GATE ME 2012 Official Paper (Previous Year Paper)

As q = total number of quantities produced.

Total cost = 5q²

Total sales = 50q

Profit (P) = total cost – total sales

P = 5q² – 50q

Differentiating w.r.t q

\(\frac{{dP}}{{dq}} = 10q - 50 = 0\)

\(q = \frac{{50}}{{10}} = 5\)

\(\frac{{{d^2}P}}{{d{q^2}}} = 10 > 0\)

∴ At q = 5, profit (P) is maximum.

The correct answer is option 1- that

Explanation 

To answer this question, we need to understand the meaning of defining and non-defining clauses. 

• A defining clause (also called an essential clause or a restrictive clause) gives information essential to the meaning of the sentence. We always use 'that' with defining clauses.

Example: My bike that has a broken seat is in the garage.

• Unlike defining clauses, non-defining clauses (also called nonessential or nonrestrictive clauses) don’t limit the meaning of the sentence. You might lose interesting details if you remove them, but the meaning of the sentence wouldn’t change. We always use 'which' with non-defining clauses.

Example: The goat was standing on the roof of the house which was abandoned.

In the given sentence, if we consider the first clause, the information given is incomplete and it can only be completed via the introduction of the second clause. Hence, we have to use 'that' in this sentence.

The correct answer is option 4. 

Explanation 

As we can see, the question has been framed to test our grammatical knowledge regarding the use of the adjective 'less'. 

• 'Less' has been used correctly in option 1. Here, 'less' means 'minus'.
• 'Less' has been used correctly in option 2 as well. It has been used here to compare the country's expenditure to that of Bangladesh.
• 'Less' has been used correctly in option 3 as well. If we don't use the word 'than' after 'less', we have to use the comparative form of less- 'lesser'.
• The usage of 'less' is wrong in option 4. 'Less' is generally used as a way of comparing two quantities or amounts. Here, the word 'less' needs to be replaced with the words 'little' or 'low'.

The correct answer is option 1- Diminish Explanation

Mitigate: make (something bad) less severe, serious, or painful; lessen the gravity of (an offence or mistake)

Diminish: make or become less; decrease in size, degree etc.

Thus we can see that among the options, 'diminish' is closest in meaning to the word 'mitigate'. They are near-synonyms.

The meaning of the other words:

• Divulge: make known (private or sensitive information)
• Dedicate: devote (time or effort) to a particular task or purpose
• Denote: be a sign of, indicate; stand as a name or symbol for

The correct answer is option 2- setbacks

Explanation   

Setback: something that happens that delays or prevents a process from developing

From the given context, it is clear that the mission finally resolved the conflict after many failed attempts and obstacles. Thus, 'setback' is the correct word for the given blank.

The meaning of the other words:

• Attempt: an effort to achieve or complete a difficult task or action
• Meeting: an assembly of people for a particular purpose, especially for formal discussion
• Delegation: a group of people who have been sent somewhere to have talks with other people on behalf of a larger group of people.

You may be confused regarding the use of 'attempts' and 'setbacks'. Attempt does not have any negative connotation to it, whereas setback has one. As the sentence is specifically referring to fruitless attempts taken in the past, 'setbacks' would be a more apt choice.

Explanation:

• Clearly, in the given question there is nothing related to gender is mentioned in the advertisement, eliminating options 1)
• Xenophobic means having or showing a dislike of or prejudice against people from other countries. There, is no such thing related to people of other countries is mentioned in the advertisement, eliminating it too.
• The requirements that are mentioned for the post is designed to make the post attractive, Transportation paid, expenses reimbursed are mentioned. So option 4) is the most suitable answer.

The given sequence is:

Explanation:

Let E(x) represent mean

Then by the property of mean

E(Cx) = cE(x) where c is constant

And E(x + c) = E(x) + E(c) = E(x) + c

∴ Options P and R always holds true.

Now,

Let Var (x) represent the variance

Then var (x + c) = var (x) + var (c)

 var (x + c) = var (x)    {∵ var (c) = 0}

∴ S.D (x + c) = S.D (x)

Now,

var (cx) =(c²)var (x)

S.D (cx) = (c)S.D (x)

∴ Options Q and S are incorrect.

Calculation: 

Let 100 shock absorbers are supplied

so, X supplies = 60 (60% of 100)

Y supplies = 40 (40% of 100)

Reliable supply by X = 96% of 60 = 57.6

Reliable supply by Y = 72% of 40 = 28.8

Required probability \(P\left( Y \right) = \frac{{28.8}}{{57.6 + 28.8}} = 0.33\)

Given:

y = 2x – 0.1x²

differentiating w.r.t x

\(\frac{{dy}}{{dx}} = 2 - 0.2x = 0\)

\(x = \frac{2}{{0.2}} = 10\)

\({\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)_{x = 10}} = - 0.2 < 0\)

∴ At x = 10, y will have a maximum value.  

y_max = y(10) = (2 × 10) - 0.1(10²) = 10

Aggregate production planning:

• It is planning in which the general level of employment and output is planned to balance supply and demand for a period of fewer than 18 months.
• It is intermediately ranged capacity planning, used to establish employment levels, output rates, inventory levels, subcontracting and backorders for products that are aggregated.
• It is not specifically focussed on the individual product but deals with products in aggregate.

Concept:

Milling time (t) can be calculated by:

\(t = \frac{{distance}}{{feed;rate}} \)

Calculation:

Given:

OB = 100 mm, AB = 100 mm

\(Feed;rate;\left( f \right) = 50\frac{{mm}}{{min}} = \frac{{50}}{{60}}\frac{{mm}}{{sec}} = 0.833;\frac{{mm}}{{sec}}\)

Distance travelled by tool (OA)

\(OA = \sqrt {O{B^2} + A{B^2}} = \sqrt {{{100}^2} + {{100}^2}} = 141.421;mm\)

Milling time (t) is:

\(t = \frac{{distance}}{{feed;rate}} = \frac{{141.421}}{{0.833}} = 169.7;seconds \simeq 170;seconds\)

Concept:

In a forging process

Volume before forging = Volume after forging

In the case of cylinder 

πd₁²h₁ = πd₂2h₂

where d₁ = initial diameter, d₂ = final diameter, h₁ = initial height, h₂ = final height

Calculation:

Given:

d1 = 100 mm, h1 = 50 mm, h2 = 25 mm

Volume before forging = Volume after forging

\(π d_1^2;{h_1} = π d_2^2;{h_2}\)

\({d_2} = {d_1} \times \sqrt {\frac{{{h_1}}}{{{h_2}}}}\)

\({d_2} = 100 \times \sqrt {\frac{{50}}{{25}}} = 141.42\)

\({\rm{Percentage;change;in;diameter}} = \frac{{{{\rm{d}}_2} - {{\rm{d}}_1}}}{{{{\rm{d}}_1}}} \times 100 = \frac{{141.42 - 100}}{{100}} \times 100 = 41.42{\rm{;% }}\)

Concept:

Degree of Reaction (R) can be calculated by:

\(R = \frac{{change;in;pressure;energy;inside;runner}}{{change;in;total;energy;inside;runner}}\)

\(Change;in;total;energy;\left( {{\rm{\Delta }}TE} \right) = \frac{{V_1^2 - V_2^2}}{{2g}} + \frac{{U_1^2 - U_2^2}}{{2g}} + \frac{{W_2^2 - W_1^2}}{{2g}}\)

where

\(\frac{{{\rm{V}}_1^2 - {\rm{V}}_2^2}}{{2{\rm{g}}}} = {\rm{kinetic;energy;head}}\)

\(\frac{{{\rm{U}}_1^2 - {\rm{U}}_2^2}}{{2{\rm{g}}}} = {\rm{centrifugal;energy;head}}\)

\(\frac{{{\rm{W}}_2^2 - {\rm{W}}_1^2}}{{2{\rm{g}}}} = {\rm{relative;velocity;head}}\)

\(\frac{{{\rm{U}}_1^2 - {\rm{U}}_2^2}}{{2{\rm{g}}}} + \frac{{{\rm{W}}_2^2 - {\rm{W}}_1^2}}{{2{\rm{g}}}} = {\rm{change;in;pressure;energy}}\)

Calculation:

Given:

U₁ = U₂, V₂ = W₁ and V₁ = W₂

\({\rm{Change;in;pressure;energy}} = {\rm{;}}\frac{{{\rm{U}}_1^2 - {\rm{U}}_2^2}}{{2{\rm{g}}}} + \frac{{{\rm{W}}_2^2 - {\rm{W}}_1^2}}{{2{\rm{g}}}} = \frac{{{\rm{W}}_2^2 - {\rm{W}}_1^2}}{{2{\rm{g}}}}\) {∵ U₁ = U₂}

\(Change;in;total;energy;\left( {{\rm{\Delta }}TE} \right) = \frac{{V_1^2 - V_2^2}}{{2g}} + \frac{{U_1^2 - U_2^2}}{{2g}} + \frac{{W_2^2 - W_1^2}}{{2g}}\)

\({\rm{\Delta }}TE = ;\frac{{V_1^2 - V_2^2}}{{2g}} + ;\frac{{W_2^2 - W_1^2}}{{2g}}\)   {∵ U1 = U2}

\({\rm{\Delta }}TE = ;\frac{{W_2^2 - W_1^2}}{{2g}} + \frac{{W_2^2 - W_1^2}}{{2g}}\)   {∵ V2 = W1 and V1 = W₂}

∴ \({\rm{\Delta }}TE = ;\frac{{W_2^2 - W_1^2}}{g}\)

\(R = \frac{{change;in;pressure;energy;inside;runner}}{{change;in;total;energy;inside;runner}}\)

\(R = \frac{{\frac{{W_2^2 - W_1^2}}{{2g}}}}{{\frac{{W_2^2 - W_1^2}}{g}}} = \frac{1}{2} = 0.5\)

Explanation:

Effectiveness (ϵ)

Effectiveness is defined as the ratio of heat transfer rate with a fin to heat transfer rate without fin.

\(\epsilon = \frac{{{{\dot Q}_{with;fin}};}}{{{{\dot Q}_{without;fin}}}}\)

For a long fin

\(\epsilon = \frac{{{{\dot Q}_{with;fin}};}}{{{{\dot Q}_{without;fin}}}} = \sqrt {\frac{{KP}}{{h{A_c}}}} \)

where K = thermal conductivity of fin, P = perimeter of fin

h = heat transfer coefficient, A_c = cross-section area of fin

From the above formula, we can conclude:

Case 1: If \(\frac{P}{{{A_c}}}\) increases, the effectiveness of fin increases.

• A_c must be small (thin fin) and fins must be closed (not too close).
• If fins are too closed it will abstract flow of air which will decrease the effectiveness of fin.

Case 2: if K increases, the effectiveness of fin increases.

• If K is large, the temperature drop along the length will be less.
• And the temperature drop between fin and surrounding will be more and heat transfer will be more.

Case 3: if h decreases, the effectiveness of fin increases.

• Fins are more effective under free convection ( h is less).

Explanation:

Entropy change for an ideal gas can be calculated by:

\({S_2} - {S_1} = m{c_p}\ln \left( {\frac{{{T_2}}}{{{T_1}}}} \right) - mRln\left( {\frac{{{P_2}}}{{{P_1}}}} \right)\)

where m = mass of the gas, c_p = specific heat at constant pressure, R = gas constant

T₁ = initial temperature, T₂ = final temperature, P₁ = initial pressure, P₂ = final pressure

 As the process is isothermal i.e. T₁ = T₂

∴ \({S_2} - {S_1} = m{c_p}\ln \left( 1 \right) - Rln\left( {\frac{{{P_2}}}{{{P_1}}}} \right)\)  

\({S_2} - {S_1} = Rln\left( {\frac{{{P_1}}}{{{P_2}}}} \right)\)     {∵ ln(1) = 0} 

Entropy change for an ideal gas can also be calculated by:

\({S_2} - {S_1} = m{c_v}\ln \left( {\frac{{{T_2}}}{{{T_1}}}} \right) + mRln\left( {\frac{{{V_2}}}{{{V_1}}}} \right)\)

where c_v = specific heat at constant volume, V₁ = initial volume, V₂ = final volume

Given:

Angular velocity of disc (ω) = 1 rad/s, PO = 50 mm, PS = 45 mm, OQ = 25 mm, SO = 5 mm

In ΔPQO

PO² = PQ² + OQ² ⇒ PQ2 = PO2 - OQ2

\(PQ = \sqrt {{{\left( {50} \right)}^2} - {{\left( {25} \right)}^2}} =43.3\ mm\)

As ΔPQO ~ ΔSRO

\(\frac{{PQ}}{{SR}} = \frac{{PO}}{{SO}}\)

\(\frac{{43.3}}{{SR}} = \frac{{50}}{{5}} ⇒ SR=4.33\ mm\)

As point Q and Point R are situated on same link

∴ Velocity of Q = velocity of R

\({V_Q} = {V_R} = SR \times ω = 4.33 \times 1 = 4.33\ m/sec.\)

Angular velocity of PQ

\({ω _{PQ}} = \frac{{{V_Q}}}{{PQ}} = \frac{{4.33}}{{43.3}} = 0.1\ rad/sec.\)

Concept:

Flywheel

 A flywheel is used to control the variation in speed during each cycle of the engine.

The kinetic energy of flywheel can be calculated by:

K.E \(= \frac{1}{2}I{w^2}\)

where I = moment of inertia of the flywheel, ω = angular velocity of the flywheel

Calculation:

Given:

Mass of flywheel (m) = 20 kg, radius (r) = 200 mm = 0.2 m, N = 600 rpm

\(I = \frac{{m{R^2}}}{2} = \frac{{20 ~\times ~{{0.2}^2}}}{2} = 0.4;kg - {m^2}\)

\(w = \frac{{2\pi N}}{{60}} = \frac{{2~ \times~ \pi ~\times~ 600}}{{60}} = 62.83;rad/s\)

∴ \({\rm{K}}.{\rm{E}} = \frac{1}{2}{\rm{I}}{{\rm{\omega }}^2} = \frac{1}{2} \times 0.4 \times {\left( {62.83} \right)^2} = 789.5{\rm{;J}} \simeq 790{\rm{;J}}\)

Concept:

\({EI}\frac{{{d^2}y}}{{d{x^2}}} = M\)

Calculation:

By integrating the above equation we get,

\(EI\frac{{dy}}{{dx}} = Mx + {C_1}\)

Once again integrating, we get

\(EI;y = \frac{{M{x^2}}}{2} + {C_1}x + {C_2}\)

For cantilever beam at \(x = 0,\frac{{dy}}{{dx}} = 0;& ;x = 0,~y = 0\) (x taken from fixed-end).

From this we get C₁ = C₂ = 0,  Hence

\(y = \frac{{M{x^2}}}{{2EI}}\)  

Deflection will be maximum when x = L

∴ Maximum deflection is

 \({y_{max}} = \frac{{M{L^2}}}{{2EI}}\)

Concept:

Buckling load:

The load at which column buckle is termed as buckling load. Buckling load is given by:

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

where E = Young’s modulus of elasticity, I_min = Minimum moment of inertia, and L_e = Effective length

Calculation:

Given:

Critical Buckling load for column fixed at both ends (Pcr₁) \(= \frac{{4{\pi ^2}EI}}{{\begin{array}{*{20}{c}} {{L^2}} \end{array}}}\)

Critical Buckling load for a column hinged at both ends (Pcr₂) = \(\frac{{{\pi ^2}EI}}{{{L^2}}}\)

\(\therefore\frac{(P_{cr})_1}{(P_{cr})_2}=4\)

The function f(x) = x³ + 1 has a point of inflection at x = 0, since in the graph sign of the curvature (i.e., the concavity) is changed.

Alternate solution:

Given:

f(x) = x³ + 1

f^'(x) = 3x² = 0 ⇒ x = 0 (critical point)

f^"(x) = 6x

f"(0) = 0

Now, f^'''(x) = 6

f'''(0) = 6  (non zero)

As the first non zero derivative value occurs at third derivative which is an odd number.

∴ function f(x) has point of inflection at x = 0.

