Mechanical Engineering  >  GATE Mechanical (ME) 2023 Mock Test Series  >  Practice Test: Mechanical Engineering (ME)- 12 Download as PDF

Practice Test: Mechanical Engineering (ME)- 12


Test Description

65 Questions MCQ Test GATE Mechanical (ME) 2023 Mock Test Series | Practice Test: Mechanical Engineering (ME)- 12

Practice Test: Mechanical Engineering (ME)- 12 for Mechanical Engineering 2023 is part of GATE Mechanical (ME) 2023 Mock Test Series preparation. The Practice Test: Mechanical Engineering (ME)- 12 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Practice Test: Mechanical Engineering (ME)- 12 MCQs are made for Mechanical Engineering 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Practice Test: Mechanical Engineering (ME)- 12 below.
Solutions of Practice Test: Mechanical Engineering (ME)- 12 questions in English are available as part of our GATE Mechanical (ME) 2023 Mock Test Series for Mechanical Engineering & Practice Test: Mechanical Engineering (ME)- 12 solutions in Hindi for GATE Mechanical (ME) 2023 Mock Test Series course. Download more important topics, notes, lectures and mock test series for Mechanical Engineering Exam by signing up for free. Attempt Practice Test: Mechanical Engineering (ME)- 12 | 65 questions in 180 minutes | Mock test for Mechanical Engineering preparation | Free important questions MCQ to study GATE Mechanical (ME) 2023 Mock Test Series for Mechanical Engineering Exam | Download free PDF with solutions
1 Crore+ students have signed up on EduRev. Have you?
Practice Test: Mechanical Engineering (ME)- 12 - Question 1

Mr. Rammurthy starts a business investing 250000 in year 1998. In year 1999 he invested an additional amount of 100000 and Mr. Siddharth Rau joined him with an amount of 350000. In year 2000. Mr. Rammurthy invested another additional amount of 100000 and Mr. Rajan Babu joined them with an amount of 350000. What will be Siddharth Rai’s share in the profit of 1500000 earned at the end of three years from the start of the business in 19988?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 1 Rammurthy’s investments for 3 yr

= (250000 × 3 + 100000 × 2 + 100000 × 1)

=1005000

Siddharth’s investment for 3 yr

= 350000 × 2 =700000

Ranhan’s investment for 3 yr

=350000 × 1 = 350000

Ratio = 1005000:700000:350000

= 3 :2 :1

Siddhartha’s profit

Practice Test: Mechanical Engineering (ME)- 12 - Question 2

What is the difference between the compound interest and simple interest on Rs. 4000 at 5% per annum for 2 years?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 2 Difference = Principal (r/100)2

Practice Test: Mechanical Engineering (ME)- 12 - Question 3

In the following question, out of the five alternatives, select the word similar in meaning to the given word.

Sedition

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 3 Sedition (n.): conduct or speech inciting people to rebel against the authority of a state or monarch.

Allegiance (n.): loyalty or commitment to a superior or to a group or cause.

Homage (n.): special honour or respect shown publicly.

Adulation (n.): excessive admiration or praise.

Allaying (v.): diminish or put at rest (fear, suspicion, or worry).

Insurrection (n.): a violent uprising against an authority or government.

So, the correct answer is option D.

Practice Test: Mechanical Engineering (ME)- 12 - Question 4

Two pipes A and B can fill a tank in 36 min and 45 min respectively. A waste pipe C can empty the tank in 30 min. If all the pipes are opened together then the tank will be full in how much time?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 4 Part filled by pipe A in 1 min = 1/36

Part filled by pipe B in 1 min = 1/45

∴ Total water dropped in the tank in 1 min = 1/36 + 1/45

Water leaked out in 1 min by C = 1/30

∴ Total tank filled in 1 min

Hence, the tank will be filled in 60 min.

Practice Test: Mechanical Engineering (ME)- 12 - Question 5

Direction: In each question, a sentence with four words printed in bold are given. These are numbered as (1), (2), and (3). One of these four words printed in bold may be either wrongly spelt or inappropriate in the context of the sentence. Find out the word, if any, which is wrongly spelt or inappropriate. The number of that word is the answer. If all the words printed in bold are correctly spelt and appropriate in the context of the sentence, mark (4), i.e., 'All correct' as the answer.

No incidence of violence (1)/ occurred during the protests (2)/ against the Government’s (3)/ All correct (4)

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 5 Replace ‘No incidence of violence’ by ‘No incident of violence’, which means no serious event, such as a crime or an attack. Hence option A is correct.
Practice Test: Mechanical Engineering (ME)- 12 - Question 6

In the following question, select the related word pair from the given alternatives.

