Practice Test: Mechanical Engineering (ME)- 13 - Mechanical Engineering MCQ

# Practice Test: Mechanical Engineering (ME)- 13 - Mechanical Engineering MCQ

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## 65 Questions MCQ Test GATE Mechanical (ME) 2024 Mock Test Series - Practice Test: Mechanical Engineering (ME)- 13

Practice Test: Mechanical Engineering (ME)- 13 for Mechanical Engineering 2024 is part of GATE Mechanical (ME) 2024 Mock Test Series preparation. The Practice Test: Mechanical Engineering (ME)- 13 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Practice Test: Mechanical Engineering (ME)- 13 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)- 13 below.
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Practice Test: Mechanical Engineering (ME)- 13 - Question 1

### A 300-meter-long train passes a 450-meter-long platform in 5 sec. If a man is walking at a speed of 4 m/sec along the track and the train is 100 m away from him, how much time will it take to reach the man?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 1
The train can cover (300 + 450) m distance in 5 sec.

The speed of the train = 150 m/sec

Relative speed of the man and the train is 154 m/sec or 146 m/sec

To cover the distance of 100 m, in either case, it will take less than 1 second.

Hence, (C) is the correct option.

Practice Test: Mechanical Engineering (ME)- 13 - Question 2

### A boat takes half the time moving a certain distance downstream than upstream. The ratio of the speed of the boat to that of the current is

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 2
Let,

Boats speed = x kmph

Current speed = y kmph

Time taken in upstream = t hours

Therefore, (x - y)*2t = (x+y)*t

⇒ 2x - 2y = x + y

⇒ x = 3y ⇒ x : y = 3 : 1

Practice Test: Mechanical Engineering (ME)- 13 - Question 3

### Which of the following words is opposite to the word “connivance”?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 3

Connivance- willingness to allow or be secretly involved in an immoral or illegal act

Conspiracy- a secret plan by a group to do something unlawful or harmful.

Sufferance - capacity to endure pain, hardship, etc.; endurance

Complot - a plot involving several participants; conspiracy

Intolerance - unwillingness or refusal to tolerate or respect persons of a different social group, especially members of a minority group.

Practice Test: Mechanical Engineering (ME)- 13 - Question 4

Direction: A number is wrong in the following number series. Select the wrong number.

64, 130, 264, 536, 1076

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 4
The series is

64 × 2 + 2= 130

130 × 2 + 4= 264

264 × 2 + 6= 534

534 × 2 + 8 = 1076

Practice Test: Mechanical Engineering (ME)- 13 - Question 5

Direction: Each statement has a blank followed by four options. Select the most appropriate word for the blank. Guru was always able to maintain a _______ face when he said something silly, and that contrast made everyone laugh.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 5
The statement means to state that Guru had an ability to maintain a serious face while saying something silly. The biggest hint in the question is the word ‘contrast’. Hence, c.
Practice Test: Mechanical Engineering (ME)- 13 - Question 6

A supermarket launched a scheme that if a customer purchases two CDs, one extra CD will be free and if he purchases 3 Mobiles he will get one extra Mobile free. If the cost price of 3 CD and 4 Mobile be Rs.6700 and Rs.232500 respectively. If a customer purchase 2 CD and 3 Mobile as per scheme he availed 1 product free of each category, then at what price these product should be sold so that the agency can get overall profit of 17.5%

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 6
CP of 2 CD and 3 Mobile = 6700 + 232500 = 300000

Now, since we required 17.5% profit,

So, SP = 300000 × 117.5/100 = Rs.352500.

Practice Test: Mechanical Engineering (ME)- 13 - Question 7

In the following question, two/three statements are given followed by four conclusions. You have to consider the statements to be true even if they seem to be at variance from commonly known facts. You have to decide which of the given conclusions, if any, follow from the given statements.

Statements:

I. Some cats are owls.

II. Some owls are elephant.

Conclusion:

I. Some cats are elephant.

II. All owls are elephant.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 7
Here two possibilities emerge based on statement I and II.

Conclusion I:- It cannot be said with certainty that some cats are elephants.

Conclusion II: - It cannot be said with certainty that all owls are elephants.

Hence, neither conclusion I nor II follows.

Practice Test: Mechanical Engineering (ME)- 13 - Question 8

In the following question, a set of labelled sentences is given. Out of the four alternatives, select the most logical order of the sentences to form a coherent paragraph.

(P) The over 11-km Mundka-Bahadurgarh section of Delhi Metro’s Green Line, connecting the capital of Haryana, was inaugurated by the Prime Minister.

(Q) The rapidly growing industry in the area was waiting for Metro connectivity, said Mr.PM, adding that it would create new employment opportunities and benefit students.

(R) Built at a cost of Rs 2,028 crore, the Mundka-Bahadurgarh elevated section, with seven stations, is the third Metro extension between Delhi and Haryana.

