ESE (ME) Paper II Mock Test - 2 Free Online Test 2026

Concept:
Circular pitch is the distance measured along the pitch circle between two similar points on adjacent teeth.
P_C = πD/T
where D is pitch diameter and T is the number of teeth on the gear.
Pitch diameter (D):
Pitch diameter or pitch circle diameter is the diameter of the pitch circle. The size of the gear is usually specified by the pitch circle diameter.
Calculation:
Given:
Pitch diameter (D) = 200 mm and Teeth (T) = 20
P_C = πD/T = π x 200/20 = 31.4 mm

• The automatic expansion valve maintains a constant pressure in the evaporator.
• The constant pressure characteristics of the valve result from the interaction of two opposing forces, the evaporator pressure, and the spring pressure.
• The evaporator pressure exerted on one side of the bellows or diaphragm acts to move the valve in a closing direction, whereas the spring pressure, acting on the opposite side of the bellows or diaphragm acts to move the valve in an opening direction. When the compressor is running, the valve functions to maintain the evaporator pressure in equilibrium with the spring pressure.

Concept:
Homogeneous transformation Matrix:
In 3-D space, a physical point is located and if we want to change from one coordinate to another
frame then we need to use 4 X 4 homogenous transformation matrix.
It consists of four sub-matrices,
Generally given as
, where

• 3 × 3 submatrix R is a rotation matrix.
• 3 × 1 column vector P is a translation matrix.
• 1 × 3 ηT is a perspective vector.
• Scalar σ is a non-zero scale factor set to unity.

In case of pure translation , it takes the form of

The new vector will be
V_N = T × V_o;
Where V_o - old vector, V_N - new vector;
Calculation:
Given vector is V_o = 5i + 2j + 2k, x = 5, y = 4, z = 2;
Now the transformation matrix will be

Now the new vector will be

Concept:

For solving this question, you should have clear idea of T-S and p-h diagram, for vapour compression refrigeration cycle.

Figure: Cycle 1-2-3-4’-1; Ideal vapour compression refrigeration cycle (T-S plot)
Processes
1-2: Isentropic compression
2-3: Heat rejection
3-4: Isentropic expansion
3-4’: Isenthalpic expansion (Irreversible process)
4-1: Evaporation (Heat absorption)
We know that area under any process in T-S plot represents heat interaction
i.e.
When expansion process is isenthalpic (3-4’), heat absorption (refrigeration effect) is represented by area 4’-5’-6-1.
When expansion process is isentropic (3-4), refrigeration effect is represented by area 4-5-6-1.

…1)

Since, work input for both cycles is same (Process 1-2).
Area (4-5-6-1) > Area (4’-5’-6-1)
⇒ RE (Isentropic expansion) > RE (Isenthalpic expansion)
⇒ COP (isentropic expansion) > COP (isenthalpic expansion), {From 1)}
So, option 1) is correct
This can also be proved by drawing p-h chart

Figure: Pressure – enthalpy (p-h) chart for ideal VCR cycle
1-2: Isentropic compression
2-3: Heat rejection
3-4’: Isenthalpic expansion
3-4: Isentropic expansion
4-1: Heat addition (Evaporation)
COP = RE/W
For original cycle: (Isenthalpic expansion:3-4’)
RE = h₁ - h'₄, W = h₂ - h₁ ⇒ COP|_original = h₁ - h'₄/h₂ - h₁  …3)
For modified cycle: (Isentropic expansion) (3-4)
RE = h₁ - h'₄, W = h₂ - h₁ ⇒ COP|_Modified = h₁ - h'₄/h₂ - h₁  …4)

{∵ h₁ – h₄ > h₁ – h₄’}
⇒ COP|_Modified > COP|_original
Key Points:
Remember T-S and p-h plot for vapour compression refrigeration cycle. Also understand the nature of constant entropy lines in p-h chart. Study the effect of various parameters like evaporator pressure, condenser pressure on COP.

