Mechanical Engineering Exam  >  Mechanical Engineering Tests  >  RRB JE Mock Test Series for Mechanical Engineering (ME) 2025  >  RRB JE ME (CBT II) Mock Test- 2 - Mechanical Engineering MCQ

RRB JE ME (CBT II) Mock Test- 2 - Mechanical Engineering MCQ


Test Description

30 Questions MCQ Test RRB JE Mock Test Series for Mechanical Engineering (ME) 2025 - RRB JE ME (CBT II) Mock Test- 2

RRB JE ME (CBT II) Mock Test- 2 for Mechanical Engineering 2024 is part of RRB JE Mock Test Series for Mechanical Engineering (ME) 2025 preparation. The RRB JE ME (CBT II) Mock Test- 2 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The RRB JE ME (CBT II) Mock Test- 2 MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for RRB JE ME (CBT II) Mock Test- 2 below.
Solutions of RRB JE ME (CBT II) Mock Test- 2 questions in English are available as part of our RRB JE Mock Test Series for Mechanical Engineering (ME) 2025 for Mechanical Engineering & RRB JE ME (CBT II) Mock Test- 2 solutions in Hindi for RRB JE Mock Test Series for Mechanical Engineering (ME) 2025 course. Download more important topics, notes, lectures and mock test series for Mechanical Engineering Exam by signing up for free. Attempt RRB JE ME (CBT II) Mock Test- 2 | 150 questions in 120 minutes | Mock test for Mechanical Engineering preparation | Free important questions MCQ to study RRB JE Mock Test Series for Mechanical Engineering (ME) 2025 for Mechanical Engineering Exam | Download free PDF with solutions
RRB JE ME (CBT II) Mock Test- 2 - Question 1

A submerged body will be in stable equilibrium if the centre of gravity is

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 1
Stability of unconstrained Submerged Bodies in Fluid

The equilibrium of a body submerged in a liquid requires that the body’s weight acting through its centre of gravity should be collinear with an equal hydrostatic lift acting through the centre of buoyancy.

  • Stable Equilibrium: If the body returns to its original position by retaining the originally vertical axis as vertical
  • Unstable Equilibrium: If the body does not return to its original position but moves further from it
  • Neutral Equilibrium: If the body neither returns to its original position nor increases its displacement further, it will simply adapt its new position

The relative position of the centre of gravity (G) and centre of buoyancy (B) of a body determines the stability of a submerged body.

Stable Equilibrium: B is above G

Unstable Equilibrium: B is below G

Neutral Equilibrium: B coincides with G

RRB JE ME (CBT II) Mock Test- 2 - Question 2

The sum of pressure head and elevation head is known as ________.

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 2

Bernoulli's equation along streamlines

Total Head (H) = Pressure head (P/ρg) + Kinetic head (v2/2g) + Potential head (z)

Total Dynamic Head (H) = Static Head ((P/ρg) + z) + Dynamic Head (v2/2g)

1 Crore+ students have signed up on EduRev. Have you? Download the App
RRB JE ME (CBT II) Mock Test- 2 - Question 3

Hydraulic grade line as compared to the centre line of conduct ______.

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 3

Bernoulli's equation along streamlines

Total head (H) = Pressure head (P/ρg) + Kinetic head (v2/2g) + Potential head (z)

Total head (H) = Hydraulic gradient line ((P/ρg) + z) + Kinetic head (v2/2g)

The line representing the sum of pressure head, datum head, and velocity head with respect to some reference line is known as total energy line (T.G.L).

The line representing the sum of pressure head and potential head or datum head with respect to some reference line is the hydraulic gradient line (H.G.L) which is always above compared to the centre line of conduct.

RRB JE ME (CBT II) Mock Test- 2 - Question 4

In a steady flow:

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 4
Streamlines are a family of instantaneously tangent curves to the velocity vector of the flow. These show the direction in which a massless fluid element will travel at any point in time.

Streaklines are the loci of points of all the fluid particles that have passed continuously through a particular spatial point in the past. Thus, dye steadily injected into the fluid at a fixed point extends along a streak line.

Pathlines are the trajectories that individual fluid particles follow. These can be thought of as "recording" the path of a fluid element in the flow over a certain period. The fluid streamlines will determine the direction the path takes at each moment in time.

For steady flow, path lines, streamlines, and streaklines coincide.

RRB JE ME (CBT II) Mock Test- 2 - Question 5

In a flow field, at the stagnation point _______.

