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Mechanical Engineering Exam  >  Mechanical Engineering Tests  >  UPSSSC JE Mechanical Mock Test Series 2024  >  UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Mechanical Engineering MCQ

UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Mechanical Engineering MCQ


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30 Questions MCQ Test UPSSSC JE Mechanical Mock Test Series 2024 - UPSSSC JE Mechanical Paper 2 Mock Test - 5

UPSSSC JE Mechanical Paper 2 Mock Test - 5 for Mechanical Engineering 2024 is part of UPSSSC JE Mechanical Mock Test Series 2024 preparation. The UPSSSC JE Mechanical Paper 2 Mock Test - 5 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The UPSSSC JE Mechanical Paper 2 Mock Test - 5 MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for UPSSSC JE Mechanical Paper 2 Mock Test - 5 below.
Solutions of UPSSSC JE Mechanical Paper 2 Mock Test - 5 questions in English are available as part of our UPSSSC JE Mechanical Mock Test Series 2024 for Mechanical Engineering & UPSSSC JE Mechanical Paper 2 Mock Test - 5 solutions in Hindi for UPSSSC JE Mechanical Mock Test Series 2024 course. Download more important topics, notes, lectures and mock test series for Mechanical Engineering Exam by signing up for free. Attempt UPSSSC JE Mechanical Paper 2 Mock Test - 5 | 150 questions in 120 minutes | Mock test for Mechanical Engineering preparation | Free important questions MCQ to study UPSSSC JE Mechanical Mock Test Series 2024 for Mechanical Engineering Exam | Download free PDF with solutions
UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 1
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The first law of thermodynamics deals with ___.

Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 1
In physics and chemistry, the law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time.
UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 2
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The amount of radiation depends upon—

Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 2
Radiation is a heat transfer method that does not rely upon any contact between the heat source and the heated object. It depends on the nature of the body, its temperature, and the kind & extent of its surface.
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UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 3
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The thickness of oil film between gauge block is of the order of

Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 3
The thickness of the ring film is about 25 nanometers (0.98 μin). The gauge's nominal length is also known as the interferometric length. In use, the blocks are removed from the set, cleaned of their protective coating (petroleum jelly or oil), and wrung together to form a stack of the required dimension.
UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 4
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Flank wear occurs mainly on the
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 4

Crater wear occurs on the rake face of the tool, while flank wear occurs on the relief (flank) face of the tool.

UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 5
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The molecular mass expressed in gram (i.e. 1 gmole) of all gases, at N. T. P., occupies a volume of
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 5

According to Avogadro's law ,the volume of a gram mole of all gases at the pressure of 760 mm of Hg and temperature of 0 degree centigrade is same and is equal to 22.4 litres.

UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 6
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In which of the following operations performed on the lathe machine, chips do not occur?
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 6

Knurling: Knurling is a manufacturing process whereby a visually attractive diamond‐shaped (crisscross) pattern is cut or rolled into metal. This pattern allows human hands or fingers to get a better grip on the knurled object than would be provided by the originally smooth metal surface.

Boring: Boring always involves enlarging an existing hole, which may have been made by a drill or maybe the result of a core in a casting.

Reaming: Reaming removes a small amount of material from the surface of holes. It is done for two purposes: to bring holes to a more exact size and to improve the finish of an existing hole.

Threading: It is the process of making internal or external threads on the workpiece.

Out of the above four processes except knurling, all others involve the chips. So Knurling is the right option.

UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 7
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One kg of air (R=287 J/kg-K) goes through an irreversible process between two equilibrium states, 1 (30°C, 1.2 m3) and state 2 (30°C, 0.8 m3). What is the change in entropy (in J/kg-K)?
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 7
Entropy changes for a closed system

For constant volume:

s2−s1 = cv lnT2/T1

For constant pressure:

s2−s1 = cp lnT2/T1

For constant temperature process:

s2−s1 = R ln v2/v1

For adiabatic process (Reversible):

dQ = 0 ⇒ dS = 0

Calculation:

Here T1 = T2

So the process is isothermal.

s2−s1 = R ln v2/v1

UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 8
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The ratio of the depth of the bucket for a Pelton wheel to the diameter of a jet is of the order of _____.
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 8
The width of the bucket for a Pelton wheel is generally five times the diameter of the jet. The depth of the bucket for a Pelton wheel is generally 1.2 times the diameter of the jet. The number of buckets on the periphery of a Pelton wheel is given by (15+D/2d), Where D is the pitch diameter of the wheel and d is the diameter of the jet.
UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 9
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For the given two statements:

I. The performance of an SI engine can be improved by increasing the compression ratio.

II. Fuels of higher-octane numbers can be employed at a higher compression ratio.
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 9
The compression ratio of the engine also strongly affects knock. (The performance of an SI engine cannot be improved by increasing the compression ratio) The higher the compression ratio, the better the thermal efficiency but, the greater the tendency for knock to occur.

