GATE ME 2020 Official Paper: Shift 2 (Previous Year Paper)

Explanation:

In the first blank, the word ‘to’ is the correct word because according to the rule with the words like proposal, idea, suggestion, view, opinion etc. we have to use ‘to’ preposition only.

In the second blank, ‘with’ must be the correct option because while referring to some person ‘with’ is used only by rule.

The correct answer is option 1 i.e. increase.

Key Points

• After reading the sentence we can easily comprehend that recent measures have been taken in order to improve the output, now we can ask ourselves how can we increase output? and then let's go through each option one by one.
• Here "would' is a modal so the word in the blank must be in base form (V₁). It is quite obvious that to increase the output we must increase production(output and production are directly proportional) hence, 'decrease' 'speed' and 'equalise' are inappropriate choices.

Complete Sentence: The recent measures to improve the output would increase the level of production to our satisfaction.

The correct answer is option 2 i.e. Lightening.

From the first part of the analogy, it becomes clear that a noun is needed after its corresponding adjective. White is an adjective (color) and it is related to Whitening (noun) that means the act or process of making or becoming white. Similarly, light is also an adjective that means having little weight and it is related to 'lightening' i.e. to reduce in weight or quantity.

The meaning of the rest of the options are given below:

• Lightning: the flashing of light produced by a discharge of atmospheric electricity.
• Lighting : illumination or ignition.
• Enlightening: providing or tending to provide knowledge, understanding, or insight.

Explanation:

Squared means ‘making the score even/equal’ and the meaning given in the option is incorrect.

“Upped the tempo” means to increase the interest toward the game by playing so long.

“Saw him finish” means playing so long and making so many runs results in making 135 runs and equals the five matches series.

Concept:

Price paid by customer at end = initial cost × (increased percentage)ⁿ

Where,

n = no. of levels

Calculation:

Given:

Cost of product at first level P = 120 Rupees.

Also,

It is increased by 25% in each level.

Hence,

Initial cost = 120 Rupees.,

Increased percentage = 25%,

n = 5

Now,

Price paid by customer at end = 120 × 1.25⁵ Rupees

∴ Price paid by customer at end = 366.21 Rupees.

The correct answer is Option 3 i.e. Mitigation deals with actions taken to reduce the use of fossil fuels.

After reading the above paragraph one can easily comprehend that mitigation deals with the steps or action taken to reduce the consumption of non-renewable energy resources like fossil fuels whereas, Adaptation deals with reducing the vulnerability of climate change aspects, now let's go through each option one by one.

• Option 1 states that- "Mitigation deals with consequences of climate change.", this option is incorrect because it has already been mentioned in the passage that it deals with the steps or action taken to reduce the consumption of non-renewable energy resources like fossil fuels.
• Option 2 states that- "Adaptation deals with causes of climate change.", this option is incorrect, it has been clearly mentioned in the passage that adaptation deals with reducing the vulnerability of climate change aspect.
• Option 4 states that- "Adaptation deals with actions taken to combat green-house gas emissions.", adaptation specifically deals with the action taken to reduce vulnerability on climate change aspects, it has nothing to do with green-house gas emissions. Hence, this option is incorrect.

Assign all the corner letters numbers as.

a = 1, b = 2, c = 3 and so on.

Assume “n = 4” which is at the centre     (by looking at the problem)

5 + 4 = 9

Now, t = 20 and x = 24

∴ x = t + n

⇒ 24 = 20 + n

⇒ n = 4   

∴ Our assumption is correct.

Now, h = 8

Case 1

h + n = unknown

8 + 4 = 12 = L (not in the option)

Case 2

Unknown + n = h

⇒ Unknown + 4 = 8

⇒ Unknown = 4 = d

Concept:

M₁ × D₁ = M₂ × D₂

Where,

M₁ = Number of men in initial condition, M₂ = Number of men in final condition

D₁ = Number of days required in the initial condition,

D₂ = Number of days required in the final condition

Calculation:

Given:

M₁ = 52,

D₁ = 10, M₂ = 52 - 12 = 40,

D₂ = ?

52 × 10 = 40 × D­₂

∴ D₂ = 13 days

Now,

Numbers of days more than the original estimate is equal to:

Number of days required in the final condition - Number of days required in the initial condition

∴ Numbers of days more than the original estimate = D_{2 –} D₁

∴ Numbers of days more than the original estimate = 13 – 10

∴ Numbers of days more than the original estimate = 3 days

As line passes through the origin

(X × Y) = mZ (where m is the slope of the line)

Case 1:

For constant Z,

X × Y = Constant

\({\rm{X}} = \frac{{{\rm{Constant}}}}{{\rm{Y}}}\)

\(\therefore {\rm{X}} \propto \frac{1}{{\rm{Y}}}\)

X is inversely proportional to Y.

Case 2:

For constant Y,

X × constant = constant × Z

X = constant × Z

∴ X ∝ Z

X is directly proportional to Z

Case 3:

For constant X,

constant × Y = constant × Z

Y = constant × Z

∴ Y ∝ Z

Y is directly proportional to Z

Case 4:

(X × Y) = mZ

(X × Y) = constant × Z

\(\therefore \frac{{{\rm{X}} \times {\rm{Y}}}}{{\rm{Z}}} = {\rm{Constant;}}\)

Total number of students in the University = 5000

Total number of girls = 1500

Total number of students in arts (students) _arts = 0.2 × 5000 = 1000

Total number of girls in arts (girls) _arts = 0.3 × 1500 = 450

∴ Total number of boys in arts (boys) _arts = (students)_arts – (girls)_arts

⇒ (boys)_arts = 1000 – 450

⇒ (boys)_arts = 550

Total number of girls in management (girls) _management = 0.15 × 1500

⇒ (girls) _management = 225

\(\therefore \frac{{{{\left( {{\rm{boys}}} \right)}_{{\rm{arts}}}}}}{{{{\left( {{\rm{girls}}} \right)}_{{\rm{management}}}}}} = \frac{{550}}{{225}} = \frac{{22}}{9} = 22;:9\)

Explanation:

Normal Sum Theorem:

According to the normal sum theorem, two statistically independent normal variables sum to another normal variable, 

\({N_1}\left( {{m_1},;a_1^2} \right) + {N_2}\left( {{m_2},a_2^2} \right) = N\left( {{m_1} + {m_2},a_1^2 + a_2^2} \right)\)

When \({N_1}\left( {{m_1},;a_1^2} \right)\) and \({N_2}\left( {{m_2},a_2^2} \right)\) are statistically independent. The mean and variance of sum of a statistically independent random variable is the sum of the individual mean and variances.

Concept:

Every square matrix is expressed as the sum of symmetric and skew-symmetric matrix. Here, S is symmetric matrix and V is skew-symmetric matrix.

∴ P = S + V

Calculation:

\(P = ;\left[ {\begin{array}{*{20}{c}} { - 4}&4&2\ 4&3&{\frac{7}{2}}\ 2&{\frac{7}{2}}&2 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 0&{ - 2}&3\ 2&0&{\frac{7}{2}}\ { - 3}&{ - \frac{7}{2}}&0 \end{array}} \right]\)

\(\therefore P = ;\left[ {\begin{array}{*{20}{c}} { - 4}&2&5\ 6&3&7\ { - 1}&0&2 \end{array}} \right]\)

Given:

\(I = \mathop \smallint \nolimits_{x = 0}^1 \mathop \smallint \nolimits_{y = 0}^{{x^2}} x{y^2}dydx\)

0 ≤ y ≤ x² (this is represented by vertical strip)

And x varies from 0 to 1.

Now if we change the order of integration, we have to draw a horizon strip.

After changing the order of Integration

\(\sqrt y \le x \le 1\)

And, 0 ≤ y ≤ 1

∴ \(I = \mathop \smallint \nolimits_{y = 0}^1 \mathop \smallint \nolimits_{x = \sqrt y }^1 x{y^2}dxdy\)

Concept:

Few necessary Laplace transform properties are given below,

L[f"(t)] = s²L[f(t)] - sf(0) - f'(0) 

L[f'(t)] = sL[f(t)]- f(0)

Calculation: 

Here, the given equation is \(\frac{{{d^2}y}}{{d{t^2}}} - y = 1,\)

⇒ L (y" – y) = L(y") - L(y) = L (1)

\(\frac{{{d^2}y}}{{d{t^2}}} - y = 1,\)

The above equation can be written as:

f"(y) – y = 1

Now for calculation the laplace domain of the above equation:

L (y" – y)

∴ L(y") - L(y) = L (1)

[s²L(y) - sy(0) - y'(0)] – [L(y)] = L(1)

\(\therefore {s^2}\left[ {L\left( y \right)} \right] - L\left( y \right)] = \frac{1}{s};\)

\(\therefore {\rm{L}}\left( {\rm{y}} \right){\rm{;}}\left[ {{{\rm{s}}^2}{\rm{;}}-{\rm{;}}1} \right]{\rm{;}} = \frac{1}{s};\)

\(\therefore {\rm{L}}\left( {\rm{y}} \right) = {\rm{;}}\frac{1}{{s\left( {{s^2} - 1} \right)}}\)

\(\therefore {\rm{L}}\left( {\rm{y}} \right) = \frac{1}{{s\left( {s + 1} \right)\left( {s - 1} \right)}}\)

Explanation:

When the roller is just about to climb the step then, 

• Static friction at point A is zero because when the cylinder is about to make out of the curb, it will lose its contact at A, the only contact will be at point B.
• At point B, the roller will be in a state of pure rolling so even the surfaces are rough there will be no friction at point B.
• Force F, contact force at point B, and weight of the roller W will be concurrent at point C.

