GATE ME 2019 Official Paper: Shift 1 (Previous Year Paper)

The correct answer is option 3 i.e. eminent.

As the article used before the blank is 'an', it is clear that the word in the blank cannot begin with a consonant, thus eliminating option 2 and 4.

Moreover, looking at the meaning of the words:

Imminent: Approaching

Eminent: Famous and respected.

Prominent: Well-known; Projecting.

Dominant: Having power or influence over others.

It is clear that Eminent is the only correct choice here.

The correct answer is option 3 i.e. Had, would.

The phrase "Had I...", in simple language means "If I had...".

There is no such meaning behind "Have I" and therefore it would have been grammatically incorrect to use it in the sentence.

Moreover, the general rule regarding question tags suggests that they use the negative form of the helping verb used in the sentence.

For example:

1. He is adorable, isn't he?

2. He isn't well, is he?

Now, since the helping verb in the clause containing the question tag is 'wouldn't', the question tag would be 'would'.

Therefore, the answer is 'had, would'.

For hour hand;

360° = 12 hours

⇒ 1° = 12/360 hours

225° = 12/360 × 225 = 7.5 hours

Let x and y be any two integers:

x + y = 26

xy = 165

⇒ x(26 - x) = 165

26x – x² = 165

x² – 26x + 165 = 0

$x = \frac{{26 \pm \sqrt {{{\left( {26} \right)}^2} - \left( 4 \right)\left( {165} \right)} }}{2} = \frac{{26 \pm 4}}{2} = 15,;11$ 

When x = 15, y = 11

⇒ x – y = 4

Option 2 is the correct answer.

skirting means attempting to ignore or avoiding to attempt.

The first sentence states that the minister is avoiding something.

So, skirting is correct in this context.

The other words don't fit in the context of the sentence.

It is possible for the judge to determine the truth or falsity of the statements in options 1, 2 and 4 and hence, the judge would not set the prisoner free.

But the judge cannot decide whether the statement 'I will be shot' is true or false and the likely decision would be to release the prisoner.

Hence, option 3 is the correct answer.

Let Rs. 100,000 is divided into two parts x and y respectively.

Part I: Scheme – 1 (10% profit)

After profit in Scheme – 1, he will get 1.10x rupees

Scheme – 2 (12% profit)

After profit in Scheme – 2, he will get 1.12y rupees

Total amount that he will get from Scheme 1 and Scheme 2 after applying profits of 10% and 12% respectively = 1.10x + 1.12y     …1)

Part II: When profit percentages are interchanged

Scheme – 1: (12% profit), Scheme – 2: (10% profit)

Total amount after applying these profit percentages = 1.12x + 1.10y     …2)

Given that,

1.10x + 1.12y – 1.12x – 1.10y = 120

0.02y – 0.02x = 120

2y – 2x = 12000

y – x = 6000     …3)

y + x = 100000     …4)

Solving 3) and 4)

2y = 106000

Y = 53000, x = 47000

Ratio between the investments = x/y = 47/53

Explanation:

The language of origin of the term 'Nzadi o Nzere' cannot be determined from the given paragraph and hence option 2 is incorrect.

As there is no hint of any Portuguese intervention in the paragraph, the statement in option 3 cannot be inferred and hence is incorrect.

It is given that Mobuto was a dictator of Congo and not Portuguese and hence the statement in option 4 cannot be inferred. 

Despite the change in name, the new name Zaire still has Portuguese origin and hence it can be inferred that Mobuto was not completely successful in his attempt to Africanise the name of his country.

Hence option 1 is the correct answer.

Total employees in 2010 = 600

Type ‘P’ employees = 20/100 × 600 = 120

Type ‘Q’ employees = 25/100 × 600 = 150

Type ‘R’ employees = 25/100 × 600 = 150

Type ‘S’ employees = 25/100 × 600 = 150

Type ‘T’ employees = 5/100 × 600 = 30

Total employees at level P, Q and R = 120 + 150 + 150 = 420

Total employees in 2016 = 600 + 600 × 15/100 = 690

Type ‘S’ employees in 2016 $= 150 + \frac{{150 \times 40}}{{100}} = 210$

⇒ Type ‘T’ employees in 2016 = 690 – 210 – 420 = 60

Using the above symbols we get the following family tree:

From the tree diagram, we can see that R and W are the parents of Y as Y is a legitimate child of W.

Therefore it is possible that R is the father of Y and W is the wife of R. Hence the statements in options 2 and 3 are not necessarily false.

Also, from the tree diagram, we can conclude that M and N are the grandparents of Y and hence the statement in option 1 is not necessarily false.

As Y is a legitimate child of W, W is the spouse of R and not P.

Hence the statement 'W is the wife of P' is necessarily false.

Concept:

Eigenvalues of upper triangles matrix or lower triangular matrix are just the diagonal elements of the matrix.

Calculation:

In the given matrix below,

\(P = \left[ {\begin{array}{*{20}{c}} 1&1&0\ 0&1&1\ 0&0&1 \end{array}} \right]\)

Diagonal elements are 1, 1, and 1, so there is only one distinct eigenvalue i.e. λ = 1.

We can also calculate this by applying |A - λI| = 0, where A is any square matrix. In this case,

|P - λI| = 0 yields

\(\left| {\begin{array}{*{20}{c}} {1 - \lambda }&1&0\ 0&{1 - \lambda }&1\ 0&0&{1 - \lambda } \end{array}} \right| = 0 \Rightarrow \left( {1 - \lambda } \right)\left( {1 - \lambda } \right)\left( {1 - \lambda } \right) = 0\)

⇒ λ = 1 is the only distinct eigenvalue

Key points:

Please do remember all the properties of eigenvalues and eigenvectors. Don’t apply the formula. It will be time-consuming.

