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RRB JE ME CBT 2 Full Test 1 - Railways MCQ


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30 Questions MCQ Test RRB JE (Railways Junior Engineer) CBT Mock Test Series 2025 - RRB JE ME CBT 2 Full Test 1

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RRB JE ME CBT 2 Full Test 1 - Question 1

Work done is zero for the following process

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 1

For constant volume process
W = ∫PdV
Since
dV = 0
so W = 0 for constant volume process.
For free expansion work done is zero. W = 0
The process in which high-pressure fluid is converted to low pressure by using a throttle valve is Throttling. In the throttling process enthalpy remains constant, work done is zero.

RRB JE ME CBT 2 Full Test 1 - Question 2

In a reversible adiabatic process the ration  is equal to:

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 2

Relationship between P - V - T for a reversible adiabatic process:

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RRB JE ME CBT 2 Full Test 1 - Question 3

Which of the following engine can be associated with heterogeneous combustion?

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 3

IC Engines may be classified based on the state of air-fuel mixture present at the time of ignition in the engine cycle, the type of ignition employed and the nature of combustion process after ignition of the air-fuel mixture.

Physical State of Mixture

  • Homogeneous Charge:
    • Premixed outside (conventional gasoline and gas engines with fuel inducted in the intake manifold)
    • Premixed in-cylinder: In- cylinder direct injection and port fuel injection
  • Heterogeneous Charge (Diesel Engine)

Ignition Type

  • Positive source of Ignition e.g., spark ignition
  • Compression ignition

Mode of Combustion

  • Flame propagation
  • Spray combustion

The combustion process in a diesel engine is heterogeneous—that is, the fuel and air are not premixed prior to initiation of combustion. Because of heterogeneous material low speed and low pressure is generated as compared to SI engine where homogenous mixture of fuel is used

RRB JE ME CBT 2 Full Test 1 - Question 4

A frictionless heat engine can be 100% efficient only if the exhaust temperature is

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 4


Efficiency of a Reversible Heat Engine:

η = 1 when TL = 0 for any TH
A heat engine can work on 100% efficiency when there is only heat input and no heat loss.
As W = QH – QL
All the heat supplied will be used in producing the work and there will be no losses.
All the temperature scale is taken in Kelvin.

RRB JE ME CBT 2 Full Test 1 - Question 5

In an Otto cycle, the heat addition and heat rejection take place at

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 5

Otto cycle: It is the standard air cycle used in the spark ignition (SI) engines or petrol engines.

Different processes of the Otto cycle are as follows:
Process 1-2: Isentropic compression
Process 2-3: Constant volume heat addition
Process 3-4: Isentropic expansion
Process 4-1: Constant volume heat rejection. 

RRB JE ME CBT 2 Full Test 1 - Question 6

The rate of heat transfer through a hollow cylinder of inner and outer radii r1 and r2, respectively, depends on

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 6

For hollow cylinder:


The rate of heat transfer through a hollow cylinder of inner and outer radii r1 and r2, respectively, depends on ratio of radii (r2/r1).

Note:

RRB JE ME CBT 2 Full Test 1 - Question 7

Which of the following quantity is not a property of a system?

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 7

Any identifiable/observable characteristic of a system by which the physical condition of the system may be described is called property of the system.

Some familiar properties are pressure, temperature, volume and mass.

Properties may be divided into two categories i.e. extensive (dependent of mass) and intensive (independent of mass) properties.

RRB JE ME CBT 2 Full Test 1 - Question 8

In a closed system, a gas undergoes a quasi-equilibrium process as per the law P = (-4V + 10) N.m2 and the volume of the gas, V changes from 1 m3 to 2 m3. The work done will be

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 8


As W is positive to it is work output.

Note:

  • Work done on the system is negative
  • Work done by the system is positive
RRB JE ME CBT 2 Full Test 1 - Question 9

Maximum work by an expansion of a gas in a closed system is possible when the process takes place at constant

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 9


As a work done is the area under the PV diagram represents work done which is maximum for isobaric (P = c) process.

RRB JE ME CBT 2 Full Test 1 - Question 10

An ideal air compressor cycle (with clearance) on p-v diagram can be represented by ________ processes.

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 10

In practical design of compressors, some clearance is required between the cylinder and piston to prevent hitting of piston to crown of the cylinder. A compressor cycle with clearance is consists of two adiabatic processes and two isobaric processes.

