GATE-2016 Mechanical Set-1 Full Solution

Page 1

GATE 2016 MECHANICAL SET - 1 DETAILED SOLUTION & ANSWER KEY

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Mechanical eryc GATEMENTOR

GATE 2016

General Aptitude - GA Set-1

Q. 1 – Q. 5 carry one mark each. Q.1

Which of the following is CORRECT with respect to grammar and usage? Mount Everest is ____________. (A) the highest peak in the world (B) highest peak in the world (C) one of highest peak in the world (D) one of the highest peak in the world

Q.2

The policeman asked the victim of a theft, “What did you (A) loose

Q.3

(B) lose

(C) loss

(B) availability --- used (D) acceptance --- proscribed

R O

T N

E M

In a huge pile of apples and oranges, both ripe and unripe mixed together, 15% are unripe fruits. Of the unripe fruits, 45% are apples. Of the ripe ones, 66% are oranges. If the pile contains a total of 5692000 fruits, how many of them are apples?

E T

(A) 2029198

Q.5

(D) louse

Despite the new medicine’s ______________ in treating diabetes, it is not ______________widely. (A) effectiveness --- prescribed (C) prescription --- available

Q.4

?”

(B) 2467482

(C) 2789080

(D) 3577422

Michael lives 10 km away from where I live. Ahmed lives 5 km away and Susan lives 7 km away from where I live. Arun is farther away than Ahmed but closer than Susan from where I live. From the information provided here, what is one possible distance (in km) at which I live from Arun’s place?

A G

(A) 3.00

(B) 4.99

(C) 6.02

(D) 7.01

Q. 6 – Q. 10 carry two marks each. Q.6

A person moving through a tuberculosis prone zone has a 50% probability of becoming infected. However, only 30% of infected people develop the disease. What percentage of people moving through a tuberculosis prone zone remains infected but does not show symptoms of disease? (A) 15

(B) 33

(C) 35

(D) 37

1/2

GATE 2016

Q.7

General Aptitude - GA Set-1

In a world filled with uncertainty, he was glad to have many good friends. He had always assisted them in times of need and was confident that they would reciprocate. However, the events of the last week proved him wrong. Which of the following inference(s) is/are logically valid and can be inferred from the above passage? (i) His friends were always asking him to help them. (ii) He felt that when in need of help, his friends would let him down. (iii) He was sure that his friends would help him when in need. (iv) His friends did not help him last week. (A) (i) and (ii)

Q.8

(B) (iii) and (iv)

(C) (iii) only

(D) (iv) only

Leela is older than her cousin Pavithra. Pavithra’s brother Shiva is older than Leela. When Pavithra and Shiva are visiting Leela, all three like to play chess. Pavithra wins more often than Leela does.

R O

Which one of the following statements must be TRUE based on the above?

T N

(A) When Shiva plays chess with Leela and Pavithra, he often loses. (B) Leela is the oldest of the three.

E M

(C) Shiva is a better chess player than Pavithra. (D) Pavithra is the youngest of the three.

Q.9

If đ?‘žđ?‘ž −đ?‘Žđ?‘Ž

=

1 đ?‘&#x;đ?‘&#x;

A G

(A) (đ?‘&#x;đ?‘&#x;đ?‘&#x;đ?‘&#x;đ?‘&#x;đ?‘&#x;)−1 Q.10

E T

and đ?‘&#x;đ?‘&#x; −đ?‘?đ?‘?

=

1 đ?‘ đ?‘

and đ?‘ đ?‘ −đ?‘?đ?‘?

(B) 0

=

1

đ?‘žđ?‘ž

, the value of abc is

.

(C) 1

(D) r+q+s

P, Q, R and S are working on a project. Q can finish the task in 25 days, working alone for 12 hours a day. R can finish the task in 50 days, working alone for 12 hours per day. Q worked 12 hours a day but took sick leave in the beginning for two days. R worked 18 hours a day on all days. What is the ratio of work done by Q and R after 7 days from the start of the project? (A) 10:11

(B) 11:10

(C) 20:21

(D) 21:20

END OF THE QUESTION PAPER

2/2

GATE 2016

SET-1

Mechanical Engineering (ME)

Q. 1 – Q. 25 carry one mark each. Q.1

The solution to the system of equations

2 2 5 đ?‘Ľ = −30 −4 3 đ?‘Ś

is (A) 6, 2

Q.2

(C) −6, −2

(C)

∞ đ?‘ đ?‘Ą đ?‘’ 0 ∞ đ?‘–đ?‘ đ?‘Ą đ?‘’ 0

đ?‘“ đ?‘Ą dđ?‘Ą

(B)

đ?‘“ đ?‘Ą dđ?‘Ą

(D)

∞ −đ?‘ đ?‘Ą đ?‘’ đ?‘“ đ?‘Ą dđ?‘Ą 0 ∞ −đ?‘–đ?‘ đ?‘Ą đ?‘’ đ?‘“ đ?‘Ą dđ?‘Ą 0

đ?‘“ đ?‘§ = đ?‘˘ đ?‘Ľ, đ?‘Ś + đ?‘– đ?‘Ł đ?‘Ľ, đ?‘Ś is an analytic function of complex variable đ?‘§ = đ?‘Ľ + đ?‘– đ?‘Ś where đ?‘– = −1. If đ?‘˘ đ?‘Ľ, đ?‘Ś = 2 đ?‘Ľđ?‘Ś, then đ?‘Ł đ?‘Ľ, đ?‘Ś may be expressed as