S)Mr. Prime Minister said Bahadurgarh was known as the Gateway of Haryana and would now be the Gateway of Development with the advent of the Metro.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 8
The given passage talks about the plan of a metro, hence P should be the first sentence as it states about the initiation of the plan. R gives another basic information about the plan, i.e. its cost etc. Thus, R should be the second sentence. Sentence S must follow next because Q contains the abbreviation which is mentioned in sentence R. Hence, option A is correct.
Practice Test: Mechanical Engineering (ME)- 13 - Question 9

Study the information given below and answer the questions based on it.

Eight friends, A, B, C, D, E, F, G and H are sitting around a circle (not necessarily in the same order) facing the centre. B sits third to the left of F. E is an immediate neighbour of both B and H. Only one person sits between A and H. C and G are immediate neighbours of each other. Neither C nor G is an immediate neighbour of B. Only one person sits between C and D.

Who amongst the following is an immediate neighbour of both A and H?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 9
From the given seating arrangements, we can conclude as follows

Hence F is an immediate neighbour of both A and H.

Practice Test: Mechanical Engineering (ME)- 13 - Question 10

Direction: Study the line graph carefully and answer the given questions.

The graph shows the total number of students who attended the exams and the total number of students who passed the exam in different years of school.

Note:

1. The total number of students enrolled in each year= The total number of students who appeared in the exam + The total number of students who didn’t appear in the exam

2. The total number of students who appeared in the exam= The total number of students passed in the exam + The total number of students who failed in the exam

What is the average number of students who failed over the years?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 10

Total number of students failed over the years=120+100+80+80+60+40=480

So, required average=480/6=80

Practice Test: Mechanical Engineering (ME)- 13 - Question 11

Determine the variation in tractive effort if the driving wheel diameter is 2m of a two cylinder locomotive. Consider the following parameters

Mass of reciprocating parts per cylinder = 300 kg

Distance between cylinder centre lines = 0.55 m

Crank radius = 0.2 m

Cranks are at angles = 90°

If the velocity of hammer blow is 80km/h and take c = 0.75

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

D = 2 m

Variation in tractive effort = √2(1 - c)mω2r

ω = V/(Diameter of treads of driving wheel/2) = 22.2 m/s / (2/2) = 22.2 rad.

= √2(1 – 0.75)300 × 22.22 × 0.2

= 10454.71 N

= 10.454 KN

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

In a heat exchanger, the hot liquid enters with a temperature of 200°C and leaves at 180°C. The cooling fluid enters at 50°C and leaves at 130°C. The capacity ratio of the heat exchanger is __________.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 12
Given,

Thi = 200°C, The= 180°C

Tci= 50°C, Tce = 130°C

from energy balance,

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

⇒ Ch(20 ) = Cc (80)

Capacity Ratio, C= capacity ratio

Capacity Ratio, C = 0.25

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

Consider a signal x (t) given by , then x(t) is given by

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 13
If L[f(t)] = F[S]

Then

Given FS = logS + 5 - log5 + 6

Differentiate both side,

By property(1)

Practice Test: Mechanical Engineering (ME)- 13 - Question 14

A smaller sphere (1) of radius 50 mm is inside a larger sphere (2) of radius 100 mm. What will be the value of the radiation shape factor F22:

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 14

F11 = 0 and F12 = 1

F21 + F22 = 1

And A1.F12 = A2.F21.F21 = 0.25

It means F22 = 1 - 0.25 = 0.75

Practice Test: Mechanical Engineering (ME)- 13 - Question 15

A cast steel slab of dimension 40×20×10 cm, is poured horizontally using a side riser. The riser is cylindrical in shape with diameter and height both equal to 20 cm. Then the freezing ratio is___.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 15

Modulus MC =

= 20/7

Modulus MR = = D/6

Freezing ratio = MR / MC

=

=

= 1.167

Practice Test: Mechanical Engineering (ME)- 13 - Question 16

Determine the magnitude of the moment transferred to the center of the pulley. If a differential pulley of diameters 400 mm and 200mm respectively are used, and is subjected to belt tensions.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 16

Resultant moment due to horizontal force, (Clockwise)

MH = (250 - 75) × 0.2

MH = 35 N-m

Resultant moment due to vertical force, (anticlockwise)

MV = (270 - 80) × 0.1

MV = 19 N-m

Therefore,

Net moment = MH – MV

= 35 - 19 = 16 N-m

Practice Test: Mechanical Engineering (ME)- 13 - Question 17

The given eigen values of the following matrix are represented by λ1 and λ2

A = The value of λ12 + λ22 + λ1λ2 is ___.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 17