Therefore the free body diagram will be as follows

Concept:

For pure rolling of the disc between the plates at the point of contact, the velocity of the disc will be equal to the velocity of the plate.

The velocity of the disc at the point of contact will be equal to the product of distance of the instantaneous centre from the point of rotation and the angular velocity of the disc.

Calculation:

The instantaneous centre of rotation of the disc can be found by drawing the line perpendicular to the velocity and using the proportionality of velocity and distance from the instantaneous centre to create similar triangles

Velocity at point P, V = IP × ω

Velocity at point Q, 2V= IQ × ω

Then from similar triangle

\(\frac{V}{{2V}} = \frac{{IP}}{{IQ}}\)

\( \Rightarrow IQ = 2;IP\)

and the sum of IP + IQ is the diameter of the disc

Then IP + IQ = 2r

⇒ IP + 2 IP = 2r

\(IP = \frac{{2r}}{3}\)

Now V = IP × ω

\( \Rightarrow V = \frac{{2r}}{3} \times \omega \)

\(\omega = \frac{{3V}}{{2r}}\)

Concept:

\({\rm{Damping;factor;}} = \zeta = \frac{c}{{2\sqrt {km} }}\)

Where,

ζ = damping factor, c = coefficient of damping, k = spring constant, m = mass

Calculation:

Given:

\(\frac{{{d^2}x}}{{d{t^2}}} + 3\frac{{dx}}{{dt}} + 9x = 10\sin \left( {5t} \right)\)

By comparing this equation with \(m\ddot x + c\dot x + kx = F(t)\)

m = 1 kg, c = 3 N s/m, k = 9 N/m,

\({\rm{\zeta }} = \frac{3}{{2;\sqrt {9 \times 1} ;}}\)

\({\rm{\zeta ;}} = \frac{3}{6}\)

∴ ζ = 0.5

Explanation:

The following points are important about the inversion of a kinematic chain:

• The inversions of a kinematic chain are obtained by fixing different links one at a time.
• For a kinematic chain consisting of 4 links, the maximum possible kinematic inversions are four.
• The qualitatively distinct inversions mean that all the inversions should be able to convert the input into a distinct output.

 

A 4 revolute pair chain is called the Grashof chain if the following condition is satisfied:

l + s < p + q (Class-I four-bar linkage)

i.e. the sum of the shortest and longest link should not be greater than the other two.

If a Grashof chain with 4 revolute pairs is considered then the following three distinct inversions are possible.

Double crank:

A double-crank results when the shortest link is fixed.

In this mechanism both the driving and the driven links make a complete rotation.

Double rocker:

A double rocker results when the shortest link is made coupler.

In this mechanism, both the driving and the follower links make only oscillations and none of them makes a complete rotation.

Crank-rocker:

If the link adjacent to the shortest link is fixed then it results in a crank-rocker.

In this mechanism, the driving link makes complete rotation and the driven link makes only oscillation.

Class-II four-bar linkage:

When the sum of the lengths of the largest and the shortest links is more than the sum of the lengths of the other two links, the linkage is known as class-II, four-bar linkage i.e. l + s > p + q.

In such links, fixing of any of the links always results in a rocker-rocker or double rocker mechanisms.

In other words, the mechanism and its inversions give the same type of motion i.e. (not distinct).

Explanation:

The unconventional machining process and their characteristics and the application areas are discussed in the table below:

Heat Affected Zone (HAZ):  

• The area of the base material of metal which is affected by the heat of the welding process. Melting of the base material does not occur here only microstructure is changed.
• Heat affected zone may range from small to large depending on the rate of heat input. A process with low rates of heat input will result in a large HAZ.
• The size of HAZ also increases as the speed of the welding process decreases.

 

\({\rm{Size;of;HAZ}}; \propto \frac{1}{{speed;of;welding}}\)

So, order of welding processes in increasing speed is

Arc welding → Submerged Arc welding → MIG welding → Laser Beam welding

Therefore, the order of size of the heat affected zone in increasing sequence is

Laser Beam welding → MIG welding → Submerged Arc welding → Arc welding 

Important Points

Butt welding:  Joining of metal by its whole cross section side by side.

Phase diagram:  

• The graphical representation of the values of thermodynamic variables when system is in equilibrium. Thermodynamic variables changes as temperature changes.
• An equilibrium phase diagram presents the phases and phase changes under equilibrium conditions, but it provides no information about the rates of transformation.
• Although changes in pressure, composition or temperature can cause phase transformations, it is temperature changes that are more important.

 

Phase diagram classified as.

1. Unary diagram: If a system consists of just one component e.g. water
2. Binary diagram: If a system consists of two components. It can be either two metals (Cu and Ni), or a metal and a compound (Fe and Fe3C), or two compounds.

Rate of transformation is explained in Time-Temperature-Transformation (TTT) diagram which is also known as iso-thermal diagram. An alloy has to be cooled rapidly and then kept at a temperature to allow for respective transformation to take place.

Surface texture is the degree of finish conveyed to the machinist by a system of symbols devised by a Standards Association. These symbols provide a standard system of determining and indicating the surface finish.

Surface terms:

Flaws: Flaws are random irregularities or scratches, holes depression, seams, tear, or inclusions. These are defects caused during the machining operation.

Lay: It is the direction of the surface pattern caused by the machining and is visible to the naked eye.

Roughness: Roughness is the irregular deviation of the surface.

Roughness height: Roughness height, Ra is the deviation of the centerline.

Roughness width: Roughness width is the distance between successive roughness peaks parallel to the nominal surface.

Waviness: Waviness is the recurrent deviation form a flat surface, it is measured and described in terms of the surface between adjacent crests of the waves (waviness width) and height between the crests and valleys of the waves (waviness height).

 

Standard symbol to represent surface texture:

Explanation:

Material Requirement Planning:

It is a material control system in inventory handling to assure that the required materials are available when needed.

The major objectives of an MRP system are:

• 1. Ensure the availability of materials, components, and products for planned production activity.
• 2. Maintain the lowest possible level of inventory.
• 3. Plan manufacturing activities, delivery schedules, and purchasing activities.

So, there are various lot-sizing approaches in material requirement planning. Some are discussed below.

Explanation:

The partially submerged body resembles the floating body, where the weight of the body is balanced by the buoyancy force acting in the upwards direction. The stability of the floating is governed by the metacentre of the floating body.

Metacentre is the point about which a body starts oscillating when the body is tilted by a small angle. It is the point where the line of action of buoyancy will meet the normal axis of the body when the body is given a small angular displacement.

Stability of floating bodies:

Stable equilibrium: Metacentre is above the centre of gravity of the body, then the disturbing couple is balanced by restoring couple, the body will be in stable equilibrium.

Unstable equilibrium: Metacentre is below the centre of gravity of the body, then the disturbing couple is supported by restoring couple, the body will be in unstable equilibrium.

Neutral equilibrium: The metacentre and the centre of gravity coincides at the same point, then the body is in neutral equilibrium.

Stability of submerged bodies: In case of the submerged body the centre of gravity and centre of buoyancy is fixed, therefore the stability or unstability is decided by the relative positions of the centre of buoyancy and the centre of gravity.

Stable equilibrium: For stable equilibrium, the body the centre of buoyancy is above the centre of gravity. The disturbing couple is countered by the restoring couple.

Unstable equilibrium: For unstable equilibrium, the centre of buoyancy is below the centre of gravity of the body, the disturbing couple is supported by the restoring couple.

Neutral equilibrium: when the centre of gravity and the centre of buoyancy coincide then it is the state of neutral equilibrium.

Concept:

The barometer is an instrument used to measure the pressure of air. There are two typed of barometers, the mercury manometer and, aneroid barometer.

Mercury barometer: In a mercury barometer a tube filled with mercury is placed upright with its open end down into a container of more mercury.

Air pressure on the mercury in the container keeps the mercury from draining out of the tube.

Aneroid barometer: An aneroid barometer uses a flexible metal bellows instead of mercury to measure air pressure.

Calculation:

The gauge pressure = P_abs - P_atm

At atmospheric pressure the gauge pressure = 0

Now if we write the equation for pressure in vertical column then,

P_atm – ρ_m × g × h – P_vapour = 0

⇒ P_atm = ρ_m × g × h + P_vapour

Now Since At atmospheric pressure the gauge pressure = 0

(P_vapour­)_gauge = -ρ_m × g × h

∴ Gauge pressure of vapour will be negative.

When the vessel is at 25°C the saturation pressure of the vapour inside the vessel 3.169 kPa is much lesser than the atmospheric pressure.

When this vessel is kept inside the isothermal oven then at 80°C the saturation pressure of vapour at 80°C 47.39 kPa is again lesser than the atmospheric pressure.

Since the pressure inside the vessel is lesser than the atmospheric pressure, then if the valve is open inside the oven which is at atmospheric pressure the hot air from the oven will go inside the vessel through the Valve.

 

In stage 1:

The pressure will be less than the atmospheric pressure because of the presence of water vapour. 

In stage 2:

The water vapour receives heat and gets superheated since the vessel is placed in an oven at 80°C.

Also, we need to remember that in stage 2 the heating of water vapour takes place at constant volume since the oven has a fixed boundary. 