R O

(A) − đ?‘Ľ 2 + đ?‘Ś 2 + constant (C) đ?‘Ľ 2 + đ?‘Ś 2 + constant

Q.4

(D) 6, −2

If � � is a function defined for all t ≼ 0, its Laplace transform F(s) is defined as (A)

Q.3

(B) −6, 2

(B) đ?‘Ľ 2 − đ?‘Ś 2 + constant (D) − đ?‘Ľ 2 + đ?‘Ś 2 + constant

T N

Consider a Poisson distribution for the tossing of a biased coin. The mean for this distribution is Âľ. The standard deviation for this distribution is given by

E M

(B) đ?œ‡2

(A) đ?œ‡

(C) Âľ

(D) 1/đ?œ‡

Q.5

Solve the equation đ?‘Ľ = 10 cos(đ?‘Ľ) using the Newton-Raphson method. The initial guess is đ?‘Ľ = đ?œ‹/4. The value of the predicted root after the first iteration, up to second decimal, is ________

Q.6

A rigid ball of weight 100 N is suspended with the help of a string. The ball is pulled by a horizontal force F such that the string makes an angle of 30ď‚° with the vertical. The magnitude of force F (in N) is __________

E T

A G

30ď‚° F 100 N

ME (Set-1)

1/11

GATE 2016

Q.7

SET-1

Mechanical Engineering (ME)

A point mass M is released from rest and slides down a spherical bowl (of radius R) from a height H as shown in the figure below. The surface of the bowl is smooth (no friction). The velocity of the mass at the bottom of the bowl is

M R H

(A) đ?‘”đ??ť

Q.8

(B) 2��

(D) 0

(C) 2đ?‘”đ??ť

The cross sections of two hollow bars made of the same material are concentric circles as shown in the figure. It is given that đ?‘&#x;3 > đ?‘&#x;1 and đ?‘&#x;4 > đ?‘&#x;2 , and that the areas of the cross-sections are the same. J1 and J2 are the torsional rigidities of the bars on the left and right, respectively. The ratio J2/J1 is

R O

(A) > 1

Q.9

(B) < 0.5

E M

T N

E T

(C) =1

(D) between 0.5 and 1

A cantilever beam having square cross-section of side a is subjected to an end load. If a is increased by 19%, the tip deflection decreases approximately by (A) 19%

(B) 29%

(C) 41%

(D) 50%

A G

Q.10

A car is moving on a curved horizontal road of radius 100 m with a speed of 20 m/s. The rotating masses of the engine have an angular speed of 100 rad/s in clockwise direction when viewed from the front of the car. The combined moment of inertia of the rotating masses is 10 kg-m2. The magnitude of the gyroscopic moment (in N-m) is __________

Q.11

A single degree of freedom spring mass system with viscous damping has a spring constant of 10 kN/m. The system is excited by a sinusoidal force of amplitude 100 N. If the damping factor (ratio) is 0.25, the amplitude of steady state oscillation at resonance is ________mm.

Q.12

The spring constant of a helical compression spring DOES NOT depend on (A) coil diameter (B) material strength (C) number of active turns (D) wire diameter

ME (Set-1)

2/11

GATE 2016

Q.13

SET-1

The instantaneous stream-wise velocity of a turbulent flow is given as follows:

u( x, y, z, t )  u ( x, y, z)  u( x, y, z, t ) The time-average of the fluctuating velocity u ( x, y, z, t ) is (A) �′/2

Q.14

Mechanical Engineering (ME)

(B) − đ?‘˘/2

(D) �/2

(C) zero

For a floating body, buoyant force acts at the (A) centroid of the floating body (B) center of gravity of the body (C) centroid of the fluid vertically below the body (D) centroid of the displaced fluid

Q.15

A plastic sleeve of outer radius r0 = 1 mm covers a wire (radius r = 0.5 mm) carrying electric current. Thermal conductivity of the plastic is 0.15 W/m-K. The heat transfer coefficient on the outer surface of the sleeve exposed to air is 25 W/m2-K. Due to the addition of the plastic cover, the heat transfer from the wire to the ambient will (A) increase (B) remain the same (C) decrease (D) be zero

Q.16

R O

T N

Which of the following statements are TRUE with respect to heat and work? (i) They are boundary phenomena (ii) They are exact differentials (iii) They are path functions (A) both (i) and (ii)

E T

E M

(B) both (i) and (iii)

(C) both (ii) and (iii)

(D) only (iii)

Q.17

Propane (C3H8) is burned in an oxygen atmosphere with 10% deficit oxygen with respect to the stoichiometric requirement. Assuming no hydrocarbons in the products, the volume percentage of CO in the products is __________

Q.18

Consider two hydraulic turbines having identical specific speed and effective head at the inlet. If the speed ratio (N1/N2) of the two turbines is 2, then the respective power ratio (P1/P2) is _____________