Concept:

The gradient of a scalar function f(x,y,z) is given by:

\({\rm{grad;f}} = \frac{{\partial {\rm{f}}}}{{\partial {\rm{x}}}}{\rm{\hat i}} + {\rm{;}}\frac{{\partial {\rm{f}}}}{{\partial {\rm{y}}}}{\rm{\hat j}} + \frac{{\partial {\rm{f}}}}{{\partial {\rm{x}}}}{\rm{\hat k}}\)

Unit outward normal vector of a function f at a point P is given by:

\({\bf{\vec n}} = \frac{{{{\left( {{\bf{grad}};{\bf{f}}} \right)}_{{\bf{at}};{\bf{p}}}}}}{{\left| {{\bf{grad}};{\bf{f}}} \right|}}\)

Calculation:

Given:

Spherical surface x² + y² + z² = 1, point P is \(\left( {\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }},0} \right)\)

F(x,y,z) = x² + y² + z² – 1 = 0

\({\rm{grad;f}} = \frac{{\partial {\rm{f}}}}{{\partial {\rm{x}}}}{\rm{\hat i}} + {\rm{;}}\frac{{\partial {\rm{f}}}}{{\partial {\rm{y}}}}{\rm{\hat j}} + \frac{{\partial {\rm{f}}}}{{\partial {\rm{x}}}}{\rm{\hat k}} = 2{\bf{x\hat i}} + 2{\bf{y\hat j}} + 2{\bf{z\hat k}}\)

(grad f)_{at P} = \(\frac{2}{{\sqrt 2 }}{\rm{\hat i}} + \frac{2}{{\sqrt 2 }}{\rm{\hat j}} + 0{\rm{\hat k}} = \sqrt 2 {\rm{\hat i}} + \sqrt 2 {\rm{\hat j}} + 0{\rm{\hat k}}\)

\(\left| {{\rm{grad;f}}} \right| = \sqrt {2 + 2} = \sqrt 4 = 2\)

Unit outward normal vector of function f at point P is given by:

\(\vec n = \frac{{{{\left( {{\rm{grad;f}}} \right)}_{{\rm{at;p}}}}}}{{\left| {{\rm{grad;f}}} \right|}} = \frac{{\sqrt 2 {\rm{\hat i}} + \sqrt 2 {\rm{\hat j}}}}{2} = \frac{1}{{\sqrt 2 }}{\bf{\hat i}} + \frac{1}{{\sqrt 2 }}{\bf{\hat j}}\)

Various Metal forming process and their properties are explained in the table given below.

Explanation:

• In AJM, the material removal takes place due to the impingement of the fine abrasive particles.
• These particles move with a high speed air (or gas) stream.
• When an abrasive particle impinges on the work surface at high velocity, the impact causes a tiny brittle fracture.
• The following air (or gas) carries away the dislodged small workpiece particle.

Effect of nozzle tip distance on MRR

• When the nozzle tip distance increases, the velocity of abrasive particles impinging on the work surface increases due to acceleration after they leave the nozzle. This, in turn, increases the MRR.
• With a further increase in nozzle tip distance, the velocity reduces due to the drag of atmosphere which initially checks the increase in the MRR and finally decreases it.

Concept: 

Maximum interference: Maximum shaft size - minimum hole size

Minimum interference: Minimum shaft size - maximum hole size

Calculation:

Given:

Maximum size of shaft = 25 + 0.040 = 25.040 mm

Minimum size of hole = 25 + 0.020 = 25.020 mm

Maximum interference = Maximum shaft size - minimum hole size

Maximum interference = 25.040 - 25.020  = 0.020 mm = 20 μm

Explanation:

Various heat treatment process and their methods are described in the table below:

Concept:

Head loss through due to friction in a pipe is given by:

\({h_f} = \frac{{fL{V^2}}}{{2gD}}\)

where f is the friction factor.

Calculation:

Given:

D = 200 mm = 0.2 m, f = 0.0225, L = 500 m, Q = 0.2 m³/s

\(Area;\left( A \right) = \frac{\pi }{4}{D^2} = \frac{\pi }{4} \times {\left( {0.2} \right)^2} = 0.0314;{m^2}\)

Discharge (Q) = Area (A) × Velocity (V)

\(V = \frac{Q}{A} = \frac{{0.2}}{{0.0314}} = 6.369;m/s\)

Head loss due to friction is

\({h_f} = \frac{{fL{V^2}}}{{2gD}}\)

\({h_f} = \frac{{0.0225~ \times~ 500 ~\times ~{{\left( {6.369} \right)}^2}}}{{2 ~\times ~9.81 ~\times ~0.2}}=116.29 ~m\)

Alternate Solution:

Head loss due to friction is,

\({h_f} = \frac{{fL{V^2}}}{{2gD}} = \frac{{fL{\left(\frac{Q}{A}\right)}^2}}{{2gD}} =\frac{{fL{Q^2}}}{{\left({\frac{\pi}{4}{D^2}}\right)^2~\times ~2gD}} = \frac{{fL{Q^2}}}{{12\times{D^5}}}\)

\({h_f} = \frac{{fL{Q^2}}}{{12\times{D^5}}}\)

\({h_f} = \frac{{0.0225~ \times~ 500 ~\times ~{{\left( {0.2} \right)}^2}}}{{12 ~\times ~{\left(0.2\right)^5}}}=117.18 ~m\)

Explanation:

Q = Q_A + Q_R + Q­_T

_{\(1 = \frac{{{Q_A}}}{Q} + \frac{{{Q_R}}}{Q} + \frac{{{Q_T}}}{Q}\)}

1 = α + ρ + τ 

where

\(\alpha = \frac{{{Q_A}}}{Q} = Absorptivity\)

\(\rho = \frac{{{Q_R}}}{Q} = Reflectivity\)

\(\tau = \frac{{{Q_T}}}{Q} = Transmissivity\)

For an opaque body, transmissivity is zero i.e. τ = 0.

α + ρ = 1

For a transparent body, τ = 1.

α = ρ = 0

For a black body, α = 1.

ρ = τ = 0

Concept:

Enthalpy of the saturated mixture is given by:

h = h_f + x (h_g – h­_f)

where h_f = enthalpy of saturated liquid, h_g = enthalpy of saturated vapour, x = dryness fraction

Calculation:

Given:

Mass flow rate of steam = 10 kg/s

Enthalpy of steam at turbine entrance (h₁) = 3251 kJ/kg

At the exit, the pressure is 15 kPa, x = 0.9

At P₂ = 15 kPa, h_f = 225.94 kJ/kg, h_g = 2598.3 kJ/kg

Let the enthalpy of the saturated mixture is h_2.

h₂ = h_f + x (h_g – h­_f)

h₂ = 225.94 + 0.9(2598.3 – 225.94)

h₂ = 2361.064 kJ/kg

Power output of turbine can be calculated by:

P = ṁ(h₁ – h₂)

P = 10 × (3251 – 2361.064)

P =  8899.36 kW = 8.89 MW ≃ 8.9 MW

Concept:

Velocity ratio for helical hears can be calculated by:

\(\frac{{{N_2}}}{{{N_1}}} = \frac{{{D_1}\cos {\psi _1}}}{{{D_2}\cos {\psi _2}}}\)

where D₁ and D₂ are pitch diameters and ψ₁ and ψ₂ are respective helix angles.

Calculation:

Given:

D₁ = 80 mm, D₂ = 120 mm, ψ₁ = 30°, ψ₂ =22.5°, N₁ = 1440 rpm

Velocity ratio 

\(\frac{{{N_2}}}{{{N_1}}} = \frac{{{D_1}\cos {\psi _1}}}{{{D_2}\cos {\psi _2}}} = \frac{{80 \times \cos 30^\circ }}{{120 \times \cos 22.5^\circ }} = \frac{{80 \times 0.866}}{{120 \times 0.923}} = 0.625\)

N₂ = 0.625 × N₁ = 0.625 × 1440 = 900 rpm

Concept:

Linear velocity V is given by:

\(V=r×ω\)

where ω is angular velocity.

Tangential acceleration in the no-slip condition is

\(\frac{{dv}}{{dt}} = r\frac{{dω }}{{dt}} = r×α \)

a_t = r × α

Centripetal acceleration is given by:

a_c = r × ω²

The instantaneous velocity of the point of contact is zero.

So at the point of contact, Instantaneous tangential acceleration is also zero.

∴ Only centripetal acceleration is there at the point of contact.

