 10 years ∴ 4 [1 + (SPPWF) 5 ] = 1.1[( USPWF)10 ] i.e.

3.636 [1 + (SPPWF) 5 ] = (USPWF)10

After tally in the table 3.636 (1 + 0.56743) ≈ 5.6502 This is for i = 12% Internal rate of return = 12 %

PI-16 (a) Three categories of principles of motion economy are as follows: 1. Use of human body: nine principles are included in this category. 2. Arrangement of work place: eight principles. 3. Design of tools and equipments: five principles. (b) Complexity increases from class 1 to class 5 Class 1 2

Pivot Knuckle Wrist

Body members moved Finger Hand and finger

37

38

Elite Academy 3 4

Elbow Shoulder

Forearm, hand and finger Upper arm, forearm, hand

5

Trunk

and finger Torso, upper arm, forearm,

GA/QB/PI−III

hand and finger. PI-17

Job sequence using Johnson’s algorithm is A B C E D Job processing order is from M 1 to M2 Gantt chart

Minimum Make span = 30 PI−18. Let Q = Order quantity in units C = Unit cost of procuring an item D = Annual consumption in units S = Cost of placing an order h = Annual cost per rupee value of holding inventory Incremental cost is the cost required to hold the inventory between two successive orders. Average inventory during the period between two successive orders =

Q . 2

GA/QB/PI−III

Elite Academy

Q  Incremental cost =  × C × h  2 

per year.

1 Q  × C × h × Incremental cost /week =  50 2  Assuming 50 weeks/year. Package deal.

4 Normal   Units 1 Large 

Storage cost normal = 1 × 50 = 50 Rs / annum Storage cost large

= 2 × 50 = 100 Rs / annum

Consumption / year (Normal) = 50 × 50 = 2500 Consumption / year large = 50 × 25 = 1250 . Ordering cost is Rs 900 For package deal ordering cost Normal = Ordering cost for large =

900 × 4 = 720 Rs 5

900 × 1 = 180 Rs 5

Economic order Quantity Normal =

2 DS = h .c

2 × 2500 × 720 50

= 268 .

Large =

2 ds = hc

2 × 1250 × 180 100

= 67. In package deal economic order Quantity Normal 268,

large 67

In a year, half of the annual demand

39

40

Elite Academy

GA/QB/PI−III

(i.e 625) of large size product is obtained from the package deal. The remaining 625 units of large size have to be ordered separately. Annual demand of large product 625.

(E O Q) large = =

2 DS h.c 2 × 625 × 900 100

= 106 . Economic order Quantity for package deal.

268 (Normal) and 67 large

Economic order Quantity for replenishing the large products only, 106.

PI−19. Maximize Profit. i.e.

Z = 20 x 1 + 25 x 2 + 10 x 3 + 10 x 4 + 5x 5 + 5x 6

Subject to 0.15 x 1 + 0.3x 2 + x 3 ≤ 6000 (Almond) 0.25x 1 + 0.3x 2 + x 4

≤ 7500 (Cashew)

0.25x 1 + 0.2 x 2 + x 5

≤ 7500 (Walnut)

0.35 x 1 + 0.2 x 2 + x 6

≤ 800 (Pista)

x 1 , x 2 , x 3 , x 4 , x 5 and and all are integers .

x6 ≥ 0

GA/QB/PI−III

Elite Academy

PI−20. p = percentage defective = 2% = 0.02 n 1 = n 2 = n = 50 . m = n p = 50 × 0.02 = 1 P(r) = Probability of (r) defectives m r . e −m r!