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 5
A stagnation point is a point in a flow field where the local velocity of the fluid is zero. The Bernoulli equation shows that the static pressure is highest when the velocity is zero, and hence static pressure is at its maximum value at stagnation points. This static pressure is called stagnation pressure.
RRB JE ME (CBT II) Mock Test- 2 - Question 6

The depth of the centre of pressure (h) is given by the relation

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 6

Centre of pressure is the point of application of the total pressure force on the surface. It can be found by applying the principle of moments i.e. the sum of the moments of the resultant force about an axis is equal to the sum of the moments of the component force about the same axis.

RRB JE ME (CBT II) Mock Test- 2 - Question 7

The length of the divergent portion of Venturimeter in comparison to the ,convergent portion is

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 7
  • A venturi meter is a device used to measure the flow rate of a fluid of liquid flowing through a pipe.

  • The venturi meter always have a smaller convergent portion and a larger divergent portion.

  • This ensures a rapid converging passage and a gradual diverging passage in the direction of flow to avoid the loss of energy due to the separation.

  • In the course of flow through the converging part, the velocity increases in the direction of flow according to the principle of continuity, while the pressure decreases according to Bernoulli’s theorem.

  • The velocity reaches its maximum value, and pressure reaches its minimum value at the throat.

  • Subsequently, a decrease in the velocity and an increase in the pressure take place in the course of flow through the divergent part.

  • The angle of convergence ≈ 20°, Angle of divergence = 6° - 7° → It should be not greater than 7° to avoid flow separation.

RRB JE ME (CBT II) Mock Test- 2 - Question 8

When can a piezometer be not used for pressure measurement in pipes.

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 8
  • Piezometer is the simplest form of manometer used for measuring gauge pressure.

  • One end of this manometer is connected to the point where the pressure is to be measured, and the other end is open to the atmosphere.

  • As one end is open to the atmosphere so it can’t be used for measuring the gas pressure

  • Gas pressure cannot be measured by means of piezometers because a gas forms no free atmosphere surface and it can’t be used when large pressures in the lighter liquid are to be measured.

RRB JE ME (CBT II) Mock Test- 2 - Question 9

In centrifugal pump, cavitation is reduced by:

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 9

Cavitation is the phenomenon of the formation of vapour bubbles of a flowing liquid in a region where the pressure of the liquid falls below the vapour pressure of the fluid and sudden collapsing of these bubbles in the region of higher pressure.

In centrifugal pumps the cavitation may occur at the inlet of the impeller of the pump or at the suction side of the pumps, where the pressure is considerably reduced.

Hence if the pressure at the suction side of the pump drops below the vapour pressure of the liquid then cavitation may occur.

In order to determine whether cavitation will occur in any portion of the suction side of the pump, the critical value of Thoma’s cavitation factor (σ) is calculated.

H = Head developed by the pump

HS = Suction pressure head in m of water

HV = Vapour pressure head in m of water

hLS = Head lost due to friction in the suction pipe

If the value of σ is greater than σc (Critical cavitation factor), the cavitation will not occur in that pump.

So, the cavitation can be reduced by reducing the suction head.

RRB JE ME (CBT II) Mock Test- 2 - Question 10

The specific gravity of liquids is usually measured by means of a:

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 10
A hydrometer or areometer is an instrument that measures the specific gravity (relative density) of the liquids-the ratio of the density of the liquid to the density of water.

A hygrometer, also known as a psychrometer, is a device that is used to measure the humidity in the air.

A thermometer is a device that measures temperature or a temperature gradient.

A piezometer is a device used to measure liquid pressure in a system by measuring the height to which a column of the liquid rises against gravity.

RRB JE ME (CBT II) Mock Test- 2 - Question 11

Compression ratio in case of SI engine is in the range of

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 11

RRB JE ME (CBT II) Mock Test- 2 - Question 12

Sub-cooling in a vapour compression cycle

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 12
Subcooling is the process of cooling the liquid refrigerant below the condensing temperature for a given pressure.

The difference between the saturation temperature and the temperature of sub cooled liquid at that pressure is called the Degree of subcooling.

Subcooling is beneficial as it increases the refrigeration effect by reducing the throttling loss at no additional specific work input.

Also, subcooling ensures that only liquid enters the throttling device leading to its efficient operation.

Effect of Subcooling:

  • Increase the refrigeration effect
  • Work input remains same
  • Increase in COP

RRB JE ME (CBT II) Mock Test- 2 - Question 13

310 J of heat is required to raise the temperature of 2 mole of an ideal gas at constant pressure from 25°C to 35°C. The amount of heat required to raise the temperature of the same amount of gas through the same temperature range at constant volume is

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 13
Gas constant for ideal gas: R = 8.3 J/mol K

Heat transfer at constant pressure:

δQP = mcpΔT

310 = 2 × cp (35 – 25) = 20 cp

cp = 31/2 = 15.5 J/mol K

R = cp – cv ⇒ cp = cp – R = 15.5 – 8.3 = 7.2 J/mol K

Heat transfer at constant volume:

δQ = mcvΔT

δQ, = 2 × 7.2 × (35 – 25) = 144 J

RRB JE ME (CBT II) Mock Test- 2 - Question 14

Transient heat conduction means

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 14
Heat transfer problems are often classified as being steady (also called steady state) or transient (also called unsteady).