λ12 + λ22 + λ1λ2

= λ12 + λ22 + 2λ1λ2 - λ1λ2

= (λ1 + λ2)2 - λ1λ2

Sum of eigen values λ1 + λ2, = trace of matrix

= sum of diagonal elements

= 1 - ⅓ = ⅔

Products of eigen values, ,λ1λ2 = determinant of matrix

=

= 1/9

∴ (λ1 + λ2)2 - λ1λ2 = (⅔ )2 - 1/9

= 4/9 - 1/9

= 3/9

= 0.33

Practice Test: Mechanical Engineering (ME)- 13 - Question 18

A solid aluminium shaft, 5 cm diameter and 1 m long, is to be replaced by a tubular steel shaft of the same length and the same outside diameter (i.e. 5 cm), such that each of the two shaft could have the same angle of twist per unit torsional moment over the total length. Modulus of rigidity of steel is three times that of aluminium. The inner diameter of the tubular shaft is

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 18

Now,

= 0.9036

d1 = 5 × 0.9036 = 4.518 cm

Practice Test: Mechanical Engineering (ME)- 13 - Question 19

Threaded Bolts B1 and B2 of similar length and material are subjected to similar tensile load. If the elastic strain energy stored in bolt B1 is 4 times as compared to bolt B2, then what will be the mean diameter of bolt B2. The mean diameter of bolt B1 is 20mm.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 19

The stain energy be represented as:

U = σ2 x V / 2E

= F2 / A2 x AL / 2E

= F2L / 2AE

UB1 = 4UB2

Now as the two bolts are of similar length and material which are subjected to similar tensile load, so EB1 = EB2

LB1 = LB2

FB1 = FB2

Now

1/AB1 = 4/AB2

1/d2B1 = 4/d2B2

d2B2 = 4d2B1

dB2 = 2dB1

So, 2 x 20mm = 40mm.

Practice Test: Mechanical Engineering (ME)- 13 - Question 20

Which of the following are correct?

(P) Slenderness ratio is defined as the ratio of the effective length of the column to the minimum radius of the gyration

(Q) Euler’s formula is applicable for short columns

(R) Rankine’s constant for mild steel is 1/7500

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 20
Slenderness ratio is the ratio of the length of a column and the least radius of gyration of its cross section. Often denoted by lambda. It is used extensively for finding out the design load as well as in classifying various columns in short/intermediate/long.

Euler’s formula is applicable for long columns.

The value of Rankine's constant for different materials having hinged ends is as follows:

Cast iron = (1 / 1600)

Wrought iron = (1 / 9000)

Mild steel = (1 / 7500)

Timber = (1 / 750)

The Rankine's constant for a mild steel column with both ends hinged is 1/7500. Hence, option C is correct.

Practice Test: Mechanical Engineering (ME)- 13 - Question 21

A cantilever is subjected to a UDL the cross section of the beam is H-section placed as shown in fig. The bending stress distribution across the cross section will be

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 21
Bending stress at the centroid is always zero. In this case, the cross section is symmetrical about the X - X plane. So this eliminates the option b and d.

Now

all notations has usual means, y is the distance of top fiber from neutral axis

so

σ ∝ E × y

E will be constant as same material

σ ∝ y

Stress will vary linearly with distance so C option is correct

Practice Test: Mechanical Engineering (ME)- 13 - Question 22

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 × 10 − 6 m2/s, the thermal boundary layer thickness (in mm) at 0.5 m from the leading edge is __________

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 22

Pr = 3

T∞ = 500 k

6 = 1.5 m

T3 = 300 k, x = 0.5 m

V = 10 m/s

v = 30 × 106 m2/s

= 5 x 105

Laminar flow

= 166666.67

= 0.00612 m

Sth = δ = 0.00612 m = 6.12 mm.

Practice Test: Mechanical Engineering (ME)- 13 - Question 23

A slot is to be milled by a side and face milling cutter with 10 teeth and of diameter 150 mm. the cutting speed is 50 m/min and feed is 0.95 mm/tooth. The table feed in mm/min will be

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 23

η = 10

V = 50 m/min

D = 150 mm

Feed f = 0.95 mm/tooth

=

106 rpm

Tube feed f = f × η × N

= 0.95 × 10 × 106

= 1007 mm/min

Practice Test: Mechanical Engineering (ME)- 13 - Question 24

In the gear train of 1 : 10 as shown below, the pinion transmits 250 kW at 1800 rpm. What is the tangential load on the gear tooth?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 24

Power to pinion = Power to gear,

R / r = 10

R + r = 660

11r = 660

r = 66 mm

R = 660 mm = 0.6m

∴ for Gear

NG = 1800/10 = 180 rpm

T =

Ft x R =

Ft =

Ft = 2210.48 N

Ft = 2.2105 KN

Practice Test: Mechanical Engineering (ME)- 13 - Question 25

Four technicians are required to do four different jobs. Estimates of time to complete every job provided by the technicians are as below

The minimum time to complete all jobs is _______.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 25

Sub-step 1:

Subtract the minimum of each row from all the elements of the row, i.e.

Sub-step 2:

Since columns 2 and 3 contain no zero entry, we have to subtract minimum element of each column from all the elements of the column, i.e.

