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Practice Test: Mechanical Engineering (ME)- 11 - Mechanical Engineering MCQ


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65 Questions MCQ Test GATE Mechanical (ME) 2024 Mock Test Series - Practice Test: Mechanical Engineering (ME)- 11

Practice Test: Mechanical Engineering (ME)- 11 for Mechanical Engineering 2024 is part of GATE Mechanical (ME) 2024 Mock Test Series preparation. The Practice Test: Mechanical Engineering (ME)- 11 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Practice Test: Mechanical Engineering (ME)- 11 MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Practice Test: Mechanical Engineering (ME)- 11 below.
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Practice Test: Mechanical Engineering (ME)- 11 - Question 1

A man covered certain distance on foot at the rate of 5km/h and rest by bicycle at the rate of 12km/h. If he took 15hrs to cover a distance of 110km. Find the distance covered by him on foot.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 1
His average speed to cover 110km = 110/15 = 22/3km/h

Hence the ratio of time taken on foot/by bicycle is = 2/1.

Now, since total time taken is 15 hrs.

Therefore, journey on foot is for 10hrs.

Hence, distance covered on foot = 10×5 = 50 km.

Practice Test: Mechanical Engineering (ME)- 11 - Question 2

The compound interest compounded half yearly on a certain sum of money at 10% per annum for 1 year is Rs. 246.The simple interest on the same sum for 2.5 years at 7% per annum is:

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 2
SI = (P × R × t)/100

A = P(1+R/100)t; Cl= A-p

Given, compound interest compounded half yearly on a certain sum of money at 10% per annum for 1 year is Rs. 246.

∴ R = 10/2 = 5%

t = 1 year = 2 half years

CI = Rs. 246

P=?

A = CI + P

⇒ A = P + 246

∴ P + 246 = P (1+(5/100))2

⇒ P + 246 = P × 1.052

⇒ 246 = 1.1025P – P

⇒ 246 = 0.1025P

⇒ P = 2400

Now, SI on the same sum for 3 years at 6% per annum.

∴ R = 7%

t = 2.5 years

SI = (2400 x 7 x 2.5)/100

⇒ SI = Rs. 420

Practice Test: Mechanical Engineering (ME)- 11 - Question 3

If the world ‘MAJORITY’ is encoded as ‘PKBNXSHQ’, how will the word ‘DAUGHTER’ be encoded?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 3
Divide the given word in 2 equal parts

M A J O R I T Y and each part is rewritten in reverse sequence

O J A M Y T I R.

Position of letters in the alphabet is 15/10/1/13 and 25/20/9/18 and position of letters in the given code PKBN/XSHQ is 16/11/2/14 and 24/19/8/17. HVBEQDSG

On comparison, you notice, the first half is obtained by adding ‘1’ and second half is obtained by subtraction ‘1’.

Practice Test: Mechanical Engineering (ME)- 11 - Question 4

In the given figure, which number represents cow, lion and goat but not tiger.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 4
Circle is cow, triangle is goat, rectangle is tiger and square is lion has been shown in the given figure.

The number which represents lion, cow and goat but not tiger has been shown in blue shaded part which is 9.

Practice Test: Mechanical Engineering (ME)- 11 - Question 5

In a car showroom, the average value of three cars is 15 Lakh, where the average price of two costliest cars is double the price of the cheapest car. Find the cost of the cheapest car.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 5
Let the price of the car be x > y > z

According to question (x+y)/2 = 2z

x + y = 4z

Given, x+y+z = 15 × 3 = 45 lakh

4z + z = 45 ⇒ 5z = 45

z = 9 Lakh

Practice Test: Mechanical Engineering (ME)- 11 - Question 6

In the following question, four words are given out of which one word is correctly spelt. Select the correctly spelt word.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 6
Arguement is the incorrect spelling of the word Argument. It means an exchange of diverging or opposite views, typically a heated or angry one.

Accommodate is the correctly spelt word. It means to provide lodging or sufficient space to someone. Alterd is the incorrect spelling of the word Altered. It means to change in character or composition, typically in a comparatively small but significant way.

Aquire is the incorrect spelling of the word Acquire. It means to buy or obtain an asset for oneself. Hence, option B is the correct answer.

Practice Test: Mechanical Engineering (ME)- 11 - Question 7

A is in the south of B at a distance of 4m. C is in the east of B at a distance of 5m. D is in the west of C at a distance of 12m. E is in the north of D at a distance of 5m. How far is C from E and E is in which direction from C?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 7
We know that

We can show the given data in the following figure :

From the figure, E is in north-west direction of C.

Using Pythagoras theorem,

distance between E and C =

Practice Test: Mechanical Engineering (ME)- 11 - Question 8

In the sentence, identify the segment which contains the grammatical error.

Every student has to face a lot of pain to get selected in this competitive exam.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 8
The error is in part 'D'.