When volume is constant then pressure is directly proportional to temperature i.e

P ∝ T

where P is the pressure and T is the temperature

∴ \(\frac{{{{\rm{P}}_1}}}{{{{\rm{P}}_2}}} = \frac{{{{\rm{T}}_1}}}{{{{\rm{T}}_2}}}\)...(i)

Where P₁ = Saturated vapour pressure of the mixture at 25°C = 3.17 kPa (Standard value)

P₂ = Pressure of the mixture at 80°C, T₁ = Initial temperature of the mixture at 25°C, T₂ = Final temperature of the mixture 

T₁ = 25 + 273 = 298 K, T₂ = 80 + 273 = 353 K 

Substituting in equation (i) and simplifying we will get

\(\frac{{3.17}}{{{{\rm{P}}_2}}} = \frac{{298}}{{353}}\)

P₂ = 3.75 kPa

Note:

P₂ is the pressure of the mixture at 80°C after thermal equilibrium is attained.

Therefore, Since the pressure inside the vessel after heating is less than the atmospheric pressure (101.32 kPa) the hot air goes inside the vessel.

Explanation:

For the Diesel cycle:

Processes involved in compression engine (diesel; cycle) are:

1-2: Reversible adiabatic compression

2-3: Constant pressure heat addition

3-4: Reversible adiabatic expansion

4-1: Constant volume of heat rejection

Otto cycle:

Diesel cycle

 

Dual Cycle:

Carnot Cycle:

Concept:

The degree of reaction of a hydraulic turbomachine is defined as 

\(DOR = \left( {\frac{{Enthalpy;drop;in;rotor}}{{Enthalpy;Drop;in;rotor ;+ ;Enthalpy;drop;In;stator}}} \right)\)

Calculation:

Given, enthalpy at the stator inlet = h₁, enthalpy at the stator outlet = h₂ and enthalpies at the rotor outlet = h₃

Enthalpy drop in rotor = h₂ - h₃

Enthalpy drop in stator = h₁ – h₂

Then, \(DOR = \left( {\frac{{{h_2}; - ;{h_3}}}{{{h_1} - ;{h_2}; + ;{h_{2;}} - ;{h_3}}}} \right)\)

Given, (h₂ – h₁) = (h₃ – h₂)

⇒ h₁ – h₂ = h₂ – h₃

\( \Rightarrow DOR = ;\frac{{{h_2} - {h_3}}}{{2\left( {{h_2} - {h_3}} \right)}} = 0.5 = 50% \)

Explanation:

The 100-dimensional identity matrix will have 1 as the diagonal elements and all the other elements as zero.

That implies the eigenvalue is only 1.

Now E is the set of its distinct (no value appears more than once in E) real eigenvalues.

⇒ E = {1}

So number of distinct elements in E is 1.

Concept:

The hinge support can provide reactions in horizontal and vertical directions as well. The roller support can provide reaction in the direction perpendicular to the surface.

For the beam to remain in equilibrium following equilibrium conditions should be satisfied

ΣF_H = 0 i.e. sum of horizontal forces should be zero

ΣF_V = 0 i.e. Sum of vertical forces should be zero

ΣM = 0 i.e. the sum of all the moments about any point should be zero

Calculation:

The reaction at support Q will be as shown in the figure below,

From moment equilibrium at Hinged point P

ΣM_P = 0

R_Q × 4 - 1200 × 5 = 0

\(\Rightarrow {R_Q} = \frac{{1200 \times 5}}{4} = 1500 ~N\)

Concept:

When the fluctuating load is acting on a member then the magnitude of stress varies between maximum and minimum stress value.

If maximum stress is σ_max and minimum stress is σ_min then

Amplitude of stress \({\sigma _a} = \frac{{{\sigma _{max}}; - ;{\sigma _{min}}}}{2}\)

Mean of the stress \({\sigma _m} = \frac{{{\sigma _{max}}; + ;{\sigma _{min;}}}}{2}\)

If σ_e is the endurance limit and σ_y is the permissible yield stress in the material then

According to Soderberg line

\(\frac{{{{\rm{\sigma }}_m}}}{{{\sigma _y}}} + \frac{{{\sigma _a}}}{{{\sigma _e}}} = \frac{1}{{F.O.S}}\)

Calculation:

Given, fluctuating stress σ = σ_o cos (8πt), σ_e = 350 MPa, F.O.S = 3.5

The maximum will occur when cos (8πt) = 1

∴ σ_max = σ₀

The minimum will occur when cos (8πt) = -1

∴ σ_min = -σ₀

Amplitude of stress \({\sigma _a} = \frac{{{\sigma _o}; - \left( { - {\sigma _o}} \right)}}{2} = {\sigma _o}\)

Mean of the stress \({\sigma _m} = \frac{{{\sigma _o} - {\sigma _o}}}{2} = 0\)

Then from the Soderberg line equation

\(\frac{0}{{{\sigma _y}}} + \frac{{{\sigma _o}}}{{350}} = \frac{1}{{3.5}}\)

\(\therefore {\sigma _o} = \frac{{350}}{{3.5}} = 100;MPa\)

Note: Since mean stress is zero in the given question any of the design criteria will give the same result.

Concept:

If the length of unsupported portion of the rod is greater than 3 times the diameter of rod.

The maximum die diameter is 1.5 times the diameter of rod.

Calculation:

Given, d = 12 mm, l = 40 mm

Maximum die diameter = 1.5 × diameter of rod

∴ (d_die)_max = 1.5 × 12 = 18 mm

Concept:

This is a case of activity on node network system (AON).

Slack = latest finish time – Earliest finish time

Calculation:

Slack = latest finish time – Earliest finish time

Slack on Activity L = 42 – 40 = 2

Important Points:-

1. AON network system enables efficient scheduling and controlling. 2. AON network system eliminates the use of dummies thus logical relationship is established by using arrows.

Concept:

Heat flux through the wall is given by

\(Q = ; - kA\frac{{dT}}{{dx}}\)

Where k is the thermal conductivity, A is the area perpendicular to the direction of heat flow, dT is the temperature difference and dx is the thickness of the wall.

Heat flux flowing through wall remains same.

∴ Heat flux through Brick wall = Heat flux through insulation

Calculation:

Given,

For brick wall, k₁ = 2.5 W/m.k, dT₁ = 700 – 1100 = -400 K

For Insulation, dT­₂ = 200 – 700 = -500 K and L₂ = L₁/4

Since the heat flux through the wall remains same,

\({k_1}\frac{{d{T_1}}}{{d{x_1}}} = {k_2}\frac{{d{T_2}}}{{d{x_2}}};\)

\( - 2.5 \times \frac{{\left( { - 400} \right)}}{{{L_1}}} = - \frac{{{k_2}\left( { - 500} \right)}}{{{L_1}/4}}\)

\(\Rightarrow {k_2} = ;\frac{{2.5 \times 400}}{{4 \times 500}} = 0.5;W/m.K\)

Concept:

Coefficient of Performance of Refrigerator

\({\left( {C.O.P} \right)_R} = \frac{{{T_L}}}{{{T_H} - {T_L}}}\)

Coefficient of Performance of Heat Pump

\({\left( {C.O.P} \right)_{H.P}} = \frac{{{T_H}}}{{{T_H} - {T_L}}}\)

Where T_L is the lower temperature limit and T_H is the high-temperature limit

Calculation:

Given, T_L = -3°C = 273 – 3 = 270 K, T_H = 27°C = 273 + 27 = 300 K

COP of the refrigerator

\({\left( {C.O.P} \right)_R} = \frac{{{T_L}}}{{{T_H} - {T_L}}} = \frac{{270}}{{300 - 270}} = 9\)

COP of the Heat Pump

\({\left( {C.O.P} \right)_{HP}} = \frac{{{T_H}}}{{{T_H} - {T_L}}} = \frac{{300}}{{300 - 270}} = 10\)

Now,

\(\frac{{{{\left( {C.O.P} \right)}_R}}}{{{{\left( {C.O.P} \right)}_{H.P}};}}=\frac{9}{10} = 0.9\)

Concept:

Directional Derivative = Gradient of function × Unit direction Vector

If F = f(x,y,z) then,

Grad \(f = \left( {\hat i\frac{{\partial f}}{{\partial x}} + \hat j\frac{{\partial f}}{{\partial y}} + \hat k\frac{{\partial f}}{{\partial z}}} \right)\)

For the given direction vector \(a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\)

Unit direction vector \(= \frac{{\vec a}}{{\left| {\vec a} \right|}}\)

and \(\left| {\vec a} \right| = \sqrt {a_1^2 + a_2^2 + a_3^2} \)

Calculation:

Given, f(x,y,z) = xyz

Grad \(f = \left( {\hat i\frac{{\partial f}}{{\partial x}} + \hat j\frac{{\partial f}}{{\partial y}} + \hat k\frac{{\partial f}}{{\partial z}}} \right)\)

Grad \(f = \left( {\hat i\frac{{\partial (xyz)}}{{\partial x}} + \hat j\frac{{\partial (xyz)}}{{\partial y}} + \hat k\frac{{\partial (xyz)}}{{\partial z}}} \right)\)

\(Grad;f = yz;\hat i + xz;\hat j + xy;\hat k\)

(Grad f)_{P = (-1, 1, 3)} \( = 1 \times 3;\hat i + \left( { - 1} \right)\left( 3 \right)\hat j + \left( { - 1} \right)\left( 1 \right)\hat k\)