? : Sun :: Rain : ?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 6

Here the related word pairs are- (Light, Cloud)

Sun gives us light in same way Cloud gives us rain (so, Light : Sun :: Rain : Cloud).

Hence, option (B) is the correct response.

Practice Test: Mechanical Engineering (ME)- 12 - Question 7

72% of the students of a certain class took Biology and 44% took Mathematics. If each student took at least one of Biology or Mathematics and 40 students took both of these subjects, the total number of students in the class is?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 7

Percentage of students opting for both subjects

= 72 + 44 - 100 = 16%

If the total number of students be x, then

x = 4000/16 = 250

Practice Test: Mechanical Engineering (ME)- 12 - Question 8

Study the following paragraph and answer the given questions.

Fashion has become one of the largest fads among the youth. The amount of time wastage and expenditure on fashion is very large. What bothes however is the fact that fashion is here to stay despite countless arguments against it. What is required therefore is that strong efforts should be made in order to displace excessive craze of fashion from the minds of today's youth.

Which of the following statements finds least support by the argument made by the author in the given paragraph?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 8

Fashion usually is neither named nor noted but is simply the lens through which our society perceives itself and the mold to which it increasingly shapes itself. This hidden, powerful, mental sort of fashion is thus worth taking stock of. In spite of its great parade of intellect, its support in influential places, and its mellifluous accompaniment of self-promoting public relations, the new variety of fashion pretty much shares the creed and the limitations of the frivolous, pirouetting old variety of fashion—that of dress.

Option C is correct.

Practice Test: Mechanical Engineering (ME)- 12 - Question 9

Direction: Find out the number of students who play only cricket.

Note : Total number of students - (25 +22) = 47

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 9

The number of students who play only cricket

= 25 - 16 = 9

Practice Test: Mechanical Engineering (ME)- 12 - Question 10

If the uncle of the father of Rani is the grandson of the father of Anup and Anup is the only son of his father, then what is the relation of Anup with Rani?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 10

There are 3 lines (1 line for each generation) between Anup and Rani, So this is clear that Anup is the great grandfather of Rani.

Practice Test: Mechanical Engineering (ME)- 12 - Question 11

A copper rod 400mm long is pulled in tension to a length of 401.2 mm by applying a tensile stress of 330 MPa. Then Young’s modulus of copper is


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 11 given,

Lf = 401.2mm, Li = 400mm

σ = 330MPa

Practice Test: Mechanical Engineering (ME)- 12 - Question 12

Find optimum order quantity where annual demand is 400 and ordering cost is 250 per order Carrying cost per unit 2%.


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 12

Which is in Range, this is order quantity.

Practice Test: Mechanical Engineering (ME)- 12 - Question 13

Consider the following statements:

P: Casting yield is kept minimum while designing a gating system.

Q: Gating ratio of 4:1:1 corresponds to a pressurised gating system.

R: Possibility of air aspiration effect is more in unpressurised gating system

S: Pressurised gating system can only be used for non ferrous castings

Which of above statements are correct?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 13 Casting yield is kept as high as possible

Pressurised gating system is used for ferrous castings

Practice Test: Mechanical Engineering (ME)- 12 - Question 14

The general solution of the differential equation with C1 and C2 as constants are:

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 14 m2 - 4m + 4 = 0

m = 2, 2

Complementary function = (C1 + C2x)e2x

Particular Integral

Put D = 2 (case fails)

(case fails)

General solution = C.F+P.I

Practice Test: Mechanical Engineering (ME)- 12 - Question 15

A square bar with tapered length having circular cross section area having diameter 10 cm and 20 cm at two ends and length of the bar is 80 cm is subjected to a tensile force of 50 KN. If the bar is made up of steel having modulus of elasticity 200GPa, Find the elongation of the tapered bar (in mm) _______.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 15 Given,

P = 50 kN,

L = 80cm,

D1 = 10cm

D2 = 20 cm

Elongation of tapered bar is given as

Practice Test: Mechanical Engineering (ME)- 12 - Question 16

A stone is dropped from a balloon going up with a uniform velocity of 10 m/s. if the balloon was 40 m high when the stone was dropped, the height of balloon from the ground when the stone hits the ground is (take g = 10 m/s2)

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 16 The stone was going up with 10m/s velocity and after that it falls freely under gravity.

Considering vertically downward as +ve and upward -ve.

