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Practice Test: Mechanical Engineering (ME)- 16 - Mechanical Engineering MCQ


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30 Questions MCQ Test - Practice Test: Mechanical Engineering (ME)- 16

Practice Test: Mechanical Engineering (ME)- 16 for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Practice Test: Mechanical Engineering (ME)- 16 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Practice Test: Mechanical Engineering (ME)- 16 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)- 16 below.
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Practice Test: Mechanical Engineering (ME)- 16 - Question 1

Direction: A sentence with one blank is given, indicating that something has been omitted. Choose the word that best fits the blank appropriately.

______ you have enough swimming practice, you will be able to swim well.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 1
"Once" is the correct word that should fill up the given blank. Here, the type of the given sentence is conditional. The happening of the second event depends on the occurrence of the first one .i.e. If someone has enough swimming practice, then only he will be able to swim well. Hence option D is the correct response.
Practice Test: Mechanical Engineering (ME)- 16 - Question 2

If price of an article increases from Rs 200 to Rs 240, when quantity demanded decreases from 1,000 units to 800 units. Find price elasticity of demand?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 2
P = Rs 200, P1 = Rs 240

Q1 = Rs 1,000, Q2 = Rs 800

Price elasticity of demand =

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Practice Test: Mechanical Engineering (ME)- 16 - Question 3

With reference to India's freedom struggle consider the following statements:

1] In March 1940, the Muslim League passed a resolution committing itself to the creation of a separate nation called "Pakistan".

2] The "Salt Satyagraha" campaign began in August 1942.

3] Jayaprakash Narayan was a socialist member of the Congress and was active in the underground resistance during the Quit India Movement.

Which of the statements given above is/are correct?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 3
The correct statements are

1. In March 1940, the Muslim League passed a resolution committing itself to the creation of a separate nation called "Pakistan".

2. The "Salt Satyagraha" campaign began in 1930.

3. Jayaprakash Narayan was a socialist member of the Congress and was active in the underground resistance during Quit India Movement.

Practice Test: Mechanical Engineering (ME)- 16 - Question 4

Choose the pair of words to be placed in the 2 blanks in the given sentence so that the completed sentence becomes meaningful.

After having been close friends for more than a decade, they had a _____ last year and have not _____ each other ever since.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 4
Out of 4 given options, it is easy to eliminate options (a) and (b). Option (a) requires ‘to’ after talked in the second blank; in option (b) ‘argument’ needs ‘an’ to come before.
Practice Test: Mechanical Engineering (ME)- 16 - Question 5

A 200 × 100 × 50 mm3 steel block is subjected to a hydro-static stress of 15 MPa. The Young’s modulus and Poisson’s ratio of the material are 200 GPa and 0.3 respectively. The change in the volume of the block in mm3 is

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 5
Given,

dimension of steel block = 200×100×50mm3

σ = 15 MPa

E = 200 GPa, μ = 0.3

Bulk modulus,

E=3K(1-2μ)⇒200=3K(1-2×0.3)

K = 166.67GPa

ΔV = 90mm3

Practice Test: Mechanical Engineering (ME)- 16 - Question 6

From a group of 61 students, each student appears for at least one of the 3 papers i.e. GATE, ESE or SSC. Out of the students appearing for SSC, the number of students appearing for ONLY SSC is equal to the number of students who also appear for GATE. The number of students who appear for only GATE is 3 more than the number of students who appear for all 3; number of students who appear for ESE alone is higher than the previous number by 5. If 32 students appear for ESE and 36 students appear for exactly ONE exam, then the number of students appearing for all 3 exams are ______.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 6

Base on the diagram and information available,

c = f + g ...(i)

a + b + c = 36

⇒ d + e + f + g = 61 – 36 = 25 ...(ii)

a + b + c = 36

⇒ g + 3 + g + 8 + c = 36

⇒ 2g + c = 25 = 3g + f ...(iii)