Q.19

The INCORRECT statement about regeneration in vapor power cycle is that

A G

(A) it increases the irreversibility by adding the liquid with higher energy content to the steam generator (B) heat is exchanged between the expanding fluid in the turbine and the compressed fluid before heat addition (C) the principle is similar to the principle of Stirling gas cycle (D) it is practically implemented by providing feed water heaters

ME (Set-1)

3/11

GATE 2016

Q.20

SET-1

The “Jominy test� is used to find (A) Young’s modulus (C) yield strength

Q.21

Q.22

(B) hardenability (D) thermal conductivity

Under optimal conditions of the process the temperatures experienced by a copper work piece in fusion welding, brazing and soldering are such that (A) Twelding > Tsoldering> Tbrazing

(B) Tsoldering > Twelding > Tbrazing

(C) Tbrazing >Twelding > Tsoldering

(D) Twelding > Tbrazing > Tsoldering

The part of a gating system which regulates the rate of pouring of molten metal is (A) pouring basin

Q.23

Mechanical Engineering (ME)

(B) runner

(C) choke

(D) ingate

The non-traditional machining process that essentially requires vacuum is (A) electron beam machining (C) electro chemical discharge machining

R O

(B) electro chemical machining (D) electro discharge machining

T N

Q.24

In an orthogonal cutting process the tool used has rake angle of zero degree. The measured cutting force and thrust force are 500 N and 250 N, respectively. The coefficient of friction between the tool and the chip is _________

Q.25

Match the following:

E T

P. Feeler gauge Q. Fillet gauge

A G

R. Snap gauge

E M

S. Cylindrical plug gauge

I.

Radius of an object

II.

Diameter within limits by comparison

III. Clearance or gap between components IV. Inside diameter of straight hole

(A) P–III, Q–I, R–II, S–IV

(B) P–III, Q–II, R–I, S–IV

(C) P–IV, Q–II, R–I, S–III

(D) P–IV, Q–I, R–II, S–III

Q. 26 – Q. 55 carry two marks each. Q.26

Consider the function đ?‘“ đ?‘Ľ = 2đ?‘Ľ 3 − 3đ?‘Ľ 2 in the domain [−1, 2]. The global minimum of đ?‘“(đ?‘Ľ) is ____________

Q.27

If đ?‘Ś = đ?‘“(đ?‘Ľ) satisfies the boundary value problem đ?‘Ś ′′ + 9 đ?‘Ś = 0, đ?‘Ś 0 = 0, đ?‘Ś đ?œ‹/2 = đ?‘Ś đ?œ‹/4 is ________

ME (Set-1)

2, then

4/11

GATE 2016

Q.28

SET-1

The value of the integral

Mechanical Engineering (ME)

∞

−∞

đ?‘Ľ2

sin đ?‘Ľ dđ?‘Ľ + 2đ?‘Ľ + 2

evaluated using contour integration and the residue theorem is (B) −đ?œ‹ cos(1)/e

(A) – đ?œ‹ sin(1)/e

Q.29

(C) sin(1)/e

(D) cos(1)/e

Gauss-Seidel method is used to solve the following equations (as per the given order): đ?‘Ľ1 + 2đ?‘Ľ2 + 3đ?‘Ľ3 = 5 2đ?‘Ľ1 + 3đ?‘Ľ2 + đ?‘Ľ3 = 1 3đ?‘Ľ1 + 2đ?‘Ľ2 + đ?‘Ľ3 = 3 Assuming initial guess as đ?‘Ľ1 = đ?‘Ľ2 = đ?‘Ľ3 = 0, the value of đ?‘Ľ3 after the first iteration is __________

Q.30

A block of mass m rests on an inclined plane and is attached by a string to the wall as shown in the figure. The coefficient of static friction between the plane and the block is 0.25. The string can withstand a maximum force of 20 N. The maximum value of the mass (m) for which the string will not break and the block will be in static equilibrium is ____________ kg. Take cosď ąď€ ď€˝ď€ ď€°ď€Žď€¸ď€ and sinď ąď€ ď€˝ď€ ď€°ď€Žď€śď€Ž

R O

T N

Acceleration due to gravity g = 10 m/s2

E M

E T

Q.31

A G

A two-member truss PQR is supporting a load W. The axial forces in members PQ and QR are respectively L

P

Q

30ď‚° 60ď‚°

R

W

(A) 2đ?‘Š tensile and 3đ?‘Š compressive (B) 3đ?‘Š tensile and 2đ?‘Š compressive (C) 3đ?‘Š compressive and 2đ?‘Š tensile (D) 2đ?‘Š compressive and 3đ?‘Š tensile

ME (Set-1)

5/11

GATE 2016

Q.32

SET-1

Mechanical Engineering (ME)

A horizontal bar with a constant cross-section is subjected to loading as shown in the figure. The Young’s moduli for the sections AB and BC are 3E and E, respectively.

For the deflection at C to be zero, the ratio P/F is ____________

Q.33

The figure shows cross-section of a beam subjected to bending. The area moment of inertia (in mm4) of this cross-section about its base is __________

10 R4

R4

R O

All dimensions are in mm

8 10

T N

10

Q.34

E M

A simply-supported beam of length 3L is subjected to the loading shown in the figure. P P L L L

E T

A

A G

It is given that P = 1 N, L = 1 m and Young’s modulus E = 200 GPa. The cross-section is a square with dimension 10 mm × 10 mm. The bending stress (in Pa) at the point A located at the top surface of the beam at a distance of 1.5L from the left end is _____________ (Indicate compressive stress by a negative sign and tensile stress by a positive sign.)