This is initial basic feasible solution. Checking if optimal assignment can be made. Examine rows successively until a row with exactly one unmarked zero is found. Mark ( ) this zero, indicating that an assignment will be made there. Mark (X) all other zeros in the same column showing that they cannot be used for making other assignments, i.e.

Next examine column for single unmaked zeroes, making them ( ) and also making (X) any other zeroes in the rows. Keep repeating the same procedure for rows and columns i.e.

Since there is one assignment in each row and in each column, the optimal assignment can be made in the current solution.

Optimum assignment is

A→1,B→2,C→4,D→3

Total work time = 20 + 34 + 18 + 35 = 107 hours.

Practice Test: Mechanical Engineering (ME)- 13 - Question 26

An orifice meter consisting of 0.1m diameter orifice in a 0.25m diameter pipe has a coefficient equal to 0.65. The pipe delivers oil of specific gravity 0.8. The pressure difference on the two sides of the orifice plate is measured by the Mercury-Oil differential manometer. If the differential gauge reads 0.8 meter of Mercury, the rate of flow in cubic meters per second is

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 26

H = 12.8 m of oil;

a1 = 0.049m2;

ao = 0.00785m2;

Cd = 0.65

Discharge through orifice meter,

Q =

Q = 0.0819 m3/sec.

Practice Test: Mechanical Engineering (ME)- 13 - Question 27

Oil having density of 800 Kg/m3 and viscosity of 0.2Ns/m2 is flowing through a pipeline of 50mm diameter at an average velocity of 2 m/s. The Darcy friction factor for the flow is

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 27

Re = 400

For laminar flow,

f = 64 / 100

f = 0.16

Practice Test: Mechanical Engineering (ME)- 13 - Question 28

What is directional derivative of ∅ =xy2 + yz3 at P (2, -1,1) in the direction of a normal to the surface x log z - y2 = -4 at A (-1, 2, 1)?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 28
we have to find the directional derivative of ∅ = xy2 + yz3 at P (2, -1,1) in the direction of a normal to the surface

x log z - y2 = -4 at A (-1, 2, 1)

The vector to the surface

f = x log z - y2 + 4 =0 is

n→ = grad f

= (log z) i – 2yj + x/z k

At A (-1, 2, 1)

Now, directional derivative =

=

Practice Test: Mechanical Engineering (ME)- 13 - Question 29

The pressure outside of the water droplet of diameter 0.06mm is 12 N/cm2. Calculate the pressure within the droplet. The surface tension is given as 0.08N/m.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 29

Diameter of droplet d = 0.06mm

σ = 0.08N/m

Pressure outside the droplet P= 12 N/cm2

the pressure inside the droplet in excess of outside pressure is given by

p = 4 × σ/d

= 0.533 N/cm2

Pressure inside the droplet = p + pressure outside the droplet

= 0.533 + 12 = 12.533 N/cm2

Practice Test: Mechanical Engineering (ME)- 13 - Question 30

For exchanging notes in a bank 50 customers arrive in 60 min which follows Poisson's distribution and service time follows exponential distribution. Due to lack of notes there is a single window for exchange and serves as a first come first serve basis. Probability of more than 20 customers in the system is 0.3 then, how much mean time is required for exchanging notes for a customer on the counter?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 30
Here 50 customers arrive in hour i.e. γ = 50/Hr

Now probability of more than k customers is P (n ≥ k + 1) = ρk+1

Where ρ = no of servers = γ/ 0.3 = ρ20+1

μ = mean service rate ρ = 0.944

μ = 50/0944 = 52.95 = 53 /hr

Now

Mean service time = 1/53 × 60 = 1.13 minutes for a customer on the counter

Practice Test: Mechanical Engineering (ME)- 13 - Question 31

A manometer contains a liquid having a specific gravity of 1.5. The gauge pressure at pt. A is -15kN/m2, determine the pressure at point C (air column between point B and C)

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 31
Given

Gauge pressure, pg = -15 kN/m2

Pressure at pt. B;

pB = -15 + (1.5 × 9.81 × 0.5)

= -7.6425 kN/m2

Air column is present between points B & C,

Therefore pB = pC

Pressure at pt. C;

pC= -7.6425 kN/m2

Practice Test: Mechanical Engineering (ME)- 13 - Question 32

Which of the following is True for this matrix A?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 32

=

(b) False

=

Thus orthogonal

(a) True

|A| ≠ 0 Thus A–1 exits

(c) False

A–1 = AT Thus –AT ≠ A–1

(d) False

Practice Test: Mechanical Engineering (ME)- 13 - Question 33

The power required to pump cooling water to the condenser is 20kW. Cooling water at 20℃ is flowing through CI pipe that is 0.8km long and 0.4m diameter at the rate of 0.07m3/s. Calculate average velocity of flow.

