GATE ME 2017 Official Paper: Shift 2 (Previous Year Paper)

Probability = $\frac{{No;of;favourable;cases}}{{No;of;possible;cases}}$

∴ The probability that both children are boys if the order one is a boy = (1/2)

P is married and P looks at Q. Q looks at R but marriage status of Q is unknown. R is not married at all.

As we do not know the maritial status of Q hence here there are two possible cases.

Case 1: Q is married

In this case the answer will  be one pair because in the question it has been asked to find the number of pairs of people in which a married person is looking at an unmarried person. And here P is married and looking at Q who is also married so this pair would not be counted but now Q is looking at R and R is unmarried and hence the number of pairs will be 1.

Case 1: Q is unmarried

Similarly from the above case the pair will be only 1 in this case too as P is looking at Q and Q is not married.

Hence the final answer will be 1 (Including both cases whether the Q is married or not married.)

If we choose one option the second option becomes unavailable to us. This is best expressed by the word ‘forgo’. Also, why we need to forgo the second option is because the two plans cannot exist simultaneously. Thus ‘exclusive’ is the correct word to fit the second blank.

The subject of the sentence ‘the ways in which this game’ is plural and hence the blank takes the plural verb ‘are’.

According to the given relation of a – b = even, there is a possibility of odd-odd (or) even-even is equal to even.

From the options, Option (D) is correct.

Since, odd × odd – odd (or) even × even – even → is always even number.

Given number of days required that X bullocks and Y tractors to plough a field = 8 days

(I.e. X + Y→ 8D ⇒ 8X + 8Y → 1 day) No of days required that X/2 bullocks and 2Y tractors to plough field = 5

[i.e, $\frac{X}{2} + 2Y \to 5D \Rightarrow \left( {\frac{5}{2}} \right){\rm{X}} + 10{\rm{Y}} \to 1{\rm{D}}]$

From above equations,

Y = 11X/4 and from 1^st equation, X = 30 days

‘Cleaving’ means separating. Thus the correct word that expresses this meaning is ‘splitting’.

The difference between the maximum and the minimum pollutant concentrations

(i) in winter = 8 - 0 = 8ppm,

(ii) in summer = 10.5 - 1.5 = 9ppm

∴ (i) is false & (ii) is correct from the graph.

If P is paired with y; they

Q has 4 choices

R has 3 choices

S has 2 choices

Total 24 choices

(Or)

If Q is paired with y;

Then

P has 3 choices

R has 3 choices

S has 2 choices

Total 18 choices

(Or)

If R is paired with y; then

P has 3 choices

Q has 3 choices

S has 2 choices

Total 18 choices

(Or)

If S is paired with y; then

P has 3 choices

Q has 3 choices

S has 2 choices

Total 18 choices

∴ Total number of ways = 24 + 18 + 18 + 18 = 78

Here let’s check each option,

a) Both P and Q are knights → If both are knights then both must say the truth.

P says “Both of us are knights”, Q says “None of us are knaves”.

Thus can be true.

b) P is a knight; Q is a knave → If P is a knight then P must say the truth.

P says “Both of us are knights”, but Q says “None of us are knaves”.

This cannot be true.

c) Both P and Q are knaves → If both are knaves then both must lie.

P says “Both of us are knights”, Q says “None of us are knaves”.

Thus can be true.

Thus both P and Q are knights or both P and Q are knaves.

Thus identities of P, Q cannot be determined.

Alternate Solution

Essentially both P and Q are saying the same thing. But we cannot infer the truth of the statements. If both are knights, then both are telling the truth. However, if both are knaves then they are lying. Thus the identities of P and Q cannot be determined.

Explanation:

Measures of central tendency provide us with a summary that describes some central or middle point of the data. 

There are five important measures of central tendency, viz.,

i) arithmetic mean,

ii) median,

iii) mode,

iv) geometric mean, and

v) harmonic mean. 

Out of these the last two measures, viz., geometric mean and harmonic mean, have very specific uses and thus less frequently used.

Mode: The word mode has been derived from the French word “la Mode” which signifies the most fashionable values of distribution because it is repeated the highest number of times in the series. The mode is the most frequently observed data value. It is denoted by Mo.

• The mode is seldom used and its computation is easy, but it is highly unstable and may change with minor shifts in the frequencies from one interval to another.
• However, there are situations in which the only mode can be used.
• For example, if a shoe company wants to how which size of shoe it should produce more, it would use mode as a measure of central tendency. The most frequently sold size of the shoes is the mode.
  