Q.35

A slider crank mechanism with crank radius 200 mm and connecting rod length 800 mm is shown. The crank is rotating at 600 rpm in the counterclockwise direction. In the configuration shown, the crank makes an angle of 90 with the sliding direction of the slider, and a force of 5 kN is acting on the slider. Neglecting the inertia forces, the turning moment on the crank (in kN-m) is __________ 200 mm

800 mm 90

5 kN

ME (Set-1)

6/11

GATE 2016

Q.36

SET-1

Mechanical Engineering (ME)

In the gear train shown, gear 3 is carried on arm 5. Gear 3 meshes with gear 2 and gear 4. The number of teeth on gear 2, 3, and 4 are 60, 20, and 100, respectively. If gear 2 is fixed and gear 4 rotates with an angular velocity of 100 rpm in the counterclockwise direction, the angular speed of arm 5 (in rpm) is 4 2

3 5

Q.37

(A) 166.7 counterclockwise

(B) 166.7 clockwise

(C) 62.5 counterclockwise

(D) 62.5 clockwise

A solid disc with radius a is connected to a spring at a point d above the center of the disc. The other end of the spring is fixed to the vertical wall. The disc is free to roll without slipping on the ground. The mass of the disc is M and the spring constant is K. The polar moment of inertia for the disc about its centre is J = ��2 /2.

R O

T N

M, J

K d

E M

a

E T

The natural frequency of this system in rad/s is given by

Q.38

2 đ??ž (đ?‘Ž+đ?‘‘)2 3 đ?‘€ đ?‘Ž2

A G

(A)

(B)

2đ??ž 3đ?‘€

(C)

2 đ??ž (đ?‘Ž+đ?‘‘)2 đ?‘€ đ?‘Ž2

(D)

đ??ž (đ?‘Ž+đ?‘‘)2 đ?‘€ đ?‘Ž2

The principal stresses at a point inside a solid object are ď ł1 = 100 MPa, ď ł2 = 100 MPa and ď ł3 = 0 MPa. The yield strength of the material is 200 MPa. The factor of safety calculated using Tresca (maximum shear stress) theory is nT and the factor of safety calculated using von Mises (maximum distortional energy) theory is nV. Which one of the following relations is TRUE? (A) nT = ( 3/2)nV (B) nT = ( 3)nV (C) nT = nV (D) nV = ( 3)nT

ME (Set-1)

7/11

GATE 2016

Q.39

SET-1

Mechanical Engineering (ME)

An inverted U-tube manometer is used to measure the pressure difference between two pipes A and B, as shown in the figure. Pipe A is carrying oil (specific gravity = 0.8) and pipe B is carrying water. The densities of air and water are 1.16 kg/m3 and 1000 kg/m3, respectively. The pressure difference between pipes A and B is __________kPa. Acceleration due to gravity g = 10 m/s2.

Q.40

Q.41

R O

Oil (kinematic viscosity, νđ?‘œđ?‘–đ?‘™ = 1.0 Ă— 10−5 m2/s) flows through a pipe of 0.5 m diameter with a velocity of 10 m/s. Water (kinematic viscosity, νđ?‘¤ = 0.89 Ă— 10−6 m2/s) is flowing through a model pipe of diameter 20 mm. For satisfying the dynamic similarity, the velocity of water (in m/s) is __________

E M

T N

A steady laminar boundary layer is formed over a flat plate as shown in the figure. The free stream velocity of the fluid is Uo.. The velocity profile at the inlet a-b is uniform, while that at a downstream location c-d is given by đ?‘˘ = đ?‘ˆđ?‘œ 2

E T

A G

b

đ?‘Ś

ď ¤

đ?‘Ś 2

−

ď ¤

ď€

.

y Uo

d

ď€ ď ¤

Uo

a

c

The ratio of the mass flow rate, đ?‘šđ?‘?đ?‘‘ , leaving through the horizontal section b-d to that entering through the vertical section a-b is ___________

Q.42

ME (Set-1)

A steel ball of 10 mm diameter at 1000 K is required to be cooled to 350 K by immersing it in a water environment at 300 K. The convective heat transfer coefficient is 1000 W/m2-K. Thermal conductivity of steel is 40 W/m-K. The time constant for the cooling process đ?œ? is 16 s. The time required (in s) to reach the final temperature is __________

8/11

GATE 2016

Q.43

SET-1

Mechanical Engineering (ME)

An infinitely long furnace of 0.5 m Ă— 0. 4 m cross-section is shown in the figure below. Consider all surfaces of the furnace to be black. The top and bottom walls are maintained at temperature T1 = T3 = 927 oC while the side walls are at temperature T2 = T4 = 527 oC. The view factor, F1-2 is 0.26. The net radiation heat loss or gain on side 1 is_________ W/m. Stefan-Boltzmann constant = 5.67 Ă— 10−8 W/m2-K4

Q.44

Q.45

R O

A fluid (Prandtl number, Pr = 1) at 500 K flows over a flat plate of 1.5 m length, maintained at 300 K. The velocity of the fluid is 10 m/s. Assuming kinematic viscosity, ν = 30 Ă— 10−6 m2/s, the thermal boundary layer thickness (in mm) at 0.5 m from the leading edge is __________