∴ (Grad f)_P \( = 3\hat i - 3\hat j - \hat k\)

Direction vector \(\left( {\vec a} \right) = \hat i - 2\hat j + 2\hat k\)

Unit direction vector \( = \frac{{\vec a}}{{\left| {\vec a} \right|}} = \frac{{\widehat {i;} - ;2\widehat {j;} + ;2\hat k}}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( { - 2} \right)}^2} + {{\left( 2 \right)}^2}} }} = \frac{{\hat i + 2\hat j + 2\hat k}}{3};\)

∴ Directional Derivative \( = \left( {3\hat i - 3\hat j - \hat k} \right) \times \frac{{\hat i - 2\hat j + 2\hat k}}{3}\)

\(D.D = \frac{{3 \times 1 + \left( { - 3 \times - 2} \right) + \left( { - 1 \times 2} \right)}}{3} = \frac{7}{3}\)

Concept:

f(z) = u + iv

u = real part

v = imaginary part

If f(z) is an analytic function

\(\left. {\begin{array}{*{20}{c}} {\frac{{\partial u}}{{\partial x}} = \frac{{\partial v}}{{\partial y}}}\ {\frac{{\partial v}}{{\partial x}} = - \frac{{\partial u}}{{\partial y}}} \end{array}} \right} \to C.R;equation\)

\(\therefore dv = \frac{{\partial v}}{{\partial x}}dx + \frac{{\partial v}}{{\partial y}}dy\)

\(dv = \frac{{ - \partial u}}{{\partial y}}dx + \frac{{\partial u}}{{\partial x}}dy\) (This is an exact differential equation)

Calculation:

Given,

u = x³ – 3xy²

\(\frac{{\partial u}}{{\partial x}} = 3{x^2} - 3{y^2}\)

\(\frac{{\partial u}}{{\partial y}} = - 6xy\)

\(dv = \frac{{ - \partial u}}{{\partial y}}dx + \frac{{\partial u}}{{\partial x}}dy\)

\(dv = 6xy;dx + \left( {3{x^2} - 3{y^2}} \right)dy\)

It is an exact differential equation the solution is obtained by treating y as constant in the first term and in the second term only that part is integrated which is not containing x.

Integrating the above equation

\(v = 3{x^2}y - {y^3} + constant\)

Concept:

The deflection of the beam at the free end due to load P is \(\Delta = \frac{{P{l^3}}}{{3EI}}\)

Since the deflection in beam and spring are equal, therefore stiffness of both i.e. beam and spring are in parallel connection.

For parallel combination:

k_eq = k₁ + k₂

where k₁ = stiffness of spring and k₂ = stiffness of beam.

Calculation:

Given:

 

Deflection of the beam due to point load P if spring is not there \(\Delta = \frac{{P{l^3}}}{{3EI}}\)

Stiffness \({k_2} = \frac{P}{\Delta }\)

\({k_2} = \frac{P}{{\frac{{P{l^3}}}{{3EI}}}} = \frac{{3EI}}{{{l^3}}}\)

\(\therefore ;{k_{eq}} = k + \frac{{3EI}}{{{L^3}}} = \frac{{k{L^3};+;3EI}}{{{L^3}}}\)

∴ Transverse deflection under the load

\(\Rightarrow \Delta_{net}= \frac{P}{{{K_{eq}}}} = \frac{{P{L^3}}}{{K{L^3} + 3EI}}\)

\( \Rightarrow \Delta_{net} = \frac{{P{L^3}}}{{3EI}};\left( {\frac{1}{{\frac{{K{L^3}}}{{3EI}};+;1}}} \right)\)

\(\Rightarrow \Delta = \frac{{P{L^3}}}{{3EI}};\left( {\frac{{3EI}}{{K{L^3};+;3EI}}} \right)\)

Shortcut Method:

Deflection of beam = Deflection of spring

\(\frac{{\left( {P;-;R} \right){L^3}}}{{3EI}} = \frac{R}{K}\)

\( \Rightarrow \left( {\frac{1}{K} + \frac{{{L^3}}}{{3EI}}} \right)R = \frac{{P{L^3}}}{{3EI}}\)

\(R = \frac{{P{L^3}}}{{3EI}} \times \frac{{3EIK}}{{3EI;+;{L^3}}}\)

\(\Delta_{net} = \frac{R}{K} = \frac{{P{L^3}}}{{3EI}};\left( {\frac{{3EI}}{{3EI;+;K{L^3}}}} \right)\)

Concept:

• When two gears are in contact they have the same module (m).
• The module is the ratio of pitch circle diameter (or simply diameter) of gear to the number of teeth of the gear.

\(m = \frac{{diameter}}{{number;of;teeths}} = \frac{D}{T}\)

• When gears are connected externally they have opposite direction or revolution while when gears are connected internally they have the same direction of revolution.
• The velocity ratio between two mating gears is given by:

\(\frac{{{\omega _2}}}{{{\omega _1}}} = \frac{{{T_1}}}{{{T_2}}}\)

where ω is the angular velocity and T is the number of teeth on the gears.

Calculation:

Given:

Number of teeth on the sun (T_s) = Number of teeth on the planet (T_p), Angular velocity of sun = ω_s (anticlockwise), Angular velocity of ring = ω_r (anticlockwise)

Let Angular velocity of planet = ω_p

The radius of the sun, planet, and ring is R_s, R_p, and R_r respectively.

The module of each gear is m.

Now,

R_s + 2R_p = R_r      ----(1)

\(m = \frac{{diameter}}{{number;of;teeths}} = \frac{D}{T}\)

⇒ D = mT

⇒ R = 2mT

Putting the value of R in equation (1)

2mT_s + 4mT_p = 2mT_r

⇒ T_s + 2T_p = T_r

⇒ 3T_p = T_r                   (T_p = T_s)

Let anticlockwise motion be positive.

\(\begin{array}{l} \frac{{{\omega _p}}}{{{\omega _s}}} = - \frac{{{T_s}}}{{{T_p}}}\ \Rightarrow \frac{{{\omega _p}}}{x} = - \frac{{{T_P}}}{{{T_p}}}; \Rightarrow ;{\omega _p} = - x \end{array}\)

And,

\(\begin{array}{l} ;\frac{{{\omega _r}}}{{{\omega _p}}} = \frac{{{T_p}}}{{{T_r}}}\ \Rightarrow \frac{{{\omega _r}}}{{\left( { - x} \right)}} = \frac{{{T_p}}}{{3{T_p}}} \Rightarrow {\omega _r} = \frac{{ - x}}{3} \end{array}\)

Now, After giving arm +y

x + y = ω_s      ----(2)

\(- \frac{{\rm{x}}}{3}{\rm{;}} + {\rm{;y;}} = {\rm{;}}{{\rm{\omega }}_{\rm{r}}}\)     ----(3)

Subtracting equation (3) from equation (2)

\(\begin{array}{l} \Rightarrow x + \frac{x}{3} = {\omega _s} - {\omega _r}\ \Rightarrow \frac{{4x}}{3} = {\omega _s} - {\omega _r}\ \Rightarrow x = \frac{3}{4}({\omega _s} - {\omega _r}) \end{array}\)

Putting the value of x in equation (2)

\(\begin{array}{l} \Rightarrow y = {\omega _s} - ;\frac{3}{4}({\omega _s} - {\omega _r})\ \Rightarrow y = {\omega _s} - \frac{{3{\omega _s}}}{4} + \frac{{3{\omega _r}}}{4}\ \Rightarrow y = \frac{{{\omega _s}}}{4} + \frac{{3{\omega _r}}}{4} \end{array}\)

Concept:

When a thin-walled cylinder is subjected to internal fluid pressure then there are two stresses acting upon the cylinder

Hoop stress in the circumferential direction \({\sigma _h} = \frac{{pr}}{t}\)

and longitudinal stress \({\sigma _l} = \frac{{pr}}{{2t}}\)

Where p is the internal pressure of the fluid in the cylinder above atmospheric pressure, r is the radius of the cylinder and t is the thickness of the cylinder.

If the cylinder is open from the sides then longitudinal stress is zero.

Strain due to these stress

\({\epsilon_h} = \frac{1}{E}\left( {{\sigma _h} - \nu {\sigma _l}} \right)\)

\({\epsilon_l} = \frac{1}{E}\left( {{\sigma _l} - \nu {\sigma _h}} \right)\)

ν is Poisson’s ratio.

Thermal strain due to temperature rise is given by

ϵ_th = αΔT

α is the coefficient of thermal expansion, ΔT is the change in temperature

Calculation:

It is given that both the ends of the cylinder are open therefore the longitudinal stress acting will be equal to zero. 

The longitudinal stress, σ_long = 0

The strain in the longitudinal direction will be due to the hoop stress only.

Now the temperature of the cylinder is increased by ΔT, then in order to avoid the leakage, the sum of strains due to increased temperature and the strain due to hoop stress (in the longitudinal direction) should be equal to zero.