Assume f (coefficient of friction) =0.02

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 33

Power P = ρgQhf

20 x 103 = 103 x10× 0.07 × hf

hf = 28.57 m

hf =

28.57 =

v = 1.88 m/s.

Note:- f is identified as friction coefficient and F is Darcy Friction Factor.

Practice Test: Mechanical Engineering (ME)- 13 - Question 34

A steel cube contracted on all six faces is heated so that the temperature rises uniformly by 40 ºC. If the thermal coefficient of the steel is 12 × 10-6 per oC, Young’s modulus is 210GPa and the Poisson’s ratio is 0.28, the thermal stress developed in the cube due to heating is

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 34

All the six faces of the cube are contracted

Temperature rise(ΔT) = 40oC

Volumetric strain in the cube(εv) = εx + εy + εz --------(1)

Also Volumetric strain(εv)

1-2μ/E= (σx + σy + σz) -------(2)

From (1) & (2)

Practice Test: Mechanical Engineering (ME)- 13 - Question 35

A agro-industry has six different jobs a, b, c, d, e, f in process with promised deadlines of days 60, 72, 67, 72, 65, 70 but today is day 60 still work remaining for days of 2, 2, 2, 4, 5, 6 respectively. So, find the correct job sequence.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 35

Here deadline of job and work remain given i.e. sequence of find by critical ratio

Critical ratio = date required - today date / work remain

Correct sequence of job = a - e - f - d - c - b

Practice Test: Mechanical Engineering (ME)- 13 - Question 36

At a point in a material there are two direct stresses of 60 MPa tensile and 50 MPa compressive acting at right angles to each other. The greatest principal stress in the material is limited to 125 MPa in tension. To what shearing stress may the material be subjected on the given plane.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 36

According to the given conditions

125 =

Practice Test: Mechanical Engineering (ME)- 13 - Question 37

For a 50 mm shaft and hole pair designated H7 S8 the basic size lies in the range of 50 – 80 mm.

Date:-For-Tolerance

-H7-16i

-S8-25i

The fundamental deviation for ‘S ’ shaft is (IT7+0.4D) microns.

The allowance in mm will be

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 37

D = √50 × 80 = 63.24 mm

Fundamental tolerance,

i- = 0.45 D1⁄3 + 0.001D

i- = 0.45 (63.24)1⁄3 + 0.001 (63.24)

i- = 1.85 microns

Hole → max-= 50.029 mm

min-= 50 mm

shaft → max- = 50.10 mm

min- = 50.054 mm

Allowance- = 0.054 + 0.046 = 0.10 mm.

Practice Test: Mechanical Engineering (ME)- 13 - Question 38

Two venturi meters, with the same manometers, of different area ratios are installed in the same pipeline. The reading of 1st manometer is h and the area ratio is 2. If the 2nd manometer reads 8h, its area ratio is

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 38
Since both the manometers are installed on a same pipeline, their discharge is same

ar2 = 5

Practice Test: Mechanical Engineering (ME)- 13 - Question 39

Match the items in columns I and II.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 39

So the correct pairs are, P-4, Q-3, R-1, S-2.

Practice Test: Mechanical Engineering (ME)- 13 - Question 40

A water sprinkler has an 8 mm diameter nozzle at either end of the rotating arm. It is discharging water at right angle to the rotating arm in opposite direction at a velocity of 10 m/s. If the axis of rotation is at a distance of 0.2 m from one end and 0.24 m from other end, then the angular velocity of arm is (neglect frictional effect)

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 40

Given, R1 = 0.2 m, R2 = 0.24 m

Suppose the angular velocity of sprinkle is ω rad/s then the absolute velocity of flow through nozzle at both ends (discharge water velocity and sprinkle velocity will be in opposite direction)

V1 = 10 – 0.2 ω

V2 = 10 – 0.24 ω

There is no external torque, so total angular momentum will be conserved,

ρ Q [V1R1 + V2R2] = 0

ρ Q ≠ 0

V1R1 + V2R2 = 0

(10 – 0.2 ω ) × 0.2 + (10 – 0.24 ω) × 0.24 = 0

2 – 0.04 ω + 2.4 – 0.0576 ω = 0

ω = 45.082 rad/s

Practice Test: Mechanical Engineering (ME)- 13 - Question 41

The value of the function is

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 41
We know

ab = ebtna

=

By applying L’ Hospital rule,

=

= e0

= 1

Practice Test: Mechanical Engineering (ME)- 13 - Question 42

A cantilever beam of length L and negligible mass, carries a mass (m) at the free end. If the same beam needs to be replaced with a spring mass system, the equivalent stiffness of the spring under the same mass (m) will be?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 42

For cantilever beam

Deflection

Stiffness k load/deflection =

Practice Test: Mechanical Engineering (ME)- 13 - Question 43

Maximum fluctuation of kinetic energy in an engine has been calculated to be 2600 J. Assuming that the engine runs at an average speed of 200 rpm, the polar mass moment of inertia (in kg. m2) of a flywheel to keep the speed fluctuation within ±0.5% of the average speed is______________?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 43

∆E = Iω2Cs

⇒ 2600 =

⇒ I = 595

Practice Test: Mechanical Engineering (ME)- 13 - Question 44

If A is then |A121 - A120| is?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 44

=

=

= A120 x 0 = 0

Practice Test: Mechanical Engineering (ME)- 13 - Question 45

Two dice are thrown. What is the probability that the sum of the numbers on the two dice is eight?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 45

Two dice are thrown.