T N

E M

For water at 25 ď‚°C, dđ?‘?s /dđ?‘‡s = 0.189 kPa/K (đ?‘?s is the saturation pressure in kPa and đ?‘‡s is the saturation temperature in K) and the specific volume of dry saturated vapour is 43.38 m3/kg. Assume that the specific volume of liquid is negligible in comparison with that of vapour. Using the Clausius-Clapeyron equation, an estimate of the enthalpy of evaporation of water at 25 ď‚°C (in kJ/kg) is __________

E T

A G

Q.46

An ideal gas undergoes a reversible process in which the pressure varies linearly with volume. The conditions at the start (subscript 1) and at the end (subscript 2) of the process with usual notation are: đ?‘?1 = 100 kPa, đ?‘‰1 = 0.2 m3 and đ?‘?2 = 200 kPa, đ?‘‰2 = 0.1 m3 and the gas constant, R = 0.275 kJ/kg-K. The magnitude of the work required for the process (in kJ) is ________

Q.47

In a steam power plant operating on an ideal Rankine cycle, superheated steam enters the turbine at 3 MPa and 350 oC. The condenser pressure is 75 kPa. The thermal efficiency of the cycle is ________ percent. Given data: For saturated liquid, at P = 75 kPa, h f  384.39 kJ/kg, v f  0.001037 m3/kg, sf = 1.213 kJ/kg-K At 75 kPa, hfg = 2278.6 kJ/kg, sfg = 6.2434 kJ/kg-K At P = 3 MPa and T = 350 oC (superheated steam), h  3115.3 kJ/kg, s  6.7428 kJ/kg-K

ME (Set-1)

9/11

GATE 2016

Mechanical Engineering (ME)

A hypothetical engineering stress-strain curve shown in the figure has three straight lines PQ, QR, RS with coordinates P(0,0), Q(0.2,100), R(0.6,140) and S(0.8,130). 'Q' is the yield point, 'R' is the UTS point and 'S' the fracture point.

Engg. Stress (MPa)

Q.48

SET-1

160 140 120 100 80 60 40 20 0

(0.6, 140) (0.8, 130)

R

(0.2, 100)

S

Q

P

(0, 0)

0

0.2

0.4

0.6

0.8

1

Engg. Strain (%) The toughness of the material (in MJ/m3) is __________

Heat is removed from a molten metal of mass 2 kg at a constant rate of 10 kW till it is completely solidified. The cooling curve is shown in the figure.

Temperature (K)

Q.49

1100 1000 900 800 700 600 500

(0s, 1023K)

R O

T N

(20s, 873K)

(10s, 873K)

E M

E T 0

10

20

(30s, 600K)

30

40

Time (s)

Assuming uniform temperature throughout the volume of the metal during solidification, the latent heat of fusion of the metal (in kJ/kg) is __________

Q.50

A G

The tool life equation for HSS tool is �� 0.14 � 0.7 �0.4 = Constant. The tool life (T) of 30 min is obtained using the following cutting conditions: V = 45 m/min, f = 0.35 mm, d = 2.0 mm If speed (V), feed (f) and depth of cut (d) are increased individually by 25%, the tool life (in min) is (A) 0.15

Q.51

ME (Set-1)

(C) 22. 50

(D) 30.0

A cylindrical job with diameter of 200 mm and height of 100 mm is to be cast using modulus method of riser design. Assume that the bottom surface of cylindrical riser does not contribute as cooling surface. If the diameter of the riser is equal to its height, then the height of the riser (in mm) is (A) 150

Q.52

(B) 1.06

(B) 200

(C) 100

(D) 125

A 300 mm thick slab is being cold rolled using roll of 600 mm diameter. If the coefficient of friction is 0.08, the maximum possible reduction (in mm) is __________ 10/11

GATE 2016

Q.53

SET-1

Mechanical Engineering (ME)

The figure below represents a triangle PQR with initial coordinates of the vertices as P(1,3), Q(4,5) and R(5,3.5). The triangle is rotated in the X-Y plane about the vertex P by angle  in clockwise direction. If sin = 0.6 and cos = 0.8, the new coordinates of the vertex Q are

(A) (4.6, 2.8)

(B) (3.2, 4.6)

(C) (7.9, 5.5)

(D) (5.5, 7.9)

Q.54

The annual demand for an item is 10,000 units. The unit cost is Rs. 100 and inventory carrying charges are 14.4% of the unit cost per annum. The cost of one procurement is Rs. 2000. The time between two consecutive orders to meet the above demand is _______ month(s).