∵ b + d + e + g = 32

⇒ g + 8 + d + e + g = 32

⇒ d + e + g + g = 24

⇒d + e + f + g + g = 24 + f = 25 + g

⇒ f = g + 1

∵ 3g + f = 25

⇒ 4g = 24

∴ g = 6

Practice Test: Mechanical Engineering (ME)- 16 - Question 7

Raja and Joseph start from their school to walk towards their hostel. The distance covered by Raja in 3 steps is identical to the distance covered by Joseph in 4 steps while the time taken by both of them to cover a step is identical. At a particular instance of time when both of them had covered 800 steps each, the distance between them was 300 m. Now, Raja cuts down the length of his step by 1/3rd and Joseph increase the length of his step by 20% with time taken to cover a step by both remaining identical. Both will be at the same spot at the same time in ______ more steps.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 7
Let R and J be the length of each step of Raja and Joseph respectively.

We can write the following equations based on the given information:

3R = 4J

and 800R – 800 J = 300 m

Solving, we get R = 1.5 m = 150 cm and J = 112.5 cm

Revised value for length of each step of Raja R1 =

and for Joseph, J1 = 112.5 + (20% of 112.5) = 135 cm

Once again, difference in the length of a step of Raja and Joseph = 35 cm with Joseph’s step being longer. In order to catch up with Raja who is ahead of him by 300 m, the number of steps required = 300/ 0.35 = 857.14 = 858

since step can’t be written in fraction so we can write it & the result is found is more than 857 so the answer would be 858.

Practice Test: Mechanical Engineering (ME)- 16 - Question 8

In metals, resistivity is composed of two parts: one part is characteristic of the particular substance. The other part is due to

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 8
Resistivity is composed of two components:-

(i) Thermal Resistivity: It is due to lattice vibration and is characteristic of the particular substance

(ii) Residual Component: This component arises due to impurity and defect present in the material. This is independent of temperature.

Practice Test: Mechanical Engineering (ME)- 16 - Question 9

Direction: In the given questions, two statements are given followed by two conclusions I and II. You have to consider the two 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 :

(1) Best performance in Olympics fetches a gold medal.

(2) Player ‘X’ got gold medal but later was found to be using a prohibited drug.

Conclusions :

I. ‘X’ should be allowed to keep the gold medal.

II. Gold medal should be withdrawn.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 9

Statements :

(1) Best performance in Olympics fetches a gold medal.

(2) Player ‘X’ got gold medal but later was found to be using a prohibited drug.

Conclusions :

I. ‘X’ should be allowed to keep the gold medal. (not follows, as ‘X’ was found to be using a prohibited drug, hence ‘X’ has no right to keep the gold medal)

II. Gold medal should be withdrawn. (follows, as ‘X’ was found to be using a prohibited drug, hence ‘X’ has no right to keep the gold medal hence Gold medal should be withdrawn and given to the next person)

Practice Test: Mechanical Engineering (ME)- 16 - Question 10

They are expert statisticians and help the Black Belts in case of issues.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 10
The Six Sigma Belts(Green, Black and Master Black) denotes the different levels an individual can achieve in Six Sigma.

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

A pitot tube measures a differential pressure of 0.008 bar. What is the speed of the plane, if it is flying at a height of about 12000 m.? Take density of air at this height 0.31 kg/m3

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 11
Pitot tube fixed to the plane, measures the speed of air with respect to plane.

pressure difference= 0.008×105 = 800 Pa

Velocity of plane,

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

Suppose a toy car is made by a group of students for analysing the effect of various forces in running a car. The car works with help of a slider crank mechanism fitted inside the toy car. The car moves with a constant angular velocity of 10 rad/s and the crank radius is 60 mm. If the mass of the slider is 0.2 kg. Find the hammer blow acting on the car that tends to lift the car vertically upward.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 12
Hammer blow is the maximum primary unbalanced force acting on the car in vertical direction

hammer blow=mrω2

for r=60mm, m=0.2kg, w=10rad/s

hammer blow=0.2×0.06×102

=1.2 N.

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

The area of the region bounded by the curves y = |x-2|, x = 1, x = 3 and the x-axis is __________.