Concept:
Pressure in static fluid:

• For incompressible fluids, the variation of pressure in the vertical direction in a static fluid is given by dp/dh = pg
• In pressure measurement, we consider the downward direction as positive.​
• Absolute pressure is the sum of atmospheric and gauge pressure P_m(abs) = P_atm + ρgh

Calculation:
Given:
h = 3 m, ρ_liquid = 1.53 x 10³ kg/m³
h_atm = 750 mm of mercury (Hg) = 0.75 m,
specific gravity ,S_Hg = 13.6, ρ = 13.6 x 1000 kg/m3
Patm = ρgh_atm = 13.6 x 1000 x 9.81 x 0.75 = 100,062 N/m²
P_abs = P_atm + P_gauge
P_gauge = (ρgh)_liquid = 1.53 x 10³ x 9.81 x 3 = 45027.9 N/m².
Pabs = Patm + Pgauge = 100062 + 45027.9 = 145090 N/m².

The pressure drop in a laminar flow through a circular pipe is given by:

where μ = viscosity of fluid, L = length of pipe, D = diameter of pipe, V = average velocity of flow
After rearranging

for a small length ∂x, the pressure gradient is
∴ Equation 1 becomes,

Average velocity is:

Concept:

The main components of a solar water heater are the solar collector, storage, and heat distribution.

Several configurations differ on the heat transport between the solar collector and the storage tank, as well as on the type of freeze protection.

The most successful solar heaters are the integrated collector and storage (ICS), Thermosiphon, drain-back, and drain-down systems

The ICS and thermosiphon are passive solar water heaters where fluid circulation occurs by natural convection.

In open-loop systems, the water that is pumped through the collectors is the same hot water to be used. These systems are not recommended for sites where freezing occurs. These active open-loop systems are called drain-down systems and they can operate in either manual or automatic mode

Drain-back system is a closed loop active solar heater.

Drain-down refers to draining the collector fluid out of the system; drain-back refers to draining the collector fluid back into the storage tank.

Concept:
Hydrostatic Forces on Submerged Surface

I_G = bd³/12 (For rectangular plate)
I_G = πd⁴/64 (For circular plate)
Here,
A = Area of surface touching fluid = b × d
I_G = Area moment of inertia about centroidal axis.
= Vertical distance of C.O.G. of body from free surface
ω = Specific weight
θ = Angle at which the surface is inclined with horizontal
Calculation:

The hydrostatic press. Distribution on gate AB,

Magnitude of hyd. Force (F_H) 

 = 0.75 m (CG of sq. plate)
F_H = 1000 × 9.81 × (1.5)² × 0.75 = 1687.5 × 9.81 N
Force will act at a distance of 

Taking moment about hinge A
F_H × 0.5 = F × 0.75
⇒ 1687.5 × 9.81 × 0.5 = F × 0.75

Concept:
When density of the body is less than that of the water, it floats. According to Archimedes' principle the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces.
Weight of the body = Immersed volume x Density of water.
Calculations:

Weight of cube = Buoyancy Force
Buoyancy (B) = Immersed volume × density of water
Buoyancy = Area of cross section × Immersed depth × 10 kg/m³ = 2 m × 2 m × 1 m × 10 = 40 kN
Weight of cube = 40 kN

If the water is not supposed to freeze then its enthalpy should be higher than that of 0°C and it should also carry the heat which it is losing into the atmosphere to ensure that the water is not freeze

Energy Equation:

Concept:
Domestic electrolux refrigerator:

• This type of refrigerator is also called the “Three-fluids absorption system”. The three fluids used in this system are ammonia, hydrogen, and water.
• The “ammonia” is used as a refrigerant because it possesses most of the desirable properties.
• The “hydrogen” being the lightest gas, is used to increase the rate of evaporation (the lighter the gas, the faster is the evaporation) of the liquid ammonia passing through the evaporator.
• The hydrogen is also non-corrosive and insoluble in water. This is used in the low-pressure side of the system.
• The “water” is used as a solvent because it has the ability to absorb ammonia readily.
• The ammonia liquid leaving the condenser enters the evaporator and evaporates into the hydrogen at the low-temperature corresponding to its low partial pressure.

Explanation:
For statement 1:
Role of Hydrogen:
By the presence of hydrogen, it is possible to maintain uniform total pressure throughout the system and at the same time permit the refrigerant to evaporate at low temperature in the evaporator corresponding to its partial pressure.
Thus, the condenser and evaporator pressures of the refrigerant are maintained as below:

• In the condenser only ammonia is present, and the total pressure is the condensing pressure.
• In the evaporator hydrogen and ammonia are present. Their relative masses are adjusted such that the partial pressure of ammonia is the required evaporator pressure.
• Thus, the partial pressures of the refrigerant vapor (ammonia) must be low in the evaporator and higher in the condenser.