'pains' in place of 'pain' because the meaning of pain is suffering. Some nouns change their meaning if they are used in plural.

E.g. Pain (physical) Pains (sufferings/tension)

Colour (hue) Colours (hues/ flag)

Practice Test: Mechanical Engineering (ME)- 11 - Question 9

Arrange the given words in the sequence in which they occur in the dictionary.

1. Culmination, 2. Conclusion, 3. Climax, 4. Cultivation

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 9
In this question, we show that all word arranged alphabetically order as in a dictionary

3. Climax, 2. Conclusion, 1. Culmination, 4. Cultivation.

So the correct sequence is 3, 2, 1, 4.

Practice Test: Mechanical Engineering (ME)- 11 - Question 10

Gustavo, Pablo and Gaviria are working together on a project. They can finish it together in a day. Gustavo alone can finish it in 2 days and Pablo in 3 days. How much time will Gaviria take to finish the project alone?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 10
Let the total amount of work to be done be W

Let x be the number of days required by Gaviria to finish the project alone

Gustavo’s speed = W/2

Pablo’s speed = W/3

Gaviria’s speed = W/x

Combined speed = W/6

Gaviria will take 6 days to finish alone.

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

Let A ϵ M3 (R) be such that det (A-I) = 0, where I denotes the 3 x 3 identity matrix. If the trace of A = 13 and det A = 32, then the sum of squares of the eigen values of A is _________.


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 11
Given that

Λ1 + λ2 + λ3 = 13

Λ1⋅λ2⋅λ3 = 32 and det⁡(A−I) = 0

∴ (λ1−1) (λ2−1) (λ3−1) = 0

From (i), (ii) and (iii), we get

From (i), (ii) and (iii), we

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

The value of surface integral ndS over the surface of the sphere x2 + y2 + z2 = 16 is


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 12
According to Gauss Divergence Theorem,
Practice Test: Mechanical Engineering (ME)- 11 - Question 13

A cargo ship weighs 50 MN and the ships is tilted through an angle of 5 due to water of weighing 500 KN is being filled on ship’s boats. The mean distance of water (in mm) from the centre of the ships is 12 m. The metacentric height is

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 13
Given,

w = weight of filled water = 500 KN

W = weight of cargo ships + weight of filled water = 50000 + 500 = 50500 KN

θ = tilted angle = 5

x = mean distance of water from the centre of the ships = 12 m

Ship’s boat after filled water

The metacentric height,

Practice Test: Mechanical Engineering (ME)- 11 - Question 14

A dam is constructed across river for the purpose of irrigation. The depth of water (H) stored on the upstream side of dam is 5 m. The total pressure exerted on the upstream face for 1 m length of dam (in KN) is (assume g = 10 m/s2)

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 14
Total pressure =

Where, ρ = Density of water = 103kg/m3

A = Area of upstream face of dam = H×1= 5m2x¯= Centroid of the surface at a vertical depth below free surface of liquid = H/2 = 2.5m Total pressure = 103 × 10 × 5 × 2.5 = 125000N = 125KN

Centre of pressure, h = x¯+1G/AR IC = area moment of inertia about axis passing through the centroid of area and parallel to water surface level

Practice Test: Mechanical Engineering (ME)- 11 - Question 15

Three kg of oxygen gas is heated from 15oC to 90oC. , the change in enthalpy( in KJ) is

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 15
Given, m = 3 Kg , molecular weight of oxygen, M = 32

T1 = 15 +273 = 288

T2 = 90 + 273 = 363

So, change in enthalpy is dH = mCP(T2 – T1)

dH = 3 x 0.909 x (363- 288)

dH = 204.525KJ

Practice Test: Mechanical Engineering (ME)- 11 - Question 16

Which of the following curve represent a non-strain hardening material

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 16

Stress-strain curve for perfectly plastic of non-strain hardening materials is shown below

Practice Test: Mechanical Engineering (ME)- 11 - Question 17

A thin cylindrical pressure vessel of internal diameter 800 mm and wall thickness 10 mm is given a coating which cracks when the strain reaches to 0.0002. The internal pressure (in MPa) at which cracking start in coating is (Take E = 210 GPa, Poisson ratio = 0.25)


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 17
Given, d = 800 mm = 0.8 m, t = 0.01 m, μ = 0.25

As the longitudinal stress is less than the hoop stress, the cracks will develop along the longitudinal axis firstly.

f2 = longitudinal stress = pd/4t

f1 = circumferential stress = pd/2t

First cracks will develop along the longitudinal axis as f1 = 2× f2

Practice Test: Mechanical Engineering (ME)- 11 - Question 18

The effect of forced convection can be neglected if

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 18

When GrL/Re2L = 1, the effect of both free and forced convection should be considered.