Q.55

Maximize Z=15X1 + 20X2

R O

12X1 + 4X2 ≥ 36 12X1 − 6X2 ≤ 24

E T

X1, X2 ≥ 0

T N

E M

subject to

The above linear programming problem has (A) infeasible solution (C) alternative optimum solutions

A G

ME (Set-1)

(B) unbounded solution (D) degenerate solution

END OF THE QUESTION PAPER

11/11

GATE MECHANICAL BOOKS

le b a il Ava All At

g n i d a e L k o o B s e r Sto

GATE MECHANICAL BOOKS

bl e a l i Ava At

GATE – 2016 Set - 1

ANSWER KEYS Set 1 - Aptitude section 1 A 2 B 3 A 4 A 6 C 7 B 8 D 9 C Set 1 - technical section 1 D 12 B 23 A 34 0 2 B 13 C 24 0.5 35 1 3 A 14 D 25 A 36 C 4 A 15 A 26 -5 37 A 5 1.56 16 B 27 -1 38 C 6 57.735 17 14.28 28 A 39 2.2or-2.2 7 C 18 0.25 29 1.55 40 22.25 8 A 19 C 30 5 41 0.3333 9 D 20 B 31 B 42 42.22 10 200 21 D 32 4 43 24530.69 11 20 22 C 33 21439.056 44 6.124

5 10

C C

45 2443.248 46 15 47 26.017 48 0.85 49 50 50 B 51 D 52 1.92 53 A 54 2 55 B

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GATE – 2016 Set - 1

APTITUDE/ENGLISH 1.(A)

Not diseased  0.5  0.7

2.(B)

 0.35

3.(A)

 35%

4.(A)

7.(B) Number of apples  x

8. (D) order of age

Number of oranges  y

Shiva

x  y  5692000

Leela

Unripe fruits

Pavithra 9.(C)

 0.15  x  y 

1 1 1 1 1 1  ,  ,  qa r r b s sc q

 853800

Ripe fruits

 q a  r , r b  s, s c  q  a ln q  ln r , b ln r  ln s, c ln s  ln q ln r ln s ln q  a ,b  ,c  ln q ln r ln s  abc  1

 0.85  x  y   4838200

Unripe Number of apples

 0.45 833800  10.(C)

 384210

Ripe number of applies

 1  0.66 4838200  1644988

Number of applies

Q  25 days  12hr / day  

1 per day 25

R  50days  12hr / day  

1 per day 50

Work by Q  512 

x  384210  1644988 x  2029198

1 25

Work by R  7 18 

5.(C) Only possible answer is C which is between 5 km and 7 km. 6.(C) Probability of infected but

1 50

60 Q 60 20  25  2  R 7 18 7 18 21 50

TECHNICAL SECTION 2x  5 y  2

…(i)

u v u v  ,  y x x y

4 x  3 y  90

…(ii)

u  2 xy v  ?

1.(D)

Solving (i) and (ii), x  6

y  2

2.(B)

u v  x y 2y 

L F  t    e st f  t dt 

0

3.(A) For analytic function

v y

u  y2  F  x

u  F ' x x

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…(i)

GATE – 2016 Set - 1

v  F ' x y

F 100  sin150 sin120 F  57.73 N

2 x  F '  x  7. (C)

F  x    x 2  constant

Now energy balance

From (i)

PE1  KE2

v  y  x  constant 2

2

mgH1 

4.(A) For poission distribution

v  2 gH

Mean   variance  

8. (A)

Standard deviation  

A2  A1

r

2 4

5. (1.5641)

F  x   x  10cos  x 

x1  x0 

xo 

F  xn 

2 2 2 2 J 2  r4  r3  r4  r3   2 2 2 J1  r2  r1  r2  r12 

F '  xn 

F  x0 

F '  xo 

2 2 J 2  r4  r3   J1  r22  r12 

 4

F  xo  

r3  r1 and r4  r2

   10cos   4 4

F '  xo   1  10sin

 r42  r32  r22  r12

F '  xo   8.07106

4

r42  r32 1 r22  r12

J2 1 J1

9. (D)

from equation (i)

4

F  xo   6.2856

x1 

 r32    r22  r12 

r4  r4 J 2 32  4 3    4 4 J1  r2  r1  32

F '  x   1  10sin x

xn 1  xn 

1 mv2 2

Deflection 

6.2856 8.07106

Pl 3 3EI

 2 I1 a14 1    1 I 2 a24 1.19 4

x1  1.5641

2  0.4986 1

6.(57.73)

Percentage change in deflection 1200 600

F

   2  1 100   2  1 100  50.14 1   1 

 2  0.4986  1

900 100N

So

1 decreased by 50%

Sine – rule

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GATE – 2016 Set - 1

10.(200)

15.(A)

Gyroscopic couple “c”

C  I . p V 20 2   r 100 10

p 

k 0.15   6mm h 0.25

rc 

Q

  100 rad / sec I  10 Kg / m2

rc

C  I  p  10 100 

r

2  200 Nm 10

ro  1mm

11. (20)

Since r0  rc

Steady state amplitude, X static Xm  2 1  r 2    2 r 2

X static 

So increasing r will increase the heat Q for upto 6mm radius.

F 100   0.01 m, K 10 103

Work and heat are path function and are inexact differential and boundary phenomena .

  0.25 r  1 for resonance  Xm 

17.(14.28%)

0.01

1  1    2  0.25 1 2 2

2

X m  20 mm

For complete combustion of fuel (stoichiometric):-

C3 H8  5O2  3CO2  4H 2O But here 10% deficit of O2 than

12.(B) K

16.(B)

9 C3 H8  O2  2CO2  4H 2O  CO 2

4

G.d 8D3 N

So % of CO in exhaust will be

k  d4

 100 

1 D3 1 k N k

1  14.28% 7

18.(0.25) Specific speed

13.(C)

NS 

N P H

Fluctuation

5 4

,

N1 2 N2

Now

N S1  N S 2 Vavg

u  xyz   const

T

o

u  xyzt  1

time

 zero

14.(D) Centroid of the displaced fluid is also known as centre of buoyancy.