=

P(0) =

1o × e −1

= e −1 = 0.3678

11 . e −1 = e −1 = 0.3678 P(1) = 1! P(2) =

12 × e −1 e −1 = = 0.1839 2! 2

P(a) = Probability of acceptance = P(0) + P(1) × P(0) + P(1) × P(1) + P(2) × P(0) = 0.3678 + (0.3678) 2 + (0.3678) 2 + 0.1839 × 0.3678 = 0.706. ∴

P–21. Net Work

AOQ = P × P(a ) = 2 × 0.706 = 1.412 %

41

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Elite Academy

TE =

GA/QB/PI−III

To + 4 TM + TP 6

A Mean Time TE Standard deviation σ Variance = σ 2

10.33 1 1

B 4.83 5 6 25 36

C 12.16 5 6 25 36

D 8.83 2 3 4 9

TE = TL = 10.33

TE = 0

TE = TL = 22.5

TL = 0 Expected time of completion of project = 31.33 lest variance path = 1 – 3 – 4

TE = TL = 31.33

GA/QB/PI−III

Elite Academy

43

PI–22. Symbols used in system flow charts BASIC OPERATION INSPECTION

COMBINATION OPERATION AND INSPECTION INSPECTION DURING DELAY

TRANSPORT

INSPECTION DURING STORAGE

DELAY

INSPECTION DURING TRANSPORT AND OTHER COMBINATIONS

STORAGE

*******************

44

Elite Academy

GA/QB/PI−III

2002 PI : Production and Industrial Engineering Duration: Three hours

Maximum marks: 150

Read the following instructions carefully. 1. All answers must be written in ENGLISH. 2. This question paper contains TWO SECTIONS: ‘A’ and ‘B’. 3. Section A consists of two questions of the multiple choice type. Question 1 consists of TWENTY FIVE sub−questions of ONE mark each and Question 2 consists of TWENTY FIVE sub−questions of TWO marks each. 4. Answer Section A only on the special machine−gradable OBJECTIVE RESPONSE SHEET (ORS). Questions in Section A will not be graded if answered elsewhere. 5. Write your name, registration number and name of the Centre at the specified locations on the right half of the ORS for Section A. 6. Using a HB pencil, darken the appropriate bubble under each digit of your registration number. 7. Questions of Section A are to be answered by darkening the appropriate bubble (marked (A), (B), (C) or (D)) using a HB pencil against the question number on the left−hand side of the ORS. In case, you wish to change an answer, erase the old answer completely using a good soft eraser. 8. The ORS will be collected after 120 minutes from the start of the examination. In case you finish Section A before the expiry of 120 minutes, you may start answering Section B. 9. There will be NEGATIVE marking in Section A. For each wrong answer to 1– and 2– mark sub–questions, 0.25 and 0.5 marks will be deducted respectively. More than one answer marked against a question will be deemed as an incorrect response and will be negatively marked. 10. Answer question in Section B in the answer book.

Section B consists of TWENTY

questions of FIVE marks each. ANY FIFTEEN out of them have to answer. If more numbers of questions are attempted, score of the answers not to be evaluated, else only the first fifteen unscored answers will be considered.

GA/QB/PI−III

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45

11. Answer for each question in Section B should be started on a fresh page. Question numbers must be written legibly and correctly in the answer book. 12. In all 5 mark questions (Section B), clearly show the important steps in your answers. These intermediate steps will carry partial credit.

SECTION − A (75 Marks) PI 1

Given below are TWENTY FIVE sub–questions (1.1–1.25) of ONE mark each. For each of these sub–questions, four possible answers ((A), (B), (C) and (D)) are given, out of which one is correct or most appropriate. Answer each sub–question by darkening the appropriate bubble on the OBJECTIVE RESPONSE SHEET (ORS) using a HB pencil. Do not use the ORS for any rough work. You may to use the Answer Book for any rough work, if needed.

1.1

[1 × 25 = 25]

The expected value of f (x), for the density function of a random variable x given by, f (x) (A) zero

1.2

= x2 / 2

0<x<2

= 0

otherwise, is (B) 2

(C) 4

(D) 8

The distribution related to the probabilities of events which are extremely rare, but which have a large number of independent opportunities for occurrence, is

1.3

(A) Bernoulli distribution

(B) Gaussian distribution

(C) Weibull distribution

(D) Poisson distribution

One of the roots of the equation x 3 − 3x − 7 = 0 , is closer to (A) 7

(B) 5

(C) 2.5

(D) 1

46 1.4

Elite Academy

GA/QB/PI−III

The temperature of points in space is given by τ (x, y, z) = x 2 + y 2 − z . The direction in which a mosquito located at (1, 1, 2) should fly to get warmed up quickly is