Total possible number of outcomes = 6 × 6 = 36

When sum of numbers on the two dice is eight

(2,6), (3,5), (4,4), (5,3), (6,2)

i.e., Total 5 possible outcomes.

So, Probability = 5/36

Practice Test: Mechanical Engineering (ME)- 13 - Question 46

An aeroplane moving with a velocity of 300 kmph negotiates a turn along a circular path of radius 600 m. If the angular velocity of the rotating parts of the plane is 120 rad/s and the moment of inertia of the rotating parts is 60 kg.m2, the gyroscopic couple is?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 46
A gyroscope is a device for measuring or maintaining orientation, based on the principles of angular momentum. Mechanically, a gyroscope is a spinning wheel or disk in which the axle is free to assume any orientation. Although this orientation does not remain fixed, it changes in response to an external torque much less and in a different direction than it would without the large angular momentum associated with the disk's high rate of spin and moment of inertia. After a short interval of time t, let the disc be spinning about the new axis of spin OX’ (at an angle δθ ) with an angular velocity (ω +δω). Using the right hand screw rule, initial angular velocity of the disc ω is represented by vector ox; and the final angular velocity of the disc (ω +δω ) Gyroscopic Couple Consider a disc spinning with an angular velocity ω rad/s about the axis of spin OX, in anticlockwise direction when seen from the front. Since the plane in which the disc is rotating is parallel to the plane YOZ therefore it is called the plane of spinning.

Gyroscopic couple = Iωωc

ωc = 120 rad/sec

ω = u/r = 300 × 5/18 ÷ 600 = 0.1388

I = 60 kgm2

Gyroscopic couple = 60 x 120 x 0.1388 = 1000 N.m

Practice Test: Mechanical Engineering (ME)- 13 - Question 47

Two metallic sheets, each of 2.0 mm thickness, are welded in a lap joint configuration by resistance spot welding at a welding current of 10 kA and welding time of 10 milliseconds. A spherical fusion zone extending up to the full thickness of each sheet is formed. The properties of the metallic sheets are given as: ambient temperature = 293 K

melting temperature = 1793 K

density = 7000 kg/m3

latent heat of fusion = 300 kJ/kg

specific heat = 800 J/kg K Assume:

(i) contact resistance along sheet – sheet interface is 500 micro–ohm and along electrode – sheet interface is zero;

(ii) no conductive heat loss through the bulk sheet materials; and

(iii) the complete weld fusion zone is at the melting temperature.

The melting efficiency (in %) of the process is ?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 47

Heat generated due to current,

H1 = I2RT

= 10 × 1032 × 500 ×10-6 ×10 × 10-3

= 500 J

Heat required for melting,

H2 = 300 × 103 × 7000 × volume + 800 × 1793 - 293 × 7000 × volume

As fusion zone is spherical in shape,

So, volume

π = 43 πR3

Where , r = t = thickness of sheet = 2mm

⇒ Melting efficiency = 70.35 %

Practice Test: Mechanical Engineering (ME)- 13 - Question 48

The drum of a simple band brakes at 400 rpm. The steel band is wrapped around 3/4th of the drum diameter as shown below. If 30 kW power is to be absorbed then the maximum tension in the band is ____ N. (take μ = 0)

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 48
Tension ratio,

T1/T2 = ew

Torque equation T = (T1 - T2)

⇒ T1 - T2 = 716.197N

From eq. (i) and (ii) we get-

T1 = 1173.54 N

T2 = 457.34N

So, maximum tension = T1 = 1173.54 N

Practice Test: Mechanical Engineering (ME)- 13 - Question 49

Consider the following Linear Programming Problem (LPP):

Maximize z = 3x1 + 2x2

Subject to x1 ≤ 4

x2 ≤ 6 (CO - OR)

3x1 + 2x2 ≤ 18

x1 ≥ 0, x2 ≥ 0

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 49

Objective function

Maximize z = 3x1 + 2x2 ,

Constraints, x1 ≤ 4............(i)

x2 ≤ 6..........(ii)

3 x1+ 2x2 ≤ 18.........(iii)

x1 ≥ 0, x2 ≥ 0............(iV)

All the equations are plotted on a graph and find the common area.

Slope of constraint (iii) and objective function are same hence, the objective function will have the multiple solutions as at points B and C the value of objective function is same.