 H1  H 2 N1 P1  N2 P2

N12  P1  N22  P2 P1  N 2    P2  N1 

2

P1 1  P2 4

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GATE – 2016 Set - 1

m

19.(C) Principle is not similar to the principle of Stirling gas cycle, Regeneration is constant pressure in vapor power cycle but constant volume in stirling cycle.

2

 9  0

m  3i

Now

y  C1 cos3x  C2 sin 3x

20.(B)

y '   3C1 sin 3x   3C2 cos3x

Jominy test is used for finding hardenability of test material such as ability up to how much depth hardness can be achieved.

From equation (i)

y  0  0

21.(D)

0  C1

Twelding > Tbrazing > Tsoldering

From equation (i)

22.(C)

  y   2 2

The part of a gating system which regulates the rate of pouring of molten metal is choke area (minimum area).

 3  2  C2 sin    2 

23.(A)

C2   2

Vacuum is required in Electron beam machining operation to have non dispersive medium for electron propagation.

Putting value of C1 and C2 in equation (i) y   2 sin 3x

24.(0.5)

FT  tan     0.5 Fc

…(ii)

From equation (ii),   y     1 4

25.(A) 26.(-5)

28.(A)

F  x   2 x  3x 3

 sin 1

2

x   1, 2

e

29.(1.55)

F '  x   6 x2  6 x

F ' x  0  x  0 x  1

Highest coefficient should calculation of x1, x2, x3

At

x1  2 x2  3x3  5

…(i)

x  0 F  0  0

2 x1  3x2  x3  1

…(ii)

x  1, F 1  1

3x1  2 x2  x3  3

…(iii)

Now at F  1  5  minimum

Largest coefficient of x1 in the above equation

F  2  4  maximum Hence minimum

value

of F  x  is -5 in

x   1, 2

be

taken

for

is 3 which is in equation (iii) then from equation (iii),

 x1 1 

1 3  2  x2 0   x3 0  3 1 3

 x1 1  3  2  0  0  1

27.(-1) y " 9 y  0,

D

2

A.E.,

y  0  0

 9 y  0

  y   2 3

In the left equation Largest coefficient of x2 in the above equation is 3 which is in equation (ii) then from equation (ii),

 x2 1 

1 1  2  x1 1   x3 0  3

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GATE – 2016 Set - 1

1 3

 x2 1  1  2 1  0  

 PL   PL      0  AE  AB  AE  BC

2 3

 F  P L

Only left equation is equation (i), then

A  3E

 x3 1 

1 5   x1 1  2  x2 1  3

P  4.F

 x3 1 

1  2   14 5  1  2     1.55  3  3  9

P 4 F

FL 0 AE

30.(5) m

os gc

20N

33.(21439.05) Parallel axis theorem

mg

s

in 

mg

mg

cos

cos  0.8

20

sin   0.6

8

x

1   m 10  cos   m10sin   20 4

x

10

m  5kg

X’

x'

31.(B)

10 TPQ 30

270

I xx 

60

10  203       84   6465.6 mm4 3  64 

M.O.I of inertia about base,

TRQ

w

I x ' x '  I x ' x '  A 10 

TPQ TRQ W   sin 30 sin 60  sin 90

 2   I x ' x '  6465.6   10  20     82    10   4    4 I x ' x '  21439.05 mm

From above equation we get , 34.(0)

TPQ  3W  tensile

P

TRQ  2W  compressive

3

P

l

32.(4) A

l

A C

B

2

P l P 3

F

P

MA 

P 3 l  l  P 3 2 2

FAB  F  P

MA 

Pl Pl  2 2

Force in BC,

MA  0

FBC  F

Moment At A = 0, stress is also = 0

Force in AB,

 c  0  given 

 AB   BC  0

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GATE – 2016 Set - 1

35.(1)

37. (A) 5KN 

800mm 90

k  a  d 

I

a

 d 

d

5KN

c a

Ft  r  5000  0.2 (Turning moment)

 1000 Nm  1 KNm

ma 2 3  m  a 2  ma 2 2 2

I o  I c  mr 2 

36.(C)

Now

Taking cw  ve and ccw  -ve

I o  K  a  d    a  d   0 2 5

4

3

100=T4

T2=60

Condition

Ar m 5

Into

Gear 2 1

Gear 3

meq

Gear 4

60 20  3 

20 100  0.6  3 

x

3x

0.6x

y

x y

3x  y 0.6x  y

K a  d  3 2 ma 2

2k  a  d 

n 

0

Add  y

K eq

n 

T3=20

Arm  Fix 0 Gear 2  +1rev

x

3 2  2 ma    K  a  d    0 2 2

2

3ma 2

38.(C)

1  100 MPa

 2  100 MPa

3  0 S yt  200 MPa Maximum shear stress,

N2  0 x y 0

 max  …(i)

N4  100  C.C.W 

Solving equation (i) and (ii)