S = ut+1/2gt2

S = 40m

u = −10m/s

40 = − 10t + 1/2 × 10 × t2

t2 − 2t − 8 = 0

t2 − 4t + 2t − 8 = 0

t = −2 (no significance in this problem)

t = 4s

Distance covered by balloon in 4 s in upward direction

= 10t = 40m

Total height of balloon = 40 + 40 = 80m

Practice Test: Mechanical Engineering (ME)- 12 - Question 17

A banner hoarding having dimensions 2m × 2m is kept vertical by a 5m long pillar. Air blowing at 5 m/s is striking directly to the banner. If the maximum moment the pillar can withstand is 100 Nm. To reduce the drag force holes need to be cut out from the hoarding with a punch of 10 cm diameter. If the air drag coefficient is 0.5, then the total number of holes need to be cut out are

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 17 Bending moment at the bottom of the pillar.

M= F × (5 + 1)

⇒ F= 100/6 = 16.67

That much amount of force can be applied by air, hence it should be equal to drag force,

F= 1/2 CDρAV2

16.67=1/2 (0.5)(1.2) × A × (5)2

A = 2.223 M2

The excess area need to cut from the hoarding

Area of hoarding – A

= 2 × 2 – 2.223

= 1.777 m2

Number of holds need to cut,

Practice Test: Mechanical Engineering (ME)- 12 - Question 18

An double slider chain mechanism with the slotted link of length 5 units FIXED. A point on the link traces a curve which has a slope dy/dx = -2√3 eat the point (√3/2,1). Find the equation of the curve.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 18 We know in this elliptical trammel, a point on the link will have a locus of ellipse

x2/a2 + y2/b2 = 1

Since the point (√3/2,1) lies on a curve, substituting in the above equation we get

(3/4a2) + (1/b2) = 1

Again differentiating we get 2x/a2 + 2y/b2dy/dx = 0

Putting point (x, y) = (√3/2,1) and dy/dx = -2√3

We get b2 = 4a2

Solving both we get a = 1,b = 2

The curve traced by a point on the link is 4x2 + y2 = 4

Practice Test: Mechanical Engineering (ME)- 12 - Question 19

A slender bar of 100 mm2 cross-section is subjected to loading as shown in the figure. If the modulus of elasticity is taken as 200 × 109 Pa, then the elongation produced in the bar will be ____ mm.


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 19 Elongation of the bar

= 0

Practice Test: Mechanical Engineering (ME)- 12 - Question 20

The resultant of two vectors when acting at right angles is 10kN. If they act at 600 their resultant is 5√6 kN. Find the magnitude of individual vectors.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 20 Resultant of two vectors is

Practice Test: Mechanical Engineering (ME)- 12 - Question 21

Two coins are tossed once, where E :no tail appears, F : no head appears. Find P(E/F).


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 21 Here, E: = no tail appears' and F: 'no head appears'

i.e., E = {H H} and F = {T T}

⇒ E ∩ F = { } = ϕ

Hence, P (E ∩ F) = 0 and P(F) = 1/4

Practice Test: Mechanical Engineering (ME)- 12 - Question 22

For a beam simply supported at ends, carrying UDL, the ratio of maximum principal stress at extreme fibre to maximum principal stress at neutral axis is___. The beam is square and length is 10 times the side of square.


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 22

Maximum stress on extreme fibre will be on point of maximum bending stress.

Max principal stress at N.A = τmaxNA

At point of maximum

Practice Test: Mechanical Engineering (ME)- 12 - Question 23

A body is said to be provided optimum amount of streamlining when ______.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 23 A streamlined body is a shape that lowers the friction drag between a fluid, like air and water, and an object moving through that fluid.
Practice Test: Mechanical Engineering (ME)- 12 - Question 24

At a production machine, parts arrive according to a Poisson process at the rate of 0.35 parts per minute. Processing time for parts have exponential distribution with mean of 2 min. What is the probability that a random part arrival finds that there are already 8 parts in the system (in machine + in queue)?


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 24 Given that

λ = 0.35 per minute

µ = 1 /2 = 0.5 per minute

Hence,

ρ = λ/ µ = 0.7

The probability that a random part arrival finds that there are already 8 parts in the system is Pn = ρn (1−ρ) = 0.78 (1−0.7) = 0.01729

Practice Test: Mechanical Engineering (ME)- 12 - Question 25

A Circular beam, 50 mm in diameter, is welded to the support by means of circumferential fillet weld as shown in the diagram. It is subjected to a eccentric load of 20kN at a distance of 100mm parallel to support and at distance of 200mm away from the support. Thickness of the weld is 10mm. Find Maximum principal stress at weld on X-X plane.


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 25

Primary shear will act upward

Secondary shear will act anti-clockwise so right side of the section equivalent shear will be the summation of primary and secondary shear.