Properties of Eigenvalues:

• The sum of Eigenvalues of a matrix A is equal to the trace of that matrix A
• The product of Eigenvalues of a matrix A is equal to the determinant of that matrix A
• If λ is an eigenvalue of a matrix A, then λn will be an eigen value of a matrix An.
• If λ is an eigenvalue of a matrix A, then kλ will be an eigenvalue of a matrix kA where k is a scalar

Concept:

Volume of the solid of rotation obtained by rotating the shaded area by 360° around the x-axis is asked,

there is a direct relation for this;

 ${\forall _1} = \smallint \pi {y^2}dx$…1)

Calculation:

Given area;

Using (1); Volume of solid of rotation can be calculated by:

\({\forall _1} = \mathop \smallint \nolimits_0^1 \pi xdx\)

\( = \pi \mathop \smallint \nolimits_0^1 xdx = \pi \left{ {\frac{{{x^2}}}{2}} \right}_0^1 \)

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

${\forall _1} = \frac{\pi }{2}units;$

Key points:

In the given problem, the volume is generated by revolving the area by 360° about the x-axis.

But if rotation/revolution is about the y-axis, then the volume of solid of rotation is calculated by:

${\forall _2} = \smallint \pi {x^2}dy$ …2)

So, always be careful about which axis rotation is asked.

Depending upon that, you should use either 1) or 2).

Concept:

For solving first order, first-degree differential equations always first inspect with variable separation method.

Calculation:

Given the differential equation is,

$\frac{{dy}}{{dx}} + 7{x^2}y = 0 \Rightarrow \frac{{dy}}{{dx}} = - 7{x^2}y$, separating variables

$\frac{{dy}}{y} = - 7{x^2}dx$, integrating both sides;

$\smallint \frac{{dy}}{y} = - 7\smallint {x^2}dx;lny = - 7\frac{{{x^3}}}{3} + lnA$

Where A is a constant.

$\ln y - \ln A = - \frac{{7{x^3}}}{3} \Rightarrow \ln \left( {\frac{y}{A}} \right) = - \frac{{7{x^3}}}{3}$

$\frac{y}{A} = {e^{ - \frac{7}{3}{x^3}}} \Rightarrow y = A{e^{ - \frac{7}{3}{x^3}}}$ …(1)

Use condition $y\left( 0 \right) = \frac{3}{7}$ in (1)

$\Rightarrow \frac{3}{7} = A;use;in;(1) \Rightarrow y = \frac{3}{7}{e^{ - \frac{{7{x^3}}}{3}}}$

Key Points

Practice all the methods of solving first order first-degree differential equations.

In these questions, options are very confusing. So, study all options carefully.

Concept:

Let X be the generalized normal random variable having mean (μ) and standard deviation (σ).

Convert this generalized normal random variable to standardized normal random variable (Z) having mean (μ) 0 and S.D. (σ) 1.

This transformation is done through,

i.e. x{μ, σ} ⇒ z{0, 1}

Now, p{X₁ < X < X₂} ⇒ p{Z₁ < Z < Z₂}

Then, required probability can be calculated by;

p{Z₁ < Z < Z₂} = p(Z₂) – p(Z₁)

Figure: Standardized normal distribution (Z) having mean 0 and S.D. 1

Calculation:

Given data

Mean (μ) = 440 mm, S.D. (σ) = 1 mm

X₁ = 438 mm, X₂ = 441 mm

\(\left. {\begin{array}{*{20}{c}} {{Z_1} = \frac{{{X_1} - \mu }}{\sigma } = \frac{{438 - 440}}{1} = - 2}\ {{Z_2} = \frac{{{X_2} - \mu }}{\sigma } = \frac{{441 - 440}}{1} = + 1} \end{array}} \right}\)

Now, p{Z₁ < Z < Z₂} = p{Z₂} – p{Z₁}

⇒ p{-2 < Z < +1} = p{1} – p{-2}

Always remember some standard results, mentioned below,

p{-1 < Z < 1} = 68.27% …1)

This is known as 1σ limits

p{-2 < Z < 2} = 95.44% …2)

This is known as 2σ limits

p{-3 < Z < 3} = 99.74% …3)

This is known as 3σ limits

⇒ p{-2 < Z < 1} = p{-2 < Z < 0} + p{0 < Z < 1} …4)

\(\Rightarrow p\left{ { - 2 < Z < 0} \right} = \frac{{95.44}}{2} = 47.72% \)

\(\Rightarrow p\left{ {0 < Z < 1} \right} = \frac{{68.27}}{2} = 34.135% \) use in 4)

p{-2 < Z < 1} = 47.72 + 34.135 = 81.855%

⇒ Correct option is A) 81.85%

Keypoints:

For solving such questions, always convert given normal variable (X) to standardized normal variable (Z).

Always remember standard results like 1σ, 2σ and 3σ limits as expressed by equations 1), 2) and 3).

Be careful while entering the value of probability in decimals/percentage in numerical answer type questions.

Concept:

A flat faced follower is driven by a circular eccentric cam,

Let cam rotates through angle θ.

Vertical displacement of follower is given by (From above figure);

y(t) = e – e cos θ …1)

$\omega = \frac{\theta }{t} \Rightarrow \theta = \omega t$

y(t) = e(1 – cos ωt) …2)

Keypoints:

Apart from displacement, velocity and acceleration of follower can also be asked; they can be calculated by;

$\nu = \frac{{dy}}{{dt}} = e\left[ {0 + \sin \omega t \times \omega } \right] = e\omega \sin \omega t$

$a = \frac{{d\nu }}{{dt}} = e\omega \cos \omega t \times \omega = e{\omega ^2}\cos \omega t$

Concept:

Springs in series and parallel

Case – I: When springs are in series

k = spring stiffness, δ = spring deflection

In series connection, load P is same far both springs 1) and 2) but total deflection will be sum of individual spring deflections.

i.e. δ = δ₁ + δ₂ …1)

Let k_e be equivalent spring stiffness of the combination.