For statement 2:

• Due to the presence of H₂ in the low-pressure side of the system, the pressure of NH₃ will be below that of NH₃ on the condenser side. Thus, the NH₃ can evaporate at low pressure. So, statement 2 is correct.

The most important property of nano material is that they have large surface to volume ratio due to this most of the atoms lies on the surface of nano material. Since they have high surface to volume ratio, they are highly reactive.

→ Nano material have reduced imperfection compare to bulk material due to small size.

Application of Nano-material:

1) Fuel cell

2) Catalysis

3) Medicine

4) Biomedical engineering

5) Sensors

6) Rechargeable batteries

7) Next generation computer chips

From a functional and safety point of view following demands are put on the behaviour of friction materials-

• High coefficient of friction
• Stable coefficient of friction irrespective of temperature, velocity, pressure, humidity, wear, corrosion, and water spraying
• The low wear rate of the friction material and long life
• The low wear rate of the disc
• Smooth braking without noise and vibrations
• Regeneration of original properties after severe braking and with ageing
• Environment-friendly raw materials
• Low cost

A binder holds different components of a friction material together. Thermosetting phenolic resins are usually used as binders, with the addition of rubber for increased damping properties. The binder resin strongly affects the fade and wear resistance of friction material. The low bonding strength between constituent particles can improve the wear characteristics of the brake system.

The wear rate decreases with an increase in contact pressure between the brake surfaces under dry conditions while it increases with an increase in contact pressure under wet conditions.

On the other hand, the wear rate should be low at low temperatures but with an increase in temperature wear rate also increases because asperity present in contacting surfaces gets distributed with the influence of temperature and pressure. These deformed and broken asperity produce wear debris causing more heat generation on the surface.

In the case of the sliding velocity of the brake surface interface, the wear rate increases continuously with increasing sliding velocity. The rate of wear is slow at the medium velocity of sliding off the surface of the brake.

ABC analysis -

• A-B-C analysis is a basic analytical management tool, which enables any store manager to expand its effects and energy where the results will be best. This analysis is purely based on the Selective Management Principle or Pareto's Law.
• It is also popularly known as "Always Better Control" or the alphabetical approach and has universal application in many areas of human endeavour.
• A-B-C analysis does not depend on the unit cost of the items but only on their annual usage and not on their importance because all items are important. Analysts commonly classify inventory into three categories 'A', 'B', and 'C'.
• The 'A' items have high annual usage in terms of money investment. 'B' items are average, while 'C' items have a low value of usage.

Every item in inventory is ranked and listed in order of items' annual value of usage. At the top of the list would be the high-value items, followed by a longer list of medium-value items, and finally a long list of Iow-value items.
The normal items in most organizations show the following pattern:
1) 10 percent to 15 percent of the top number of items account for about 70 percent of the total consumption value. These items are called 'A' items.
2) 20 percent to 25 percent of the number of items account for 20 percent of the total consumption value. These items are called 'B' items.
3) The remaining 65 percent to 70 percent of items account for the balance of 10 percent of the total issue value. These items are called 'C' items.

VED analysis -

• The limitation of the A-B-C analysis is that it is based only on monetary value and the rate of consumption of the items. Sometimes, particularly in a hospital, an item of low monetary value and consumption (e.g. Injection Adrenaline, Anti-Snake Venom, etc.,) may be very vital or even life-saving. Their importance cannot be overlooked simply because they do not appear in A category of inventory.
• V-E-D analysis is based on critical values and shortage costs of the item. Based on their criticality, the items could be classified into three categories -Vital, Essential, and Desirable.
• V-E-D classification is applicable to a large extent in spare parts management. Stocking a spare part is based on strategies different from those of raw materials.

1) Vital items: There are several vital items in the inventory of a hospital, which could make difference between life and death. Such items should always be stocked in sufficient quantity to ensure their constant availability. Top management should control this group of items.

2) Essential items: The shortage of such items can be tolerated for a short period. If these items are not available for a few days or a week, the functioning of the hospital can be adversely affected (drugs like Antibiotics, etc.). Top middle level management ' should preferably control these items.