Arithmetic mean: 

• The arithmetic mean is the most commonly used measure of central tendency.
• The mean represents the central tendency. It is defined as the sum of the values of all observations divided by the number of observations and is usually denoted by X.
• In general if there are N observations as X1, X2, X3, ..., XN, then the Arithmetic Mean is given by
• Mean= ${X1 + X2 + X3 +...+ Xn}\over n$.
• This will be written in simpler form without the index i.
• Thus mean = N ∑ $X\over N$ where, ΣX = sum of all observations and N = total number of observations. 
  

Median is that positional value of the variable which divides the distribution into two equal parts, one part comprises all values greater than or equal to the median value and the other comprises all values less than or equal to it. 

• The Median is the “middle” element when the data set is arranged in order of magnitude. Since the median is determined by the position of different values, it remains unaffected if, say, the size of the largest value increases.
• The median can be easily computed by sorting the data from smallest to largest and finding out the middle value
  

Calculation:

17, 18, 17, 17, 13, 18, 5, 5, 6, 7, 8, 9, 20, 17, 3

Mode refers the most frequently appeared value (which is 17 in this case).

Concept:

Laplace transform of ${t^n} \cdot {e^{at}} u(t) = \frac{{n!}}{{{{\left( {s - a} \right)}^{n + 1}}}}$

Now $L\left( {t{e^t}} u(t)\right) = \frac{{1!}}{{{{\left( {s - 1} \right)}^{1 + 1}}}} = \frac{1}{{{{\left( {s - 1} \right)}^2}}}$

Explanation:

Springs are in Parallel so k_eq. = k + k = 2k

$\omega = \sqrt {\frac {k_{eq}}{m}}=\sqrt {\frac{{2k}}{m}}$

Concept:

${M_{x - x}} = P \cdot x$

\(Strain;energy = \mathop \smallint \limits_0^L \frac{{{{\left( {{M_{x - x}}} \right)}^2}dx}}{{2EI}}\)

\({U_{X - X}} = \mathop \smallint \limits_0^L \frac{{{{\left( {Px} \right)}^2}}}{{2EI}}dx = \frac{{{P^2}}}{{2EI}}\mathop \smallint \limits_0^L {x^2}dx\)

$= \frac{{{P^2}}}{{2EI}}\left| {\frac{{{x^3}}}{3}} \right|_0^L = \frac{{{P^2} \cdot {L^3}}}{{2EI \times 3}} = \frac{{{P^2}{L^3}}}{{6EI}}$

Concept:

Impulse Turbine: If at the inlet of the turbine, the energy available is only kinetic energy, the turbine is known as impulse turbine. e.g. a Pelton wheel turbine.

Reaction Turbine: If at the inlet of the turbine, the water possesses kinetic energy as well as pressure energy, the turbine is known as a reaction turbine. e.g. e Francis and Kaplan turbine.

Tangential flow turbines: In this type of turbines, the water strikes the runner in the direction of the tangent to the wheel. Example: Pelton wheel turbine

Radial flow turbines: In this type of turbines, the water strikes in the radial direction. accordingly, it is further classified as

• Inward flow turbine: The flow is inward from periphery to the centre (centripetal type); Example: old Francis turbine
• Outward flow turbine: The flow is outward from the centre to periphery (centrifugal type); Example: Fourneyron turbine

Axial flow turbine: The flow of water is in the direction parallel to the axis of the shaft. Example: Kaplan turbine and propeller turbine

Concept:

$W = -\smallint PdV$ (-Ve Sign taken because work is done on the system)

For an ideal gas, PV = mRT = const (c)

⇒ PV = C;             ∴ P = C/V

\(W = -\mathop \smallint \limits_{{V_1}}^{{V_2}} \frac{C}{V}dV \Rightarrow W = -C\ln \left( {\frac{{{V_2}}}{{{V_1}}}} \right)\)

$W =- PV\ln \left( {\frac{{{V_2}}}{{{V_1}}}} \right)$

$W = -{P_1}{V_1}\ln \left( {\frac{{{V_2}}}{{{V_1}}}} \right)$

When work is done on the system (compression), V₂ < V₁ and P₂ > P₁

${P_1}{V_1} = {P_2}{V_2} \Rightarrow \frac{{{V_2}}}{{{V_1}}} = \frac{{{P_1}}}{{{P_2}}}$

$W = -{P_1}{V_1}\ln \left( {\frac{{{P_1}}}{{{P_2}}}} \right)$

Concept:

A psychrometric chart is represented as shown in the figure.

Calculation:

Given, that moist air is contained in CLOSED VESSEL, so we can say that mass of moist air is constant.

$w = specific;humidity = \frac{{{m_v}}}{{{m_{dry;air}}}} = constant$

So specific humidity is also same.