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

Required area = Area of ΔLAB + Area of ΔMBC

Practice Test: Mechanical Engineering (ME)- 16 - Question 14

The solution of the differential equation is

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 14

⇒ m2 + m + 1 = 0

Practice Test: Mechanical Engineering (ME)- 16 - Question 15

The demand for an item is 900 units per month and the lead time is 10 days. In the past two month, the maximum demand observed is 50 units per day. For a re-ordering system based on inventory level, the re-order level (in units) is

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 15
The re-order level (ROL) (when the demand varies and the lead time is constant) is given as

ROL = average DDLT + buffer stock (BS)

Average consumption per day = = 30 units per day

Average demand during lead dime (DDLT) = 30×10 = 300

BS = (maximum DDLT) - (Average DDLT)

= 50×10-30×10

= 200 units

ROL = 300+200 = 500 units.

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

The solidification time for a spherical mould was 205 sec. If the radius of the mould is 5 cm, determine the mould constant.

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

Solidification time = 205 sec

Area A = 4πr2

Volume V = 4/3 πr3

Where T = solidification time

V = volume of the cavity

SA = surface area of the cavity

S = mould constant

For sphere

So,

S = 0.738 s/mm2

Practice Test: Mechanical Engineering (ME)- 16 - Question 17

The areas of ram and plunger of a hydraulic press are 40mm2 and 2 mm2 respectively. The plunger applies a force of 500N, determine the weight lifted based on the intensity of pressure measured at ram.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 17
Given: AR = 0.04 m2, AP = 0.002m2

F = 500N

Intensity of pressure due to plunger,

P = F/AP = 500 / 0.002

P = 2.5 × 105 N/m2

Since the intensity of pressure will be equally transmitted, therefore intensity of pressure at ram will be same. (by Pascal’s law)

Intensity of pressure at ram = W / AR

2.5 × 105 = W / 0.04

W = 10 kN

Practice Test: Mechanical Engineering (ME)- 16 - Question 18

A Quick return mechanism is shown below. The value of α will be _________ (in degrees) stroke=60 cm Length of slotted lever = 50 cm

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 18

stroke=PP'

but, PP'=DD'

OD=50cm

∴ OM2+MD2 = OD2

OM2 + OD2 - MD2

OM + 502 - 302

OM=40cm

tan(β)= 30/40

β=36.87°

θ = 180 - (90+36.869)

=53.14°

2θ=106.28°

α=360-106.28

α=253.72°

Practice Test: Mechanical Engineering (ME)- 16 - Question 19

The state of stress is showing as a particular point in a stressed body. The diameter of the Mohr’s circle for this state of stress is.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 19
Radius of Mohr’s circles is given by

σx = 100 MPa

σy = 50 MPa

τxy = 50 MPa

Radius = 90.138 MPa

and

Diameter = 2 × radius

= 2 × 90.138

= 180.27 MPa

Practice Test: Mechanical Engineering (ME)- 16 - Question 20

Two blocks are on the frictionless surface as shown in fig. having mass 1 kg and 2 kg. The 1 kg block moves with a speed 5 m/s towards other block kept at rest. The spring is attached to 2 kg block and has a spring constant 60 N/m. The maximum compression of spring (in m) is

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 20

The maximum compression of spring will take place when both the block moves with same velocity. Let the final velocity of both blocks is V m/s. As there is no external force acting on the system (combined spring and blocks as a system), so the total linear momentum will be conserved.

Initial linear momentum = Final linear momentum

1×5 + 2×0 = 1×V + 2×V

Using conservation of energy

Change in kinetic energy of blocks = Energy stored by spring

Where, K is spring constant and x is maximum compression of spring

12.5 - 4.167 = 30 x2

x = 0.52 m.

*Answer can only contain numeric values
Practice Test: Mechanical Engineering (ME)- 16 - Question 21

A heat exchanger is used to heat cold water (CP = 4.18 kJ/kg-K) entering at 15 °C at a rate of 1.2 kg/s by hot air (CP = 1.005 kJ/kg-K) entering at 90 °C at a rate of 2.5 kg/s. The highest rate of heat transfer in the heat exchanger is _____ kW.


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 21
Maximum rate of heat transfer is given by

Qmax=CminTh1-Tc1

Ch=mhcph=2.5×1.005=2.5125 kW/K

Cc=mccpc=1.2×4.18=5.016 kW/K

Cmin =Ch = 2.5125 kW/K

Qmax=2.5125×90-15=188.44 kW

Practice Test: Mechanical Engineering (ME)- 16 - Question 22

What will be the transmission ratio of gear train?