The critical compression ratio can be defined as the compression ratio when knock starts. So higher fuel octane quality is required (Fuels of higher-octane number can be employed at a higher compression ratio).

Important Point:

Knock occurs when the unburnt gases ahead of the flame front (the end gases) spontaneously ignite, causing a sudden rise in pressure accompanied by a characteristic pinging sound. This results in a loss of power and can lead to damage to the engine.

The antiknock value of an SI engine fuel is determined by comparing its antiknock property with a mixture of two reference fuels, iso-octane (C8H18) and normal heptane (C7H16).

Iso-octane, which has a very high resistance to knock and therefore is given an octane number of 100. Normal heptane, which has a very poor antiknock quality and is therefore given a zero value.

The octane number of fuels is defined as the percentage, by volume, of iso-octane in a mixture of iso-octane and normal heptane. A blend of 10% n-heptane and 90% iso-octane by volume has an octane number of 90.

Engines with low compression ratios can use fuels with lower octane numbers, but high compression engines must use high-octane fuel to avoid self-ignition and knock.

UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 10
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Which process follows heat treatment?
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 10

Grinding

In the past, many scientists investigated the heat dissipation in grinding and the resulting influences on the surface integrity of the workpieces. Under abusive grinding conditions, the formation of a heat-affected zone was observed, which damaged the ground surface of hardened steels. This paper aims to introduce a new surface heat treatment process, namely by making use of the heat flux generated in grinding. In this case, the grinding conditions have to be optimized to induce martensitic phase transformations in the surface layers of annealed or tempered steels, which is achieved by other surface strengthening processes. The fundamentals of this new heat treatment method called grind-hardening are given in this paper.

UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 11
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What is the value of the angle (degree) between streamlines and equipotential lines at the point of intersection in the flow net?
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 11
The angle between streamlines and equipotential lines at a point of intersection in the flow net is always intersect at 90°

UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 12
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In arc welding, penetration is minimum for
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 12
In arc welding, penetration is minimum for DC – Electrode Positive.

Types and importance of polarity in arc welding

Polarity indicates the direction of current flow in the welding circuit.

Kinds of polarity

  • Straight polarity or electrode negative (DCEN)
  • Reverse polarity or electrode positive (DCEP)

Straight polarity: In straight polarity, the electrode is connected to the negative and the work to the positive terminal of the power source.

Electrode Negative Or straight polarity (DCEN)

Reverse Polarity: In reverse polarity, the electrode is connected to the positive and the work to the negative terminal of the power source.

Electrode Positive Or Reverse Polarity (DCEN)

DCSP (Electrode Negative)—Maximum penetration.

AC—Moderate penetration.

DCRP (Electrode Positive)—Minimum penetration

UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 13
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Which of the following statements is wrong?
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 13
Knocking in a CI engine occurs because of an ignition lag in the combustion of fuel between the time of injection and the time of actual burning.

As the ignition lag increases, the amount of fuel accumulated in the combustion chamber increases. When combustion actually takes place, an abnormal amount of energy is suddenly released, causing an excessive rate of pressure rise, which results in a knock.

The CI engine knock can be controlled by reducing the delay period. The delay is reduced by the following :

a. High charge temperature

b. High fuel temperature

c. Good turbulence

d. Injection of fuel just before TDC

To decrease the tendency of knock, it is necessary to start the actual burning as early as possible after the injection begins. In other words, it is necessary to decrease the ignition delay and thus decrease the amount of fuel present when the actual burning of the first few droplets starts.

The following are the differences in the knocking phenomenon of the S.I. and C.I. engines:

  • In the S.I engine, the detonation occurs near the end of combustion, whereas in the C.I engine detonation occurs near the beginning of combustion

  • The detonation in the S.I engine is of a homogeneous charge causing a very high rate of pressure rise and very high maximum pressure.

  • In the C.I. engine, the fuel and air are imperfectly mixed, and hence the rate of pressure rise is normally lower than that in the detonating part of the charge in the S.I engine.

  • In the C.I. engine, the fuel is injected into the cylinder only at the end of the compression stroke; there is no question of pre-ignition as in S.I. engine.

  • In the S.I. engine, it is relatively easy to distinguish between knocking and non-knocking operations as the human ear easily finds the distinction.

UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 14
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At the triple point of a pure substance, the number of degrees of freedom is
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 14

UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 15
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Cutting and forming operations can be performed in a single operation in a _______ die.
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 15
Simple dies are also known as single operation dies as a single operation is performed for each stroke of the die press.

Progressive dies perform two or more operations at different stages every time the ram descends.

Compound dies differ from a progressive die in that it performs two or more cutting operations during one stroke of the press at one station only. These die allow simultaneous cutting of internal and external part features in a single stroke in some cases.