Spring stiffness is defined as $\left( k \right) = \frac{{Load\left( P \right)}}{{Deflection\left( \delta \right)}}$

⇒ P = kδ; 1) becomes

$\frac{P}{{{k_e}}} = \frac{P}{{{k_1}}} + \frac{P}{{{k_2}}} \Rightarrow \frac{1}{{{k_e}}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}}$ …2)

Case – II: When springs are in parallel

In this case load is divided between two springs, but deflection of both springs will be same.

i.e. P = P₁ + P₂, δ₁ = δ₂ = δ

k_eδ = k₁δ₁ + k₂δ₂

⇒ k_eδ = k₁δ + k₂δ

⇒ k_e = k₁ + k₂ …3)

Remember 2) and 3) for further calculations.

Calculation part

Given configurations are;

Figure: Springs in series

Using 2) $\frac{1}{{{k_e}}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}}$ , k₁ = k₂ = k

$\Rightarrow \frac{1}{{{k_e}}} = \frac{2}{k};{k_e} = \frac{k}{2}$ …4)

Natural frequency of spring mass system is given by,

${\omega _n} = \sqrt {\frac{{{k_e}}}{M}}$

For above case; ${\omega _I} = \sqrt {\frac{k}{{2M}}}$ …5)

Figure: Springs in parallel

For parallel combination,

k_e = k₁ + k₂ = 2k

$\Rightarrow {\omega _{II}} = \sqrt {\frac{{{k_e}}}{M}} = \sqrt {\frac{{2k}}{M}}$ …6)

$\frac{{{\omega _I}}}{{{\omega _{II}}}} = \frac{{\sqrt {\frac{k}{{2M}}} }}{{\sqrt {\frac{{2k}}{M}} }} = \frac{{\sqrt k }}{{\sqrt 2 \sqrt M }} \times \frac{{\sqrt M }}{{\sqrt 2 \sqrt k }} = \frac{1}{2}$

$\Rightarrow \frac{{{\omega _I}}}{{{\omega _{II}}}} = \frac{1}{2}$

Key Points

A special configuration is generally given;

Here springs are in PARALLEL NOT IN SERIES.

Equivalent stiffness $\left( {{k_e}} \right) = {k_1} + {k_2},;NOT;\frac{1}{{{k_e}}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}}$

Concept:

The force which driving tooth exerts on the driven tooth is along a line from the pitch point to point of contact of the two teeth. This line is also the common normal at the point of contact of the two mating gears and is known as the line of action or pressure line. This is shown below.

This force along the pressure line can be resolved into two components. First is tangential component which helps in transmitting power. Other component is radial component. This is shown below.

F_T = F_N cos ϕ …1)

F_R = F_N sin ϕ …2)

$\frac{{{F_R}}}{{{F_T}}} = \tan \phi$

⇒ F_R = F_T tan ϕ …3)

Calculation:

Given data; Power (P) = 20 kW = 20 × 10³ W

Angular velocity (ω) = 200 rad/s

$P = Tω \Rightarrow T = \frac{P}{ω } = \frac{{20 \times 1000}}{{200}} = 100;Nm$

⇒ Torque (T) = 100 Nm

Pitch circle diameter (D) = 100 mm

Radius (R) = 50 mm

$T = {F_T}R \Rightarrow {F_T} = \frac{{100}}{{50}} \times 1000 = 2000N$

So, tangential force (F_T) = 2000 N

But we need radial force (F_R) =?

From 3)

F_R = (2000) tan (20°) = 727.94 N = 0.72794 kN

Hence, magnitude of radial force (F_R) = 0.72794 kN ≈ 0.73 kN

Keypoints:

Be careful about calculation of forces.

There are 3 forces i.e.

1. Normal force (F_N)

2. Tangential force (F_T) = F_N cos ϕ

3. Radial force (F_R) = F_N sin ϕ

Power transmitting component is tangential force (F_T).

Always see carefully which force component is asked to calculate out of 3 components and calculate accordingly.

Concept:

The first law of thermodynamics

For a closed system/non-flow system undergoing a process, (1 - 2)

Q_1-2 = ΔE + W_1-2 …1)

E = Stored energy of a system

This stored energy can be viewed as the sum of microscopic and macroscopic energies.

⇒ Q_1-2 = Δ (U + KE + PE) + W_1-2

⇒ Q_1-2 = ΔU + ΔKE + ΔPE + W_1-2 …2)

For a non-flow or closed system at equilibrium, ΔKE and ΔPE are negligible,

So, these 2 terms can be neglected.

⇒ Q_1-2 = ΔU + W_1-2 …3)

Also, for a perfect gas, the internal energy is a function of temperature only.

i.e. dU = mC_νdT …4)

Calculation:

Given equation is

Q_1-2 = W_1-2 …5)

But first law states that; Q_1-2 = ΔU + W_1-2 …6)

Comparing 5) and 6)

⇒ ΔU = 0 …7)

But for perfect gas; dU = mC_νdT, integrating both sides

\(\mathop \smallint \nolimits_1^2 dU = \mathop \smallint \nolimits_1^2 m{C_\nu }dT\)

⇒ U₂ – U₁ = mC_ν(T₂ – T₁) {Assuming constant m, C_ν}

ΔU = mC_ν(ΔT) …8)

Comparing 7) and 8)

⇒ ΔT = 0 {∵ m ≠ 0, C_ν ≠ 0}

⇒ T₂ = T₁ = Constant = Isothermal process

Key Points

Remember the properties of perfect gases and apply these directly instead of writing first law.

Study all the basic processes in detail like the adiabatic process, Isobaric, isochoric etc.

Concept:

Always remember standard results mentioned below;

Part – I For constant surface heat flux (q_s = constant);

Hydrodynamically and thermally fully developed laminar flow through a circular pipe of constant cross section

Nu_q = 4.36      …1)

Part – II For constant wall temperature (T_w = constant)

For this case

Nu_T = 3.66     …2)

Calculation:

Comparing 1) and 2)

Nu_q > Nu_T ⇒ Option A is correct.