1.5

(A)

( 2I − 2J − K ) 3

(B)

( 2I + 2J − K ) 3

(C)

( − 2 I − 2J − K ) 3

(D)

( − 2I − 2J + K ) 3

1 4 The Eigen values of the matrix   are 3 2 (A) 1, 4

1.6

(B) 3, 2

(C) 5, − 2

(D) 3, 4

In iron−carbide diagram, pearlite is (A) eutectoid mixture of austenite and ferrite (B) eutectoid mixture of cementite and ferrite (C) eutectic mixture of austenite and cementite (D) eutectic mixture of austenite and ferrite

1.7

A box weighing 600 N is kept in a lift.

If the lift moves down with an

acceleration of 6 m / s 2 , the reaction experienced by the lift is (A) 233 N 1.8

(B) 600 N

(C) 967 N

(D) 1200 N

A lift strikes a stopper spring made of helical coil placed in a telescopic guide and breaks it equally into two halves. The ratio of the resultant stiffness to the original stiffness is (A) 1/4

1.9

(B) 1/2

(C) 1

(D) 2

When a particular system is subjected to a reversible process from state 1 to state 2, its entropy increases by 3 kJ/K. If the same system is now subjected to an irreversible process from state 2 to state 1, its entropy change (S1 − S 2 ) is

GA/QB/PI−III

1.10

Elite Academy

(A) + 3 kJ/K

(B) − 3 kJ/K

(C) between − 3 kJ/K and + 3 kJ/K

(D) zero

47

The capacity of a material to withstand plastic deformation m compression without fracture is known as (A) toughness

1.11

(B) ductility

(C) malleability

(D) resilience

A sheet is subjected to plane−strain stretching. If the true strain on the surface of the sheet is + 0.15, the true strain in the thickness direction is (A) zero

1.12

(B) − 15

(C) + 0.15

(D) + 0.30

Normalizing of steel is done by (A) leaving in the open yard for six months to two years (B) heating below critical temperature followed by air cooling (C) heating above critical temperature followed by air cooling (D) heating above critical temperature followed by furnace cooling

1.13

1.14

1.15

The appropriate process in manufacturing polyimide seals is (A) thermoforming

(B) pultrusion

(C) foam moulding

(D) compression moulding

Plastic bottles are made using (A) blow moulding

(B) injection moulding

(C) preform moulding

(D) slush moulding

Parts having thin section and complex geometry are joined by (A) gas welding

(B) submerged arc welding

(C) friction welding

(D) dip brazing

48 1.16

Elite Academy Spline−holes are machined by (A) milling

1.17

GA/QB/PI−III

(B) boring

(C) drilling

(D) broaching

Machining of complex shapes on CNC machines requires (A) simultaneous control of x, y and z axes (B) simultaneous control of x and y axes (C) independent control of x and y axes (D) independent control of x, y and z axes

1.18

The assignment problem is (A) integer valued at the optimum when solved by simplex method (B) not solvable by simplex method (C) the dual of the transportation problem (D) the dual of the maximum flow problem

1.19

Dynamic programming is a method (A) based on a complete enumeration of all solutions (B) based on partial enumeration of all possible solutions (C) for solving only the problems that have decisions occurring sequentially over time (D) to solve problems easily as the number of state variables increase

1.20

1.21

The formula for economic order quantity does not contain (A) order cost

(B) carrying cost

(C) cost of the item

(D) annual demand

exponential smoothing model is (A) a regression model of all data points (B) a simple moving average model considering all data points (C) a weighted average model with unequal weights to all data points

GA/QB/PI−III

Elite Academy

49

(D) a weighted average model with equal weights to all data points. 1.22

Concurrent engineering is an approach (A) where product design proceeds at the same time as process design (B) where product design and facilities layout are carried out at the same time (C) to reduce variety in manufacturing a group of parts. (D) to eliminate complex design features with a view to reduce costs.