Practice Test: Mechanical Engineering (ME)- 13 - Question 50

The humidity ratio of atmospheric air at 27 °C dry bulb temperature and 760 mm of Hg is 0.015 kg/kg of dry air. The vapour density (in kg/m3 of dry air) is?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 50

0.015 =

⇒ Pv = 173896 mm of Hg

Vapour of density,

=

= 0.0172kg/m2 of d.a.

Practice Test: Mechanical Engineering (ME)- 13 - Question 51

A block of weight W1 = 250N rests on a horizontal surface and supports on top of it another block of weight W2 = 75N. The block W2 is attached with a vertical wall by a string AB as shown in the figure. Find the force P (in N) for impending slipping.(Take μ = 0.3)

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 51

Here,

tan⁡α = 3/4,sin⁡α = 3/5, and cos⁡α = 4/5

Considering FBD of W2

∑H = 0

μN2 = Scos⁡α

S = μN2 / cos⁡α

∑nV= 0

N2 + Ssin⁡α − W2 = 0

N2 + μN2 / cos⁡α x sin⁡α − W2 = 0

N2(1 + μtan⁡α) = W2

N2 =

Considering FBD of W1

∑V = 0

N1 = N2 + W1 = 61.2245 + 250 = 311.2245

∑H = 0

P = μN1 + μN2 = 0.3(61.2245 + 311.2245) = 111.7347

P = 111.7347N

Practice Test: Mechanical Engineering (ME)- 13 - Question 52

What should be the angle between the two cylinder V line engines such that all the primary and secondary forces as well as couples are perfectly balanced.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 52

• V line engine is the only type of engine which can be completely balanced

• All primary and secondary forces can be balanced when the angle between the two cylinder in v-line engine is 90 degree

*Answer can only contain numeric values
Practice Test: Mechanical Engineering (ME)- 13 - Question 53

Calculate the value of poisson's ratio if bulk modulus(K) and modulus of rigidity(G) are 300MPa and 400MPa respectively. upto 3 decimal places

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 53

We know that,

E = 2G (1 + µ)

E = 3K (1 - 2µ)

2G (1 + µ) = 3K (1 - 2µ)

µ(2G + 6K) = 3K - 2G

µ =

= 0.0385

Practice Test: Mechanical Engineering (ME)- 13 - Question 54

In a spot welding process, a current of 40,000 amperes is released for 0.015 sec. The effective resistance of the joint is 120 micro ohm. The joint can be considered as a cylinder of 6 mm diameter and 3mm height. The heat required for the melting of material is 10 J/mm3. Calculate the heat transferred to environment (in kcalories) from the joint where 4.184 joules = 1 calories

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 54

Heat Supplied =I2RT

=(40000)2 × 120 × 10-6 × 0.015

=2880 Joules

Heat required = π/4d2 × H × e

= π/4 × 62 × 3 × 10

= 848.23

Heat escape to surrounding = 2880 − 848.23

= 2031.76 joules = 485.60 calories = 0.485 kcal

Practice Test: Mechanical Engineering (ME)- 13 - Question 55

An electric light fixture is held with the arrangement shown in the figure. If the weight of the fixture is 1N and the hinge is an ideal one, Then axial force in the bar PR and string PQ is

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 55

∑V = 0 S1sin⁡(30) − 1 = 0

S1 = 1 / sin⁡(30) = 2N

∑H = 0

S2 + S1cos⁡(30) = 0

S2 = −S1cos⁡(30) = −2 × 32 =−1.7321N

Practice Test: Mechanical Engineering (ME)- 13 - Question 56

The diameters of inner and outer tubes of a Double pipe heat exchanger are 24mm and 44mm. The thickness of the inner tube is;

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 56

For a double pipe heat exchanger,

2δ = do – di

δ = (44 – 24) /2

δ = 10 mm

Practice Test: Mechanical Engineering (ME)- 13 - Question 57

A point mass having mass M is moving with a velocity V at an angle θ to the wall as shown in the figure. The mass undergoes a perfectly elastic collision with the smooth wall and rebounds. The total change (final minus initial) in the momentum of the mass is

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 57

Let w is the velocity after collision

V sinθ j^

−wsin ϕj^

After collision

E = velocity of separation / velocity of approach = 1

{∴ e = 1,or perfectly elastic collision}

= w sin⁡∅j = Vsin⁡∅j _________(1)

Changes in momentum = final momentum - initial momentum

ΔP = {−wsin⁡∅j − (Vsin⁡∅j)}M

= −2MVsin⁡∅j (∵ wsin⁡∅j = Vsin⁡∅j)

Practice Test: Mechanical Engineering (ME)- 13 - Question 58

Which parameter should be increased to avoid built-up-edge formation?

P: Rake Angle

Q: Velocity

R: Feed

S: Depth of cut

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 58

High rake angle means less cutting force and less dissipation of heat.

High cutting speed means less chip tool contact time which will lead to insufficient time to form a bond between chip and tool.