T  fos by Tresca 

v  Fos by MDE

N4  100 0.6 x  y  100

100  0  50 Mpa 2

…(ii)

Now by Tresca theory

T 

x  62.5, y  62.5

S yt 2  max S yt

Speed of arm

 max 

N5  y

Now by MDE

N5  62.5 N5  62.5  C.C.W 

T 

200 2 2  50

2.F .O.S

S yt

   2   1 2 2 1

3

200 100  1002  1002 2

Therefore,

T  v

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2

GATE – 2016 Set - 1

39. (–2.2 or 2.2)

43.(24530.7)

Pressured left T3

Column = Pre. At right column

PA  air g   0.08  oil  g   0.2  water  g  0.38  PB

800k

0.4

T4

 PA  PB   2.2 KPa

Or  PB  PA   2.2 KPa

800k T2(527)0C

0.5

40. (22.25)

oil  1105 m2 / s   0.5 m V  10 m / s

F12  0.26

water  0.89 106 m2 / s d  0.02m

Now

Satisfy the dynamic similarities 

 Re1  Re2 

10  0.5 V  0.2  105 0.89 106

V  22.15 m / s

Q1 2  F12  A1    T14  T24 

Q1 2  12265.34 w / m Due to symmetry,

 Q12  Q14 

41.(0.33) Mass out flux  b   t Vo

   b (displacement thickness) Vo      y   y 2        b   1   2        dy   V0    o            

Total heat transfer,

Q1  Q1 2  Q1 4   2  Q1 2   24530.68 w / m 44.(6.12) T  =500 k

Pr=1

10m/S

      b  Vo          b   Vo 3 3  mass  bd  

1200k

  b    vo 3

mass  ab     b  Vo

Vo m  bd    A  3 ratio of   0.33 m  ab    A  V0

300k=TS 1.5m

v  300 106 m2 / s

x  0.5m

Re 

h 

42.(42.22)

d  10mm To  1000K T  350K

T  300K

h  1000 w / m2  K

K  40 w / mk  *  16sec

T 

.V.L Vx   166.66 103  v 5x Re

 6.12mm

h 1

 pr  3

 6.12 mm

45.(2443)

vg  43.38 m3 / kg T  298k

 T  T  e x To  T

 T  T   ln   16  To  T 

  42.22sec

dPs Kpa  0.189 dTs k Clausius equation

h fg dPs  dTs Ts.V fg 0.189 

h fg 298  43.38

h fg  2443 kj / kg

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GATE – 2016 Set - 1

mass  L.H  10 10  at the time of phase change

46.(15)

L.H  50 KJ / kg 50.(B)

2

200 KPa

VT 0.14 F 0.7 d 0.4  C

P

T1  30min d1  2.0 mm V1  45 m / min

100 KPa

f1  0.35 mm

1 0.1m3 V

Now 25% increase individually

0.2m3

V2  56.25 m / min

Area under curve is =15 kj

f 2  0.4375 mm

d2  2.5 mm

47. (26)

Now

V1T10.14 F10.7 d10.4  V2T20.14 F20.7 d20.4 3MPa h1 75KPa

350 h4 h3

T2  1.055min 51.(D)

h2

Part

100mm

h1  3115.3 KJ / kg h3  384.39 KJ / kg

h4  h3  v f dp

200mm

h4  387.42 KJ / kg S1  S2  6.7428  1.213  x  6.2434 x  0.8856

h2  2402 KJ / kg



net Qadd

Or  

V  R2 H  A 2 RH  2 R 2

V R.H   25 A 2H  2R

V  M R &er     A R

 h1  h2    h4  h3  h1  h4

H

b

 0.26  26%

48.(.85)

Riser

Area under the stress – strain diagram is toughness:-

1 1    0.2 102  100  100  140  2 2   0.4  102 

1 140  130   0.2 102 2

 0.85 MJ / m3 49.(50)

mass  2 kg Q  10 kw  10kJ / sec

 D2  H D 4    2 5  DH  D 4

Modulus method

M Riser  c  M casting c is ratio of solidification time of riser to casting. As c is not given it can be assumed 1

D  1 25 5 D  125 mm

Time for phase change = 10 sec

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GATE – 2016 Set - 1

52.(1.92)

EOQ 

H max   R   0.08  03  1.92mm 2

2

2  D  CO 2 10, 000  2000  Cc 14.4

EOQ  1666.66 unit

53.(A) Q

l 

P (1,3)

13

No. of order = 6 per year (4,5)

Month total in year = 12 Time between two order 

2

33.690

55.(B)

(4,3)

36.87 3.18

12  2 month 6

max z  15x1  20 x2

x  3 .6

Subjected to l  13

y  0 .2

9

Unbounded Solution

Q (4.6,2.8)

Rotation by  ,

sin   0.6    36.87

2

3

4

Refer above figure,

sin 3.18  cos 3.18 

y 13 x 13

-4

 y  0.2  x  3.6

Coordinate of point Q,

Q  1  3.6,3  0.2  Q   4.6, 2.8 54.(2) D  10,000 per year

C  100 per unit Cc  0.144 100  14.4

12 x1  4 x1  36

…(i)

12 x1  6 x2  24

…(ii)

x1 x2  0 Since there is no limit in maximum value of Z, therefore solution is unbounded.

Co  2000 per order

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