The polar moment of Inertia of circular beam

As at X-X plane, there will not be any bending stress

So max principal stress = T1 + T2 = 63.66 MPa

Practice Test: Mechanical Engineering (ME)- 12 - Question 26

The turbine rotor of a ship has a mass of 5000 kg and its radius of gyration is 500 mm. The turbine rotor rotates at 1500 rpm in counter-clockwise direction (view from stern or rear end). If the ship travels at a speed of 90 km/h and steer to the left in a curve of 100 m then the gyroscopic couple on the ship and its effect respectively are

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 26 Given, m = 5000kg,k = 0.5m,N = 1500rpm, V = 90

km/h = 25m/s,R = 100m

The gyroscopic couple, C = IωωP

I = mk2 = 5000 × 0.52 = 1250kg − m2

ωp = ∨/R = 25/100 = 0.25m/s

C = 1250 × 157.08 × 0.25 = 49087.5N − m = 49.0875

kN−m

As rotor rotates in anticlockwise direction, when view from rear end and steering is towards left, so, the effect of gyroscopic couple will be to lower the bow and raise the stern end.

Practice Test: Mechanical Engineering (ME)- 12 - Question 27

Consider following statements:

I) A gas is said to be an ideal gas when intermolecular forces are zero.

II) A gas behaves as an ideal gas under high temperature and high pressure.

III For an ideal gas, the temperature variation w. r. t. pressure during throttling condition is zero.

IV) A gas cools during expansion only when the Joule Thomson coefficient is negative.

Which of the following statements is/are correct?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 27 A gas behaves as an ideal gas under high temperature and low pressure and a gas cools during expansion only when the Joule Thomson coefficient is positive. If negative then upon expansion it heats (refer inversion curve).

Practice Test: Mechanical Engineering (ME)- 12 - Question 28

Two castings of the same metal have the same surface area. One castings is in the form of a sphere and the other is a cube. What is the ratio of the solidification time for the sphere to that of the cube?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 28 As surface area are equal 4πr2 = 6a2

Practice Test: Mechanical Engineering (ME)- 12 - Question 29

The uniform rod in figure weighs 400 N, length of the rod is 12 m

Determine the angle ɵ in degrees.


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 29 From the figure-

Fram 1 & 2

cos3 θ = 2/3

θ = 29.12

Practice Test: Mechanical Engineering (ME)- 12 - Question 30

The system of equations x - 4y + 7z = 12 , 3x + 8y - 2z = 10, 26 z - 8y = 6

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 30

Here, Rank [A:B] = Rank⁡[A] = 3 = n

And n = 3

Therefore, the system has unique solution.

Practice Test: Mechanical Engineering (ME)- 12 - Question 31

A bar of length 2.1m and diameter 38mm is subjected to pure torsion. If the shear modulus of the material is 41GPa, calculate the torsional stiffness of the bar.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 31 We know that Torsional Stiffness

So, Torsional stiffness = 3996.67Nm

Practice Test: Mechanical Engineering (ME)- 12 - Question 32

Two shafts of same material and same length are connected in series and subjected to a torque of 20 kN-m. If the ratio of their diameters is 1:3 then the ratio of their angles of twist is:

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 32 In case of series connection torque remains same. Thus, by torsion equation

θ = TL/GJ

Where:

J = Polar moment of Inertia

Practice Test: Mechanical Engineering (ME)- 12 - Question 33

The temperature of two thermometers A and B are tA and tB. The reading t a and tB agree at the ice point and steam point and are related as

Where, a,b and c are constants and tA and tB are in C The thermometers A and B, bath are immersed in a fluid and reads 45C

and 47C respectively. The thermometer reading A, when the thermometer B reads 22C

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 33

At ice point, tA = tB = 0C

a = 0

At steam point, tA = tB = 100C

100 = 0 + b × 100 + c × 1002

b + 100c = 1(11)

At tA = 45, tB = 47

45 = 0 + b × 47 + c × 472

b + 47c = 0.957 (iii)

Solving equation (ii) and (iii)

b = 0.919

c = 8.11 × 10−4

So, finally eq" (i) becomes

At tB = 22,

tA = 0.919 × 22 + 8.11 × 10-4 × 222 = 20.61°C

Practice Test: Mechanical Engineering (ME)- 12 - Question 34

A shaft compressed all round by a hub will have a Mohr’s circle as __________.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 34 When a shaft compressed all round by a hub will have the same impact as a hydrostatic stress acting all around the hub. That will be a point, since it is compression, a point on the negative x-axis.
Practice Test: Mechanical Engineering (ME)- 12 - Question 35

Steam flows at the rate of 5m3/s at a particular section in a reaction turbine where the diameter of blade is 2m and the velocity of flow is 6m/s. What is the blade height of section _________mm.