RRB JE ME CBT 2 Full Test 1 - Question 11

Polytropic index n is given

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 11


RRB JE ME CBT 2 Full Test 1 - Question 12

The rate of radiation of a black body at 0°C is E J/sec. The rate of radiation of this black body at 273°C will be

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 12

The radiation energy emitted by a black body per unit time and per unit surface area is given by
Eb = σT4 (W/m2)
where Stefan – Boltzmann constant, σ = 5.67 × 10-8 W/m2 K4, T is absolute temperature of the surface in Kelvin.
Eb = Total emissive power of a black body
For a non-black surface having an emissivity ϵ, it follows that
Eb = ϵ σT4
Calculation:


E2 = 16 E1 = 16 E

RRB JE ME CBT 2 Full Test 1 - Question 13

Refrigeration is based on _________

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 13

Refrigeration is based on Clausius statement of second law of thermodynamics. This law states that it is impossible to develop a device which works on a cycle and transfers heat from lower temperature to higher temperature without any external energy input. On the other hand, heat engine works on the Kelvin Planck’s statement of second law of thermodynamics.

RRB JE ME CBT 2 Full Test 1 - Question 14

A Carnot cycle refrigerator operates between 250 K and 300 K. Its coefficient of performance is

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 14



For Carnot cycle refrigeration:

RRB JE ME CBT 2 Full Test 1 - Question 15

A body subjected to coplanar non-concurrent forces will remain in a state of equilibrium if

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 15

Equations of equilibrium for Non-concurrent force System: A non-concurrent force systems will be in equilibrium if the resultant of all forces and moments is zero.

  • ΣFx = 0, ΣFy = 0 and ΣM = 0

Equations of equilibrium for concurrent force System: For the concurrent forces, the lines of action of all forces met at a point and hence the moment of those force about that point will be zero or ΣM = 0 automatically.

  • ΣFx = 0 and ΣFy = 0
RRB JE ME CBT 2 Full Test 1 - Question 16

The maximum frictional force, which comes into play, when a body just begins to slide over the surface of the other body, is known as

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 16

It has been observed that when a body, lying over another body, is gently pushed, it does not move because of the frictional force, which prevents the motion.

If the applied force exceeds this limit, the force of friction cannot balance it and the body begins to move, in the direction of the applied force. This maximum value of frictional force, which comes into play, when a body just begins to slide over the surface of the other body, is known as limiting friction.

Static friction is the friction experienced by a body when it is at rest or it is the friction when the body tends to move.

RRB JE ME CBT 2 Full Test 1 - Question 17

The loss of kinetic energy, during inelastic impact of two bodies having masses m1 and m2, which are moving with velocity v1 and v2respectively, is given by

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 17
  • Momentum is conserved in all collisions.
  • In elastic collision, kinetic energy is also conserved.
  • In inelastic collision, kinetic energy is not conserved. In perfectly inelastic collision, objects stick together after collision.


Conservation of momentum:
m1v1 + m2v2 = (m1 + m2) vf

Kinetic energy before collision:

Kinetic energy after collision:



Change in kinetic energy:
 


 
Loss of kinetic energy:

RRB JE ME CBT 2 Full Test 1 - Question 18

Forces P, 3P, 2P and 5P act along the sides taken in order of a square. The magnitude of the resultant is

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 18


ΣFx = 3P – 5P = -2P
ΣFx = P – 2P = -P
 

RRB JE ME CBT 2 Full Test 1 - Question 19

A block A is released from the top of smooth inclined plane and slides down the plane. Another block B is dropped from the same point and falls vertically downwards. Which one of the following statements will be true if the friction offered by air is negligible?

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 19

Consider a block of mass m lying on a frictionless inclined plane of length AB = L, height = h and angle of inclination θ.

When the block is released, it moves down the plane under a force mg sin θ.
Hence the acceleration of the block down the plane is
a = g sin θ
If the block starts from rest from point A, then its velocity when it reaches the bottom B is given by
v2 – u2 = 2as
v2 – 0 = 2aL
 
For the block which falls vertically downward of height, h:

So, speed of both the blocks will be same.

RRB JE ME CBT 2 Full Test 1 - Question 20

Two circular discs have masses in the ratio 1 : 2 and radii in the ratio 2 : 1. The ratio of their moment of inertia about the diameter is

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 20


Moment of inertia of a disc about its diameter:

Moment of inertia of a disc about its centre:

Calculation:

Two circular discs have masses in the ratio 1: 2 and radii in the ratio 2: 1. The ratio of their moment of inertia about the diameter is

RRB JE ME CBT 2 Full Test 1 - Question 21

A thin circular ring of mass 100 kg and radius 2 m resting on a smooth surface is subjected to a sudden application of a tangential force of 300 N at a point on its periphery. The angular acceleration of the ring will be

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 21


Consider a mass m rotating around an axis, a distance r away.
Newton’s second law: F = ma
We know a = αr
(a = tangential acceleration; α = angular acceleration)
F = ma = mrα
Calculation:
F = ma = m r α
300 = 100 × 2 × α
α = 1.5 rad/s2

RRB JE ME CBT 2 Full Test 1 - Question 22

Polar moment of inertia of an equilateral triangle of side ‘x’ is given by

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 22


Consider an isosceles triangle:

For an equilateral triangle:

 

RRB JE ME CBT 2 Full Test 1 - Question 23

If one of the walls moves in the direction of flow with uniform velocity while the other wall is stationary, then the resulting flow between parallel walls is called ______.