The term steady implies no change with time at any point within the medium, while transient implies variation with time or time dependence.

Therefore, the temperature or heat flux remains unchanged with time during steady heat transfer through a medium at any location, although both quantities may vary from one location to another.

During transient heat transfer, the temperature varies typically with time as well as position.

RRB JE ME (CBT II) Mock Test- 2 - Question 15

The inside and outside heat transfer coefficient of a fluid across a brick wall of 150 cm thickness and thermal conductivity of 0.10 W/mK are 30 W/m2K. The overall heat transfer coefficient (W/m2K) will be closer to

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 15

Given:

h1 = h2 = 30 W/m2K

L = 15 cm = 0.15 m

k = 0.10 W/mK

The overall heat transfer coefficient, U:

U = 15/226 = 0.07 W/m2K

U ≈ 0.1 ≈ k

RRB JE ME (CBT II) Mock Test- 2 - Question 16

Which of the following is incorrect regarding the first law of thermodynamics?

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 16

The first law of thermodynamics is a restatement of the law of conservation of energy. It states that energy cannot be created or destroyed in an isolated system; energy can only be transferred or changed from one form to another.

When heat energy is supplied to a gas, two things may occur:

  • The internal energy of the gas may change
  • The gas may do some external work by expanding

 

According to the first law of Thermodynamics:

δQ = δW + ΔU

When a system executes a process, the change in stored energy of the system is numerically equal to the net heat interaction minus the network interaction during the process:

ΔU = δQ – δW where U is the internal energy which is introduced by this law.

For the cyclic process: ΔU = 0

The first law of thermodynamics states that for a cyclic process, the cyclic integral of heat is equal to the cyclic integral of work.

The second law of thermodynamics introduces the concept of entropy.

RRB JE ME (CBT II) Mock Test- 2 - Question 17

A Carnot engine whose sink is at 300 K has an efficiency of 40%. How much should the source’s temperature be increased for the same sink to increase its efficiency by 50% of the original efficiency?

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 17

Concept:

The efficiency of Carnot Engine:

Calculation:

Case 1:

η1 = 40% = 0.4, TL1 = 300 K

Case 2:

If we need to increase the efficiency by 50% of original, efficiency = 40% + (50/100) × 40% = 40% + 20% = 60%

η2 = 60% = 0.6, TL2 = TL1 = 300 K

Increase in source temperature:

ΔT = TH2 – TH1 = 750 – 500 = 250 K

RRB JE ME (CBT II) Mock Test- 2 - Question 18

The internal energy of a gas obeying van der Waals, equation (p + a/V2)(V - b) = RT depends on its

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 18

internal energy: U = u (T, V)

For a real gas, the internal energy if a function of both the temperature and the specific volume.

That is,

U = f(T,u)

For ideal gas (∂U/∂V)T = 0

∴ U = U (T) only

Important Points

For an ideal gas (no intermolecular interactions and no molecular volume), appropriate equation of state would be: PV = nRT ⇒ V = nRT/P i.e. V = f(T,P,n)

There are many equations of state describing real gases. These equations take in consideration molecular volume and interactions. The most well-known such equation is the Van der Waals equation.

The internal energy of an ideal gas is a function of temperature only and is independent of pressure and volume. u = u(T)

RRB JE ME (CBT II) Mock Test- 2 - Question 19

The tendency of knocking in CI engines is reduced by

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 19
Knocking in CI Engine

Knocking in CI engines 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; and 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.

Hence a good CI engine fuel should have a short ignition lag so that it will ignite more easily.

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

In order 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.

Important Point :

  • In the SI engine, knocking occurs near the end of combustion whereas in the CI engine, knocking occurs near the beginning of combustion.

RRB JE ME (CBT II) Mock Test- 2 - Question 20

The ratio of radii of two spheres of metal is 1 : 2. The ratio of the amounts of radiation emitted by them at same temperature will be

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 20

The radiation energy emitted by a body per unit time is given by

Eb = ϵAσT4

Where ϵ is emissivity of the body.