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 35 Flow rate = Area × Velocity of flow

Q = A × V

5 = πdh × Vf

Practice Test: Mechanical Engineering (ME)- 12 - Question 36

A conical valve of weight 3 kN is used to stop the flow of water from bottom the tank as shown in the figure. Find the force F (in kN) required to stop the flow of water from the bottom of the tank.


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 36 The FBD of the cone

Now the radius of the cone is , r = 1 × tan 300 = 0.577 m

Fv = Fv1 + Fv2 = (πr2 * 2 – πr2*1/3) ρg

Fv = 17.12 kN

Fv + 3 = F

F = 17.12 + 3 = 20.12 kN

*Answer can only contain numeric values
Practice Test: Mechanical Engineering (ME)- 12 - Question 37

Far the Froude model law 1/25 model of a pipe is to be tested. If the discharge is 300m3/s, then the discharge at which the model must be tested is ……….m3/s?


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 37 For the model

Practice Test: Mechanical Engineering (ME)- 12 - Question 38

The pressure on the fluid is 3 MPa. The Young’s modulus is 210 GPa and the Poisson’s ratio is 0.25, the volumetric compressive strain produced is

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 38

Given- P = 3MPa

E = 210 x 103 MPa

μ = 0.25

E = 3K(1 - 2μ)

εy = 21.42 x 10-6

Practice Test: Mechanical Engineering (ME)- 12 - Question 39

If there are two materials A & B, for material “A”, Young’s modulus is twice the Shear modulus & for “B” Young’s modulus is thrice the bulk modulus, then μA & μB respectively are

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 39

For A

E = 2G(1 + μ)

For B

E = 3K(1 − 2μ)

E = 2G for "A",μA = 0

E = 3K for "B",μB = 0

Practice Test: Mechanical Engineering (ME)- 12 - Question 40

Consider a stepped shaft subjected to a twisting moment applied at B & C as shown in the figure. Assume shear modulus, G = 80 GPa. The angle of twist at D (in degrees) is ______ (corrected up to 3 decimal places)


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 40 θ = Angle of twist = TL/GJ

Tatal twist at C will be equal ta D

θe = θAB + θBC = θD

θD = 0.00673 radian

= 0.386 degree

Practice Test: Mechanical Engineering (ME)- 12 - Question 41

In an epicyclic gear train, an arm carries two gears A and B having 36 and 45 teeth respectively. If the arm rotates at 150 r.p. m. in the anticlockwise direction about the centre of the gear A which is fixed, determine the speed of gear B. If the gear A instead of being fixed, makes 300 r.p.m. in the clockwise direction, what will be the speed of gear B?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 41

Let

NA = Speed of gear A

NB = Speed of gear B, and

NC = Speed of arm C

Assuming the arm C to be fixed, speed of gear A relative to arm C

= NA − NC

and speed of gear B relative to arm

C = NB − NC

since the gears A and B revolve in opposite directions, therefore

Speed of gear B when gear A is fixed

When gear A is fixed, the arm rotates at 150 r.p.m. in the anticlockwise direction, i.e.

NA = 0, and NC = +150r⋅p⋅m

or NB = −150 × −0.8 + 150 = 120 + 150 = 270r⋅p⋅m⋅ Ans.

Speed of gear B when gear A makes 300 r.p.m. clockwise

since the gear A makes 300 r.p.m. clockwise, therefore

NA = −300r⋅p⋅m

NB = -450 x -0.8 + 150 = 360 + 150 = 510 r.pm. Ans

Practice Test: Mechanical Engineering (ME)- 12 - Question 42

The cost parameters and other factor for a production inventory system of automobile factory are

Demand / year = 6000 units

Unit cost = Rs. 40/-

Setup cost (order cost) = Rs. 500/ set up

Production rate/ year = 36000 units

Holding cost/unit/year = Rs. 8/-

Shortage cost/unit/year = Rs. 20/-

The optimal lot size would be _____ units.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 42

D = 6000 units

C0 = Rs. 500, P = 36000 units

Cn = Rs. 8, Cb = Rs. 20

= 1122.49 = 1123units

Practice Test: Mechanical Engineering (ME)- 12 - Question 43

The number of degrees of freedom for the mechanism shown will be.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 43

The mechanism consists a cam pair. Therefore, its degree of freedom must be found from Grubler’s criterion.