When GrL/Re2L ≫ 1, effect of forced convection neglected

When GrL/Re2L ≪ 1, effect of free convection neglected

Practice Test: Mechanical Engineering (ME)- 11 - Question 19

Consider the following linear programming problem

Maximize X= 3y1 + 2y2

Subject toy1 ≤ 4

y2 ≤ 6

6y1 + y2 ≤ 18

y1 ≥ 0, y2 ≥ 0

The optimal value of the objective function for the above problem is


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 19
Range: 18 to 18

Area OABC is the area of feasible solution.

So value of objective function on feasible area points

At O(0,0), X = 0

At A(0,6), X = 2(6)=12

At B(2,6), X = 3(2)+2(6)=18

At C(3,0), X =3(3)=9

So the optimal value of X = 18

Practice Test: Mechanical Engineering (ME)- 11 - Question 20

In a single-channel queuing model, the customer arrival rate is 12 per hour and the waiting time in the queue is 2.5 minute. The proportion of time that a server actually spends with customers is

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 20
Given, arrival rate, λ = 12 per hr

Waiting time in queue, Wq = 2.5min = 1/24hr

Wq = p2(1−ρ)λ = 1/24

ρ = traffic intensity or the proportion of time that a server actually spends with customers

2p2 + p − 1 = 0

2p2 + 2p − p − 1 = 0

ρ = −1(not possible) ρ = 1/2 = 0.5 = 50%

Practice Test: Mechanical Engineering (ME)- 11 - Question 21

A rotating bar made of forged steel 50 mm in diameter (Sut = 600 N/m2), is subjected to completely reverse bending stress. The corrected endurance limit (in MPa) of the bar is 100 N/mm2. The fatigue strength of the bar for the life of 25000 cycles is


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 21
Given, Sut = 600 N/m2, Se = 100 N/mm2, N = 25000 cycles

Step I. Construction of S-N diagram

log10 ⁡(0.9Sut) = 2.73

log10 ⁡(Se) = 2

log10⁡ (N) = 4.4

The S-N curve for the rotating bar is shown.

Step II. Fatigue strength (Sf)

The equation of line AB

log10 ⁡Sf − 2.73 =

Sf = 245.094MPa

Practice Test: Mechanical Engineering (ME)- 11 - Question 22

Which of the following is a multi point cutting tool?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 22
Single point cutting tool = turning, shaping

Double point cutting tool = drilling

Multi point cutting tool = broaching, milling etc.

Practice Test: Mechanical Engineering (ME)- 11 - Question 23

Mechanism of material removal for ECM machining

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 23
In electrochemical machining electrolytes are used which causes ionic dissolution, to remove the material.
Practice Test: Mechanical Engineering (ME)- 11 - Question 24

A 6 mm thick sheet is rolled to 3 mm with a pair of rollers having diameter 300mm. The friction coefficient at the work roll interface is 0.1, the maximum number of passes required ____


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 24
Δhmax = µ2R

Δhmax = .12 × 150 = 1.5mm

Minimum no of passes n = Δhrequired/Δhmax = 3/1.5

Minimum no of passes n = 2

*Answer can only contain numeric values
Practice Test: Mechanical Engineering (ME)- 11 - Question 25

A 140 mm diameter billet is extruded to 70 mm, at the working temperature of 800ºC. The average flow stress of material is 350 MPa, the force required for extrusion (in MN) is _____


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 25

P = 15394.804 × 350 × 2 × 10

d0/df

P = 15394.804 × 350 × 2 × 1n

140/70

P = 7.46 MN

Practice Test: Mechanical Engineering (ME)- 11 - Question 26

The steam is expanded isentropically in steam turbine from 20 bars, 350∘C to 0.08 bars and then it condensed to saturated liquid water in the condenser. The pump feeds back the water into the boiler. Neglect losses in the processes, the cycle efficiency (in percentage) are

Given data: At p = 0.08 bar, hfg = 2403.1 kJ/kg, sfg = 7.6361 kJ/kg K, vf = 0.001008 m3/kg

h1 =3159.3 kJ/kg, s1 = 6.9917 kJ/kg K, h3 = 173.88 kJ/kg, s3 = 0.5926 kJ/kgK,

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 26
?cycle = Wnet/Q1

Wnet = WT – WP

WT = h1 – h2

As, s1 = s2 = s3 + xsfg = 0.5926 + 7.6361 x

x = 0.838

h2 = h3 + x hfg = 173.88 + 0.838 × 2403.1 = 2187.68 kJ/kg

WT = h1 – h2 = 3159.3 – 2187.68 = 971.62 kJ/kg

WP = vf (p1−p2) = 0.001008 × (20 – 0.08) × 102 = 2 kJ/kg

h4 – h3 = 2

h4 = 175.88 kJ/kg

Wnet = 971.62 – 2 = 969.62 kJ/kg

Heat added, Q1 = h1 – h4 = 2983.42 kJ/kg

?cycle = 969.62/2983.42 = 0.325 = 32.5 %

Practice Test: Mechanical Engineering (ME)- 11 - Question 27

In the figure shown below, the relative velocity of link 1 with respect to link 2 is 15 m/s. The link 2 rotates at a constant speed of 150 rpm. The magnitude of the coriolis component of acceleration of link 1 is .....