∴ Net acceleration of the point of contact is

ac = r × ω2

Concept:

Hoop stress (σ_h) for a thin spherical shell is given by:

\({{σ _h}} = \frac{{Pd}}{{4t}}=\frac{{Pr}}{{2t}}\)

Hoop stress (σh) for thin cylindrical shell is given by:

\({{σ _h}} = \frac{{Pd}}{{2t}}=\frac{{Pr}}{{t}}\)

Calculation:

Given:

Now r^' = 1.01r, t^' = 0.99t

\({{σ _h'}} = \frac{{Pd'}}{{4t'}}=\frac{{Pr'}}{{2t'}}= \frac{{P\times 1.01r}}{{2\times 0.99t}}=\frac{{1.01}}{{0.99}}\times \frac{{P r}}{{2 t}}=1.0202 ;σ_h\)

Percentage change in hoop stress is:

\(% ;change = \frac{{\sigma _h' - {\sigma _h}}}{{{\sigma _h}}} \times 100 = \frac{{1.0202\sigma _h - {\sigma _h}}}{{{\sigma _h}}} \times 100 = 2.02% \)

The given curves are y = x and y = x²

Solving the equations, we get

x  = 0, x = 1

\(Area = \mathop \smallint \limits_0^1 \left( {x - {x^2}} \right)dx\)

\(= \left( {\frac{{{x^2}}}{2} - \frac{{{x^3}}}{3}} \right)_0^1 = \frac{1}{2} - \frac{1}{3}\)

\(= \frac{1}{6}sq;units\)

Alternate Solution:

Concept:

Area of a region can be calculated by:

\(\int!!!\int dxdy\)

Calculation:

Solving equation y = x² and we get y = x

We get intersection points i.e. (0,0) and (1,1)

Area of the region:

\(\mathop \smallint \limits_{x = 0}^{x = 1} \mathop \smallint \limits_{y = {x^2}}^{y = x} dy.dx\)

\(\mathop \smallint \limits_{x = 0}^{x = 1} \left[ y \right]_{{x^2}}^x.dx;;\)

\(\mathop \smallint \limits_{x = 0;}^{x = 1} \left( {x - {x^2}} \right)dx\)

\(\left[ {\frac{{{x^2}}}{2} - \frac{{{x^3}}}{3}} \right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}\)

Concept:

A function f(x) is continuous at x = a if,

Left limit = Right limit = Function value = Real and finite

A function is said to be differentiable at x =a if,

Left derivative = Right derivative = Well defined

Calculation:

Given:

f(x) = |x|

|x| = x for x ≥ 0

|x|= -x for x < 0

At x = 0

Left limit = 0, Right limit = 0, f(0) = 0

As

Left limit = Right limit = Function value = 0

∴ |X| is continuous at x = 0.

Now

Left derivative (at x = 0) = -1

Right derivative (at x = 0) = 1

Left derivative ≠ Right derivative

∴ |x| is not differentiable at x = 0

Concept:

We know 

⇒ 1 - cos x = 2 sin²(x/2)​

⇒ \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1\)

Calculation:

\(\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{1 - {\rm{co}}{{\rm{s}}x}}}{{{x^2}}} = \mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{2{{\sin }^2}\left( {\frac{x}{2}} \right)}}{{{x^2}}}\)

Multiply and divide the denominator by 4

\(\mathop {\lim }\limits_{x \to 0} \frac{{2 \times {{\sin }^2}\left( {\frac{x}{2}} \right)}}{{4 \times \left( {\frac{{{x^2}}}{4}} \right)}}\)= \(\frac{1}{2}\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{\sin \left( {\frac{x}{2}} \right)}}{{\left( {\frac{x}{2}} \right)}}} \right)^2} = \frac{1}{2} \)

Concept:

Diameter of punch for blanking operation can be calculated by:

Diameter of punch = (diameter of Blank) – (2 × radial clearance) – (die allowance)

Calculation:

Given:

diameter of Blank = 25 mm, radial clearance = 0.06 mm, die allowance = 0.05 mm

Diameter of punch = Diameter of Blank - 2 × radial clearance – die allowance

Diameter of punch = 25 - 2 × 0.06 - 0.05 = 24.83 mm

Concept:

In a rolling operation change in thickness can be calculated by:

ΔH = D(1 – cosα)

where ΔH = change in thickness, D = roll diameter, α = bite angle

Calculation:

Given:

ΔH = 10 % = 0.1, D = 410 mm

Now,

ΔH = D(1 – cosα)

\(cosα = 1 - \frac{{{\rm{\Delta }}H}}{D} = 1 - \frac{{0.1 \times 8}}{{410}}\)

α = cos^-1(0.998) = 3.57°

\(α = (3.51 \times \frac{\pi }{{180}})~radians\)

α = 0.062 radians

Concept:

\(\frac{{{V_{arc}}}}{{OCV}} + \frac{I}{{SCC}} = 1\)

where OCV = open-circuit voltage, SCC = short circuit current

Calculation:

Given:

V_arc  = 20 + 5L

At L = 5 mm

V_arc = 20 + 5(5) = 45 V and current (I) = 500 A

At L = 7 mm  

V_arc = 20 + 5(7) = 55 V and current (I) = 400 A

Using:

\(\frac{{{V_{arc}}}}{{OCV}} + \frac{I}{{SCC}} = 1\)

\(\frac{{45}}{{OCV}} + \frac{{500}}{{SCC}} = 1;;;;;;;;\left( {equation;1} \right)\)

\(\frac{{55}}{{OCV}} + \frac{{400}}{{SCC}} = 1;;;;;;;;\left( {equation;2} \right)\)

Multiplying equation 1 by '4' and equation 2 by '5' and subtracting equation 1 from equation 2.

We get,

\(\frac{{275}}{{OCV}} - \frac{{180}}{{OCV}} = 1\)

OCV = 275 – 180 = 95 V

Putting the value of OCV in equation 1

We get,

\(\frac{{45}}{{95}} + \frac{{500}}{{SCC}} = 1\)

SCC = 950 A

Explanation:

Total pressure at point 1

P₁ = ρ₁gh₁ + ρ₂gh₂ + ρ₃gh₃

Applying Bernoulli’s equation between point 1 and point 2

\(\frac{{{P_1}}}{{{\rho _3}g}} + \frac{{{V_1^2}}}{{2g}} + {Z_1} = \frac{{{P_2}}}{{{\rho _3}g}} + \frac{{V_2^2}}{{2g}} + {Z_2}\)

Here,

Z₁ = Z₂, V₁ = 0, P₂ = 0

Putting the values, we get

\(\frac{{{\rho _1}g{h_1} + {\rho _2}g{h_2} + {\rho _3}g{h_3}}}{{{\rho _3}g}} + 0 + 0 = 0 + \frac{{V_2^2}}{{2g}} + 0\)

\(\frac{{{\rho _1}}}{{{\rho _3}}}{h_1} + \frac{{{\rho _2}}}{{{\rho _3}}}{h_2} + {h_3} = \frac{{V_2^2}}{{2g}}\)

\(V_2^2 = 2g\left( {\frac{{{\rho _1}}}{{{\rho _3}}}{h_1} + \frac{{{\rho _2}}}{{{\rho _3}}}{h_2} + {h_3}} \right)\)

\({V_2} = \sqrt {2g\left( {\frac{{{\rho _1}}}{{{\rho _3}}}{h_1} + \frac{{{\rho _2}}}{{{\rho _3}}}{h_2} + {h_3}} \right)} \)

\({V_2} = \sqrt {2g{h_3}\left( {\frac{{{\rho _1}{h_1}}}{{{\rho _3}{h_3}}} + \frac{{{\rho _2}{h_2}}}{{{\rho _3}{h_3}}}; + 1} \right)} \)

Concept:

The effectiveness of a heat exchanger can be calculated by:

\(ϵ= \frac{{{{\rm{Q}}_{{\rm{actual}}}}}}{{{{\rm{Q}}_{{\rm{max}}}}}}\)

\({{\rm{Q}}_{{\rm{actul}}}} = {{\rm{̇ m}}_{\rm{h}}}{{\rm{c}}_{{\rm{ph}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{h}}2}}} \right) = {{\rm{̇ m}}_{\rm{c}}}{{\rm{c}}_{{\rm{pc}}}}\left( {{{\rm{T}}_{{\rm{c}}2}} - {{\rm{T}}_{{\rm{c}}1}}} \right)\)

\({{\rm{Q}}_{{\rm{max}}}} = {{\rm{C}}_{{\rm{min}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} - {{\rm{T}}_{{\rm{c}}1}}} \right)\)

\(ϵ= \frac{{{{\rm{Q}}_{{\rm{actual}}}}}}{{{{\rm{Q}}_{{\rm{max}}}}}} = \frac{{{{\rm{C}}_{\rm{h}}}\left( {{{\rm{T}}_{{\rm{h}}1}} ;- ;{{\rm{T}}_{{\rm{h}}2}}} \right)}}{{{{\rm{C}}_{{\rm{min}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} ;- ;{{\rm{T}}_{{\rm{c}}1}}} \right)}} = \frac{{{{\rm{C}}_{\rm{c}}}\left( {{{\rm{T}}_{{\rm{c}}2}} ;-; {{\rm{T}}_{{\rm{c}}1}}} \right)}}{{{{\rm{C}}_{{\rm{min}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} ;- ;{{\rm{T}}_{{\rm{c}}1}}} \right)}}\)