F = 3(l-1)-2j-h

l→ Number of links

j→ Number of binary points

h→ Number of higher pairs

here, l = 7

j = 8

h = 1

F = 3(7 - 1) -2 × 8 - 1

F = 1

Practice Test: Mechanical Engineering (ME)- 12 - Question 44

A laminar fully developed fluid (k = 0.175 W/m-K) flow inside a 6 mm internal diameter tube under uniform wall temperature boundary conditions. Compute the heat transfer rate between the tube walls and the fluid for a length of 8 m if the mean temperature difference between the wall and the fluid is 50 °C.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 44

Nu = kD/k = 3.66

(for fully developed laminar flow under

uniform wall temperature condition)

or

= 106.75 W/m2-K

Now, Q = h(πDL)∆T

= (106.75)(π × 0.006 × 8) × 50

= 804.87 W

Practice Test: Mechanical Engineering (ME)- 12 - Question 45

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 45

Since the function has 0/0 from now, we can apply L’ Hopital rule,

Applying limit now,

Practice Test: Mechanical Engineering (ME)- 12 - Question 46

A grinding operation requires 5 kW of power to be carried by the motor. The material removal rate is given as 23.65 mm3/sec and the grinding ratio is given as 12. Find the wear rate of the grinding grains of the grinding wheel.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 46

We know that

=1.97 mm3/sec.

Practice Test: Mechanical Engineering (ME)- 12 - Question 47

In the figure shown, the input M has a 48 teeth, B and D constitute a compound planet having 60 and 36 teeth respectively.

If all the gears are of same pitch, find the ratio of speeds of gear M to gear L, assuming N to be fixed.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 47

TM = 48 teeth

TB = 60 teeth

TD = 36 teeth

Pitch diameters of wheels are proportional to the number of teeth on them.

TN = TM + 2TB = 168 teeth

TL = TM + TB + TD

TL = 144 teeth

From the given condition

Practice Test: Mechanical Engineering (ME)- 12 - Question 48

A large turbine generates a power of 1000 kW under a head of 50 m. A 1/5th model of a hydraulic turbine is tested against a head of 10 m. The power developed (kW) and the discharge flow rate (m3/s) by the model are

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 48

Given

Pp = 1000 kW

Hp = 50 m

Pp = ρgQpHp

We know that head coefficient,

and the power coefficient

and distance flow rate, Qm

Pm = ρgQmHm

Practice Test: Mechanical Engineering (ME)- 12 - Question 49

Two streams of fluids of unit constant specific heat and unit mass flow rate exchange thermal energy in an adiabatic heat exchanger. The inlet temperature of hot and cold fluid streams is 300 0C and 30 0C respectively. The effectiveness of the heat exchanger if the outlet temperature of both the fluid are equal is

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 49

For zero entropy condition, outer temperature of both the fluid must be same,

Hence, mhCh (300-t) = mhCh (t-30)

t = 105° C

And this condition is possible only in case of counter flow heat exchanger.

Practice Test: Mechanical Engineering (ME)- 12 - Question 50

Oil of viscosity 0.085 Ns/m2 is filled between two 120 cm long concentric cylinders having inner and outer radius as 5cm and 6cm. If the inner cylinder is fixed and outer cylinder, I rotate at 900 rpm. If the inner cylinder is also rotating at 500 rpm in the same direction as that of outer cylinder. Then percentage reduction in shear stress is

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 50

We know that shear stress is given by

When inner cylinder is fixed.

= 2883.98N/m2

When outer cylinder is rotated at 500 rpm in same direction,

the net rotation of outer cylinder = 900 – 500

= 400 rpm

=1281.769N/m2

% age change in shear stress

= 55.56%

Practice Test: Mechanical Engineering (ME)- 12 - Question 51

The following table shows the unit cost of transportation associated from ith source to jth destination. The demand and supply at different location are also given in table. The initial basic feasible solution using Vogel’s approximation method is

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 51

Total demand = 3 + 4 + 6 + 7 + 10 = 30

Total supply = 5+13+12 = 30

As, Total supply = Total demand, i.e., problem is balanced

No. of filled cell = 7

The required no. of cell = m + n - 1 = 5 + 4 - 1 = 8

If, No. of filled cell < required="" no.="" of="" cell="" degenerate="" />

If, No. of filled cell = required no. of cell non degenerate solution

IIf, No. of filled cell > required no. of cell redundant solution

As, no. of filled cell < required="" no.="" of="" cell,="" so="" the="" solution="" is="" degenerate="" and="" the="" corresponding="" transportation="" cost="" is="" given="" />

Total transportation cost = 2×4 + 2×1 + 2×3 + 6×3 + 4×6 + 7×3 +10×1 = 89

*Answer can only contain numeric values
Practice Test: Mechanical Engineering (ME)- 12 - Question 52