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 27
ac = 2ωV

= 2 × 2πN/60 × V

= 2×2×3.14×150×15/60

= 471m/s2

Practice Test: Mechanical Engineering (ME)- 11 - Question 28

The annual demand of the company is 10000 units per year. The ordering cost per order is Rs 100 per year and the carrying cost is Rs10 per unit per year and the shortage is allowed. The shortage cost is Rs 5 per unit per year. Maximum inventory is_______.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 28
Here we will use shortage model of inventory control,

The EOQ in shortage model is given by

Q* = EQQ QUANTITY

D = DEMAND PER YEAR

C0 = ORDERING COST PER YEAR

CC = CARRYING COST PER UNIT PER YEAR

CS = SHORTAGE COST PER UNIT PER YEAR

= 516.39

= MAXIMUM INVENTORY = Q*- S* = 258.193

Practice Test: Mechanical Engineering (ME)- 11 - Question 29

What is equal to?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 29

Applying L-Hospital Rule

Practice Test: Mechanical Engineering (ME)- 11 - Question 30

A cantilever beam of 5m length supports a triangularly distributed load over its entire length, the maximum of which is at the free end. The total load is 21 N. What is the bending moment at fixed end?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 30
Mx = -w/2*x2+w/6L*x3

maximum bending moment will be at x = L

ML = wL2/3

total load is given which is W = wL/2 (area of triangular distributed load)

So

Bending moment at fixed end = wl2/3

given total load 21 kN

Total load ⇒ wl/2 = 21

wl = 42

= 42 × 5 × 103/3

= 70,000 Nmm

Practice Test: Mechanical Engineering (ME)- 11 - Question 31

Consider the following data for a product

Demand= 1000 units/ year

Order = Rs. 40/ order

Holding cost = 10% of the unit cost/ unit year

Unit cost = Rs. 500

With a policy of ordering every month, the total annual cost would be _____.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 31

D = 1000 units

C0 = Rs.40

Cn = 10% of 500 = Rs.50

Ordering quantity per month

100/12 = 83.33

100/12 = 83.33(TIC)83.33 = 1000/83.33 × 40 + 83.33/2 × 50

= 480+2083.25 = 2563.25

Total cost = 1000 × 500 + 2563.25

= 502563.25

Practice Test: Mechanical Engineering (ME)- 11 - Question 32

For a particular product, the following information is given:

Selling price per unit: Rs. 100

Variable cost per unit: Rs. 60

Fixed cost: Rs. 1000000

Due to inflation the variable cost has increased by 10% while fixed costs have increased by 5%. If the break even quantity is to remain constant, the percentage increase in sales price would be ____.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 32
Selling price/unit = Rs. 100 = S.P.

Variable cost/unit = Rs. 60 = V.C.

Fixed costs = Rs. 10,00, 000 = F.C.

Break even quantity, QB = F.C./S.P-V.C.

= 1000000/100-60 = 25000

Given that variable cost have increased by 10%

V.C1 = 60 + 10/100 × 60 = Rs.66

Fixed cost have increased by 5%

∴ F.C. = 10,00,000 × 5100 + 10,00,000

= 1050000

But break even quantity is to remain constant

25000 = 10,50,000/SP1-66

SP1 = 108

Percentage increase in selling price = (108-100/100) × 100

= 8%

Practice Test: Mechanical Engineering (ME)- 11 - Question 33

For the following assignment problem find the minimum cost of assignment

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 33
Applying Hungarian method

Finding minimum element in row and subtracting from other element

Finding minimum element in column and subtracting from other element

Assigning the job

Cost = 12+4+5+1 = 22

Practice Test: Mechanical Engineering (ME)- 11 - Question 34

Find the reactions at A and D of a truss if it is hinged at one end while the other end is inclined at an angle of 450.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 34

Since the truss is supported on rollers at D, therefore the reaction at this support will be normal to the support i.e. inclined at 45° with the vertical (because the support is inclined at an angle of 45° with the horizontal) as shown in Fig. The reaction at A will be the resultant of vertical and horizontal forces.

Let RA = Reaction at A, and

RD = Reaction at D.