1.23

Scientific management principles were enunciated by (A) F.W. Taylor

1.24

1.25

PI 2

(B) F.W. Harris

(C) W. Stewhart

(D) F. Gilbreth

KANBAN is used in (A) Synchronous Manufacturing

(B) Materials Requirements Planning

(C) Total Quality Management

(D) Just−In−Time Manufacturing

A Management Information System does not require (A) database

(B) hardware and software

(C) procedures for analysis

(D) Bill of Materials

Given below are TWENTY FIVE sub−questions (2.1− 2.25) of TWO marks each. For each of these sub−questions, four possible answers ((A), (B), (C) and (D)) are given, out of which one is correct or the most appropriate. Answer each sub− question by darkening the appropriate bubble on the OBJECTIVE RESPONSE SHEET (ORS), using a HB pencil. Do not use the ORS for any rough work. You may use the Answer Book for any rough work, if needed.

2.1

(2 × 25 = 50)

Solution to the equations 2x + 2y + 2z = 1; x + z = − 1 and − x + 3y − 2z = 7 is (A) (9/2, 1/6, − 1//2)

(B) 9/2, 1/2, − 11/2)

50

Elite Academy (C) (1.6, 9/2, 11/2)

2.2

2.3

1 1  (B)   3 4

The inverse transform of (A) 2 e − 4 t + e 2 t

2.4

(D) (9/2, 11/6, 1/6)

 1 −1  If (P + Q) =   and (P − Q) = 3 0  4 1 (A)   4 4

GA/QB/PI−III

3 s2

+ 2s − 8

(B) 2 e 4 t + e 2 t

3 1  1 4  then PQ is    −2 −2  (C)    0 −6 

 3 −1  (D)   1 0 

is given by (C) 2e − 4 t + e − 2 t

(D) 2e 4 t + e − 2 t

The work done to propel a ship with a displacement of V for a distance S in time t is proportional to (S 2 N 2 / 3 / t 2 ) . The percentage increase of work necessary when the displacement is increased by 1%, the time is reduced by 1% and the distance is decreased by2% is : (A) 2

2.5

1 3

(B) − 1

1 3

(C) − 2

1 3

(D) 1

1 3

An incompresible medium is flowing steadily through a pipe. The possible expressions for the velocity components u and v are :

2.6

(A) u = 4xy + y 2 , v = 6xy + 5x

(B) u = 2 x 2 − y , v = x 2 − 3xy

(C) u = 2 x 2 + y 2 , v = − 4xy

(D) u = 3xy + y 2 , v = 4xy − 3x

An aluminium rod of 28diameter is inserted into a steel tube of 50 mm outer diameter and 30 mm inner diameter as shown in figure. One end of the 100 mm long assembly is fixed and the other end is welded to a flange. The Young's

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moduli of steel and aluminium are 200 GPa and 70 GPa respectively. Under a pull of 100 kN, the elongation of the assembly at the flange end is

Figure : (All dimensions are in mm) (A) 0.051 mm 2.7

(B) 0.042 mm

(C) 0.034 mm

(D) 0.016 mm

A rivet joint in a structure shown in Fig. has to support a load of 2 kN. If the primary and secondary shear stresses are 6.37MPa and 13.50 MPa respectively, the maximum stress developed in the rivet is

Fig : (All dimensions are in mm). (A) 8.5 MPa 2.8

(B) 18.6 MPa

(C) 28.5 MPa

(D) 38.5 MPa

A single disc clutch operating under dry sliding condition transmits 150 Nm. The average coefficient of friction is 0.3 and maximum operating pressure is 0.5 MPa. The ratio between the outside radius and inner radius is 1.73. Under uniform

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wear conditions, if the outside radius of the clutch is to be minimum, the inside radius will have to be (A) 64.4 mm 2.9