For a balanced heat exchanger:

\({{\rm{̇ m}}_{\rm{h}}}{{\rm{c}}_{{\rm{ph}}}} = {{\rm{̇ m}}_{\rm{c}}}{{\rm{c}}_{{\rm{pc}}}} ⇒ {{\rm{C}}_{\rm{h}}} = {{\rm{C}}_{\rm{c}}} = {{\rm{C}}_{{\rm{min}}}} = {{\rm{C}}_{{\rm{max}}}}\)

\({\rm{ϵ }} = \frac{{{{\rm{C}}_{\rm{h}}}\left( {{{\rm{T}}_{{\rm{h}}1}} ;- ;{{\rm{T}}_{{\rm{h}}2}}} \right)}}{{{{\rm{C}}_{{\rm{min}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} ;- ;{{\rm{T}}_{{\rm{C}}1}}} \right)}} = \frac{{{{\rm{T}}_{{\rm{h}}1}} ;- ;{{\rm{T}}_{{\rm{h}}2}}}}{{{{\rm{T}}_{{\rm{h}}1}} ;- ;{{\rm{T}}_{{\rm{c}}1}}}}\)

\({\rm{ϵ }} = \frac{{{{\rm{C}}_{\rm{c}}}\left( {{{\rm{T}}_{{\rm{c}}2}} ;- ;{{\rm{T}}_{{\rm{c}}1}}} \right)}}{{{{\rm{C}}_{{\rm{min}}}}\left( {{{\rm{T}}_{{\rm{h}}1}} ;- ;{{\rm{T}}_{{\rm{C}}1}}} \right)}} = \frac{{{{\rm{T}}_{{\rm{c}}2}}; -; {{\rm{T}}_{{\rm{c}}1}}}}{{{{\rm{T}}_{{\rm{h}}1}}; - ;{{\rm{T}}_{{\rm{c}}1}}}}\)

Log mean temperature difference for the counter-flow heat exchanger is:

\({LMTD} = \frac{{{Δ T_1};-;{Δ T_2}}}{{\ln \left( {\frac{{{Δ T_1}}}{{{Δ T_2}}}} \right)}}\)

where

\({Δ T_1} = {T_{{h1}}} - {T_{{c2}}};and;{Δ T_2} = {T_{{h2}}} - {T_{{c1}}}\)

Calculation:

Given:

T_h1 = 80 °C, T_c1 = 30 °C, ṁ_h = 0.5 kg/sec, c_ph = 4.18 kJ/kg-K, ṁ_c = 2.09 kg/sec, c_pc = 1 kJ/kg-K.

C_h = m_hc_h = 0.5 × 4.18 = 2.04 W/K 

C_c = m_cc_c = 2.09 × 1 = 2.09 W/K

∵ C_h = C_c = 2.09 W/K ⇒ balanced heat exchanger.

Effectiveness (ϵ):

\({\rm{ϵ }} = \frac{{{{\rm{T}}_{{\rm{h}}1}} ;- ;{{\rm{T}}_{{\rm{h}}2}}}}{{{{\rm{T}}_{{\rm{h}}1}} ;- ;{{\rm{T}}_{{\rm{c}}1}}}}\)

\(0.8 = \frac{{\left( {80; - ;{T_{{h2}}}} \right)}}{{\left( {80; -; 30} \right)}}\)

\({T_{{h2}}} = 40;^\circ C\)

\({m_h}{c_{{_h}}}\left( {{T_{{h1}}} - {T_{{h2}}}} \right) = {m_c}{c_{{_c}}}\left( {{T_{{c2}}} - {T_{{c1}}}} \right)\)

0.5 × 4.18 × (80 - 40) = 1 × 2.09 × (T_c2 - 30)

∴ T_c2 = 70 °C

ΔT₁ = T_h1 - T_c2 = 80 - 70 = 10 °C

ΔT₂ = T_h2 - T_c2 = 40 - 30 = 10 °C

In case of counter-flow heat exchanger if ΔT₁ = ΔT₂ then by L. hospital rule

ΔT1 = ΔT2 = LMTD

∴ LMTD = 10 °C

Explanation:

For a solid cube, strain in x, y and z-axis are:

\(\begin{array}{l} {\epsilon_x} = \frac{{{\sigma _x}}}{E} - \frac{{v\left( {{\sigma _y} + {\sigma _z}} \right)}}{E}\ {\epsilon_y} = \frac{{{\sigma _y}}}{E} - \frac{{v\left( {{\sigma _x} + {\sigma _z}} \right)}}{E}\ {\epsilon_z} = \frac{{{\sigma _z}}}{E} - \frac{{v\left( {{\sigma _x} + {\sigma _y}} \right)}}{E} \end{array}\)

From the symmetry of cube, \(\epsilon_x =\epsilon_y=\epsilon_z=\epsilon\)

and \(\sigma_x=\sigma_y=\sigma_z=\sigma\)

Because of the rise in temperature, the cube will try to expand (tensile strain), and as the walls are constrained so a compressive force will develop on the walls of the cube.

So, \(\sigma_x=\sigma_y=\sigma_z=- \sigma\)

As the walls are constrained so total strain should be zero.

Tensile strain due to temperature rise + Compressive strain due to compressive force = 0

Volumetric strain of the cube,

\( \epsilon_v= \frac{{\left( {1 - 2v} \right)}}{E} \times \left(- \sigma_x - \sigma_y - \sigma_z\right) =- \frac{{\left( {1 - 2v} \right)}}{E} \times3\sigma \)

Volumetric thermal strain

\(\epsilon_v=3\alpha. \Delta T\) (Thermal tensile Strain)

\(- \frac{{\left( {1 - 2v} \right)}}{E} \times3\sigma = 3\alpha. \Delta T\)

∴ \(\sigma = \frac{{ - \alpha .{\rm{\Delta }}T.E}}{{\left( {1 - 2v} \right)}}\)

Concept:

Maximum permissible shear stress can be calculated using:

\({τ _{max}} = \frac{T}{{{Z_p}}}\)

where Z_p = polar section modulus, T = torque applied

Given:

T = 50 Nm, τ_allowable = 140 MPa

F0S = 2

\({τ _{max}} = \frac{{{τ _{allowable;}}}}{{FOS}} = \frac{{140}}{{2}}=70~MPa\)

Polar section modulus for the circular road is:

\({Z_P} = \frac{{\pi {d^3}}}{{16}}\)

Now maximum shear stress is

\({τ _{max}} = \frac{T}{{{Z_p}}}\)

⇒ \({Z_p} = \frac{T}{{{τ _{max}}}}\)

\(\frac{{\pi {d^3}}}{{16}} = \frac{T}{{{τ _{max;}}}}\)

\(d = \sqrt[3]{{\frac{{16T}}{{\pi {τ _{max}}}}}} = \sqrt[3]{{\frac{{16 \times 50}}{{\pi \times 70 \times {{10}^6}}}}}\)

d = 0.01537 m = 15.37 mm ≃ 16 mm

Concept:

For a brake:

\(\frac{{{T_1}}}{{{T_2}}} = {e^{\mu \theta }}\)

where T₁ = tension on tight side, T₂ = tension on slack side, μ = coefficient of friction, θ = wrapping angle

And braking torque can be calculated by:

BT = (T₁ – T₂) × R

where R is the radius of the brake drum.

Calculation:

Given:

As the drum is rotating in the anti-clockwise direction, T1 will be tight side & T2 will be slack side.