∴ Horizontal component of reaction at D,

RDH = RDV = RDcos 45° = 0.707 RD

Now taking moments about A and equating the same,

RDV × 12 – RDH × 6 = (4 × 4) + (3 × 8)

(0.707RD × 12) – (0.707 RD × 6) = 40

or R = 40/4.242 = 9.43kN

∴ RDH = RDV = 9.43 × 0.707 ≈ 6.67 kN

Now vertical component of reaction at A,

RAV = (4 + 3) – 6.67 = 0.33 kN

and horizontal component of reaction at A

RAH = RDH = 6.67 kN

Practice Test: Mechanical Engineering (ME)- 11 - Question 35

What will be the height between the bottom of the sprue and free metal surface of the casting, having a gating ratio of 1 : 1.5 : 2. The height of metal during pouring in the basin is 100mm above the cope, where cope box has a height of 200 mm.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 35

Height of cope = 200mm

Height of molten metal in pouring basin = 100mm

ht = 200 + 100 = 300mm

Practice Test: Mechanical Engineering (ME)- 11 - Question 36

In a rolling process, a sheet of 25 mm thickness is rolled to 20 mm thickness. Roll is of diameter 1200 mm and it rotates at 100 rpm. What is the roll strip contact length?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 36

Initial thickness

h1 = 25 mm

Final thickness

h2 = 20 mm

Roll diameter = 1200 mm

Speed of rolls = 100 RPM

As we know that contact length of roll strip is given by

Where R = roll radius = 1200/2 = 600 mm

And Δh =

h1-h2 = (25 – 20) = 5 mm

Put these values we have

L = 54.77 mm

Practice Test: Mechanical Engineering (ME)- 11 - Question 37

For a fluid flow through a divergent pipe of length 2m having inlet and outlet diameters of 0.6m and 0.4m respectively and a constant flow rate of 0.5 m3/sec, assuming the velocity to be axial and uniform throughout, the acceleration at the exit is;

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 37
Given : R1 = 0.3 m R2 = 0.2 m

Q = 0.5 m3/sec L = 2 m

Acceleration at exit;

Practice Test: Mechanical Engineering (ME)- 11 - Question 38

The value of

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 38

Simplifying

Applying shifting theorem,

Practice Test: Mechanical Engineering (ME)- 11 - Question 39

Manufacturing a product requires processing on four machines A, B, C, D in the order A – B – C – D. The capacities of four machines are A = 100, B = 110, C = 120 and D = 130 units per shift. If the expected output is 85% of the system capacity, then what is the expected output?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 39
In the sequence of A – B – C – D only minimum output has to be calculated. Other machines will be in an empty position.

∴ Output = η × 100 = 0.85 ×100 = 85units

Practice Test: Mechanical Engineering (ME)- 11 - Question 40

The eigenvectors of the matrix are written in the form and What is a + b?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 40

Let A = are the eigenvalue of matrix.

therefore the characteristic matrix is

|A = λI| = 0

(1-λ)(2-λ) = 0 ⇒ λ = 1&2

Therefore eigen vector corresponding to λ = 1 is,

0 × x + 2y = 0 so x = any value and y = 0

a = 0

for λ = 2

x = 2y = k(anyvalue)

x = k,y = k/2

the sum of a & b = a + b = 0 + 1/2 = 1/2

Practice Test: Mechanical Engineering (ME)- 11 - Question 41

Ratio of the buckling strength of the hollow steel column to that of solid square column of the same material, same cross section area, same length. The ratio of outer to inner diameter of the hollow column is 2.


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 41

Same material ⇒ ρH = ρs and EH = Es

Same area ⇒ AH = As = B2

B = 1.085D

D/B = 0.9216

P is directly proportional to

Now, = 0.398

Practice Test: Mechanical Engineering (ME)- 11 - Question 42

A hollow cylinder 10 cm internal diameter and 20 cm outer diameter has an inner surface and outer surface temperatures are 3000c and 2000c. The thermal conductivity of cylinder material is 80 W/Mk. The area of the cylinder per meter length such that cylinder can be transform into an equivalent slab is

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 42
Given, di = 10cm, d0 = 20cm, L = 1m = 100cm

The area of the cylinder that can be used to transform a a

cylinder into an equivalent slab called Log mean area of the cylinder and it is given as

Where, Ao = πdOL = π × 20 × 100 = 2000πcm2

Ai = πdiL = π × 10 × 100 = 1000πcm2

*Answer can only contain numeric values
Practice Test: Mechanical Engineering (ME)- 11 - Question 43

A 0.8 mm diameter nickel- alumel thermocouple is used to measure the temperature difference of a fluid. The sensitivity of the thermocouple (in s) is

Given data:

Convective heat transfer coefficient = 500 W/mK

Specific heat of fluid = 350 J/kgK

Density of thermocouple material = 9500 Kg/m3


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 43
Given, d = 0.8mm, h = 500W/mK,C = 350J/kgK, ρ = 9500Kg/m3

The sensitivity of the thermocouple is the time required by a thermocouple to reach 63.2% of the initial temperature difference. I.e.,

T−T∞ = 0.632(T0 − T∞)(1)