(B) 84.6 mm

(C) 54.3 mm

(D) 74.5 mm

A specimen with a 12.5 mm diameter and 50 mm gauge length is used in a tension test of an aluminium alloy. At the load of 250 N, the total elongation is 2 mm. True stress and true strain are (A) 2.037 N / mm 2 and 0.04 respectively (B) 2.119 N / mm 2 and 0.039 respectively (C) 1.956 N / mm 2 and 0.04 respectively (D) 2.119 N / mm 2 and 0.017 respectively

2.10

In a non−pressurized gating system, an aluminium alloy is poured at a flow rate of 0.8× 10 − 3 m 3 / min through a sprue with a base cross sectional area of 400 mm 2 . If the friction losses in the sprue are neglected, the height of the sprue is (A) 200mm

2.11

(B) 150 mm

(C) 250 mm

(D) 300 mm

Atomized iron powder having an apparent density of 2350 kg / m 3 has a compression ratio of 3 at compact pressure of 600 MPa. To compact 1000 bushes of 20 mm outside diameter, 10 mm inside diameter and 14 mm log, the mass of the powder required is (A) 32.3 kg

2.12

(B) 23.3 kg

(C) 20.6 kg

(D) 16.6 kg

In spot pulsed later welding of aluminium plates (density = 2700 kg / m 3 , specific heat = 896 J/kg, melting temperature = 933 K, latent heat of melting =398 kJ/kg) at a temperature of 30 ºC, a pulse with energy of 0.5 J is focussed onto an area of 0.05 mm 2 . If the entire energy is coupled into the material, what will be the depth of weld assuming the cross sectional area of the weld is circular and is

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uniform throughout its depth and only heat conduction in the direction of penetration (A) 5.34 mm 2.13

(B) 2.15 mm

(C) 4.23 mm

(D) 3.85 mm

A slab of 50 mm width and 200 mm length is rough machined by a symmetrical face milling operation. The cutter diameter is 75 mm. The length of travel of the cutter is (A) 275.00 mm

2.14

(B) 237.50mm

(C) 219.54 mm

(D) 209.55 mm

Two different tools A and B having nose radius of 0.6 mm and 0.33 mm respectively are used to machine C−45 steel employing feed rates of 0.2 mm/rev and 0.1 mm/rev respectively. The tool that gives better finish and the value of ideal surface roughness are

2.15

(A) tool A and 4.166 µm respectively

(B) tool B and 4.166 µm respectively

(C) tool A and 8.333 µm respectively

(D) tool B and 8.333 µm respectively

In plunge surface grinding, a steel specimen of 2.5 mm wide and 200 mm long is ground with AA60 K5 V8 grinding wheel of 300 mm diameter and 25 mm wide. The wheel has undergone a radial wheel wear of 20 microns after removing material to a depth of 5 mm. The grinding ratio is (A) 53.05

2.16

(B) 26.53

(C) 13.26

(D) 106.10

A hole is specified on φ35 H7and size of tolerance zone is 0.025 mm. The axis of the hole is to be controlled with reference to A as shown in Fig. A shoulder−pin is used as a gauge for checking the geometrical relation. The pin size is

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Figure : (All dimensions are in mm).

2.17

(A) φ34.925 mm

(B) φ34.075 mm

(C) φ35.05 mm

(D) φ 34.95 mm

To introduce a new product that has a life of 3 years, a company spends Rs. 30 lakhs on machinery. The expected annual profit is Rs. 12 lakhs per year. For an interest rate of 15%, at breakeven, the salvage value of the machinery at the end of the third year is (A) Rs. 6.00lakhs (B) Rs. 4.76 lakhs (C) Rs. 5.42 lakhs (D) Rs. 3.95 lakhs

2.18

Consider a single server infinite population queuing model with Poisson arrival (λ = 4 per hour) and exponential service (µ = 10 minutes). The probability that there are at least two people in the system is (A) 2/3

2.19

(B) 1/3

(C) 4/9

(D) 5/9

Given a transportation problem with the following data : Two supply points with S1 =30and S 2 = 40; two demand points with D1 = 30 and D 2 = 40 Cost of transportation : S1 to D1 = Rs. 3 /unit, S1 to D 2 = Rs. 4/unit, S 2 to D1 = Rs. 5/unit and S 2 to D 2 = Rs. 4 /unit, which statement among the following is correct. (A) North west Corner rule gives a degenerate basic feasible solution (B) North West Corner rule gives an objective function value of Rs. 240 (C) North West Corner rule gives an infeasible solution. (D) North West Corner rule cannot be used to obtain a basic feasible solution in this example.