Therefore,

∴ ϵ_th + ϵ_l = 0      

\({\epsilon_{th}} = \alpha \left( {{\rm{\Delta }}T} \right)\)                   

\({\epsilon_l} = \frac{1}{E}\left( {0 - \nu \frac{{pr}}{{t}};} \right)\)

\(\therefore \alpha \left( {{\rm{\Delta }}T} \right) + \left( {\frac{{ - vpr}}{{Et;}};} \right) = 0;;;\)

\({\rm{\Delta }}T = \frac{{vpr}}{{atE}}\)

Concept:

k = spring constant = \(\frac{W}{y}\)

y = deflection of spring = \(\frac{{\delta \left( U \right)}}{{\delta W}}\)

Where U is the strain energy stored in the spring, W is the load applied at the axis of the spin and y is the deflection caused by the load W.

As the strain energy of spring is given by,

\({\rm{SE}}\left( {\rm{U}} \right) = \frac{1}{2} \times T \times \theta \)

Where, torque at the end of spring  \(T = W \times \frac{D}{2}\)

Maximum angle of twist \(\theta = \frac{{TL}}{{GJ}}\)

Where, W = load applied at end, D = diameter of spring, length of spring L = πDn,

n = number of coils, G = torsional rigidity of spring material, J = polar moment of inertia = \(\frac{\pi }{{32}} \times {d^4}\)  

d = diameter of wire

\(\therefore U = \frac{1}{2} \times \left( {W \times \frac{D}{2}} \right) \times ;\left( {\frac{{32 \times WD \times \pi Dn}}{{2 \times G \times \pi \times {d^4}}}} \right) = \frac{{4{W^2}{D^3}n;}}{{G{d^4};}}\)

Now, y = deflection of spring 

y = \(\frac{\delta }{{\delta W}};\left( {\frac{{4{W^2}{D^3}n}}{{G{d^4};}}} \right) = ;\frac{{8W{D^3}n}}{{G{d^4};}}\)   and 

k = spring constant = \(\frac{W}{y} = \frac{{G{d^4}}}{{8{D^3}n}}\)

Calculation:

Given, d₂ = 2d, D₂ = 2D, n₂ = 2n

\(\therefore \frac{{{k_2}}}{{{k_1}}} = \frac{{{{\left( {2d} \right)}^4}}}{{{{\left( {2D} \right)}^3} \times \left( {2n} \right)}} \times \frac{{{D^3}n;}}{{{d^4}}} = \frac{{{d^4}}}{{{D^3}n}} \times \frac{{{D^3}n}}{{{d^4}}} = 1\)

∴ k₂ = k₁   

∴ k₂ = k                      (∵ k₁ = k)

Let us draw both the blocks in one V-groove,

Now if we draw the line parallel to the edge of the V-groove, then sin α can be written as

\(\sin \alpha = \frac{{PQ}}{{PR}}\)

From the geometry PQ can be written as

\(PQ = \frac{{{D_1} - {D_2}}}{2}\)

Now, for PR again if we observe the geometry 

\(PR = \left( {{H_1} - \frac{{{D_1}}}{2}} \right) - \left( {{H_2} - \frac{{{D_2}}}{2}} \right) = \left( {{H_1} - {H_2}} \right) - \left( {\frac{{{D_1} - {D_2}}}{2}} \right)\)

\( \Rightarrow \sin \alpha = \frac{{PQ}}{{PR}} = \frac{{\left( {\frac{{{D_1} - {D_2}}}{2}} \right)}}{{\left( {{H_1} - {H_2}} \right) - \left( {\frac{{{D_1} - {D_2}}}{2}} \right)}}\)

\(\sin \alpha = \frac{{\left( {{D_1} - {D_2}} \right)}}{{2\left( {{H_1} - {H_2}} \right) - \left( {{D_1} - {D_2}} \right)}}\)

Concept:

The forecast for recent data is calculated by

F_n = F_n-1 + α (D – F)_n-1   

Where, F_n = Recent Forecast value, F_n-1 = Previous period forecast value,

 D_n-1 = Previous period actual sale value, α = smoothing constant

Calculation:

For 3^rd month, F_n = 31.82, F_n-1 = 31.80, D_n-1 = 32

∴ 31.82 = 31.80 + α (32 - 31.80)

31.82 – 31.80 = 0.2 α

0.02 = 0.2 α

α = 0.1

Important Points

α value lies between 0 and 1

When α = 0 means the forecast is stable

Stability: When the forecast pattern is flat, smooth or has less fluctuations. It is preferred for old existing products.

When α = 1 means the forecast is not stable

Responsiveness: When the forecast is fluctuating and swinging pattern. It is preferred for new products.

Concept:

The change in entropy of gas at given pressure is,

\({s_2} - {s_1} = {C_p}\ln \left( {\frac{{{T_2}}}{{{T_1}}}} \right) - R\ln \left( {\frac{{{P_2}}}{{{P_1}}}} \right)\)

Where, s₂ - s₁ = change in entropy, C_p = specific heat at constant pressure, R = gas constant

T = temperature, P = pressure

Calculation:

m = 1 kg, P₁ = 1 bar, P₂ = 3 bar, T₁ = 27°C = 27 + 273 = 300 K, T₂ = 127°C = 127 + 273 = 400 K

γ = 1.4, R = 287 J/kg.K   

\({C_p} = \frac{{\gamma R}}{{\gamma - 1}} = \frac{{1.4; \times ;287}}{{1.4; - ;1}} = 1005;J/kg.K\)

\(\therefore {s_2} - {s_1} = 1005 \times \ln \left( {\frac{{400}}{{300}}} \right) - 287 \times \ln \left( {\frac{3}{1}} \right)\)

s₂ - s₁ = -26.3 J/kg.K

Concept:

\({\rm{Number;of;interval}} = \frac{{{\rm{b}} - {\rm{a}}}}{{\rm{h}}}{\rm{;}}\)

where, b is the upper limit, a is the lower limit, h is the step size

According to the trapezoidal rule

\(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \frac{{\rm{h}}}{2}\left[ {{{\rm{y}}_{\rm{o}}} + {{\rm{y}}_{\rm{n}}} + 2\left( {{{\rm{y}}_1} + {{\rm{y}}_2} + {{\rm{y}}_3}{\rm{;}} \ldots } \right)} \right]\)

And,

\(Absolute;percentage;error = \frac{{\left| {true;value - approximate;value} \right|}}{{true;value}}\)

Calculation:

Given, Number of interval = 1

a = 0, \({\rm{b;}} = {\rm{;}}\frac{\pi }{2}\)

\(\begin{array}{l} {\rm{Number;of;interval}} = \frac{{{\rm{b}} - {\rm{a}}}}{{\rm{h}}}\ \Rightarrow 1 = \frac{{\frac{{\rm{\pi }}}{2} - 0}}{{\rm{h}}}{\rm{;}}\ \therefore {\rm{h}} = \frac{{\rm{\pi }}}{2} \end{array}\)

True value

\(\mathop \smallint \nolimits_0^{\pi /2} \left( {8 + 4\cos x} \right)dx = ;\left[ {8x + 4sinx} \right]_0^{\frac{\pi }{2}}\)

\(\Rightarrow {\rm{True;value}} = \left[ {\left( {8 \times \frac{\pi }{2}} \right) + \left( {4 \times \sin \left( {\frac{\pi }{2}} \right)} \right)} \right];\)

⇒ True value = 4π + 4 = 16.56

Approximate value

By trapezoidal rule

 

\(h = \frac{{\left( {\frac{\pi }{2} - 0} \right)}}{1} = \frac{\pi }{2}\)

According to the trapezoidal rule

\(\begin{array}{l} \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \frac{{\rm{h}}}{2}\left[ {{{\rm{y}}_{\rm{o}}} + {{\rm{y}}_{\rm{n}}} + 2\left( {{{\rm{y}}_1} + {{\rm{y}}_2} + {{\rm{y}}_3}{\rm{;}} \ldots } \right)} \right]\ \mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} \left( {8{\rm{;}} + 4{\rm{cosx}}} \right){\rm{dx;}} = \frac{{\rm{\pi }}}{4}\left( {12 + 8} \right) \end{array}\)

⇒  Approximate value = 5π = 15.70

Now,

\(\begin{array}{l} Absolute;percentage;error = \frac{{\left| {true;value - approximate;value} \right|}}{{true;value}}\ Absolute;percentage;error = \frac{{\left| {16.56 - 15.70} \right|}}{{16.56}} \times 100 \end{array}\)

∴ Absolute percentage error = 5.19 %

Concept:

Binomial probability refers to probability of exactly ‘r’ success in ‘n’ number of trials of a experiment which has two possible outcomes.

If the probability of success is ‘p’ in individual trial and the probability of failure is ‘q’ then

 P(r) = ⁿC_r p^rq^n-

Where, n = no. of trials, p = probability of success of random variable, q = probability of failure of random variable, r = number of occurrence of random variable

Calculation:

Probability of 4 heads in first 10 tosses and 4 tails in next ten tosses will be given by

P (4 head in 10 tosses) × P (4 tail in 10 tosses)

Probability of 4 heads in first 10 tosses

n = 10, \({\rm{p;}} = {\rm{;}}\frac{1}{2},;q = 1 - \frac{1}{2} = \frac{1}{2}\), r = 4

P (4 head in 10 tosses) = ¹⁰C₄ × 0.5⁴ × 0.5⁶ = 0.205

Probability of 4 tails in last 10 tosses

n = 10, \({\rm{p;}} = {\rm{;}}\frac{1}{2},;q = 1 - \frac{1}{2} = \frac{1}{2}\), r = 4

P (4 tail in 10 tosses) = ¹⁰C₄ × 0.5⁴ × 0.5⁶ = 0.205

∴ Probability of 4 heads in first 10 tosses and 4 tails in next ten tosses = 0.205 × 0.205 = 0.0420

Concept:

Buoyancy force on a body is given by:

F_b = ρ × V_d × g

where, ρ = Density of liquid, V_d = Volume displaced, g = Acceleration due to gravity

A body floats when the force of buoyancy is equal to the weight of the body i.e.