3) Desirable item: The shortage of these items will not adversely affect patient care or hospital function even if the shortage is prolonged (Items like vitamins). Middle or lower-level management should control desirable items.

Concept:
Volume Flow Rate in case of Laminar Flow through Parallel Plate is given by

where ΔP = Pressure Difference, h = Distance between parallel plates, b = Width of Plate, L = Length of Plate,  μ = Viscosity
Calculation:
Given:
ΔP = 0.3 N/cm² = 0.3 × 10⁴ N/m2, h = 50 mm = 0.05 m, b = 200 mm = 0.2 m, L = 1.2 m,  μ = 10 Poise = 1 Pa.s

Q = 5.208 × 10^-3 m³/sec = 5.208 L/s

Concept:
Sensible heat (S.H):

• Sensible heat is the heat which increases or decreases the temperature of the body.
• Mathematically, it is expressed as:

S.H = m × c × ΔT

Where m = mass, c = specific heat, ΔT = Temperature difference

Latent Heat (L.H):

• Latent heat is the heat which is absorbed or released during phase change.​
• Mathematically, it is expressed as:

L.H = m × LH

Where LH = Latent heat

Total heat, Q = SH + LH

And refrigeration capacity (in KW) = Q/Time
Calculation:
Given:
m = 10 tons = 10 × 1000 Kg, C_ice = 1.94 KJ / Kg-K, Latent heat of water = 335 KJ / Kg, C_water = 4.1868 KJ / Kg-K, time, t = 24 hours = 86400 seconds

Sensible heat = m × c × ΔT = 10 × 10³ × [4.186 × (30 - 0) + 1.94 × {0 - (-10)}] = 10 × 10³ × 144.98 KJ = 1449.8 MJ

Latent heat = m × LH = 10 × 10³ × 335 KJ = 3350 MJ

∴ Total heat, Q = SH + LH = 1449.8 + 3350 = 4799.8 MJ

So, Refrigeration capacity

Concept:

In a fully developed laminar flow through circular pipes velocity distribution according to Hagen-Poiseuille’s equation is

V is maximum at r = 0,

Average velocity is given by

⇒ V_avg = V_max/2

Concept:

Transmissibility is defined as the ratio of force transmitted to the shaking force.

where, F_t = Transmitted force, F_o = shaking force

Calculation:

Given:

m = 35 kg

N = 480 rpm

ω = 2πN/60 = 50.27 rad/s

As Damper is not used

ζ = 0

F_t = (0.1) F_o

The three springs will be in parallel, therefore K_eq = 3K

For zero damping, the Transmissibility equation gets reduced to:

K = 2677.76 N/m = 2.67 N/mm ≃ 2.7 N/mm

Concept:

The Maximum Shear Stress Theory was given by Guest and Tresca,

i.e. Maximum shear stress τ_max = larger of |(σ₁−σ₂),(σ₂−σ₃),(σ₃−σ₂)| ≤ S_yt/N

Tensile yield strength (Proportionality limit) = S_yt = 250 MPa

Calculation:

Given:

The three principal stresses are σ₁ = 120 MPa, σ₂ = 60 MPa and σ₃ = -30 Mpa

= larger of |(120  − 60),(60 − (−30)),((−30) − 120)| ≤ 250/N MPa

= larger of |60, 90, −150| ≤ 250/N MPa

⇒ 150 ≤ 250/N

Therefore, Factor of safety, N = 1.7

Note:

Maximum shear stress theory is used for Ductile materials and it is an oversafe and expensive.

The best theory for Ductile material used in market is Maximum Distorsion Energy theory and it is safe and economical.