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The lead time demand for an item follows the distribution given below : Demand (Number of units)

Probability

10

0.5

20

0.3

25

0.2

For a safety stock of 20 units, the reorder level is (A) 35 2.21

(B) 55

(C) 36

(D) 75

The demands for an item for six consecutive periods are 101, 99, 102, 100, 99 and 101. Which statement among the following is correct for the forecast of the seventh period : (A) Using a five period moving average is greater than that using a three period moving average. (B) Using a three period moving average is greater than the arithmetic mean. (C) Using a five period moving average is greater than a arithmetic mean. (D) Using a three period moving average is greater than that using a four period moving average.

2.22

An assembly of a product comprises of eight work elements with assembly times 7, 9, 6, 9, 7, 9, 8 and 9 seconds. Taking into consideration the precedence constraints, elements 1, 2 and 3 are assembled in the first station, elements 4, 5 and 7 in the second and elements 6 and 8 in the third station. The efficiency of the line is (A) 81.48%

(B) 92.59%

(C) 94.44%

(D) 88.88%

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For a single item, cumulative production and cumulative demand for four months are given below : Month

Cumulative production

Cumulative demand

1

100

80

2

180

180

3

250

260

4

320

300

Cost of inventory is Rs. 2 /unit /month and cost of backordering is Rs.10 /unit /month. Assuming that inventory cost is calculated based on ending inventory values; the sum of inventory and backorder costs is (A) Rs. 50 2.24

(B) Rs. 180

(C) Rs. 420

(D) Rs. 600

The following failure rates were observed for bulbs in a building : End of month

:

1

2

3

4

Cumulative probability of failure :

0.2

0.4

0.8

1.0

Assume 100 bubbles and an individual replacement policy, the expected number of bulbs to be replaced at the end of the second month is (A) 70 2.25

(B) 50

(C) 30

(D) 34

A time study was conducted in a machine shop for a manual operation. The times for five observations (in minutes) were 20, 22, 21, 24 and 25. The performance rating for the operator is 120%. The allowance factor is 20%. The standard time is (A) 22.4 minutes

(B) 25 minutes

(C) 26.88 minutes (D) 33.6 minutes

SECTION – B (75 marks) This section consists of TWENTY questions of FIVE marks each. Answer any FIFTEEN questions in the ANSWER BOOK provided. Answer for each question should start on a

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fresh page. Question numbers must be written legibly and correctly. If more numbers of questions are attempted, cancel the answers not to be evaluated, else only first fifteen [15 × 5 = 75]

answers would be considered strictly. PI 3

Fig. B1 shows a circle of Diameter D in which a chord of length 2L is drawn. The cordal height is x.

Figure: B1 (A) Write the relation among D, L and x. (B) For dx /dD = - 1, find the value of L /D. PI 4

Shown in Fig. B2 is the mechanism of a crank-press. With the help of velocity and acceleration diagrams, determine the velocity and acceleration of the ram for the crank position shown in the figure.

Fig. B2: (All dimensions are in mm).

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Forces acting on the belts are shown in Fig. B3. The yield strength of the shaft material is 400 MPa. Neglecting bearing friction and using maximum shear stress theory, determine the diameter of the shaft for a factor of safety of 2.

Fig. B3: (All dimensions are in mm), PI 6

A frictionless piston cylinder device contains initially 0.01 kg of air at a pressure of 1 bar and a specific volume of 0.87 m 3 /kg. The air is now slowly compressed according to pv n = constant, where n = 1.35, until the pressure of air is doubled. The specific internal energy of air is given by u = 0.0025 pv, where u is in kJ/kg, p is in N / m 2 and v is in m 3 /kg. Determine the magnitude of heat transfer during this process.