Key Points

Go through both derivations (i.e. q_s = constant and T = constant) and remember graphs of both cases.

Concept:

The three hydraulic turbins can be divided based upon head and flow rate (discharge). This is shown below.

 

Calculation:

Based upon above table, descending order of flow rate/discharge will be;

Kaplan > Francis > Pelton ⇒ option A) is correct

Keypoints:

If descending order of available head had been asked, then answer would be;

Pelton > Francis > Kaplan

Concept:

Fourier’s law of heat conduction states that;

$\dot Q = kA\frac{{\left( {{T_1} - {T_2}} \right)}}{L} = \frac{{{\rm{\Delta }}T}}{{{R_{cond}}}}$ …1)

DIAGRAM

${R_{cond}} = \frac{L}{{kA}} = Conduction;resistance;of;plane;wall$    …2)

Calculation:

Given configuration

Heat transfer in radial direction is negligible, contact resistance is also negligible. We can express given composite bar by equivalent thermal circuit as →

Heat transfer rate can be given by,

${\dot Q_{cond}} = \frac{{{T_1} - {T_2}}}{{\frac{L}{{{k_1}A}} + \frac{L}{{{k_2}A}}}} = \frac{{{\rm{\Delta }}T}}{{\frac{L}{A}\left[ {\frac{1}{{{k_1}}} + \frac{1}{{{k_2}}}} \right]}} = \left( {\frac{{{k_1}{k_2}}}{{{k_1} + {k_2}}}} \right)\left( {\frac{A}{L}} \right)\left( {{\rm{\Delta }}T} \right)$ …3)

Now, we have to replace the composite bar by a bar of length 2L, area A and thermal conductivity k_eq.

Figure: Equivalent bar having thermal conductivity k_eq

${\dot Q_{cond}} = {k_{eq}}\frac{{A\left( {{T_1} - {T_2}} \right)}}{{2L}}$ …4)

Heat transfer through both the bars must be same.

Comparing 3) and 4)

$\frac{{{k_1}{k_2}}}{{{k_1} + {k_2}}}.\frac{A}{L}{\rm{\Delta }}T = {k_{eq}}A\frac{{\left( {{\rm{\Delta }}T} \right)}}{{2L}}$ 

$\Rightarrow {k_{eq}} = \frac{{2{k_1}{k_2}}}{{{k_1} + {k_2}}}$ 

Key Points:

In such questions, confusion arises due to cylindrical rod.

We generally confused whether should we use

$\dot Q = \frac{{\left( {2\pi kL} \right){\rm{\Delta }}T}}{{{\rm{ln}}\left( {{r_2}/{r_1}} \right)}};or;\dot Q = \frac{{kA\left( {{\rm{\Delta }}T} \right)}}{L}$ 

If geometry is cylindrical and heat transfer is in radial direction, then use $\dot Q = \frac{{\left( {2\pi kL} \right)\left( {{\rm{\Delta }}T} \right)}}{{{\rm{ln}}\left( {{r_2}/{r_1}} \right)}}$

If geometry is cylindrical and heat transfer is in axial direction than use $\dot Q = kA\left( {\frac{{{\rm{\Delta }}T}}{L}} \right)$

Concept:

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

Figure: Cycle 1-2-3-4’-1; Ideal vapour compression refrigeration cycle (T-S plot)

Processes

1-2: Isentropic compression

2-3: Heat rejection

3-4: Isentropic expansion

3-4’: Isenthalpic expansion (Irreversible process)

4-1: Evaporation (Heat absorption)

We know that area under any process in T-S plot represents heat interaction

i.e. \({Q_{1 - 2}} = \mathop \smallint \nolimits_1^2 TdS\)

When expansion process is isenthalpic (3-4’), heat absorption (refrigeration effect) is represented by area 4’-5’-6-1.

When expansion process is isentropic (3-4), refrigeration effect is represented by area 4-5-6-1.

$COP = \frac{{Refrigeration;effect}}{{Work;input}} = \frac{{RE}}{W}$    …1)

Since, work input for both cycles is same (Process 1-2).

Area (4-5-6-1) > Area (4’-5’-6-1)

⇒ RE (Isentropic expansion) > RE (Isenthalpic expansion)

⇒ COP (isentropic expansion) > COP (isenthalpic expansion), {From 1)}

So, option 1) is correct

This can also be proved by drawing p-h chart

Figure: Pressure – enthalpy (p-h) chart for ideal VCR cycle

1-2: Isentropic compression

2-3: Heat rejection

3-4’: Isenthalpic expansion

3-4: Isentropic expansion

4-1: Heat addition (Evaporation)

$COP = \frac{{RE}}{W}$

For original cycle: (Isenthalpic expansion:3-4’)

$RE = {h_1} - h_4',;W = {h_2} - {h_1} \Rightarrow {\left. {COP} \right|_{original}} = \frac{{{h_1} - h_4'}}{{{h_2} - {h_1}}}$       …3)

For modified cycle: (Isentropic expansion) (3-4)

$RE = {h_1} - {h_4},;W = {h_2} - {h_1} \Rightarrow {\left. {COP} \right|_{Modified}} = \frac{{{h_1} - {h_4}}}{{{h_2} - {h_1}}}$      …4)

\(\frac{{{{\left. {COP} \right|}{Modified}}}}{{{{\left. {COP} \right|}{Original}}}} = \frac{{{h_1} - {h_4}}}{{{h_1} - h_4'}} > 1\) {∵ h₁ – h₄ > h₁ – h₄’}

\(\Rightarrow {\left. {COP} \right|{Modified}} > {\left. {COP} \right|{Original}}\)

Key Points:

Remember T-S and p-h plot for vapour compression refrigeration cycle. Also understand the nature of constant entropy lines in p-h chart. Study the effect of various parameters like evaporator pressure, condenser pressure on COP.

Vertical channel through which molten metal flows downward from pouring basin to runner is called SPRUE.