Five jobs are to be assigned at a particular firm. Their work (processing) time and due dates are given in the following table. If the sequence of processing is according to earliest due date (EDD) then the average no. of jobs in the system is


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 52 Table: according to EDD rule

Average no. of jobs in system

Practice Test: Mechanical Engineering (ME)- 12 - Question 53

The principal stresses at a point inside a solid object are σ1 = 100 MPa, σ2 = 100 MPa and σ3 = 0MPa. The yield strength of the material is 200 MPa. The factor of safety calculated using Tresca theory is nT and the factor of safety calculated using Von-Mises theory is nV. Which one of the following relations is TRUE?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 53

Given that, σ1 = 100Mpa, σ2 = 100Mpa

σ3 = 0,σy = 200Mpa

Naw, maximum shear stress is given by

Hence, Tmax = 50Mpa.

Applying Tress's theory

⇒ FosT = 2

⇒ nT = 2……(1)

Applying Yon-mises theory.

Using N = nt

and salving using the given values, we get nt = 2

Sa nv = nt = 2

*Answer can only contain numeric values
Practice Test: Mechanical Engineering (ME)- 12 - Question 54

The inner and outer diameters of a multiple plate clutch disc are 100 mm and 200 mm respectively. The friction coefficient and permissible intensity of pressure is 0.2 and 1 MPa. The no. of pair of contacting surfaces if the torque carrying capacity of clutch at 750 rpm is 1.1 KN-m (assume uniform pressure theory)


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 54 The torque carrying capacity of multiple clutches is given by Mt = ZμPRm

Where, Z= No. of pair of contacting surfaces μ = 0.2

P = Operating farce p = 1MPa

r = 100/2 = 50mm

R = 200/2 = 100mm

Rm = mean radius

For uniform pressure theory

Far uniform wear theory

P = 7T [R2 - r2] x p = π [1002 - 502] x 1

P = 23561.95N

M t = Z x 0.2 x 23561.95 x 77.77N - m m = 0.366 x ZKN - m

1.1 = 0.366 x Z

Z = 3

*Answer can only contain numeric values
Practice Test: Mechanical Engineering (ME)- 12 - Question 55

A close-coiled helical spring, made out of a 10 mm diameter steel wire, has 20 complete coils each of mean diameter of 100 mm. A weight of 10 kg is dropped onto the spring, compresses it 60 mm, then height from which the weight is dropped (in mm) will be (take G = 85000 MPa)


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 55 Applying the deflection equation of close coiled helical spring

Now, by energy balance

⇒ h = 37.48mm

*Answer can only contain numeric values
Practice Test: Mechanical Engineering (ME)- 12 - Question 56

In a machining experiment, the tool life was found to vary with cutting speed in following manner

If tool changing time is 3 minutes, then the optimum value of cutting speed (in m/min) for maximum productivity is


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 56 Range: 103.5 to 105.5

Using Taylar's taal life equation,

VTn = C

100 × 10n = C…….(1)

75 × 30n = C………

From equation (1) and (2) , we get

n = 0.2616 and C = 182.6

Now applying the maximum productivity criteria

*Answer can only contain numeric values
Practice Test: Mechanical Engineering (ME)- 12 - Question 57

A sprue with gate diameter of 5 cm is used to fill a 80 cm diameter and 50 cm high cylindrical casting. If the head available for filling metal is 75 cm, then the minimum time taken in seconds to fill up the casting is


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 57 Range: 33 to 34

Top gating always takes less time than bottom gating. So, the minimum filling time will be in top gating.

Velocity at gate,

Then time taken to fill the cavity is given by

Practice Test: Mechanical Engineering (ME)- 12 - Question 58

In a wire drawing operation the 12 mm dia. wire is reduce to 8 mm. Mean flow stress is taken 500 MPa, drawing load and stress required for wire drawing operation respectively.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 58

P = 20.378KN

Drawing stress:

Practice Test: Mechanical Engineering (ME)- 12 - Question 59

A shaft has a dimension of the respective values of upper deviation, lower deviation and mean deviation.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 59

0.04,-0.02,0.01 upper deviation = max. size − basic size

= 20.04 − 20 = +0.04

Lawer deviation = min size - basic size

= 19.98 − 20 = −0.02

Mean deviation

Practice Test: Mechanical Engineering (ME)- 12 - Question 60

A CNC part programming on a block is given by

M020 G02 X45.0 Y25.0 R5.0

The tool motion during the execution of the part programme will be

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 60

Given:- MO20 GO2 X45.0 Y25.0 R5.0

Here term ×45.0Y25.0R5.0 will produce circular motion

because radius is consider in this term and GO2 will

produce clockwise motion of the tool.