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 23

Couette flow is the flow of a viscous fluid in the space between two surfaces, one of which is moving tangentially relative to the other. The configuration often takes the form of two parallel plates or the gap between two concentric cylinders.

In plug flow, the velocity of the fluid is assumed to be constant across any cross-section of the pipe perpendicular to the axis of the pipe. The plug flow model assumes there is no boundary layer adjacent to the inner wall of the pipe.

Stokes flow or creeping flow is a type of fluid flow where advective inertial forces are small compared with viscous forces. The Reynolds number is low i.e. Re≪1. This is a typical situation in flows where the fluid velocities are very slow, the viscosities are very large.

RRB JE ME CBT 2 Full Test 1 - Question 24

The resultant upward pressure of the fluid on an immersed body is called

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 24

Buoyancy or upthrust, is an upward force exerted by a fluid that opposes the weight of an immersed object. Archimedes' Principle states that the magnitude of the buoyant force on an object is equal to the weight of the fluid it displaces.

RRB JE ME CBT 2 Full Test 1 - Question 25

Mercury is considered as a superior barometric fluid due to its

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 25

Barometer is used to determine the local atmospheric pressure.

Mercury is employed in the barometer because its density is sufficiently high for a relative short column to be obtained and also because it has very small vapour pressure at normal temperature. High density scales down the pressure head (h) to represent same magnitude of pressure in a tube of smaller height.

Considered a small diameter glass tube with mercury at normal temperature is inverted within a small tank of mercury, an equilibrium will be reached.

PB = ρgh + Pv

Pv is the pressure exerted on the free surface of the mercury within the glass tube.

The vapour pressure of mercury at normal temperature and atmospheric pressure is only 0.16 Pa while the quantity ρgh is equal to approximately 105 Pa i.e. an insignificant fraction can be neglected.

Due to its small vapour pressure mercury used to be the standard fluid for measuring absolute pressures.

RRB JE ME CBT 2 Full Test 1 - Question 26

The velocity of fluid particle at the centre of the pipe section is:

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 26


Velocity distribution of a flow of viscous fluid through circular pipe:

The velocity is maximum when r = 0 i.e. at the centre of pipe.

Average velocity:

RRB JE ME CBT 2 Full Test 1 - Question 27

The concept of stream function which is based on the principle of continuity is applicable to

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 27

It is defined as the scalar function of space and time, such that its partial derivative with respect to any direction gives the velocity component at right angles to that direction.

It is denoted by ψ and defined only for two-dimensional flow.

 

Properties of Stream function:

  • If stream function exists, it is a possible case of fluid flow which may be rotational or irrotational
  • If the stream function satisfies the Laplace equation i.e.  it is a case of irrotational flow

Recommended Similar Important Concept:

Velocity Potential Function:

It is defined as the scalar function of space and time, such that its negative derivative with respect to any direction gives the velocity in that direction.

It is denoted by ϕ and defined for two-dimensional as well as three-dimensional flow.

 

Properties of Stream function:

  • If velocity potential function exists, the flow should be irrotational
  • If the velocity potential function satisfies the Laplace equation i.e. it is a case of steady incompressible irrotational flow
RRB JE ME CBT 2 Full Test 1 - Question 28

 

All the terms of energy in Bernoulli’s equation  have dimension of

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 28

From the law of dimensional homogeneity, the constant must have the same dimensions as the other additive terms in the equation.
Thus P/γ and v2/2g will have same dimension as z i.e. Length.

All the terms of Bernoulli’s equation:

P/γ = P/ρg = Pressure energy per unit weight of fluid or pressure head
v2/2g = Kinetic energy per unit weight or kinetic head
z = Potential energy per unit weight or potential head

RRB JE ME CBT 2 Full Test 1 - Question 29

A reservoir containing water has two orifices of the same size at depths of 4 m and 9 m below the free surface of water. The ratio of discharges through these orifices is

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 29



For same orifice at different height:
Q ∝ √h

RRB JE ME CBT 2 Full Test 1 - Question 30

Shear stress for a general fluid motion is represented by where n and A are constants. A Newtonian fluid is given by:

Detailed Solution for RRB JE ME CBT 2 Full Test 1 - Question 30



For Newtonian Fluid: A = 0 and n = 1 (Example: Air, Water, Glycerin)

For Bingham Plastic: A = τ0 and n = 1(Fluid does not move or deform till there is a critical stress. Example: Toothpaste)

For Dilatant: A = 0 and n > 1 (Fluid starts ‘thickening' with increase in its apparent viscosity. Example: starch or sand suspension or shear thickening fluid)

For Pseudo plastic: A = 0 and n < 1 (Fluid starts ‘thinning' with increase in its apparent viscosity. Example: Paint, polymer solutions, colloidal suspensions or shear thinning fluid)

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