σ = The Stefan – Boltzmann constant = 5.67 × 10-8 W m-2K-4

Calculation:

For same material and same temperature: E ∝ A

RRB JE ME (CBT II) Mock Test- 2 - Question 21

Arrange the thermal conductivity of the following materials in ascending order:

copper, mercury, silver, water

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 21

thermal conductivity of a material can be defined as the rate of heat transfer through a unit thickness of the material per unit area per unit temperature difference. The thermal conductivity of a material is a measure of the ability of the material to conduct heat. A high value for thermal conductivity indicates that the material is a good heat conductor, and a low value indicates that the material is a poor heat conductor or insulator. The thermal conductivities of some common materials at room temperature are given as

Kwater < Kmercury < Kcopper < KSilver

RRB JE ME (CBT II) Mock Test- 2 - Question 22

An ideal gas is compressed to half its initial volume by means of several processes. Which of the processes results in the maximum work done on the gas?

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 22
Polytropic Process is represented by

PVn = C

  • n = 0 ⇒ P = C ⇒ Constant Pressure Process (Isobaric Process)
  • n = 1 ⇒ PV = C ⇒ Constant Temperature Process (Isothermal process)
  • n = γ ⇒ PVγ = C ⇒ Adiabatic Process
  • n = ∞ ⇒ V = C ⇒ Constant Volume Process (Isochoric process)

Area under P-V diagrams represents the work done.

For compression between two volumes, we can see that area is maximum for the adiabatic process.

So work done on the gas will be maximum for the adiabatic process.

RRB JE ME (CBT II) Mock Test- 2 - Question 23

When the wavelength corresponding to maximum emission of radiation is halved, the radiant power of the body becomes

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 23

Concept:

Wien’s displacement law

For a black body emissive spectrum, the wavelength λmax giving the maximum emissive power at a temperature is inversely proportional to the absolute temperature.

λmax = C/T

λmaxT= C

λ1T1 = λ2T2

The radiation energy emitted by a black body per unit time is given by:

Eb = σAT4

The radiation energy emitted by a body per unit area per unit time is given by:

Eb = ϵσT4

Calculation:

E2 = 16 × E1

RRB JE ME (CBT II) Mock Test- 2 - Question 24

Which of the following statements is CORRECT for the modulus of resilience?

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 24
Proof resilience is defined as the maximum energy that can be absorbed within the elastic limit, without creating a permanent distortion.

The modulus of resilience is defined as the maximum energy that can be absorbed per unit volume without creating a permanent distortion (i.e. The proof of resilience per unit volume of a body is called modulus of resilience).

RRB JE ME (CBT II) Mock Test- 2 - Question 25

The equivalent stiffness of two springs of stiffness S1 and S2 joined in series is given by:

RRB JE ME (CBT II) Mock Test- 2 - Question 26

What will be the thermal stress (in MPa) developed in a rod having a diameter of 4 cm and length of 2 m. It experiences heating from temperature 50°C to 200°C. The coefficient of thermal expansion is α = 10 × 10-6/°C and young’s modulus is 250 GPa?

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 26
Thermal stresses are the stresses induced in a body due to a change in temperature.

σt = Eɑ△T = 250 x 109 x 10 x 10-6 x 150 = 375 MPɑ

RRB JE ME (CBT II) Mock Test- 2 - Question 27

Calculate the total angle of twist for a stepped shaft which is subjected to the torque (T) as shown in the figure below.

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 27

RRB JE ME (CBT II) Mock Test- 2 - Question 28

Which of the following statements is INCORRECT?

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 28

The shear force is of at point is given by the slope of the bending moment diagram at that point.

Where,

V = shear force

M = bending moment

The value of distributed load at any point is equal to the slope of the shear force diagram at that point.

RRB JE ME (CBT II) Mock Test- 2 - Question 29

A shaft is subjected to a bending moment, M and a torque, T simultaneously. The ratio of the maximum bending stress to maximum shear stress developed in a shaft is

RRB JE ME (CBT II) Mock Test- 2 - Question 30

Modulus of Rigidity is related to

Detailed Solution for RRB JE ME (CBT II) Mock Test- 2 - Question 30
The ratio of shearing stress to the corresponding shearing strain is called the shear modulus of the material and is represented by G. It is also called the modulus of rigidity.

Modulus of rigidity is associated with the change in the shape of a body.

Note:

Young’s Modulus of Elasticity is associated with the change in the length of a body.

Bulk Modulus of Elasticity is associated with the change in the volume of a body.

View more questions
158 tests
Information about RRB JE ME (CBT II) Mock Test- 2 Page
In this test you can find the Exam questions for RRB JE ME (CBT II) Mock Test- 2 solved & explained in the simplest way possible. Besides giving Questions and answers for RRB JE ME (CBT II) Mock Test- 2, EduRev gives you an ample number of Online tests for practice

Top Courses for Mechanical Engineering

Download as PDF

Top Courses for Mechanical Engineering