F_b = mg

When an equation is of the form \(\ddot x+ \frac{k}{m}x = 0\)

then the natural frequency is given by:

\({\omega _n} = \sqrt {\frac{k}{m}} \)

Calculation:

Given, Radius(r) = 20 cm, The density of water (ρ) = 1000 kg/m³

Acceleration due to gravity (g) = 10 m/s²

Let,

Mass of body = m, The density of body = ρ_s, The volume of body = V_s

The volume of fluid displaced \({{\rm{V}}_{\rm{d}}}{\rm{;}} = {\rm{;}}\frac{{{V_s}}}{2}\)

As the body is floating

mg = ρ × V_d × g

⇒ m = ρ × V_d = \(\rho \times \frac{{{V_s}}}{2}\)

\(\begin{array}{l} \Rightarrow {\rho _s} \times {V_s} = \rho \times \frac{{{V_s}}}{2}\ \Rightarrow {\rho _s} = \frac{\rho }{2} \end{array}\)

Now, displacing the ball slightly (x) inside the water creates extra buoyancy force on the body which is equal to F_b1 = ρ × V_extra × g

Since very small disturbance x is given to the ball the instantaneous volume of the ball which will dip more into the water will be

Here V_extra = area × x = πr²x

mẍ + F_b1 = 0

\( \Rightarrow \rho \times \frac{{{V_s}}}{2} \times\ddot x + \rho \times \pi {r^2}x \times g = 0\)

\( \Rightarrow \frac{{{V_s}}}{2} \times\ddot x + \pi {r^2}x \times g = 0\)

\( \Rightarrow \frac{4}{3} \times \frac{{\pi {r^3}}}{2} \times \ddot x + \pi {r^2}x \times g = 0\)

\( \Rightarrow \ddot x + \frac{{3g}}{{2r}}x = 0\)

\(\therefore {\omega _n} = \sqrt {\frac{{3g}}{{2r}}} = \sqrt {\frac{{3; \times 10}}{{2; \times 0.2}}} =8.66;rad/s\)

⇒ ω_n = 8.66 rad/s

Concept:

Buckling load: the load at which column buckle is termed as Buckling load.

Buckling load is given by,

\({P_b} = \frac{{{\pi ^2}E;{I_{min}}}}{{L_e^2}}\)

where

E = Young’s modulus of elasticity, Imin = Minimum moment of inertia, Le = Effective length

In case of both ends hinged Le = L

Calculation:

Given, L = 1 m

Stress up to the elastic limit (σ) = 100 MPa

Strain (ϵ) = 0.002 mm/mm

\({P_b} = \frac{{{\pi ^2}E{I_{min}}}}{{L_e^2}}\)     ----(1)

Young’s modulus (E) = \(\frac{\sigma }{\epsilon} = \frac{{100}}{{0.002}} = \frac{1}{2} \times {10^5};MPa = 0.5 \times {10^{11}};Pa\)

Pb = σ × A = \(\sigma \times \frac{\pi }{4}{d^2} = ;100 \times {10^6} \times \frac{\pi }{4}{d^2} = ;{10^8} \times \frac{\pi }{4}{d^2}\)

\({I_{min}} = \frac{\pi }{{64}}{d^4}\)

Putting values of E, Pb and Imin in equation (1)

\(\begin{array}{l} \Rightarrow {10^8} \times \frac{\pi }{4}{d^2} = \frac{{{\pi ^2}; \times ;0.5; \times ;{{10}^{11}}; \times ;\frac{\pi }{{64;}}; \times ;{d^4}}}{1}\ \Rightarrow d_{min}^2 = \frac{{16}}{{{\pi ^2}; \times {{10}^3} \times 0.5;}} \end{array}\)

⇒ dmin = 0.0569 m = 56.9 mm

Concept:

The maximum fluctuation of energy is given by

ΔE_max = (T – T_mean) × θ

The maximum fluctuation of energy is also given by

ΔE_max = I × (ω_mean)² × C_s

where, I = Moment of inertia, ω_mean = Average angular velocity of the flywheel, C_s = Coefficient of fluctuation of energy

Angular velocity is also given by

\(\omega = \frac{{2\pi N}}{{60}};rad/s\)

where N = speed in rpm.

Calculation:

**Given:**T_max = 3000 Nm, T_mean = 2000 Nm

\(\theta = \pi - ;\frac{\pi }{2} = \frac{\pi }{2};radian\)

C_s = 0.04, N = 1000 rpm

The maximum fluctuation of energy

ΔE_max = (T – T_mean) × θ

\(\Rightarrow {\rm{\Delta }}{E_{max}} = \left( {3000 - 2000} \right) \times \frac{\pi }{2}\)

⇒ ΔE_max = 500π J

Average angular velocity

\({\omega _{mean}} = \frac{{2\pi N}}{{60}} = \frac{{2 \times \pi \times 1000}}{{60}} = 104.71;\frac{{radian}}{s}\)

Now,

ΔE_max = I × (ω_mean)² × C_s

⇒ 500 × π = I × (104.71)² × 0.04

⇒ I = 3.58 kg.m²

Concept:

Perfectly inelastic collision in one dimension:

In a perfectly inelastic collision, when one body moving with a velocity hits a stationary body, the two bodies after the collision move together as one body.

• The total linear momentum of the system is conserved in case of Inelastic collision.
• Some kinetic energy is always lost in an inelastic collision.

Energy stored in spring = \(\frac{1}{2}k{x^2}\)

where, k = Spring constant, x = Deflection in spring

Calculation:

Given, m₁ = 10 kg, v_o = 2m/s, m₂ = 30 kg, k = 10⁵ N/m

Let v is combined velocity after collision.

Conservation of momentum

m₁ × v_o + m₂ × 0 = (m₁ + m₂) × v

\(\Rightarrow ,,V = \frac{{20}}{{40}} = 0.5;m/s\)

Now when these combined mass touches the spring the kinetic energy of the mass will get converted into potential energy of the spring

\(\begin{array}{l} ;\frac{1}{2}; \times \left( {{m_1} + {m_2}} \right) \times {v^2} = \frac{1}{2} \times k \times {x^2};\ \Rightarrow \frac{1}{2} \times \left( {10 + 30} \right) \times {0.5^2} = \frac{1}{2} \times {10^5} \times {x^2}\ \Rightarrow {x^2} = \frac{1}{{{{10}^4}}}\ \Rightarrow x = \frac{1}{{100}};m = 1;cm \end{array}\)

Concept:

In a gear assembly,

Power = Torque × Angular speed

Torque = Tangential load × radius of the gear

\(\omega = \frac{{2 \times \pi \times N}}{{60}}\)

The module of two mating gear always remains the same

\(Module = \frac{{Diameter}}{{No.;of;teeth}} = \frac{{2 \times Radius}}{{No.;of;Teeth}}\)

Calculation:

Given, for pinion m = 1.25 mm, T = 20, N = 1200 rpm, F = 50 mm, K_v = 1.26,  Y = 0.322, σ_max = 400 MPa

\(r_p=\frac{mT}{2}=\frac{1.25\times 20}{2}=12.5 ;mm\)

In order to find the power we need to find the torque acting upon the gear,

For torque, the tangential load will be required,

Lewis Formula \(\sigma = \frac{{{K_v}{W^t}}}{{FmY}}\)

\(\therefore {W^t} = \frac{{\sigma ; \times ;F; \times ;my}}{{{K_v}}}\)

\({W^t} = \frac{{400 \times 50 \times 1.25 \times 0.322}}{{1.26}} = 6388.88;N\)

\(\therefore Radius;of;pinion = \frac{{No.;of;teeth \times Module}}{2} = ;\frac{{20 \times 1.25}}{2} = 12.5;mm\)

Torque = W^t × r_p

Torque = 6388.88 N × 12.5 mm = 79.861 N-m

Power = Torque × Angular speed

\(Power = \frac{{2\pi NT}}{{60}} = \frac{{2\pi ; \times ;1200; \times ;79.86;}}{{60}} = 10.035{\rm{;kW}};\)

Concept:

Time taken to fill mould cavity

\(t = \frac{{Volume}}{{Discharge;Rate}}\)

Discharge rate (Q) = Area of the base of sprue × Velocity at the base of the sprue

Velocity at the base of sprue \(v = \sqrt {2;g{h_t}}\)

Where h_t is the length of the sprue

Calculation:

Given, V = 1200 cm³, h_t = 10 cm, A = 2 cm²,  g = 9.81 m/s² = 981 cm/s²

Velocity st the base of the sprue \(v = \sqrt {2;g{h_t}} = \sqrt {2 \times 981 \times 10} ;cm/sec\)

\(t = \frac{V}{Q} = \frac{{1200}}{{2 \times \sqrt {2 \times 981 \times 10} }} = 4.28;sec\)

Concept:

In metal cutting

Specific cutting energy \(= \frac{{Power}}{{Metal;removal;rate}}\)

\({U_c} = \frac{{Cutting;force \times Velocity}}{{Feed \times Depth;of;cut \times Velocity}}\)

\({U_c} = \frac{{{F_c}}}{{f \times ;d}}\)