Now as the container is heated, so temperature increases. On psychometry chart, it can be represented by process 1 - 2.

Along the saturation line, RH is 100%. So in this case, RH is decreasing.

Explanation:

Condition of stable equilibrium for a floating body in terms of metacentric height (GM) as follows:

• Stable equilibrium : GM > 0 (M is above G) i.e. Metacentric height is positive.
• Neutral equilibrium : GM = 0 (M coinciding with G)
• Unstable equilibrium : GM < 0 (M is below G)

Condition of stable equilibrium for a submerged body in terms of centre of buoyancy and the centre of gravity:

• Stable equilibrium: If its centre of gravity is directly below the centre of buoyancy
• Neutral equilibrium: If its centre of gravity is coincident with the centre of buoyancy
• Unstable equilibrium: If its centre of gravity is directly above the centre of buoyancy

Explanation:

General velocity profile for Laminar flow

$\frac{u}{{{u_\infty }}} = \frac{3}{2}\left( {\frac{y}{\delta }} \right) - \frac{1}{2}{\left( {\frac{y}{\delta }} \right)^3}$

\({\tau {wall}} = \mu {\left( {\frac{{\partial u}}{{dy}}} \right){y = 0}} = \frac{{3\mu }}{{2\delta }}\)

Now $\delta = \frac{{4.65x}}{{\sqrt {R{e_X}} }}$

$\delta \propto {x^{\frac{1}{2}}}$

${\tau _{wall}} \propto {x^{\frac{{ - 1}}{2}}}$

$= \tau \propto \frac{1}{{\sqrt x }}$ as the distance from the leading edge increases, shear stress decreases.

Concept:

The radiation energy emitted by a body per unit time is given by:

Eb = ϵAσT4

Where ϵ is emissivity of the body.

σ = The Stefan – Boltzmann constant = 5.67 × 10-8 W m-2K-4

Calculation:

For a given area  E α T4

$\frac{{{E_2}}}{{{E_1}}} = \frac{{T_2^4}}{{T_1^4}}$

$\frac{{{E_2}}}{{{E_1}}} = \frac{{{{\left( {2{T_1}} \right)}^4}}}{{T_1^4}} \Rightarrow {E_2} = 16E$,

E₁ = P(given)

E₂ = 16 P

Concept:

During ultrasonic machining the metal removal is achieved by hammering action of abrasive particles.

In ultrasonic machining, a tool of desired shape vibrates at an ultrasonic frequency (19 ∼ 25 kHz) with an amplitude of around 15 – 50 μm over the workpiece.

Generally, the tool is pressed downward with a feed force, F.

Between the tool and workpiece, the machining zone is flooded with hard abrasive particles generally in the form of a water-based slurry.

As the tool vibrates over the workpiece, the abrasive particles act as the indenters and indent both the work material and the tool.

The abrasive particles, as they indent the work material, would remove the work material, particularly if the work material is brittle (due to crack initiation, propagation and brittle fracture of the material).

Additional Information

(a) Wrong: In USM, the tool vibrates at high may be sonic frequency.

(b) Wrong: USM uses transducers so as to produce low amplitude vibration

(c) Wrong: USM is an excellent process for machining hard and brittle materials.

(d) In USM slurry comprising abrasive particles and water are often uses to remove material from the workpiece by abrasion or impact grinding action.

Concept:

Crystal structure of Material

FCC: FCC stands for Face Centered Cubic. In one unit cell,1 atom at each corner, and 1 atom on each face. The crystal structure is used for Ductile materials only. 

E.g. Ni, Cu, Ag, Pt, Au, Pb, Al, Austenite or γ-iron

BCC: BCC stands for Body-Centered Cubic. In one unit cell, there is one atom at centre, 1 atom at each corner. The crystal structure is used for Brittle materials only. 

E.g. V, Mo, Ta, W, Ferrite or α-iron, δ-ferrite or δ-iron

HCP: HCP stands for Hexagonal Close Packed Crystal Structure.

E.g. Mg, Zn, Ti, Cd, Co

Explanation:

For steady-state μ > λ i.e. service rate should be greater than arrival rate.

where μ = service rate and λ = arrival rate.

As λ = 12.

So for finite queue length μ = 24 is the best option.

Explanation:

Generally, Strength — Cold working > Metal Forming > Casting > Machining > Welding

Highest strength is obtained through metal forming processes because due to continuous application of force work hardening occurs.