T₁ = 400 N, R = 0.25 m, μ = 0.25

θ = 180° = π radian

Now, we know that

\(\frac{{{T_1}}}{{{T_2}}} = {e^{\mu \theta }}\)

\(\frac{{400}}{{{T_2}}} = {e^{0.25 \times \pi }} = 2.193\)

\( \Rightarrow {T_2} = \frac{{400}}{{2.193}} = 182.375;N\)

Braking torque (BT)

BT = (T₁ – T₂) × R

BT = (400 – 182.375) × 0.25

BT = 54.40 Nm

Given:

Total number of red balls = 4

Total number of black balls = 6

Three balls are selected randomly without replacement,

Three cases are possible,

Case 1 - (B, B, R)

Case 2 - (B, R, B)

Case 3 - (R, B, B)

The probability of the selected cases are:

\(P\left( {B, B, R} \right) = \frac{{{6_{{c_1}}} \times {5_{{c_1}}}\times {4_{{c_1}}}}}{{{{10}_{{c_1}}\times {9_{{c_1}}}\times {8_{{c_1}}}}}} = \frac{{ {\left( {6 \times 5 \times 4} \right)}}}{{\left( {10 \times 9 \times 8} \right)}} = \frac{1}{6}\)

\(P\left( {B, R, B} \right) = \frac{{{6_{{c_1}}} \times {4_{{c_1}}}\times {5_{{c_1}}}}}{{{{10}_{{c_1}}\times {9_{{c_1}}}\times {8_{{c_1}}}}}} = \frac{{ {\left( {6 \times 4 \times 5} \right)}}}{{\left( {10 \times 9 \times 8} \right)}} = \frac{1}{6}\)

\(P\left( {R, B, B} \right) = \frac{{{4_{{c_1}}} \times {6_{{c_1}}}\times {5_{{c_1}}}}}{{{{10}_{{c_1}}\times {9_{{c_1}}}\times {8_{{c_1}}}}}} = \frac{{ {\left( {4 \times 6 \times 5} \right)}}}{{\left( {10 \times 9 \times 8} \right)}} = \frac{1}{6}\)

The probability that the selected set contains 1 red ball and 2 black balls is = \(P\left( {B, B, R} \right) + P\left( {B, R, B} \right) + P\left( {R, B, B} \right) = \frac{1}{6} + \frac{1}{6} +\frac{1}{6} = \frac{1}{2}\)

Concept:

For different roots of the auxiliary equation, the solution (complementary function) of the differential equation is as shown below.

Calculation:

Given:

\({x^2}\frac{{{d^2}y}}{{d{x^2}}} + x\frac{{dy}}{{dx}} - 4y = 0\)

y (0) = 0, y (1) = 1

Cauchy’s D.E

\(\begin{array}{l} y = {x^m}\ \frac{{dy}}{{dx}} = m{x^{m - 1}}\ \frac{{{d^2}y}}{{d{x^2}}} = m(m - 1){x^{m - 2}} \end{array}\)

\(\begin{array}{l} {x^2}\frac{{{d^2}y}}{{d{x^2}}} + x\frac{{dy}}{{dx}} - 4y = 0\ m(m - 1) + m - 4 = 0\ {m^2} - 4 = 0\ m = \pm 2 \end{array}\)

∴ The required solution is y = C₁x^m1 + C₂x^m2 ​= C1 x2 + C₂ x^-2

Boundary condition: 

y(0) = 0, we get C2 = 0

And y(1) = 1, we get C₁ = 1

So y = x² will be the solution.

Concept:

Let

[A] is the Coefficient matrix

[A/B] be Augmented matrix

n = total number of variables

Case 1: ρ(A) = ρ(A/B) = n

In this case, the system will be consistent and will have a unique solution.

Case 2**: ρ(A) = ρ(A/B) < n**

In this case, the system will be consistent and will have infinite solutions.

Case 3**: ρ(A) < ρ(A/B)**

In this case, the system will be inconsistent and will have no solution.

Calculation:

Given:

x  + 2y + z = 4

2x + y + 2z = 5

x – y + z = 1

Here n = 3

Augmented matrix is:

\(\left[ {A/B} \right] = \left[ {\left. {\begin{array}{*{20}{c}} 1&2&1\ 2&1&2\ 1&{ - 1}&1 \end{array}} \right|\begin{array}{*{20}{c}} 4\ 5\ 1 \end{array}} \right]\)

R₂ → R₂ – 2R₁ and R₃ → R₃ – R₁

_{\(\left[ {A/B} \right] = \left[ {\left. {\begin{array}{*{20}{c}} 1&2&1\ 0&{ - 3}&0\ 0&{ - 3}&0 \end{array}} \right|\begin{array}{*{20}{c}} {;;;4}\ { - 3}\ { - 3} \end{array}} \right]\)}

R₃ → R₃ – R₂

\($\left[ {A/B} \right] = \left[ {\left. {\begin{array}{*{20}{c}} 1&2&1\ 0&{ - 3}&0\ 0&0&0 \end{array}} \right|\begin{array}{*{20}{c}} {;;4}\ { - 3}\ {;;0} \end{array}} \right]\)

ρ(A) = ρ(A/B)  = 2 < 3

∴ System will be consistent and will have infinite solutions.$

Concept:

Maximum distortion energy theory (Von mises theory)

• According to this theory, the failure or yielding occurs at a point in a member when the distortion strain energy per unit volume reaches the limiting distortion energy (i.e. distortion energy at yield point) per unit volume as determined from simple tension test.
• yield stress under triaxial condition is given by:

\({\sigma _{y}} = \sqrt {\frac{1}{2}\left{ {{{\left( {{\sigma _x} - {\rm{;}}{\sigma _y}} \right)}^2} + {{\left( {{\sigma _y} - {\sigma _z}} \right)}^2} + {{\left( {{\sigma _z} - {\sigma _x}} \right)}^2} + 6\left( {τ _{xy}^2 + {\rm{;}}τ _{yz}^2 + τ _{zx}^2} \right)} \right}} \)

Calculation:

Given:

σ_x = 10 MPa, σ_y = 20 MPa, σ_z = -10 MPa, τ_xy = 5 MPa, τ_yz = τzx = 0

putting the value in the equation

\({\sigma _{y}} = \sqrt {\frac{1}{2}\left{ {{{\left( {{\sigma _x} - {\rm{;}}{\sigma _y}} \right)}^2} + {{\left( {{\sigma _y} - {\sigma _z}} \right)}^2} + {{\left( {{\sigma _z} - {\sigma _x}} \right)}^2} + 6\left( {τ _{xy}^2 + {\rm{;}}τ _{yz}^2 + τ _{zx}^2} \right)} \right}} \)

\({\sigma _{y}} = \sqrt {\frac{1}{2}\left{ {{{\left( {{10-20}} \right)}^2} + {{\left( {20+10} \right)}^2} + {{\left( {{-10} - {10}} \right)}^2} + 6\left( {5^2 } \right)} \right}} \)

∴ σ_y = 27.839 MPa 

Shear stress at yield is

 \({τ _y} = \frac{{{\sigma _{y}}}}{{\sqrt 3 }} = 16.07;MPa\)

Concept:

Shear angle (ϕ) can be calculated by:

\(\tan \phi = \frac{{r\cos \alpha }}{{1 - r\sin \alpha }}\)

Shear strain rate can be calculated by:

\(\dot \varepsilon = \frac{{{V_s}}}{{{t_m}}}\)

where Vs = shear velocity and tm = mean chip thickness of the primary zone

And shear velocity can be calculated by:

\(\frac{{{V_s}}}{{\cos \alpha }} = \frac{{{V_c}}}{{\cos \left( {\phi - \alpha } \right)}};\)

where V_c = chip velocity

Calculation:

Given:

r = 0.4

t = 0.6 mm

α = 10°

V_c = 2.5 m/s

t_m = 25 μm

Shear angle (ϕ) can be calculated by:

\(\tan \phi = \frac{{r\cos \alpha }}{{1 - r\sin \alpha }} = \frac{{0.4 \times \cos 10^\circ }}{{1 - (0.4 \times \sin 10^\circ )}} = 0.4233\)

ϕ = tan^-1(0.4233) = 22.9°

Now, shear velocity (v_s)

\(\frac{{{V_s}}}{{\cos \alpha }} = \frac{{{V_c}}}{{\cos \left( {\phi - \alpha } \right)}};\)

\({V_s} = \frac{{(\cos 10^\circ ) \times 2.5}}{{\cos \left( {22.9^\circ - 10^\circ } \right)}} = 2.525\)

Shear strain rate

\(\dot \varepsilon = \frac{{{V_s}}}{{{t_m}}} = \frac{{2.525}}{{25 \times {{10}^{ - 6}}}} = 1.01 \times {10^5};per;second;\)

Do not confuse shear strain rate with shear strain.

shear strain can be calculated by:

ε = cot ϕ + tan (ϕ – α)

Concept:

Machining time in drilling can be calculated by:

\({T_m} = \frac{{L + AP + OR + X}}{{f \times N}}\)

where L = thickness of the plate to be drilled

AP = approach

OR = over travel

X = necessary approach

The necessary approach can be calculated using:

\(X = \frac{D}{{2 \times \tan \beta }}\)

where D = diameter of the hole to be drilled, β = half drill point angle.