Uncut chip thickness, t = f × sin λ

Where λ = Approach angle = 90 – Side cutting edge angle

In American Standards System (ASA) system the tool signature is as follows:

Back rake angle – side rake angle – end relief angle (ERA) – side relief angle – end cutting Edge angle – side Cutting Edge Angle- Nose Radius (inch)

Calculation:

Given that U_c = U_o (t₁)^-0.4

Also \({U_c} = \frac{{{F_c}}}{{f \times ;d}}\)

If both the specific cutting energy is equated to each other then

\(\frac{{{F_c}}}{{f \times d}} = {U_o} \times {\left( {{\rm{f;}} \times {\rm{;sin;\lambda }}} \right)^{ - 0.4}}\)

If other parameters are to remain same then Fc ∝ (sin λ)-0.4

For the first case, C_s1 = 30° 

For the second case C_s2 = 0° 

\({\lambda _1} = 90^\circ - {C_{{s_1}}} = 90 - 30^\circ = 60^\circ\)

\({\lambda _2} = 90^\circ - {C_{{s_2}}} = 90 - 0^\circ = 90^\circ\)

\(\therefore \left( {\frac{{{F_{{c_2}}} - {F_{{c_1}}}}}{{{F_{{c_1}}}}}} \right) \times 100% = \left[ {{{\left( {\frac{{\sin {\lambda _2}}}{{\sin {\lambda _1}}}} \right)}^{ - 0.4}} - 1} \right] \times 100%\)

\(\Rightarrow \left( {\frac{{{F_{{c_2}}} - {F_{{c_1}}}}}{{{F_{{c_1}}}}}} \right) \times 100% = \left[ {{{\left( {\frac{{\sin 90}}{{\sin 60}}} \right)}^{ - 0.4}} - 1} \right] \times 100%\)

\(\therefore \left( {\frac{{{F_{{c_2}}} - {F_{{c_1}}}}}{{{F_{{c_1}}}}}} \right) \times 100% = - 5.59%\)

Concept:

For shaper machine

Total time taken \(= \frac{{L.D\left( {1 + m} \right)}}{{Velocity}}\) 

Where L = Length of cut, D = No. of double stroke \(= \frac{{Width}}{{Feed}}\)

\(m = \frac{{Time;of;return}}{{Time;of;cutting}} = \frac{{Velocity;of;cutting}}{{Velocity;of;return}}\)

\(T = \frac{{LW\left( {1; + ;m} \right)}}{{f; \times ;v}}\)

Calculation:

For Machine S1

\({T_1} = \frac{{L{W_1}\left( {1 + m} \right)}}{{{f_1} \times v}}\)

W1 & f1 are width & feed for machine S1 other parameters are constant, then, 

For Machine S₂

W₂ = 1.10 W₁

f₂ = 0.90 f₁

\(\therefore {T_2} = \frac{{L{W_2}\left( {1 + m} \right)}}{{{f_2} \times v}} = \frac{{1.10{W_1} \times L\left( {1 + m} \right)}}{{0.9{f_1} \times v}} = 1.22\frac{{L{W_1}\left( {1; + ;m} \right)}}{{{f_1}; \times ;v}};\)

T₂ = 1.22 T₁

\(\therefore \frac{{{T_1}}}{{{T_2}}} = 0.8181\)

Concept:

No. of tool per turning \(= \frac{{Total;time;taken;in;turning}}{{Tool;life;of;single;tool}}\)

For maximum production rate

Tool life \(= \left( {\frac{{1; - ;n}}{n}} \right) \times Tool;change;time\)

Where n is the tool life exponent

Total time taken \(= \frac{L}{{fN}}\)

Optimum tool cost per piece = No. of tool per turning (at maximum production rate) × Grinding cost of single tool

Calculation:

Given, Tool life equation, VT^0.2 = 24, f = 0.2 mm/rev, T_c = 1 min, Tool regrinding Cost = 20 Rs.

For maximum production rate

Tool life \(= \left( {\frac{{1; - ;n}}{n}} \right) \times Tool;change;time\)

\(T = \left( {\frac{{1 - 0.2}}{{0.2}}} \right) \times 1 = 4;min\)

Now, from, VT^0.2 = 24

\(V = \frac{{24}}{{4^{0.2}}} = 18.188;m/min\)

Also V = πDN m/min

∴ πDN = 18.188 m/min

\(N = \frac{{18.188}}{{\pi \times \frac{{25}}{{1000}}}}\)

N = 231.576 rpm

Total time taken in turning \( = \frac{L}{{fN}} = \frac{{250}}{{0.2; \times ;231.576}} = 5.397{\rm{;min}};\) 

No. of tool per turning \( = \frac{{Total;time;taken;in;turning}}{{Tool;life;of;single;tool}} = \frac{{5.397}}{4} = 1.349\)

Optimum tool cost per piece = No. of tool per turning × Grinding cost of single tool

Optimum tool cost = 1.349 × 20 = Rs. 26.98

Concept:

In one revolution of the stepper motor the distance traversed will be the pitch of the lead screw.

Therefore Velocity will be equal to = Revolution per minute × distance traversed per revolution

Time taken to traverse the distance = \(\frac{{distance}}{velocity}\)

Since point P is moved by rapid positioning command (G00) and there are pair of stepping motor so time taken to reach the point will be different for motors in different direction. The maximum time taken by the any individual motor will be the time required to position the point ‘P’ to the point ‘Q’.

Calculation:

Given, N = 800 rpm, p = 0.5 mm, Δx = 600, Δy = 300

\(V = 800;\left( {\frac{{rev}}{{min}}} \right) \times 0.5\left( {\frac{{mm}}{{rev}}} \right) = 400\frac{{mm}}{{min}};\)

Now the time taken to traverse in the x-direction

\(T = \frac{{600}}{{400}} = 1.5;min\)

Now the time taken to traverse in the y-direction

\(T = \frac{{300}}{{400}} = 0.75;min\)

So the time taken to reach from Point P to point Q will be 1.5 minutes, the motor in the y-direction will stop after 0.75 minutes and the motor in the x-direction will run for 1.5 minutes.

Concept:

In a back order model every time after stock out period when quantity Q is received the inventory level reaches its maximum level.

Economic order quantity \( = \sqrt {\frac{{2D{C_o}}}{{{C_h}}}\left( {\frac{{{C_b} + {C_h}}}{{{C_b}}}} \right)}\) 

Maximum inventory \(= \sqrt {\frac{{2D{C_o}}}{{{C_h}}} \times \left( {\frac{{{C_b}}}{{{C_b} + {C_h}}}} \right)}\) 

Where, D is the demand per year, C_b is the backordering cost per unit per year and C_h is the carrying cost per unit per year

C_o is the ordering cost per order

Calculation:

Given, D = 10000/year, C_h = 4 /unit/year, C_o = 300/order, C_b = 25 /unit/year

\({Q_{max}} = \sqrt {\frac{{2 \times 10,000 \times 300}}{4} \times \frac{{25}}{{25 + 4}}}\)

Q_max = 1137.147 units

Concept:

For steady incompressible and inviscid flow

Bernoulli’s equation is valid i.e.

\(\frac{P}{{ρ g}} + \frac{{{V^2}}}{{2g}} + Z = Constant\)

Continuity equation for incompressible fluid,

A₁V₁ = A₂V₂

Calculation:

Given, ρ = 1.23 kg/m3

From continuity equation

A₁V₁ = A₂V₂

0.2 × V₁ = 0.02 × 50

V₁ = 5 m/s

From Bernoulli’s Equation

\(\frac{{{P_1}}}{{ρ g}} + \frac{{{V_1}}}{{2g}} + {Z_1} = \frac{{{P_2}}}{{ρ g}} + \frac{{V_2^2}}{{2g}} + {Z_2}\)

∵ Z₁ = Z₂ & P₂ = P_atm

\(\therefore \frac{{{P_1}; - ;{P_{atm}}}}{{ρ g}} = \frac{{V_2^2; - ;V_1^2}}{{2g}}\)

\({P_1} - {P_{atm}} = \frac{{1.23}}{2}\left( {{{50}^2} - {5^2}} \right)\)

P₁ – P_atm = 1522.125 Pa

P₁ – P_atm = 1.52 kPa

Concept:

Bernoulli’s equation is applicable for an incompressible flow and it states that the total head of the water along the direction of flow remains constant 

\(\frac{{{P_A}}}{{\rho g}} + \frac{{V_A^2}}{{2g}} + {Z_A} = ;\frac{{{P_B}}}{{\rho g}} + \frac{{V_B^2}}{{2g}} + {Z_B} + {h_l}\)

Where\(\frac{P}{{\rho g}}\) is the pressure head

\(\frac{{{V^2}}}{{2g}}\) is the velocity head

Z is the datum head

Calculation:

If we apply the Bernoulli’s equation between the points A and B then,

\(\frac{{{P_A}}}{{\rho g}} + \frac{{V_A^2}}{{2g}} + {Z_A} = ;\frac{{{P_B}}}{{\rho g}} + \frac{{V_B^2}}{{2g}} + {Z_B} + {h_l}\)

∵ V_A = V_B So we can avoid velocity head

Now, \(\frac{{186 \times {{10}^3}}}{{{{10}^3} \times 10}} + 24.5 = ;\frac{{260 \times {{10}^3}}}{{{{10}^3} \times 10}} + 9.1 + {h_l}\)

⇒ h_l = 43.1 – 35.1 = 8 m

Concept:

According to Wein’s displacement law, the product of wavelength (λ_max) corresponding to the maximum monochromatic emissive power and the absolute temperature of a black body (T) is constant.