Combination dies combine cutting operations with a non-cutting operation. The cutting operations may include blanking, piercing, trimming, and cut off and are combined with non-cutting operations, which may include bending, extruding, embossing, and forming.

UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 16
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A refrigerator works on a reversed Carnot cycle producing a temperature of -40°C. Work done per TR is 700 kJ per ten minutes. What is the value of its COP?
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 16
Concept

Coefficient of performance of the refrigeration system working on reversed Carnot cycle:

Unit of refrigerating effect = One tonne of refrigeration

1 TR = 3.5 kW

Calculation:

Given: RE = 1 TR = 3.5 kW

W = 700 kJ/ten minutes

COP = 3

UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 17
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Work done obtained in a free expansion process is_______.
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 17
For free expansion, work done is zero.

So δW = 0

Also, No heat interaction takes place during free expansion. Hence δQ = 0

From the first law of Thermodynamics

δQ = dU + δW

dU = 0

UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 18
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A substance is said to be volatile if it has______
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 18
Volatility is quantified by the tendency of a substance to vaporize. Volatility is directly related to a substance's vapor pressure. A substance with higher vapor pressure evaporates more readily than a substance with a lower vapor pressure at a given temperature.
UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 19
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A flapper nozzle is used in
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 19

The nozzle and flapper in pneumatic controls is a simple low maintenance device that operates well in a harsh industrial environment and does not present an explosion risk in hazardous atmospheres. They were the industry controller amplifier for many decades until the advent of practical and reliable electronic high gain amplifiers. However, they are still used extensively for field devices such as control valve positioners and I to P and P to I converters.

UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 20
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The velocity of a freely falling object is
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 20

Whether explicitly stated or not, the value of the acceleration in the kinematic equations is -9.8 m/s/s for any freely falling object. If an object is merely dropped (as opposed to being thrown) from an elevated height, then the initial velocity of the object is 0 m/s.

UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 21
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Which of the following is a medium head turbine in which water flows inwards radially?
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 21

High head turbine: In this type of turbine, the net head varies from 150m to 2000m or even more, and these turbines require a small quantity of water. Example: Pelton wheel turbine.

Medium head turbine: The net head varies from 30m to 150m, and also, these turbines require a moderate quantity of water. Example: Francis turbine.

Low head turbine: The net head is less than 30m, and also, these turbines require a large quantity of water. Example: Kaplan turbine.

UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 22
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Two particles, each of mass m, collide head-on when their speeds are 2u and u. If they stick together on impact, find their combined speed in terms of u.
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 22

Conservation of linear momentum:

m1u1 + m2u2 = (m1 + m2) V

m(2u) + m(-u) = (2m) V

mu = 2mV

V = u/2

UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 23
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In the case of power transmission through pipes, maximum efficiency is
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 23

The efficiency of power transmission is given by

For maximum efficiency

We get

ηmax = 66.66%

UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 24
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Which of the following statements is not true for diamond?
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 24

Diamond is one of the 8 allotropes of carbon.

Properties of a diamond:

1. It is the hardest known material.

2. Diamond is non - metallic.

3. It has high thermal conductivity, i.e., it is a good conductor of heat.

4. It has a very low electrical conductivity, i.e., it is a terrible conductor of electricity, or it does not conduct electricity at all, and it is used as an insulator also.

UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 25
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The two reference fuels used for cetane rating are
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 25

 Cetane and alpha-methyl naphthalene are the reference fuels taken for cetane rating.

UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 26
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Navier-Stokes equations represent:
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 26

Navier-Stokes (N-S) equations represent mass conservation or continuity equation. The mass in the control volume can be neither created nor destroyed in accordance with physical laws. The conservation of mass is also expressed as Continuity Equation.

N-S equation also represents the momentum conservation equation.

UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 27
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Atmospheric pressure at ground level is approximately:
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 27

At ground level the atmospheric pressure is 101.32kPa=1.01 ×105Pa

1kgf=9.81N

1 kgf/cm2=9.81N/cm2=9.81×104N/m2=9.81×104Pa≃1×

105Pa

UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 28
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A man is climbing up a ladder up a ladder which is resting against a vertical wall. When he wasexactly half way up, the ladder started slipping . The path traced by the man is :
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 28

UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 29
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The motion of the particle is described by the following equations. x = t2 + 8t + 4 find the velocity when t = 2 sec
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 29

UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 30
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The following system of equations :

x - 2y + 3z = 3

2x + y - z = 9

-3x - 4y + 5z = - 8 has
Detailed Solution for UPSSSC JE Mechanical Paper 2 Mock Test - 5 - Question 30

The bottom row of the final matrix corresponds to the equation 0.x+0.y+0.z=7

which has no solution.

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