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 22
When the same direction of rotation is required for both the driver and the follower, an idler wheel is used. Transmission or movement ratio can be expressed as μM = ND / NF = TF / TD

where

μM = movement ratio

ND = revolutions of driver (rpm)

NF = revolutions of follower (rpm)

TF = number of teeth on follower

TD = number of teeth on driver

Practice Test: Mechanical Engineering (ME)- 16 - Question 23

Which of the following statement is incorrect

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 23
In punching press to reduce the variation of speed flywheel is used.

For an isochronous governor, the range of speed in on-load condition is zero.

For no load isochronous governor,

For on-load isochronous governor,

*Answer can only contain numeric values
Practice Test: Mechanical Engineering (ME)- 16 - Question 24

A solar collector receiving solar radiation at the rate of 0.5KW/m2 at temperature 80C and transforms it to a fluid within heat engine. The engine rejects energy as heat at temperature of 25C. The minimum collector area (in m2) if the plant produces 2KW of useful shaft power is


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 24
Given, T2 = 25 +273 = 298 K, T1 = 80 + 273 = 353 K

Heat input to engine, Q1 = 0.5 KW/m2 = 0.5A KW

Where, A is the area of solar collector

For the minimum area A. collector the efficiency of heat engine would be maximum.

The maximum efficiency of heat engine,

Practice Test: Mechanical Engineering (ME)- 16 - Question 25

What will be the entropy change, if 1 kg of liquid water is heated from 25◦C to 100◦C. assuming constant specific heat, and compare the result with that found when using the given values.

Take: C = 4.184; sf20◦C = 0.2989 kJ/kg; sf90◦C = 1.2105 kJ/kg

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 25
For constant specific heat, from Eq. (s2 - s1 ~= C ln T2/T1)

s2 - s1 = 4.184 ln (373/298) = 0.9392 kJ/kg K

Comparing this result with that obtained by using the given values, we have

s2 - s1 = sf90◦C - sf20◦C = 1.2105 - 0.2989

= 0.9116 kJ/kg K

Practice Test: Mechanical Engineering (ME)- 16 - Question 26

Thermodynamic relation for the isothermal compressibility is

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 26

Also we know that,

compressibility k =

Then isothermal compressibility (i.e. compressibility at constant temperature) is

Practice Test: Mechanical Engineering (ME)- 16 - Question 27

In an air standard cycle, indicated power is 80 KW and mechanical efficiency is 85%. Then the frictional power will be

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 27
Given ηM = 85%, IP = 80 KW

IP = BP + FP ⇒ FP = IP − BP

= 80 − 68

FP = 12 kW

Practice Test: Mechanical Engineering (ME)- 16 - Question 28

Which of the following chemical formula is correct for the refrigerant designated as R12

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 28
Chemical formula for the refrigerants can be given as

CmHnFpClq

The designation for such refrigerant is, R(m−1)(n+1)p

Where, n+p+q=2m+2

So from the designation, R12 or R012, we have

m−1=0⇒m=1

n+1=1⇒n=0

p=2

using n+p+q=2m+2, we get, q=2

So the chemical formula for the given refrigerant will be: CF2Cl2

*Answer can only contain numeric values
Practice Test: Mechanical Engineering (ME)- 16 - Question 29

If a body of 500 kg - m2 is rotating about its axis passing through the centre of mass and work done by the rotation of body is 100 kJ. Then the angular velocity of the body in rad/s will be


Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 29
The answer is in between 19.5 and 20.5

We know that, work done = force ×displacement

In rotational motion, Work Done =

Practice Test: Mechanical Engineering (ME)- 16 - Question 30

In a steady flow device, a liquid enters with a mass flow rate of 3 kg/s with enthalpy of 3200 kJ/kg. Heat lost to the environment is 600 kW. If the leaving fluid has an enthalpy of 2300 kJ/kg, calculate the work done in kW.

Detailed Solution for Practice Test: Mechanical Engineering (ME)- 16 - Question 30

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