Concept:

for a boiler shell, the thickness of the boiler shell is given by:

where P is pressure, D is the diameter of the shell, σ_t = tensile stress in the plate, CA is corrosion allowance, η = joint efficiency

when t ≥ 8mm,

we use Unwin formula to find the diameter of the rivet (d) hole which is:

d = 6√t

Calculation:

Given:

P = 1 MPa, D = 1.5 m, η = 0.75

Allowable tensile stress in the plate σ_t = 83 MN/m­­²

Allowable compressive stress in the plate σ_c = 138 MN/m²

Allowable shear stress in the rivet τ = 55 MN/m²

thickness of the boiler shell is:

t = 14.04 mm ≥ 8 mm

Since t > 8 mm, So using Unwin’s formula we get:

The diameter of the rivet (d) is

d = 6√t

d = 6√14.04 = 22.48 mm

Concept:

The stress and strain under uniaxial load are given by

Young’s modulus is given by

Calculation:

Given l = 2m, d = 50 mm, Δl = 5 mm, P = 400 kN;

Nearest option is 82 GPa

Statement 1:
Fluidized bed combustion (FBC) is a combustion technology used to burn solid fuels.
Statement 2:
Cyclonic separation is a method of removing particulates from an air, gas or liquid stream, without the use of filters, through vortex separation. When removing particulate matter from liquid, a hydrocyclone is used; while from gas, a gas cyclone is used. Rotational effects and gravity are used to separate mixtures of solids and fluids. The method can also be used to separate fine droplets of liquid from a gaseous stream.
Statement 3:
The PFBC system combines a fluidized-bed combustion boiler with a gas turbine. The gas turbine supplies the combustion air at high pressure to the combustor, and the gas after being cleaned in cyclones drive the gas turbine. Steam is simultaneously generated in the tube bundles in the combustor and drives the steam turbine. It is a coal-fired combined cycle.

Brayton cycle:
Gas turbines operate on the Brayton cycle/Joule cycle. The Joule cycle consists of four internally reversible processes:

• Isentropic compression (in a compressor)
• Constant-pressure heat addition
• Isentropic expansion (in a turbine)
• Constant-pressure heat rejection

Rankine Cycle:
Process 1 – 2: Isentropic compression
Process 2 – 3: Isobaric heat addition
Process 3 – 4: Isentropic expansion
Process 4 – 1: Isobaric heat rejection

Though the Rankine cycle is a constant pressure cycle, it does not come under gas-turbine as the working fluid is water here.
Atkinson Cycle:
In the case of the Atkinson Cycle heat is supplied at constant volume and rejected at constant pressure.
Stirling cycle:
A Stirling cycle consists of two reversible isothermal and two reversible constant volume (isochoric) processes.

Concept:
Heat removal per unit area is:
q = Q̇/A
Calculation:
Given:
Area = 6 m², q = 50 kJ-hr^-1m^-2°C^-1, Q̇ = 5235 kJ/hr, T_room = 25°C
q = Q̇/A
Q̇ = 50 × 6 = 300 kJ-hr^-1°C^-1
Temperature difference = 5235/300 = 17.45∘C
T_ref - T_room = 17.45
T_ref = 17.45 + 25 = 42.45°C

Concept:
Bernoulli's equation:
P/ρg + v²/2g + Z = Constant
where constant is the total head available.

Since the pipe diameter is constant before nozzle, the velocity will be same throughout.
Due to the length of the pipe there will be head loss due to friction as well.
The loss of head due to friction in the pipe is given as:

Where f = friction coefficient,
Calculation:
Given:
l = 2000 m, D1 = 500 mm = 0.5 m, Constant = 60 m
D2 = 100 mm = 0.1 m, f = 0.01, g = 9.81 m/s2

Considering point 2 at the nozzle outlet, P₂ and z will be zero.

Applying continuity equation:
A₁V₁ = A₂V₂
D₁²V₁ = D₂²V₂
(0.5)² × V₁ = (0.1)² × V₂
V₁ = V₂/25

V₂ = 30.61 m/sec

Properties of Pure Bending:

When loaded by a bending force, the beam bends so that the inner surface is in compression and the outer surface is in tension. The neutral plane is the surface within the beam between these zones, where the material of the beam is not under stress (either compression or tension).

As there is no lengthwise stress force on the neutral plane, there is no strain so when the beam bends, the length of the neutral plane remains constant.

The neutral axis is an axis in the cross-section of a beam (a member resisting bending) or shaft along which there are no longitudinal stresses or strains.

If the section is symmetric, isotropic, and is not curved before a bend occurs, then the neutral axis is at the geometric centroid.

All fibers on one side of the neutral axis are in a state of tension, while those on the opposite side are in compression.