Calculation:

Given:

L = 50 mm, D = 15 mm N = 500 rpm, f = 0.2 mm/rev, AP = 2 mm, OR = 2 mm

2β = 118° ⇒ β = 59°

Necessary approach (X)

\(X = \frac{D}{{2 \times \tan \beta }} = \frac{{15}}{{2 \times \tan 59^\circ }} = 4.5;mm\)

Machining time (T_m)

\({T_m} = \frac{{L + AP + OR + X}}{{f \times N}}\)

\({T_m} = \frac{{50 + 2 + 2 + 4.5}}{{0.2 \times 500}} = 0.585;minute = 35.1;seconds\)

Concept:

View factor (F_ij) is defined as:

\({F_{ij}} = \frac{{net;radiant;heat;transfer;from;{i^{th}}surface;to;{j^{th}}surface}}{{total;radiation;emmited;from;surface;i}}\)

Reciprocity theorem

AiFij = AjFji

Summation rule

If there are n surfaces, then according to summation rule:

F11 + F12 + F13 + … + F1n = 1

F21 + F22 + F23 + … + F2n = 1

...

Fn1 + Fn2 + Fn3 + … + Fnn = 1

Calculation:

Given:

A₁ = πD₁L, A₂ = πD₂L

F₁₁ = 0 {∵ convex surface}

As F₁₁ + F₁₂ = 1

∴ F12 = 1

From reciprocity theorem

A₁F₁₂ = A₂F₂₁

⇒ \({F_{21}} = \frac{{{A_1}}}{{{A_2}}} = \frac{{\pi {D_1}L}}{{\pi {D_2}L}} = \frac{{{D_1}}}{{{D_2}}}\)

Now F₂₁ + F₂₂ = 1

⇒ \({F_{22}} = 1 - {F_{21}} = 1 - \frac{{{D_1}}}{{{D_2}}}\)

Concept:

According to the Blasius equation, for a laminar flow

Boundary layer thickness (δ) can be calculated by:

\(\delta = \frac{{5x}}{{\sqrt {R{e_x}} }}\)

where Re_x = Reynold's number = \(\frac{{\rho Vx}}{\mu }\) {V = Velocity}

\(\therefore \delta \propto \frac{1}{V}\)

Calculation:

Given:

δ₁ = 1 mm, V₂ = 4V₁

\(As,,\delta \propto \frac{1}{V}\)

\(\frac{{{\delta _1}}}{{{\delta _2}}} = \sqrt {\frac{{{V_2}}}{{{V_1}}}} \)

\(\frac{1}{{{\delta _2}}} = \sqrt {\frac{4}{1}} = 2\)

 δ₂ = 0.5 mm

Concept:

Specific humidity (ω) can be calculated by:

\(\omega = \frac{{{m_v}}}{{{m_a}}} = 0.621 \times \frac{{{P_v}}}{{{P_{atm}}~ -~ {P_v}}}\)

where m_v = mass of vapour, m_a = mass of dry air, P_v = vapour pressure, P_atm = atmospheric pressure

Relative humidity (ϕ) can be calculated by:

\(\phi = \frac{{{m_v}}}{{{m_{vs}}}} = \frac{{{P_v}}}{{{P_{vs}}}}\)

where m_vs = mass of saturated vapour, P_vs = saturated vapour pressure

Calculation:

Given:

M_a = 35 kg, m_v = 0.5 kg, temperature (t) = 25°c, P_atm = 100 kPa, P_vs = 3.17 kPa

Specific humidity (ω)

\(\omega = \frac{{{m_v}}}{{{m_a}}} = \frac{{0.5}}{{35}} = \frac{1}{{70}}\)

Specific humidity (ω) can also be calculated by:

\(\omega = 0.621 \times \frac{{{P_v}}}{{{P_{atm}} - {P_v}}}\)

\(\frac{1}{{70}} = 0.621 \times \frac{{{P_v}}}{{100 - {P_v}}}\)

100 – P_v = 43.47 × P_v

44.37 × P_v = 100

\({P_v} = \frac{{100}}{{44.37}} = 2.253;kPa\)

Relative humidity (ϕ)

\(\phi = \frac{{{P_v}}}{{{P_{vs}}}} = \frac{{2.253}}{{3.17}} = 0.71 = 71;% \)

Concept:

Strength of double transverse fillet weld can be calculated by:

P = 2 × 0.707 × (h × l) × σ_per

where h = leg size, l = length of the single weld.

Calculation:

Given:

h = 10 mm, 2l = 30 mm ⇒ l = 15 mm, τ_per = 94 MPa

Strength of double transverse fillet weld is:

P = 2 × 0.707 × (h × l) × σ_per

P = 2 × 0.707 × 0.010 × 0.015 × 94 × 10⁶

P = 19937.4 N = 19.93 kN

Do not take l = 30 mm, as it is given total weld length which is 2l.

Strength of parallel fillet weld can be calculated by:

P = 0.707 × (h × l) × τ_per

The nature of stress in the cross-section of a transverse fillet weld is complex due to the weld being subjected to normal stress, shear stress, and bending moment. So, in order to simplify design, many times shear failure is used as a failure criterion. So we can use the equation of load for parallel fillet welds also and it is not necessary to know tensile stress if only shear stress is known.

Explanation:

Displacing the rod by a small distance x.

Taking moment about '0'

Kx × 2L = mg × L

\(x = \frac{{mg}}{{2k}}\)

From similar triangle property

 \(\frac{x}{{2L}} = \frac{\delta }{L}\)

\(\delta = \frac{x}{2} = \frac{{mg}}{{4k}}\)

Using static deflection of the mass 'm'

\({\omega _n} = \sqrt {\frac{g}{\delta }} = \sqrt {\frac{{4K}}{m}} \)

Concept:

Maximum and minimum values of normal stresses occur on planes of zero shearing stress. The maximum and minimum normal stresses are called the principal stresses, and the planes on which they act are called the principal plane.

\({\sigma _{max,;min}} = \frac{{{\sigma _{xx}};+;{\sigma _{yy}}}}{2} \pm \sqrt {{{\left( {\frac{{{\sigma _{xx}};-;{\sigma _{yy}}}}{2}} \right)}^2} +;\tau _{xy}^2} \)

But the maximum shear stress planes may or may not contain normal stresses as the case may be.

\({\tau _{max}} = \frac{{{\sigma _{max}};-;{\sigma _{min}}}}{2} = \sqrt {{{\left( {\frac{{{\sigma _{xx}};-;{\sigma _{yy}}}}{2}} \right)}^2} +;\tau _{xy}^2} \)

which is equal to the radius of the Mohr's Circle.

Calculation:

Given:

\({\sigma _{xx}} = 40~MPa, ~{\sigma _{yy}} = 100~MPa,{\tau _{xy}} = 40~MPa\)

Radius of Mohr's circle:

 \(\tau_{max}= \sqrt {{{\left({\frac{{{\sigma _{xx}} - {\sigma _{yy}}}}{2}} \right)}^2} + \tau _{xy}^2} \)

\(\therefore\sqrt{\left ( \frac{40;-;100}{2} \right )^2;+;40^2}\Rightarrow50;MPa\)

Concept:

If L-1{F(s)} = f(t)

then

L-1F(s – a) = eat.f(t) and L-1{F(s + a)} = e-at.f(t)

Calculation:

\(F\left( s \right) = \frac{1}{{s\left( {s + 1} \right)}} = \frac{A}{s} + \frac{B}{{\left( {s + 1} \right)}}\)

\(\frac{1}{{s\left( {s + 1} \right)}} = \frac{{A\left( {s + 1} \right) + B\left( s \right)}}{{s\left( {s + 1} \right)}}\)

A(s + 1) + B(s) = 1

Put s = 0, we get A = 1

Put s = -1, we get B = -1

\(F\left( s \right) = \frac{1}{{s\left( {s + 1} \right)}} = \frac{1}{s} + \frac{{ - 1}}{{\left( {s + 1} \right)}}\)

f(t) = L-1(s)

\(f\left( t \right) = {L^{ - 1}}\left( {\frac{1}{s} - \frac{1}{{s + 1}}} \right)\)

f(t) = e0t – e-t    \(\left{\because {{L^{ - 1}}\left( {\frac{1}{s}} \right) = 1} \right}\)

f(t) = 1 – e-t

Explanation:

Characteristic equation

\(\left| {A - λ I} \right| = 0 ⇒ \left| {\begin{array}{*{20}{c}} {5 - λ }&3\ 1&{3 - λ } \end{array}} \right| = 0 \)

⇒ (5 - λ)(3 - λ) - 3 = 0;

⇒ λ² – 8λ + 15 – 3 = 0 ⇒ λ² – 8λ + 12 = 0

⇒ λ = 2, λ = 6

Eigenvector for λ = 2,

(A – 2I) × X = 0

At, λ = 2

\(\left( {\begin{array}{*{20}{c}} 3&3\ 1&1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} {{x_1}}\ {{x_2}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0\ 0 \end{array}} \right);\)

⇒ 3x₁ + 3x₂ = 0;

⇒ x₁ = - x₂;

The eigenvector will be \(\left[ {\begin{array}{*{20}{c}} {1}\ {-1}\end{array}} \right]\)

Hence the required vector is \(\left[ {\begin{array}{*{20}{c}} {\frac {1}{\sqrt {2}}}\ {- \frac {1}{\sqrt {2}}}\end{array}} \right]\)

Alternate solution:

Eigen vector should satisfy Ax = λx, where λ is eigen value

Taking x = \(\left[ {\begin{array}{*{20}{c}} {\frac{1}{2}}\ {\frac{{\sqrt 3 }}{2}} \end{array}} \right]\) (option 1)

Ax = \(\left[ {\begin{array}{*{20}{c}} 5&3\ 1&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\frac{1}{2}}\ {\frac{{\sqrt 3 }}{2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{{5 + 3\sqrt 3 }}{2}}\ {\frac{{1 + \sqrt 3 }}{2}} \end{array}} \right]\)≠ λx

∴ Option 1 is not correct.

Taking x = \(\left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}\ {\frac{{ - 1}}{{\sqrt 2 }}} \end{array}} \right]\)(option 2)

Ax = \(\left[ {\begin{array}{*{20}{c}} {\frac{{5 - 3}}{{\sqrt 2 }}}\ {\frac{{1 - 3}}{{\sqrt 2 }};} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{2}{{\sqrt 2 }}}\ {\frac{{ - 2}}{{\sqrt 2 }}} \end{array}} \right] = 2\left[ {\begin{array}{*{20}{c}} {\frac{1}{{\sqrt 2 }}}\ {\frac{{ - 1}}{{\sqrt 2 }}} \end{array}} \right]\)= λx

∴ Option 2 is correct.

(No need to check other options)

Given:

Area = 100 mm² = 100 × 10^-6 m²

\(\frac{{{T_1}}}{{\sin 120^\circ }} = \frac{{{T_2}}}{{\sin 135^\circ }} = \frac{F}{{\sin 105^\circ }}\)

\(\frac{{{T_1}}}{{\sin 120^\circ }} = \frac{F}{{\sin 105^\circ }}\)

T1 = 0.8965F

\(;\frac{{{T_2}}}{{\sin 135^\circ }} = \frac{F}{{\sin 105^\circ }}\)

T2 = 0.732F

Maximum force is,

F_max = Max(T₁,T₂) =  T₁ = 0.8965F

\(Maximum;stress = \frac{{{F_{max}}}}{{area}} = \frac{{0.8965F}}{{100 \times {{10}^{ - 6}}}} = \frac{{0.8965F}}{{100}};MPa\)

Now,

Maximum stress ≤ 100 MPa

\(\frac{{0.8965F}}{{100}} MPa \le 100~ MPa\)

\(F \le \frac{{100 \times 100}}{{0.8965}};\)

F ≤ 11.154 kN

Given:

F = 1 kN

Using Lame’s theorem

\(\frac{{{T_1}}}{{\sin 120^\circ }} = \frac{{{T_2}}}{{\sin 135^\circ }} = \frac{F}{{\sin 105^\circ }}\)

 \(Taking,~\frac{{{T_1}}}{{\sin 120^\circ }} = \frac{F}{{\sin 105^\circ }}\)

T1 = 0.8965F = 0.8965 × 1 = 0.8965 kN

\(Taking,;\frac{{{T_2}}}{{\sin 135^\circ }} = \frac{F}{{\sin 105^\circ }}\)

T₂ = 0.732F = 0.732 × 1 = 0.732 kN

The vertical reaction at B (R­_B)

R_B = T₂cos 30° = 0.732 × 0.866 = 0.633 kN

Concept:

Power required for the compressor (P) is given by:

P = ṁ(h₂ – h₁)

Calculation:

Given:

ṁ = 2 kg/s

Condition 1 is saturated vapour at 120 kPa. So, h₁ = 237 kJ/kg

Condition 2 is superheated vapour at 800 kPa and S₁ = S₂ = 0.95 kJ/kg.K

h₂ = 276.45 kJ/kg

Power required for the compressor (P) is:

P = ṁ(h**₂ – h₁)**

P = 0.2 × (276.45 – 237) = 7.89 kJ/s ≃ 7.9 kJ/s = 7.9 kW

Concept:

Heat is being extracted from the refrigerated space in process 4 - 1

Rate of heat extracted (Q̇)

Q̇ = ṁ(h₁ – h₄)

Calculation:

Given:

ṁ = 0.2 kg/s

Condition 1 is saturated vapour at 120 kPa. So, h₁ = 237 kJ/kg

As h₃ = h₄ and h₃ = h_f at 800 kPa

h₄ = 95.5 kJ/kg

Rate of heat extracted (Q̇)

Q̇ = ṁ(h₁ – h₄)

Q̇ = 0.2 × (237 – 95.5) = 28.3 kJ/s = 28.3 kW

Concept:

Critical path: It is the path in the project which determines the shortest time to complete the project.

In any network, the path which takes maximum time is a critical path.

Various paths in the network are:

A-D-F-H = 18 days

A-B-E-G-H = 17 days

A-C-G-H = 16 days

As path A-D-F-H takes maximum time i.e. 18 days.

∴ A-D-F-H is the critical path.

As activity F is only in the critical path.

∴ Critical path will remain same i.e. A-D-F-H

Project duration changes to 19 days.

Concept:

The steady flow energy equation betweeen two points is given by:

\({h_1} + \frac{{V_1^2}}{2} + Q = {h_2} + \frac{{V_2^2}}{2} + W\)

Calculation:

Given:

P₁ = 300 kPa, T₁ = 500 K, V₁ = 10 m/s, A₁ = 80 cm² = 80 × 10^-4 m²

P₂ = 100 kPa, V₂ = 180 m/s, C_p = 1008 J/kg.K

Using the steady flow energy equation

\({h_1} + \frac{{V_1^2}}{2} + Q = {h_2} + \frac{{V_2^2}}{2} + W\)

Here

Q = 0 {∵ adiabatic flow}

W = 0

\({h_1} - {h_2} = \frac{{V_2^2 - V_1^2}}{2}\)

\({C_p}\left( {{T_1} - {T_2}} \right) = \frac{{V_2^2 - V_1^2}}{2}\)

\(1008\left( {500 - {T_2}} \right) = \frac{{{{180}^2} - {{10}^2}}}{2}\)

500 – T₂ = 16.02

T₂ = 483.97 K ≃ 484 K

Concept:

Using continuity equation

ρ₁A₁V₁ = ρ₂A₂V₂

\(\frac{{{P_1}}}{{{RT_1}}} \times {A_1}{V_1} = \frac{{{P_2}}}{{{RT_2}}} \times {A_2}{V_2}\){∵ P = ρRT}

\(\frac{{{P_1}}}{{{T_1}}} \times {A_1}{V_1} = \frac{{{P_2}}}{{{T_2}}} \times {A_2}{V_2}\)

Calculation:

Given:

P₁ = 300 kPa, T₁ = 500 K, V₁ = 10 m/s, A₁ = 80 cm² = 80 × 10^-4 m²

P₂ = 100 kPa, V₂ = 180 m/s, C_p = 1008 J/kg.K

Using, \(\frac{{{P_1}}}{{{T_1}}} \times {A_1}{V_1} = \frac{{{P_2}}}{{{T_2}}} \times {A_2}{V_2}\)

\(\frac{{300 \times {{10}^3}}}{{500}} \times 80 \times {10^{ - 4}} \times 10 = \frac{{100 \times {{10}^3}}}{{500}} \times {A_2} \times 180\)

On solving we get,

A₂ = 13.33 × 10^-4 m² = 13.33 cm² 

Option 4 is best suited.