Riser acts as a reservoir of molten metal which feeds the casting during liquid shrinkage (Pouring temp to solidification temp) and solidification shrinkage (phase change at constant temp).

Blister and pinholes are casting defects.

Different components of a sand casting are shown below:

Concept:

The following table indicates heat intensity/power density and maximum temperature related with different welding process,

 

Out of the above options, laser beam welding having the highest heat flux > 10⁸ W/cm².

Key points:

Note down the values of temperatures also. Maximum temperature also occurs in laser beam welding.

Figure: Slab Milling

From right-angled triangle OA’B; (OB)² = (OA^’)² + (A’B)²

$\Rightarrow {\left( {\frac{D}{2}} \right)^2} = {\left( {\frac{D}{2} - d} \right)^2} + {A^2}$

$\Rightarrow \frac{{{D^2}}}{4} = \frac{{{D^2}}}{4} + {d^2} - 2\frac{D}{2}d + {A^2}$

⇒ A² = Dd – d² = d(D - d)

$\Rightarrow A = \sqrt {d\left( {D - d} \right)}$    …1)

Time for one pass (T) $= \frac{{L + A + O}}{{FZN}}$ …2)

Terms

D = diameter of wheel

d = depth of cut

N = rpm of wheel

A = approach

O = Over-travel

L = Length of workpiece

F = Feed per tooth

Z = No. of teeth

Calculation:

Given data;

d = 2 mm, D = 100 mm, L = 400 mm, F = 0.05 mm, Z = 20, N = 1200 rpm

$A = \sqrt {\left( 2 \right)\left( {100 - 2} \right)} = \sqrt {196} = 14mm$ …3)

Approach (A) = Over-travel (O)

$\Rightarrow T = \frac{{L + 2A}}{{FZN}} = \frac{{400 + \left( 2 \right)\left( {14} \right)}}{{\left( {0.05} \right)\left( {20} \right)\left( {1200} \right)}} = 0.3567;minutes$

Time (T) = 0.3567 minutes = 21.4 seconds …4)

So, approach distance (A) = 14 mm

Time taken (T) = 21.4 s

Concept:

For this we have to understand TRANSFORMATIONS.

In 2-D, these are 4 main transformations listed below:

i) Translation

ii) Scaling

iii) Rotation

iv) Reflection

i) TRANSLATION: in homogeneous coordinates:

\(\underbrace {\left[ {\begin{array}{{20}{c}} {x'}&{y'}&1 \end{array}} \right]}_{Final;coordinates} = \underbrace {\left[ {\begin{array}{{20}{c}} x&y&1 \end{array}} \right]}_{Initial;coordinates}\left[ {\begin{array}{*{20}{c}} 1&0&0\ 0&1&0\ {{t_x}}&{{t_y}}&1 \end{array}} \right] \to Translation;matrix\) …1)

Here

 translation in x-direction

${t_y}:;$translation in y-direction

ii) SCALING: Scaling transformation;

\(\underbrace {\left[ {\begin{array}{{20}{c}} {x'}&{y'}&1 \end{array}} \right]}_{Final;coordinates} = \underbrace {\left[ {\begin{array}{{20}{c}} x&y&1 \end{array}} \right]}_{Initial;coordinates}\left[ {\begin{array}{*{20}{c}} {{s_x}}&0&0\ 0&{{s_y}}&0\ 0&0&1 \end{array}} \right] \to Scaling;matrix\) …2)

iii) ROTATION: Rotation in 2-D about origin

Sign convention: In counter clockwise direction → θ: +ve

In clockwise direction → θ: -ve

\(\underbrace {\left[ {\begin{array}{{20}{c}} {x'}&{y'}&1 \end{array}} \right]}_{Final;coordinates} = \underbrace {\left[ {\begin{array}{{20}{c}} x&y&1 \end{array}} \right]}{Initial;coordinates}\underbrace {\left[ {\begin{array}{*{20}{c}} {\cos \theta }&{\sin \theta }&0\ { - \sin \theta }&{\cos \theta }&0\ 0&0&1 \end{array}} \right]}{Rotation;about;z - axis;passing} \to Rotation;matrix\)  …3)

iv) REFLECTION/MIRRORING: Reflection about x-axis

\(\underbrace {\left[ {\begin{array}{{20}{c}} {x'}&{y'}&1 \end{array}} \right]}_{} = \left[ {\begin{array}{{20}{c}} x&y&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&0&0\ 0&{ - 1}&0\ 0&0&1 \end{array}} \right] \to Reflection;matrix\)   …4)

Reflection about y-axis;

\(\underbrace {\left[ {\begin{array}{{20}{c}} {x'}&{y'}&1 \end{array}} \right]}_{} = \left[ {\begin{array}{{20}{c}} x&y&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 1}&0&0\ 0&1&0\ 0&0&1 \end{array}} \right] \to Reflection;matrix\) …5)

Calculation:

Given figure

In this case, rotation is about z-axis passing through origin in anti-clockwise direction.

Use equation 3);

\(\left[ {\begin{array}{{20}{c}} {x'}&{y'}&1 \end{array}} \right] = \left[ {\begin{array}{{20}{c}} x&y&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\cos \theta }&{\sin \theta }&0\ { - \sin \theta }&{\cos \theta }&0\ 0&0&1 \end{array}} \right]\) 

θ = +30°, x = 20, y = 10

\( \Rightarrow \left[ {\begin{array}{{20}{c}} {x'}&{y'}&1 \end{array}} \right] = \left[ {\begin{array}{{20}{c}} {20}&{10}&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\cos 30^\circ }&{\sin 30^\circ }&0\ { - \sin 30^\circ }&{\cos 30^\circ }&0\ 0&0&1 \end{array}} \right]\)

\(= \left[ {\begin{array}{*{20}{c}} {20\cos 30^\circ - 10\sin 30^\circ }&{20\sin 30^\circ + 10\cos 30^\circ }&1 \end{array}} \right]\)

\(= \left[ {\begin{array}{*{20}{c}} {12.3205}&{18.6601}&1 \end{array}} \right]\)

\(\left[ {\begin{array}{{20}{c}} {x'}&{y'}&1 \end{array}} \right] = \left[ {\begin{array}{{20}{c}} {12.3205}&{18.6601}&1 \end{array}} \right]\) 

 

⇒ x¹ = 12.32, y¹ = 18.66 ⇒ (12.32, 18.66) is correct.