As long as the stresses are under elastic range, the extension or compression does not effect the location of neutral axis, whereas if the stresses goes beyond the elastic range the shape of the beam is distorted, such that the neutral axis is also displaced.

From Pure Bending equation:


we can conclude that normal stress varies linearly with distance from the neutral axis.

Ductile Fracture:

• Most ductile metals fracture preceded by a moderate amount of necking, followed by the formation of voids, cracks and finally shear. Hence we can say that ductile fracture takes place after necking with little sound.
• This gives a characteristic cup-and-cone fracture. In this central interior region has an irregular and fibrous appearance, which signifies plastic deformation.
• Different progressive stages of ductile fracture are shown in figure.
• As the cross-sectional areas is getting reduced at subsequent stages high energy is absorbed before fracture happens.

Additional Information
Brittle Fracture:

• The brittle fracture that takes place with little or no preceding plastic deformation. It occurs, often at unpredictable levels of stress, by rapid crack propagation.
• The direction of crack propagation is very nearly perpendicular to the direction of applied tensile stress. This fracture is also said to be trans granular because crack propagates through grains.
• Thus, it has a grainy or faceted texture.
• Most brittle fractures occur in a transgranular manner. However, the brittle fracture can occur in an intergranular manner i.e. crack propagates along grain boundaries.

Plastics can be divided into two types thermoplastics and thermosetting plastics.
Thermoplastics:

• They are polymers that can be deformed on heating and can be reformed into different shapes.
• Examples: Polythene and PVC.

Thermosetting plastics:

• They are polymers that cannot be deformed on heating or remoulded into a new shape.
• Thermosetting plastics are polymers that become irreversibly rigid on heating.
• Initially, it is in liquid form or soft solid, heat provides energy for covalent bond formation, crosslinking the polymer subunits. Thus the polymer becomes rigid.
• Examples: Bakelite, epoxy resin, melamine, etc.

Copolymers:

• Copolymers are those polymers which are made by the copolymerisation of more than one monomeric species.
• They have different properties than the homopolymers.
• It contains multiple units of each monomer used in the same polymeric chain.
• They can be made either by chain growth polymerisation or by step growth polymerisation.
• Ex. butadiene-styrene copolymer, ethylene-vinyl acetate etc.

Polymerization
The process by which the monomers get linked up is called polymerization.
Types of Polymerization

• Depending upon the mode of reaction, polymerization is classified as :
  1. Addition polymerization
  2. Condensation polymerization

Addition Polymerization:

• This process involves the addition of monomer units to themselves to form a growing chain by a chain reaction mechanism.
• It is for this reason that the process is also known as chain growth polymerization.
• Addition polymerization is achieved by adding a catalyst (known as initiator), which provides some reactive species like free radicals.

Condensation Polymerization:

• In this, the monomers combine with the elimination of a small molecule like H2O, ROH or NH3 , etc.
• The reaction is called (step growth) condensation polymerization and the product formed is called condensation polymer.
• The process involves the elimination of by product molecules, therefore, the molecular mass of the polymer is not the integral multiple of the monomer units

Cause misalignment of the teeth on the pinion relative to those on the gear:

• Check the compatibility/level of gear accuracy with the specified tolerances before actual use, i.e., to assure required accuracy and quality.
• Provide an insight into the performance of the gear manufacturing process including the setup of the gear-making machine tools (misalignment), condition of the gear cutting tools, machine tool control, and basic machining practices.
• Determine the thermal distortions caused by possible heat treatment to facilitate corrective action.
• Minimize the overall cost of manufacture by controlling rejection and scrapping.

Displacement Probes: Displacement, or eddy-current, probes are designed to measure the actual movement (i.e., displacement) relative to the probe. Therefore, the displacement probe must be rigidly mounted to a stationary structure to gain accurate, repeatable data.

• Permanently mounted displacement probes will provide the most accurate data on machines with a low-relative to the casing and support structure-rotor weight.
• Displacement transducers permanently mounted at key measurement locations to acquire data for the program.
• The useful frequency range for displacement probes is from 10 to 1.000 Hz or 600 to 60.000 rpm.
• Frequency components below or above this range will be distorted and therefore unreliable for determining machine condition.
• The major limitation with displacement or proximity probes is cost . The typical cost for installing a single probe, including a power supply, signal conditioning, and so on.