Higher feed and depth of cut means more MRR and hence more heat dissipation at the edge.

Practice Test: Mechanical Engineering (ME)- 13 - Question 59

The following figure shows the velocity-time plot for a particle travelling along a straight line. The distance covered by the particle from t=0 to t=5s is ______ m.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 59
Area under V - T wave

1/2 × 1 × 1 +1 × 1 + 1/2 × (1 + 4) × (3 - 2) + 1/2 × (4 + 2) × (5 - 3)

= 0.5 + 1 + 2.5 + 6

= 10

Practice Test: Mechanical Engineering (ME)- 13 - Question 60

Block A weighing 200 kg resting on the rough floor supports block B weighing 100 kg as shown in the figure. Both the blocks are connected with a rope passing over a smooth pulley. Find the force P required to impending motion. (Take g = 9.81 m/s2 and coefficient of friction = 0.3)

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 60

Considering FBD of block B,

S = μNB = 0.3 × 100 = 30kg

Considering FBD of block A,

NA=NB + 200 = 100 + 200 = 300kg

μμP = μNA + μNB + S = 0.3 × 300 + 0.3 × 100 + 30 = 150kg

Required force P =150kgwt = 150 × 9.81 = 1471.5N

P = 1471.5N

Practice Test: Mechanical Engineering (ME)- 13 - Question 61

A uniform beam of length L is simply and symmetrically supported over two roller supports spaced l distance with each other. The ratio of L/l so that the upward deflection at each end equals the downward deflection at the mid-point due to a central point load is ______.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 61

Due to central point load W, the slope at the ends of a simply supported beam

=Wl2 / 16EI

θsupport = Wl2 / 16EI

Here angle will be very small,

So Deflection at the overhanging end = θ*support Overhanging length of one side

=

Deflection at the midspan =wl2 /48EI↓

Equating the deflection

Note-Here the formula of central deflection and slope are right as we have considered it a simply supported beam and length of the simply supported beam is I.

Practice Test: Mechanical Engineering (ME)- 13 - Question 62

The stream function for 2-D flow is given by ψ = 4xy. Calculate the velocity at point P(2,2).

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 62

= -4x

= 4y

At P(2,2), u = -8, v = 8

Resultant Velocity = √u2 + v2 = 11.31

Practice Test: Mechanical Engineering (ME)- 13 - Question 63

Air flows over a heated plate at a velocity of 40 m/s. The local skin friction coefficient at a point on the plate 0.004. Determine the local heat transfer coefficient ( in W/m2K) at this point using Reynold’s Colburn analogy

Following are property of air

ρ= 0.88 kg/m3, Cp =1001 J/kg°K, k = 0.035 W/mK, μ=2.286×10-5 Kg/ms

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 63

Here for air ρ=0.88 kg/m3, Cp = 1001 J/kg°K , k = 0.035 W/mK

μ = 2.286×10-5 Kg/ms

local skin friction coefficient at a point Cf = 0.004

Reynolds Colburn analogy with

Stanton no St

As

h = 93.62 = 94 W/m2K

Practice Test: Mechanical Engineering (ME)- 13 - Question 64

A solid copper ball of mass 500gm when quenched in a water bath at 30C, cools from 530C to 430C in 10 sec. What will be the temperature of the ball after the next 10 seconds?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 64
We know that,

This is a case of an unsteady-state heat transfer problem.

In order to obtain the temperature after t time the formula is given by,

Where T= temperature after time t, T = temperature of bath or environment, Ti = initial temperature, τ is the time, h is heat transfer coefficient, A is the area of cross-section, ρ is the density of ball, V is the volume of ball, C is the heat capacity.

Given:

Putting the value,

Now temperature of the ball after the next 10 sec means the total time is 20sec⁡(τ = 20 sec)

Note: Time is always taken from the beginning.

Put the value of hA / ρVC from the above,

⇒ T− 30 = 320

⇒ T = 350C

Hence, the correct option is (C).

Practice Test: Mechanical Engineering (ME)- 13 - Question 65

A furniture dealer deals only in two items- tables and chairs. He has Rs. 50,000 to invest and has storage space of at most 60 pieces. A table costs Rs.2500 and a chair Rs.500. He estimates that from the sale of one table, he can make a profit of Rs.250 and that from the sale of one chair a profit of Rs.75. If he buys ‘x’ and ‘y’ number of tables and chairs respectively from the available money in order to maximize his profit, the value of ‘x’ will be, assuming that he can sell all the items which he buys?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 13 - Question 65

‘x’ and ‘y’ must be non-negative.

2500x + 500y ≤ 50000 (investment constraint)

5x + y ≤ 100

x + y ≤ 60 (storage constraint)

Max Z = 250x + 75y (objective function)

Computing the values at corner points of the shaded region. The optimum solution will be obtained for(10,50).

Hence for maximum profit dealer should buy 10 tables

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