Key Points:

Practice other transformations also like translation, scaling, rotation and reflection.

Concept:

Let us understand simple moving overage method.

 

Above table indicates monthly demand of a product. Now problem is that we have to calculate demand-forecast for month 5 using 3-months moving average method:

This will be given by;

${D_5} = \frac{{B + C + D}}{3}$  …1)

Calculation:

Given data

 

Using 3-months simple moving average method, the demand-forecast of the product for September will be:

${F_{sep}} = \frac{{495 + 475 + 560}}{3}$ 

F_sep = 510

Key Points:

One more important method in forecasting is exponential smoothing method.

As per this method;

F_n = F_n-1 + α (D_n-1 – F_n-1)      …2)

Where

F_n → Required forecast for n^th month

F_n-1 → Forecast for previous month (n-1).

D_n-1 → Demand for previous month (n-1).

α → Smoothing factor; 0 ≤ α ≤ 1

Greater the value of α, more will be the weight placed on recent data.

If α = 1 ⇒ F_n = F_n-1 + D_n-1 – F_n-1; F_n = D­_n-1

i.e. latest forecast is equal to previous period demand.

If α = 0 ⇒ F_n = F_n-1, this means demand pattern is stable.

Smoothing factor is calculated by;

$\alpha = \frac{2}{{n + 1}}$     …3)

Where n = number of periods in moving average

Concept:

Non-homogeneous equation of type AX = B has infinite solutions if ρ(A : B) = ρ(A) < Number of unknowns

Calculation:

Given set of equations

x + y + z = 1

ax – ay + 3z = 5

5x – 3y + az = 6

\(\left[ {\begin{array}{*{20}{c}} 1&1&1& \vdots &1\ a&{ - a}&3& \vdots &5\ 5&{ - 3}&a& \vdots &6 \end{array}} \right]\)

R₂ → R₂ – aR₁ and R₃ → R₃ – 5R₁

~ \(\left[ {\begin{array}{*{20}{c}} 1&1&1& \vdots &1\ 0&{ - 2a}&{3 - a}& \vdots &{5 - a}\ 0&{ - 8}&{a - 5}& \vdots &1 \end{array}} \right]\)

${R_2} \to \frac{{{R_2}}}{{ - 2a}}$

~ \(\left[ {\begin{array}{*{20}{c}} 1&1&1& \vdots &1\ 0&1&{\frac{1}{2} - \frac{3}{{2a}}}& \vdots &{\frac{1}{2} - \frac{5}{{2a}}}\ 0&{ - 8}&{a - 5}& \vdots &1 \end{array}} \right]\)

R₃ → R₃ + 8R₂

~  \(\left[ {\begin{array}{{20}{c}} 1&1&1& \vdots &1\ 0&1&{\frac{1}{2} - \frac{3}{{2a}}}& \vdots &{\frac{1}{2} - \frac{5}{{2a}}}\ 0&0&{a - 1 - \frac{{12}}{a}}& \vdots &{5 - \frac{{20}}{a}} \end{array}} \right]\) ~ \(\left[ {\begin{array}{{20}{c}} 1&1&1& \vdots &1\ 0&1&{\frac{1}{2} - \frac{3}{{2a}}}& \vdots &{\frac{1}{2} - \frac{5}{{2a}}}\ 0&0&{\frac{{{a^2} - a - 12}}{a}}& \vdots &{\frac{{5a - 20}}{a}} \end{array}} \right]\)

a² – a – 12 = 0

a² – 4a + 3a – 12 = 0

a(a - 4) + 3(a - 4) = 0

(a - 4)(a + 3) = 0

a = 4, -3

When a = 4, then ρ(A : B) = ρ(A) = 2 < 3

Hence, given system of equations have infinite solutions when a = 4.

Note: here a = -3 we cannot consider because for a = -3  ρ(A : B) ≠  ρ(A) 

Key Points:

Remember the system of equations

AX = B have

1. Unique solution, if ρ(A : B) = ρ(A) = Number of unknowns.

2. Infinite many solutions, if ρ(A : B) = ρ(A) <  Number of unknowns

3. No solution, if ρ(A : B) ≠ ρ(A).

Concept:

Let w = u + iν be a function of complex variable.

Function of a complex variable is analytic, if it satisfies Cauchy-reimann equation;

$i.e.\frac{{\partial u}}{{\partial x}} = \frac{{\partial \nu }}{{\partial y}};and;\frac{{\partial u}}{{\partial y}} = - \frac{{\partial \nu }}{{\partial x}}$

Calculation:

Given:

u(x, y) = 2x² – 2y² + 4xy, ν(x, y) = ?

$\frac{{\partial u}}{{\partial x}} = \frac{{\partial \nu }}{{\partial y}}$

$\frac{{\partial u }}{{\partial x}}=4x + 4y = \frac{{\partial \nu }}{{\partial y}}$

Integrating w.r.t y keeping x constant

ν(x, y) = 4xy + 2y² + f(x)

$\frac{\partial v}{\partial x}=4y;+;f'(x)$

$\frac{{\partial u}}{{\partial y}} = - \frac{{\partial \nu }}{{\partial x}}$

$\frac{\partial u}{\partial y}= - 4y + 4x$

4y – 4x = 4y + f’(x)

$f\left( x \right) = - \frac{{4{x^2}}}{2} + C= - 2{x^2} + C$

∴ ν(x, y) = 4xy + 2y² – 2x² + C

Concept:

In such questions, show various regions represented by equations on the graph.