PI 7

A hot gas at 250 °C flows through a long pipe of 100 mm outer diameter and 10 mm thick made out of a material of thermal conductivity k = 0.04 W/mK. The outer surface of the pipe is exposed to the surroundings at 25  C. Given that the inside and outside convective heat transfer coefficients are 30 W / m 2 K and 10 W / m 2 K respectively, calculate the temperature of the outside surface of the pipe.

PI 8

For a true centrifugal casting process with the axis of rotation in horizontal direction, an acceleration of 75 g, where g is the acceleration due to gravity, is required at the rim for sound casting.

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(A) Drive the relation between the diameter of casting D (in mm) and the rotational speed N (in rev/s) for this casting. (B) Determine the mould speed (in rev/s) for casting C.I. pipes of 5 m long with outside and inside diameters of 0.52 m and 0.5 m respectively. PI 9

A cylindrical cup shown in fig. B4 is to be extruded from a cylindrical billet using a punch of 15 mm diameter. The stress-strain relation for the material is given by ρ = 60 + 40ε, where σ is in MPa.

Fig. B4: (All dimensions are in mm) (A) Find the billet dimensions before extrusion. (B) Calculate the strain, average pressure and punch force neglecting friction and redundant work. PI 10 In orthogonal machining of mild steel with cutting tool having a rake angle of 10 °, the chip thickness is measured to be 0.4 mm. The uncut chip thickness is 0.2 mm and the width of cut is 2 mm. For a cutting speed of 200 m/min and a shear stress of 400 N / mm 2 for the material, determine the specific energy (in J/ mm 3 ) in cutting. Use Merchant’s relation : 2Φ + λ - α = 90° where Φ is shear angle, λ is friction angle and α is rake angle. PI 11 Tool life in drilling steel using HSS drill is expressed as T 0.2 = 9.8 D 0.4 / Vs 0.5 where D is the drill diameter (in mm), T is the tool life (in minutes), V is the

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cutting speed (in m/min) and s is the feed (in mm/rev). The feed is set at maximum possible value of 0.4 mm/rev for a given drill diameter of 30 mm. The length of drilling is 50 mm. The machine hour rate is Rs. 60 and the cost of drill is Rs. 400. For the given conditions. (A) Express T in term of V only. (B) Express the time for drilling the workpiece. (C) Formulate the total production cost in terms of only V and find the optimum cutting speed, Vopt , neglecting the workpiece and tool changing times. PI 12 A square plate is inspected for flatness. The readings obtained, using an electronic gauge, on a grid of 1m × 1m, are as follows : A = 40.005 mm; B = 40.010 mm; C = 40.015 mm; D = 40.000 mm and M = 40.010 mm, where M is at the grid center as shown in Fig. B5. (A) Give the equation of the assessment plane passing through A, B and D with origin at A and coordinate axes along A-B and A-D. (B) Using the above assessment plane, find the deviation of C and M and arrive at the flatness error (in µm).

Fig. B5

PI 13 A slot of width 5 H7 is to be milled on the part shown in Fig. B6 as a final operation. Neatly sketch a fixture, showing the locating, clamping and setting

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block arrangements. Show also important dimensions. The part should be shown by phantom lines as positioned in the fixture. Fig. B6: (All dimensions are in mm). PI 14 A part shown in Fig. B7 is to be finished with a right-hand turning tool on a CNC lathe employing a rotational speed of 1200 rev/min. A finishing allowance of 0.4 mm is provided on the part. Employ a feed rate of 120 mm/min while moving the cutter along linear and circular contours. The formats for different NC codes are given as : G00 X – Z - ; G01 X – Z – F -; GO2/GO3 X – Z – I – K – F -. Following this format strictly, prepare the part program using the absolute method of programming. Consider the reference point shown in the figure for generating the coordinate list. Write a set of instructions in the part program for finishing the part following the path : P0 P1 P2 ...... P9 P0 .