Using Lumped capacity method, T − T∞ = (T0−T)e(−hA/ρCV)t

From (i) 0.632 (To−T) = (To−T)e(−hA/ρcv)t

From (ii) 0.752t = 0.4588

t = 0.61s

Practice Test: Mechanical Engineering (ME)- 11 - Question 44

In a counter flow heat exchanger, the water (Cp = 4200J/kgK) flawing at the rate of 1000kg/h enters at 15C and leaves at 30C. It is heated by ail (Cp = 2100J/kgK) flowing at the rate of 500kg/h from the inlet temperature of 100C ta 40C. The na. of transfer units is

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 44
Given, mc = 1000kg/h, mh = 500kg/h, cc = 4200J/kgK

ch = 2100J/kgK,Tc,i =15C, Tc,0 = 30C,Th,i =100C, Th,0

= 40∘C

Heat capacity rate, CC = mcCC = 42 x 105J/hK, Ch = mhCh

= 10.5 × 105J/hK

Cmin = 10.5 × 105J/hK

Effectiveness of heat exchanger, ϵ = Qact/Qmax =

A ∣ sO, for the counter flow heat exchanger,

Where, NTU = no. of transfer unit,

0.705−0.25×0.705exp⁡[−0.75NTU] =

1−exp⁡[−0.75NTU]

0.82375exp⁡[−0.75NTU] = 0.295

exp⁡[−0.75NTU] = 0.358

NTU = 1.37

Practice Test: Mechanical Engineering (ME)- 11 - Question 45

In compressor of a Refrigerator, air enters in the axial flow compressor & leaves the compressor axially. Calculate the degree of reaction, if the component of air responsible for whirl is leaving the compressor at one-third the average blade velocity of rotor if the flow velocity at inlet and outlet remain same.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 45
Given,

Vv2 = 1/3u2

Vf1 = Vf2

Since,

Degree of Reaction:

Practice Test: Mechanical Engineering (ME)- 11 - Question 46

Which of the following is not a hardness test.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 46
Izod test is a toughness test whereas all others are Hardness measuring test.
*Answer can only contain numeric values
Practice Test: Mechanical Engineering (ME)- 11 - Question 47

Consider the following data for an item:

Demand = 1500 units/year,

Ordering cost = Rs. 4 per order

Storage cost = 10% of unit price per year

The optimum cost (in Rs.) is


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 47

Given, D = 1500,C0 = 4

Case 1. (900 ≤ Q2)

Cn = 0.1 × 0.87 = 0.087

Optimum order quantity, Q∗2 =

Q∗2 is nat within the range. Case2. (0≤Q1 < />

Ch = 0.1 × 0.9 = 0.09

Optimum order quantity,

Q∗1 is within the range. Total inventory cost

= 1500/365.15 × 4 + 12 × 365.15 × 0.09 + 1500 × 0.9

= 1382.86

Case3. At, Q=900 Total inventory cast =1500/900 × 4 + 12 × 900 × 0.087 + 1500 × 0.87 = 1350.81 (minimum)

Practice Test: Mechanical Engineering (ME)- 11 - Question 48

Consider the following data and the corresponding statements:

Case 1:

Case 2:

I. case 1 has higher stability of sales

II. case 2 has higher stability of sales

III. case 1 has higher weightage to recent sales

IV. case 2 has higher weightage to recent sales

Which of the following statements is/are correct?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 48
Using exponential smoothing method

Ft+1 = Ft + α(At−Ft)

Forecast for February month, F2 = F1 + α(A1−F1)

Case 1:

503 = 496 + α(510−496)

α = 0.5

Case 2:

502 = 495 + α(505−495)

α = 0.7

The higher the values of smoothing constant (α), higher

will be the weightage to recent data and lower will be the stability. As the value of α in case (I) is lower than that of case (II), so stability in case (I) will be more and higher weightage to recent sales will be in case (II).

Practice Test: Mechanical Engineering (ME)- 11 - Question 49

The moment of inertia of a beam section which is 500 mm deep is 69.49 × 107 mm4. The maximum stress in the beam is not to exceed 110 N/mm2. The maximum span of the beam is _______ m, when simply supported at the ends and could carry a uniformly distributed load of 50 kN per metre length.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 49
Section modulus of section,

z = 27.796 × 105mm3

Let the maximum spam be I metre

∴ Maximum bending moment,

= 62.5 × 103FN−mm

From bending equation, M/z ≤ σmax

⇒ 1 = 6.99m

Practice Test: Mechanical Engineering (ME)- 11 - Question 50

A beam of circular section is subjected to a load of 4kN. If the diameter of the circular section is 0.1m. Calculate the maximum shear stress.


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 50
Given: F = 4kN = 4000N

D = 100mm

r = 50mm

Average shear stress, Tavg = Shearforce/Area of circular section = F/πr2

Tavg = 0.5093 N/mm2

Maximum shear stress for circular section,

Tmax = 4/3 Tavg

Tmax = 0.68N/mm2

Practice Test: Mechanical Engineering (ME)- 11 - Question 51

A company produces two types of toys: P and Q. Production time of Q is twice that of P and the company has a maximum of 2000 time units per day. The supply of raw material is just sufficient to produce 1500 toys (of any type) per day. Toy type Q requires an electric switch which is available @ 600 pieces per day only. The company makes a profit of Rs.3 and Rs.5 on type P and Q respectively. For maximization of profits, the daily production quantities of P and Q toys should respectively be?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 51
Solve this problem, by the linear programming model.