Calculation:

Given that

0 ≤ x ≤ 10

0 ≤ y ≤ 20

p {x + y ≥ 20} = ?

Required probability = Area of right angled triangle ABC/Area of rectangular region OABD

$p=\frac{\frac{1}{2}\times 10\times 10}{10\times 20}=\frac{1}{4}=0.25$

Concept:

Dynamics of rigid bodies

$\sum \vec{F}=m\vec{a}$    ...(1)

\(\sum {{M}{G}}={{I}{G}}\) α    ...(2)

Calculation:

Applying (1) in x and y direction;

∑F_x = ma_x

$\Rightarrow \sum {{F}_{x}}=\frac{w}{g}a$    ...(3)

∑F_y = ma_y; a_y = 0

⇒ ∑F_y = 0

R_r + R_f = W    ...(4)

Let’s apply inertia face (ma) in a direction apposite to acceleration so that we can convert dynamics problem to static problem.

Now, taking moment about rear wheel to zero.

∑M_r = 0

${{R}_{f}}.l+\frac{Wa}{g}h-W.\frac{l}{2}=0$

\(\Rightarrow {{R}{f}}.l=\frac{Wl}{2}-\frac{Wah}{g};{{R}{f}}=\frac{W}{2}-\frac{Wah}{gl}\)    ...(5)

From (4)

R_r + R_f = W

\({{R}{r}}=W-{{R}{f}}=W-\frac{W}{2}+\frac{Wah}{gl}=\frac{W}{2}+\frac{Wah}{gl}\)

Hence, reactions on front and rear wheels are;

\({{R}{f}}=\frac{W}{2}-\frac{Wah}{gl};{{R}{r}}=\frac{W}{2}+\frac{Wah}{gl}\)    ...(6)

Key points:

Always be careful while converting dynamics problem into static problem using D’ Alembert’s principle.

The direction of inertial force (ma) is always opposite to the acceleration.

Concept:

i) Instantaneous Centre → It is a point in common between two members where the velocities are equal in direction & magnitude

ii) Relative velocity method → Velocity of any point on a link is always perpendicular to the line joining these points on the configuration diagram

Calculation:

Given mechanism;

ω₂ = 10 $\frac{{rad}}{{sec}}$, ω₄ = ?, AB = 5 cm, CD = 2 cm, AD = 4 cm

let’s locate I – centre 24.

Extend the lines joining the I centres 14,12 and 34, 23, The point of intersection of these two lines will be i₂₄.

Now, linear velocity of I – center(24) can be calculated as;

V₂₄ = (AI) ω₂ = (DI) ω₄

\(\Rightarrow {{\omega }{4}}=\frac{AI}{DI}.{{\omega }{2}}\)    ...(1)

From similarity of triangles BAI and CDI;

$\frac{AB}{AI}=\frac{CD}{DI}\Rightarrow \frac{AI}{DI}=\frac{AB}{CD}=\frac{5}{2}$

$\Rightarrow {{\omega }_{4}}=\frac{5}{2}.10=25~rad/s$

Angular velocity of bar DC (ω₄) = 25 rad/s

Key Points

In these problems, apply geometrical relations carefully.

Concept:

Gyroscopic couple = Iωω_p

Calculation:

Given data;

I = 10 kg m²

ω = 1100 rad/s

ν = 800 km/hr = 222.22 m/s

r = 1.5 km = 1500 m

\(\nu =r{{\omega }{p}}\Rightarrow {{\omega }{p}}=\frac{222.22}{1500}=0.148148~rad/s\) 

Gyroscopic couple (CG) = (10) (1100) (0.148148) = 1629.63 Nm

Reactive gyroscopic couple will dip the nose.

Key points:

Always use reactive gyroscopic torque for identifying direction of motion of nose.

Concept:

For solving this question, draw temperature (T) versus x profile.

Flow will be thermally fully developed when slopes of both mean fluid temperature line (T_m) and wall temperature line (T_s) are same.

Calculation:

Slope of line $PQ=\frac{80-50}{0.2}=150$

Slope of line $P'{Q}'=\frac{30-20}{0.2}=50$

⇒ Flow is not thermally developed at P.

Slope of line $QR=\frac{90-80}{0.2}=50$

Slope of line ${Q}'{R}'=\frac{40-30}{0.2}=50$

⇒ Flow is developed thermally at point Q and R.

Alternate Method

In case of uniform heat bulk mean temperature varies linearly. The difference between bulk mean temperature and wall temperature is constant in thermally developed region.

so bulk temperature.

T(x) = A + Bx (i)

At x = 0, T = 0°C

A = 0°C       (ii)

At x = 1, T = 50°C

B = 50°C    (iii)

from (i), (ii) & (iii)

bulk temperature T(x) = 50x

<br>

So, Q and R in thermally developed region.

Concept:

Apply steady flow energy equation (S.F.E.E) for solving this problem.

\(\frac{\delta Q}{\delta m}+{{h}{1}}+\frac{c{1}^{2}}{2}+g{{z}{1}}=\frac{\delta W}{{{\delta }{m}}}+{{h}{2}}+\frac{c{2}^{2}}{2}+g{{z}_{2}}\)

Neglecting change in kinetic and potential energies;

\(\Rightarrow \frac{\delta Q}{\delta m}+{{h}{1}}=\frac{\delta W}{\delta m}+{{h}{2}}\)     …(1)

Calculation:

P₁ = 100 kPa, T₁ = 27°C = 300 K, V̇₁ = 15 m³/s

R = 0.5 kJ/kgK

\({{P}{1}}{{̇{V}}{1}}=̇{m}R{{T}{1}}\Rightarrow ̇{m}=\frac{{{P}{1}}{{{̇{V}}}{1}}}{R{{T}{1}}}=\frac{100\times {{10}^{3}}\times 15}{0.5\times {{10}^{3}}\times 300}=10~kg/s\)

ṁ = 10 kg/s

Ẇ = -101 × 10³ watt, Q̇ = -51 × 10³ watt

Multiply (1) with mass flow rate

⇒ Q̇ + ṁh₁ = ẇ + ṁh₂

Q̇ + ṁc_p T₁ = ẇ + ṁc_pT₂

-51 × 10³ + (10) (10³) (300) = -101 × 10³ + (10) (10³) T₂

50 × 10³ + (10⁴) (300) = 10⁴. T₂

5 + 300 = T₂

⇒ T₂ = 305K = 32°C

Key points:

Always be careful while inserting data in S.F.E.E

Take proper sign conventions and care of unit conversion also.