Practice Test: Mechanical Engineering (ME)- 12 - Question 61

Three masses m1, m2 and m3 of magnitude 50 kg, 100 kg and 200 kg are connected to a rotating shaft. The corresponding radiuses of rotation are 0.4 m; 0.2 m and 0.1 m respectively and angle between successive masses are 60 and 150. What is the magnitude of balancing mass if its radius of rotation is 0.3 m?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 61

m1 = 50kg,m2 = 100kg,m3 = 200kg

The horizontal component of centrifugal farce ∑H =

[50 × 0.4 + 100 × 0.2 × cos⁡60 + 200 × 0.1 × cos ⁡2102

= 12.68ω2

The vertical component of centrifugal farce ∑V = [100 × 0.2 × sin⁡60+ 200 × 0.1 × sin ⁡2102 = 7.32

ω2

The resultant force,

ω= angular velocity of rotating masses. Suppose the balancing mass M acting moving with angular velocity ω at a radius of rotation 0.3m and it is given as

M × 0.3 × ω2 = 14.64ω2

M = 48.8kg

The angle made by resultant with horizontal force anticlockwise, is given as

The location of M = 180+ 30= 210 ta the anticlockwise of 200 kg mass

Practice Test: Mechanical Engineering (ME)- 12 - Question 62

A mass M, of 20 kg is attached to the free end of a steel cantilever beam of length 1000 mm having a cross-section of 25 × 25 mm. Assume the mass of the cantilever to be negligible and Esteel = 200GPa. If the lateral vibration of this system is critically damped using a viscous damper, the damping constant of the damper is

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 62

Given M = 20kg,I = 1000mm = 1m, A = 25 × 25

mm2

ESteel = 200GPa = 200 × 109Pa

Mass moment of inertia of a square section is given by,

Deflection of a cantilever, Loaded with a point load placed at the free end is

Therefore, critical damping constant C0 = 2Mωn = 2 × 20 × 31.32

= 1252.8Ns/m ≃ 1250Ns/m

Practice Test: Mechanical Engineering (ME)- 12 - Question 63

In aqua- ammonia absorption refrigeration system, the temperature to be maintained in the refrigerator is −5C and the ambient temperature is 27C. If the actual COP is 50% af the maximum COP and the heat supply to generator is 120KW at 110C, then the refrigeration load is

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 63

given, QG = 120KW

Generator temperature, T1 = 110C + 273 = 383K

Condenser and absorber temperature, T2 = 27 + 273 = 300K Evaporator temperature,

TR = -5°C + 273 = 268K

(COP)max = product of ideal COP of a refrigerator warking between TR and T2 and ideal thermal efficiency of an engine working between T1 and T2

= 1.81

(COP)actual = 0.5 X 1.81 = 0.9

Refrigeration load, QE = 108KW

*Answer can only contain numeric values
Practice Test: Mechanical Engineering (ME)- 12 - Question 64

In an ideal Brayton cycle with regeneration, air is compressed from 1 bar and 300 K to 4 bar and 450 K, is heated to 750 K in the regenerator, and then further heated to 1200 K before entering the turbine. Under cold-air standard conditions, the effectiveness of the regenerator (in percentage) is _____ [Take γ = 1.3 for gas]


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 64

Given, P1 = 1 bax = P4

P2 = 4 bar = P3

T1 = 300K

T2 = 450K

T6 = 750K

T3 = 1200K

Practice Test: Mechanical Engineering (ME)- 12 - Question 65

A 4-stroke and 8 cylinder gas engine with a stroke volume of 2 litre develops 30 KW at 510 rpm. The indicated mean effective pressure is 600 KN/m2. The average no. of times each cylinder misfired in one minute is

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 12 - Question 65

Given,

Vs = 2 litre = 0.002m3

Indicated mean effective pressure (imep) = 600 KN/m2 N = 380rpm

Indicated power per cylinder

n = no. of explosion per minute No .of explosions per minute = N/2 = 510/2 = 255 No. of misfiring per minute per cylinder = 255 − 250 = 5

Use Code STAYHOME200 and get INR 200 additional OFF
Use Coupon Code
Information about Practice Test: Mechanical Engineering (ME)- 12 Page
In this test you can find the Exam questions for Practice Test: Mechanical Engineering (ME)- 12 solved & explained in the simplest way possible. Besides giving Questions and answers for Practice Test: Mechanical Engineering (ME)- 12, EduRev gives you an ample number of Online tests for practice