Concept:

f(z) = (x² + ay²) + i bxy = u(x,y) + iv(x,y)

u(x,y) = x² + ay²

v(x,y) = bxy

$\frac{{\partial u}}{{\partial x}} = 2x;;\frac{{\partial u}}{{\partial y}} = 2ay$

$\frac{{\partial v}}{{\partial x}} = by;;\frac{{\partial v}}{{\partial y}} = bx$

Using C-R equation

$\frac{{\partial u}}{{\partial x}} = \frac{{\partial v}}{{\partial y}} \Rightarrow 2x = bx \Rightarrow b = 2$

$\frac{{\partial u}}{{\partial y}} = \frac{{ - \partial v}}{{\partial x}} \Rightarrow 2ay = - by \Rightarrow 2a = - b$

$a = \frac{{ - b}}{2} = - 1$

So, a = -1, b = 2

Concept:

Eccentric Loading of Riveted Joints:

• An eccentrically loaded joint is one in which line of application of load does not pass through centre of gravity (c.g) of rivets.
• It passes away from c.g axis.
• This has two effects, primary/direct load and secondary load.

Direct Load:

• This load acts parallel to the load acting and vertically downwards.
• The magnitude for the direct load is $P'=\frac{Load;acting}{No.;of;rivets}$
• It is represented by P'.

Secondary load:

• It acts perpendicular to the line joining the centre of gravity of rivets assembly to individual rivets.
• The direction of the secondary load is the same as given by external load i.e. the moment produced due to external load is clockwise so the secondary load will also be clockwise.
• The magnitude for the secondary load is P'' = $\frac{P;\times ;e;\times;r_1}{\sqrt{r_1^2;+;r_2^2;+;r_3^2;+;r_4^2}}$ for rivet 1 and similarly for other rivets by changing r₂, r₃ and r₄ in the numerator.

Now, both the load are added vectorially.

Calculation:

Given:

$\rm {P_{{S_1}}} = Primary;shear = \frac{6}{3} = 2kN$

${P_{{S_2}}} = \frac{{{P_e}{r_c}}}{{r_A^2 + r_B^2 + r_C^2}} = \frac{{6 \times 250 \times 50}}{{{{50}^2} + {0^2} + {{50}^2}}}$

${P_{{S_2}}} = 15kN$

∴ Total resultant shear force

${P_s} = {P_{{S_1}}} + {P_{{S_2}}} = 15 + 2 = 17kN$

Concept:

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

i.e. Maximum shear stress τmax = larger of \(\left| {\left( {{{\bf{\sigma }}_1} - {{\bf{\sigma }}_2}} \right),\left( {{{\bf{\sigma }}_2} - {{\bf{\sigma }}_3}} \right),\left( {{{\bf{\sigma }}_3} - {{\bf{\sigma }}2}} \right)} \right| \le \frac{{{{\bf{S}}{{\bf{yt}}}}}}{{\bf{N}}}\)

Tensile yield strength (Proportionality limit) = Syt = 200 MPa

Calculation:

Given:

σ₁ = 60 MPa; σ₂ = 5 MPa and σ₃ = -40 MPa

σ_yt = 200 MPa

$|(60-5),(5-(-40)),((-40)-60)|\leq\frac{200}{N}$

$|55,45,(-100)|\leq\frac{200}{N}$

$100\leq\frac{200}{N}$

N = 2

Concept:

Steady state energy equation

$\dot m{h_1} + \dot Q = \dot m{h_2} + W$

Calculation:

Applying the steady state flow equation

$\dot m{h_1} + \dot Q = \dot m{h_2} + W$

$\dot m\left( {{h_1} - {h_2}} \right) + \dot Q = \dot W$

$\dot m{C_p}\left( {{T_1} - {T_2}} \right) + \dot Q = \dot W$

$\dot m \times 1 \times \left( {1100 - 400} \right) + \left( { - 300} \right) = 4.6 \times {10^3}$

$\dot m = 7\frac{{Kg}}{s};$

V = 2.87 m³; T = 300 K

P_gauge = 0.5 bar; P_atm = 1 bar; R = 287 j/Kg-K

PV = mRT (P is absolute pressure)

P_abs = P_atm + P_gauge = 0.5 + 1 = 1.5 bar = 150 kPa

\(PV = mRT \Rightarrow m = \frac{{PV}}{{RT}} = \frac{{1502.87}}{{0.287300}}\)

m = 5 Kg

Concept:

$LMTD = \frac{{{\rm{\Delta }}{T_1} - {\rm{\Delta }}{T_2}}}{{\ln \left( {\frac{{{\rm{\Delta }}{T_1}}}{{{\rm{\Delta }}{T_2}}}} \right)}}$

Calculation:

Given:

${m_c} = 1.5\frac{{Kg}}{s};{T_{{c_1}}} = 40^\circ C;{T_{{c_2}}} = 80^\circ C$

${T_{{h_1}}} = 120^\circ C;{T_{{h_2}}} = 60^\circ C;{C_c} = 4.2\frac{{kJ}}{{KgK}};;{C_h} = \frac{{2kJ}}{{kgK}}$

U = 400 W/m²K

$Q = {m_c}{C_c}\left( {{T_{{c_2}}} - {T_{{c_1}}}} \right) = 252 \times {10^3}W$

$LMTD = \frac{{{\rm{\Delta }}{T_1} - {\rm{\Delta }}{T_2}}}{{\ln \left( {\frac{{{\rm{\Delta }}{T_1}}}{{{\rm{\Delta }}{T_2}}}} \right)}} = \frac{{40 - 20}}{{\ln \left( {\frac{{40}}{{20}}} \right)}}$

ΔT_m = LMTD = 28.854°C

Q = UAΔT_m

252 × 10³ = 400 A × 28.854

A = 21.83 m²

Concept:

From the lumped parameter analysis

$\frac{{T - {T_\infty }}}{{{T_i} - {T_\infty }}} = {e^{ - \frac{{hA}}{{\rho Vc}}t}}$

Calculation:

$\frac{A}{V} = \frac{{4\pi {R^2}}}{{\frac{4}{3}\pi {R^3}}} = \frac{3}{R}$

$\frac{{T - {T_\infty }}}{{{T_i} - {T_\infty }}} = {e^{ - \frac{{hA}}{{\rho Vc}}t}}$

$\frac{{T - 20}}{{220 - 20}} = {e^{ - \frac{{200 \times 3}}{{30 \times {{10}^{ - 3}} \times 9000 \times 400}} \times 90}}$

T = 141.3°C

Initial solution by North-west corner Rule

X = 10 × 7 + 3 × 1 + 4 × 8

X = Rs 105

Optimality: As the total number of allocations are m + n -1 = 3 and at independent positions. So optimality can be performed.

Modified distribution (MODI-Method)

1) Cost matrix for allocated cells only and finding u_i & v_j value taking v₁ = 0

u₁ + v₁ = 10 ⇒ u₁ = 10

u₂ + v₁ = 3 ⇒ u₂ = 3

u₂ + v₂ = 4 ⇒ v₂ = 1

2) Develop u_i + v_j Matrix for unallocated cells by entering sum of u_i & v_j value for unallocated cells.

u_i + v_j Matrix for unallocated cells.

3) Subtract cell values of u_i + v_j Matrix from original cost Matrix to get cell evaluation Matrix.

As one cell value is still –ve so, the current solution is not optimum.

Cell Evaluation Matrix: Performing optimality we select the smallest allocation with – ve sign, to be allocated at identified cells. Smallest allocation is 7 Units, which is allocated at identified cell.

So, the new allocations are and the corresponding cost

= 7 × 7 + 3 × 8 + 4 × 1

= Rs 77

Now again we have to check whether the current solution is optimum.

1) cost Matrix for allocated cells and u_i & v_j value

2. Subtracting from original cost Matrix to get cell evaluation Matrix.

Cell Evaluation Matrix

As all the cell values are +ve in the cell evaluation Matrix.

So, the current solution is optimum.

So, Y = Rs. 77

And X- Y = 105 – 77 = Rs. 28

Concept:

$Engineering;strain = \frac{{{\rm{\Delta }}L}}{L}$

$True~strain= ln(1 + \varepsilon _E)$

Concept:

$Engineering;strain = \frac{{{\rm{\Delta }}L}}{L} = \frac{{40 - 20}}{{20}} = 1$

True strain = ln(1 + 1) = ln2

$Engineering;stain = - \left( {\frac{{40 - 10}}{{40}}} \right) = \frac{{ - 3}}{4}$

$True;strain = \ln \left( {1 - \frac{3}{4}} \right) = \ln \left( {0.25} \right)$

Total true strain = ln 2 + ln 0.25 = -0.693

Concept:

According to Taylor’s law

VTⁿ = C

V = πDN m/min

$\left( {\pi {D_1}{N_1}} \right)T_1^n = \left( {\pi {D_2}{N_2}} \right)T_2^n$

${N_1}T_1^n = {N_2}T_2^n$

400 × (20)ⁿ = 200 × 60ⁿ = 300 × Tⁿ

$\frac{{400}}{{200}} = {\left( {\frac{{60}}{{20}}} \right)^n} \Rightarrow n = 0.6309$

Now 400 × 20ⁿ = 300 × Tⁿ

${\left( {\frac{{400}}{{300}}} \right)^{\frac{1}{n}}} \times 20 = T$