Concept:

As per maximum shear stress criterion, material will fail when maximum shear stress in complex stress system reaches the yield point in simple tensile test.

\(i.e.~\frac{{{\sigma }{1}}-{{\sigma }{3}}}{2}=\frac{{{\sigma }_{yl}}}{2}\)

\(\Rightarrow {{\sigma }{1}}-{{\sigma }{3}}={{\sigma }_{yl}}\)     …(1)

σ₁ = Maximum principle stress

σ₃ = Minimum principle stress

σ_yl = yield strength

\({{\sigma }{1,3}}=\frac{{{\sigma }{x}}+{{\sigma }{y}}}{2}\pm \frac{1}{2}\sqrt{{{\left( {{\sigma }{x}}-{{\sigma }_{y}} \right)}^{2}}+4\tau _{xy}^{2}}\)    ...(2)

Calculation:

Given:

σ_x = -40 MPa, σ_y = ?, ${{\tau }_{xy}}=0$, σ_yl = 300 MPa

\(\Rightarrow {{\sigma }{1,3}}=\frac{{{\sigma }{x}}+{{\sigma }{y}}}{2}\pm \frac{{{\sigma }{x}}-{{\sigma }_{y}}}{2}\)

⇒ σ_1,3 = σ_x, σ_y

Let σ₁ = σ_x = -40, σ₃ = σ_y

⇒ -40 - σ_y = 300

⇒ σ_y = -340 MPa = 340 MPa (compressive)

Let σ₁ = σ_y, σ₃ = -40 MPa

⇒ σ_y – (-40) = 300 MPa

⇒ σ_y = +260 MPa = 260 MPa (Tensile), this is not possible because σ_y can’t be tensile (from figure).

So, σ_y = 340 MPa (compressive)

Concept:

Shear force (F_s) is calculated by;

\({{F}{s}}=\frac{\tau bt}{\sin \phi }\Rightarrow \tau =\frac{{{F}{s}}\sin \phi }{bt}\)     …(1)

Here, b = width of cut (mm)

t = uncut chip thickness (mm)

for orthogonal turning; λ = 90°

$b=\frac{d}{\sin \lambda },t=f\sin \lambda$

⇒ b = d, t = f

$\tau =\frac{{{F}_{s}}\sin \phi }{df}$    ...(2)

F_s = F_H cos ϕ - F_v sin ϕ

Where, F_H = Tangential face, F_V = Axial force/feed force

ϕ = shear angle

$\tan \phi =\frac{rcos\alpha }{1-r\sin \alpha }$    ...(3)

Calculation:

Given data;

t_c = 0.3 mm, f = 100 mm/min, N = 1000 rpm.

\(r=\frac{t}{{{t}{c}}}=\frac{f}{{{t}{c}}}=\frac{100}{\left( 100 \right)\left( 0.3 \right)}=\frac{1}{3}\) 

r = 0.333, α = 10°

$\Rightarrow \tan \phi =\frac{\left( 0.3333 \right)\cos \left( 10{}^\circ \right)}{1-\left( 0.333 \right)\sin \left( 10{}^\circ \right)}\Rightarrow \phi =19.21{}^\circ$ 

F_H = 1601 N_, F_V = 1259 N

⇒ F_S = (1601) cos(19.21°) – (1259) sin(19.21°) = 1097.62 N

\(d=5mm,f=\frac{100}{1000}=0.1mm/rev\) 

$\Rightarrow \tau =\frac{\left( 1097.62 \right)\sin \left( 19.21{}^\circ \right)}{\left( 5 \right)\left( 0.1 \right)}=722.302~MPa$ 

Key points:

Always remember relations between orthogonal machining and turning.

\(widthofcut\left( b \right)=\frac{Depthofcut~\left( d \right)}{\sin \lambda }\) 

λ = approach angle = 90° - C_S

C_S side cutting edge angle

For orthogonal turning; λ = 90° ⇒ C_S = 0°

Uncut chip thickness (t) = f sin λ

f = feed (mm/rev)

Concept: 

Clearance fit is shown in below figure.

<br>

Clearance fit is obtained when lower limit of hole is greater than or equal to upper limit of shaft.

Calculation:

Given:

$Diameter \ of\ the \ shaft = 65.00_{-0.05}^{+0.01}~mm, \ Coating \ thickness = 50\pm5\ \mu m$

Minimum hole diameter to just provide clearance will be upper limit of shaft.

Upper limit/maximum size of shaft after applying coating will be;

${{D}_{max}}=65.01+\frac{\left( 2 \right)\left( 55 \right)}{\left( 1000 \right)}=65.12~mm$

Key points:

Always multiply by 2 to thickness before adding in diameter.

If radius is given, then don’t multiply by 2 to thickness.

Concept:

Blow holes are generated when gases inside the moved cavity are not able to escape i.e. due to poor permeability. Misrun is caused due to insufficient fluidity of molten metal.

Hot tearing is due to poor collapsibility.

Wash is caused due to mould erosion caused by high velocity.

Application:

So, correct matching will be;

(P) Blow-hole → (3) Poor permeability

(Q) Misrun → (4) Insufficient fluidity

(R) Hot tearing → (1) poor collapsibility

(S) Wash → (2) Mold erosion