λ_max T = constant i.e. 2900 μm-K

i.e. λ₁T₁ = λ₂T₂

Calculation:

Given, T₁ = 3000 K, λ₂ = 1.2 λ₁

Now from Wien’s displacement law

λ₁T₁ = λ₂T₂

⇒ λ₁ × 3000 = 1.2 λ₁ × T₂

T₂ = 2500 K

Concept:

Heat transfer axially & circumferentially must be equal to heat gain by water which gives a rise in water temperature.

Heat flux × (Circumferential area + Axial area) = Heat gain by water

Heat gain by water = ṁ × C_p × [(T_m)_exit – (T_m)_inlet] _water

The heat flux at the exit will be equal to heat convected away by the water, 

Therefore According to Newton’s law of cooling at the exit of the tube, 

\(q = h\left[ {{{\left( {{T_{inner}}} \right)}_{exit;tube}} - {{\left( {{T_{exit}}} \right)}_{water}}} \right]\)

where h is the heat transfer coefficient

Calculation:

Given, For water

C_p = 4.179 kJ/kg⋅K, (T_m)_exit = 70°C and (T_m)_inlet = 10°C, ṁ = 720 kg/h = 720/3600 kg/s

Let us draw the given situation 

Then,

Circumferential area = πDL = π × 0.03 × 20

⇒ Circumferential area = 1.884 m²

Axial area \(= 2 × \frac{\pi }{4}{d^2} = 2 × \frac{\pi }{4}{\left( {0.03} \right)^2}\)

⇒ Axial area = 1.413 × 10^-3 m²

(∵ The axial area is very small so, it can be neglected in the calculation.)

Heat flux × (Circumferential area + Axial area) = Heat gain by water

\(q × \left( {1.884 + 1.413{\rm{;}} × {\rm{;}}{{10}^{ - 3}}} \right) = \frac{{720}}{{3600}} × 4.179 × \left( {70 - 10} \right)\)

q = 26.597 kW/m² = 26.597 × 10³ W/m2

According to Newton’s law of cooling

\(q = h\left[ {{{\left( {{T_{inner}}} \right)}_{exit;tube}} - {{\left( {{T_{exit}}} \right)}_{water}}} \right]\)

26.597 × 10³ = 1697 [T - 70]

\(T = 70 + \frac{{26.597{\rm{;}} × {\rm{;}}{{10}^3}}}{{1697}} = 85.67^\circ C\)

Concept:

Work interaction by air inside the cylinder = Work done by the piston + Work done on the spring to compress it.

Work done by the piston before it reaches the stops will be a through a constant pressure process if spring is not there

W_{piston ­}= P.dV

Work done on the spring is the change of potential energy of the spring = \(\frac{1}{2} \times k \times {\delta ^2}\)

where k is the stiffness of the spring and δ is the compression in the spring.

Total Work interaction = Work done by the piston at constant pressure + Work done upon the spring 

Calculation:

Given, initial pressure in the cylinder P = 105 kPa = 105 × 10³ Pa and Area of the piston A = 300 cm² = 300 × 10^-4 m² and dx = 16 cm

Work interaction = Work done by the piston at 105 kPa + Work done upon the spring 

Work done by the piston = Pdv

dv = Area of the piston × displacement of the piston

∴ Work done by the piston= P × A × dx

W_piston = 105 × 10³ Pa × 300 × 10^-4 m² × 16 × 10^-2 m = 504 J

Work done on spring = Strain energy absorbed by spring

\({W_{spring}} = \frac{1}{2} \times k \times {\delta ^2}\)

\({W_{spring}} = \frac{1}{2} \times \left( {12.5\frac{N}{{mm}} \times {{80}^2};mm^2} \right) = 40,000;N - mm = 40;N - m = 40;J\)

∴ Work done on spring = 40 J

∴ Work interaction = 504 + 40 = 544 J

Concept:

For saturated air relative humidity (ϕ) is 100% i.e. P_v = P_vs

i.e. partial pressure of water vapour in the moist air is equal to the saturation pressure of the water vapour

Specific Humidity: It is the ratio of the mass of water vapour to the mass of air at a given temperature and volume.

Specific humidity \(\omega = \frac{{{m_v}}}{{{m_a}}} \times \frac{{{P_v}}}{{{P_t} - {P_v}}}\)

where P_v is the partial pressure of water vapour and P_t is the total pressure of moist air.

Calculation:

If we find the specific humidity before the cooling coil and after the cooling coil then the difference between two will give the amount of water condensed in the cooling coil.

Before entering into the cooling coil,

ϕ = 80% = 0.8, m_v = 18 g/mol, m_a = 28.94 g/mol, P_t = 105 kPa

At 30°C P_vs = 4.24 kPa

Relative humidity (ϕ) \(= \frac{{{P_v}}}{{{P_{vs}}}}\)

∴ P_v = ϕ × P_vs = 0.8 × 4.24

⇒ P_v = 3.392 kPa

Specific humidity (ω) \(= \frac{{{m_v}}}{{{m_a}}} \times \frac{{{P_v}}}{{{P_t} - {P_v}}}\)

\({\left( \omega \right)_{moistair}} = \frac{{18}}{{28.94}} \times \frac{{{P_v}}}{{105 - {P_v}}}\)

\(\therefore {\omega _{moistair}} = \frac{{18}}{{28.94}} \times \frac{{3.392}}{{105 - 3.392}} = ;0.0207{\rm{;kg}}/{\rm{kg;of;dry;air}}\)

Now after the exit of the cooling coil

At 15°C P_vs = 1.7 kPa

For saturated air ϕ = 1

⇒ P_v = P_vs = 1.7 kPa

\({\omega _{saturated;air}} = \frac{{18}}{{28.94}} \times \frac{{1.7}}{{100 - 1.7}} = ;0.0107{\rm{;kg}}/{\rm{kg;of;dry;air}}\)

∴ Mass of water condensing = ω_{moist air} – ω_{saturated air}

⇒ Mass of water condensing = 0.0207 – 0.0107 = 0.010 kg/kg of dry air

Mass of water condensing out from duct = 10 gm/kg of dry air

Concept:

When steam is reheated, after partial expansion and then expanded further in another turbine the work done is sum of the enthalpies drop in both the turbines.

If the given system of turbines is drawn on the T-s diagram then

1-2 is the expansion in the high-pressure turbine and 3-4 is the expansion in the low-pressure turbine.

Work done by both the turbine = (h₁ – h₂) + (h₃ – h₄)

Calculation:

h₁ is the enthalpy at 10 MPa & 500°C

∴ h₁ = 3373.6 kJ/kg

h₂ is the enthalpy at 1 MPa of saturated vapour

∴ h₂ = 2778.1 kJ/kg

h₃ i.e. enthalpy at 1 MPa of superheated vapour at 500°C

∴ h₃ = 3478.4 kJ/kg

Now for h₄

The expansion process 3-4 is isentropic, for an isentropic process, entropy remains constant,

∴ s₃ = s₄

s₃ i.e. entropy at 1 MPa of superheated vapour at 500°C

s₃ = s₄ = 7.7621 kJ/kg

s₄ = s_{f at 20 KPa} + s_{fg at 20 KPa}

s_{f at 20 KPa} = 0.8319 kJ/kg

s_{fg at 20 KPa} = s_g – s_f = 7.9085 – 0.8319 = 7.0766 kJ/kg

∴ 7.7621 = 0.8319 + x₄ 7.0766

x₄ = 0.979

h₄ i.e. enthalpy of wet vapour at 20 kPa

h₄ = h₁ + x₄ h_fg

h_f  at 20 KPa = 251.38 kJ/kg

h_g at 20 KPa = 2609.7 kJ/kg

h₄ = 251.38 + 0.979 (2609.7 – 251.38)

h₄ = 2560.175 kJ/kg

∴ Work done by both turbine = (h₁ – h₂) + (h₃ – h₄)

W_T = (3373.6 – 2778.1) + (3478.4 – 2560.175)

W_T = 1513.725 kJ/kg

Concept:

For a Diesel cycle, the thermal efficiency is given by,

\({\eta _{th}} = 1 - \frac{1}{{\gamma {r^{\gamma - 1}}}} \times \frac{{r_c^\gamma - 1}}{{\left( {{r_c} - 1} \right)}}\)

where, r is the compression ratio, γ is the ratio of specific heats, r_c is the cut off ratio

Calculation:

Case (i)

r = 15, γ = 1.3 and r_c = 2

\({\eta _{th}} = 1 - \frac{1}{{1.3{{\left( {15} \right)}^{1.3; - ;1}}}} \times \frac{{{2^{1.3}} - ;1}}{{\left( {2; - ;1} \right)}}\)

η_th = 0.5008 = 50.08%

Case (ii)

r = 21, γ = 1.3 and r_c = 2

\({\eta _{th}} = 1 - \frac{1}{{1.3{{\left( {21} \right)}^{1.3; - ;1}}}} \times \frac{{{2^{1.3;}} - ;1}}{{\left( {2; - ;1} \right)}}\)

η_th = 0.5487 = 54.87%

(ηnew|CR=21) – (ηold|CR=15) = 54.87% - 50.08% = 4.79%