We have to make the constraints from the given conditions.

For production conditions

P + 2Q < />

For raw material

P+Q< />

For electric switch

Q< />

For maximization of profit, objective function

Z = 3P+5Q ......(iV)

From the equations (i), (ii) & (iii), draw a graph for toy P and Q

Line (i) and line (ii) intersects at point A, we have to calculate the intersection point.

P + 2Q = 2000

P + Q = 1500

After solving there equations, we get A(1000, 500)

For point B,

P + 2Q = 2000

Q = 600

P = 2000 - 1200 = 800

So, B(800,600)

Here shaded area shows the area bounded by the three line equations (common area)

This shaded area have five vertices.

So, for maximization of profit

P = 1000 from point (ii)

Q = 500

Practice Test: Mechanical Engineering (ME)- 11 - Question 52

For a mechanism having 6 links and 7 joints the number of instantaneous centre is? (IC is a point at which a particular link is assumed to be a rotational motion about that point)

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 52
The number of Ic is given as n(n-1)/2

= 6(6-1)/2

= 15

Practice Test: Mechanical Engineering (ME)- 11 - Question 53

Calculate the rate of heat dissipation of a bearing, if it has a rubbing velocity of 1.57 m/s. The bearing is 80mm in length. Viscosity of this oil at the oil temperature of 50℃ is 0.050kg/ms. The temperature of the bearing is 85℃ and surrounding temperature is 30℃, assume energy dissipation coefficient to be 209.34 W/m2℃ and speed of journal is 600RPM.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 53
Given: V = 1.57 m/s L = 0.08m

tb = 85℃ ta = 30℃

Kd = 209.34 W/m2℃.

Diameter of bearing:

V = πDN/60

1.57 = πD×600/60

D = 0.05 m

Rate of heat dissipation:

Hd = KdDL(t,-ta)

= 209.34 × 0.05 × 0.08(85 - 30)

Hd = 46.055 W

Practice Test: Mechanical Engineering (ME)- 11 - Question 54

Solution of the differential equation 3y dy/dx + 2x = 0 represents a family of?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 54
3y dy/dx + 2x = 0

⇒ ∫ 3y dy = ∫ 2x dx

⇒ 3y2/2 = -2x2/2 + c ( on integrating )

⇒ x2 + 3y2/2 = c or x2/1 + y2/2/3 = c

x2/c + y2/2c/3 =1

which represents a family of ellipse.

Practice Test: Mechanical Engineering (ME)- 11 - Question 55

The cooling capacity of cooling and dehumidifying coil is 128 kW. Atmospheric air at a flow rate of 3 kg/s (on dry basis) enters to the cooling coil with an enthalpy of 85 kJ/kg of dry air and humidity ratio is 20 grams/kg of dry air. The air leaves the coil with an enthalpy of 43 kJ/kg of dry air and a humidity ratio is 10 grams/kg of dry air. What will be the enthalpy (in kJ/kg d.a) of condensate water leaving the coil?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 55
Cooling capacity = 128 kW

ma = 3 kg/s

h1 = 85 kJ/kg

ω1 = 20 g/kg of d.a.

h2 = 43 kJ/kg

ω2 = 10g/kg of d.a.

h3 = enthalpy of condensate leaving the coil?

Inlet mass of water vapour

mv1 = ma × ω1

= 3 × 30 × 10-3

= 60 × 10-3 kg

Outlet mass of water vapour

mv2 = ma × ω2

= 3 × 10 × 10-3

= 30 × 10-3 kg

Required cooling capacity = Enthalpy of inlet air - Enthalpy of outlet air + Enthalpy of condensate water

Qc = mah1 - mah2 +mvh3

⇒ 128 = 3 85-43 + 30 × 10-3 × h3

⇒ h3 = 66.67 kJ/kg

Practice Test: Mechanical Engineering (ME)- 11 - Question 56

For SHM motion of cam & follower, the maximum velocity of the follower is, where, h = rise of follower,θ = angle turned by crank, Φ = crank rotation for maximum follower displacement.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 56
y = h/2(1 - cos β)

β = harmonic angle

β = πθ/ϕ

θ = ωt

y = h/2 (1 - cos πωt/ϕ)

V = h/2 πω/ϕ sin πω/ϕ

Vmax = hπω/2ϕ

Practice Test: Mechanical Engineering (ME)- 11 - Question 57

The rod AB, of length 1 m, shown in the figure is connected to two sliders at each end through pins. The sliders can slide along QP and QR. If the velocity VA of the slider at A is 2 m/s, the velocity of the midpoint of the rod at this instant is ________m/s

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 57
Given AB√2

Since Rod AB is rigid, so

Axial velocity of A & B should be same

??cos60 = ??cos 60

?? = ?? = 2?/???

? ?? ??? ????? ?? ??

Practice Test: Mechanical Engineering (ME)- 11 - Question 58

A matrix P is a symmetric matrix and a matrix Q is a skew symmetric matrix. If both the matrix are equal then,?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 58
P = PT since symmetric

Q = -QT since skew symmetric

= - PT , Since P = Q,

PT = -PT which is only possible when all the elements are zero or when it is a null matrix.

Practice Test: Mechanical Engineering (ME)- 11 - Question 59

For the system shown below, the natural frequency of the system is _______ rad/s.

Consider force F = 143N on disc and G = 100GPa

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 59
Weight of disc = 143N

[The torsional shafts are arrangement in parallel]

kωq = k1 + k2

kωq = 2.4305 × 105 Nm.rad

Practice Test: Mechanical Engineering (ME)- 11 - Question 60

In a 0.18 m, 250 m long horizontal pipe a lubricating oil of specific gravity 0.9 and dynamic viscosity 0.15 Pa-s is pumped at the rate of 0.04 m3/s. The power required to maintain the flow (in KW) is?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 60
Given,

Length of pipe, L = 250 m

Diameter of pipe, D = 0.18 m

Dynamic viscosity μ = 0.15 m3/s

Flow rate, Q = 0.04 m3/s

The pressure drop in pipe is given by

Vavg = Q/A = 4 × 0.04/π × 0.182 = 1.57 m/s

∆P = = 58148.15 N/m2

The power required to maintain the flow,

P = Q ×∆P = 0.04 × 58148.15 = 2325.926 W = 2.326 KW

Practice Test: Mechanical Engineering (ME)- 11 - Question 61

The following set of equations has

3x + 2y + z = 4

x – y + z = 2

– 2x + 2z = 5

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 61
The given system of non-homogeneous equations is

3x + 2y + z = 4

x – y + z = 2

– 2x + 2z = 5

Now, the augmented matrix

applying R2→R2-3R1, R3→R3-2R1

So Rank of [A|B] = 3 and rank of [A] = 3 = r

Hence the system is consistent and nonhomogeneous, so will have a unique solution.

Practice Test: Mechanical Engineering (ME)- 11 - Question 62

A Roller bearing has a basic dynamic load rating (C10 for 106 revolution) of 50 kN. If the equivalent radial load on the bearing in 70 kN, the expected life in (106 revolutions) is?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 62
Option (A) is correct.

Given: C = 50 KN & P = 70 KN

For roller bearing L90 = (C/P)10/3 = (5010)10/3 = 0.325

i.e. 0.325 million revolutions.

Practice Test: Mechanical Engineering (ME)- 11 - Question 63

The product of two complex numbers 1 + i and 2 – 5i is?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 63
(1 + i) (2 - 5i)

= (2 + 2i - 5i + 5) = (7 - 3i)

Practice Test: Mechanical Engineering (ME)- 11 - Question 64

In the counter flow heat exchange, for the hot fluid the specific heat is 3 kJ/kg K, mass flow rate is 6 kg/s, inlet temperature is 150 °C and outlet temperature is 100 °C. For the cold fluid, the specific heat is 3 kJ/kg K, mass flow rate is 8 kg/s, inlet temperature is 20 °C. Neglecting the heat transfer to surrounding, the outlet temperature of the cold fluid in °C is ________.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 64
Given that

Cph = 3 kJ/kg-K = 3000 J/kg K, mh = 6 kg/s

Cpc = 3 kJ/kg K = 3000 J/kg K. , mc = 8 kg/s

Ch = mh x Cph = 6 x 3000 = 18 x 103 J/K

Cc = mc x Cpc = 8 x 3000 = 24 x 103 J/K

Now, Thi = 150°C, The = 100°C, Tci = 20°C

Apply energy conservation principle:

Heat lost by hot fluid = heat lost by cold fluid

Ch (Thi - The) = Cc (Tce - Tci)

18 x 103 (150 - 100) = 24 x 103 (Tce - 20)

Tce = 20 + (150/4)

Tce = 57.5°C

Practice Test: Mechanical Engineering (ME)- 11 - Question 65

Consider the following second-order differential equation:

y``- 4y` +3y = 2t - 3t2,The particular solution of the differential equations is?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 11 - Question 65

given y``- 4y` +3y = 2t - 3t2.

⇒ (D2 - 4D+3y = 2t - 3t2)

By the definition of particular solution-

verifying options, option (a) satisfies,

(D2-4D+3) (-2-2t - 3t2) = -2 + 8 + 8t - 6 - 6t 3t 2 = 2t - 3t2

∴(A) is constant

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