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:

Favorable outcomes are (T,H), (H,T), (H,H)

Possible Outcomes are (T,T);(T,H);(H,T);(H,H)

Number of favorable outcome = 3

Number of possible outcome = 4

Probability of at least one head is = ¾ = 0.75

Explanation:

Divergence of a vector function \(F = {F_1}\hat i + {F_2}\hat j + {F_3}\hat k\)

\( = \nabla \cdot F = \frac{{\partial {F_1}}}{{\partial x}} + \frac{{\partial {F_2}}}{{\partial y}} + \frac{{\partial {F_3}}}{{\partial z}}\)

Calculation:

Vector = -yi +xj

∴ Divergence \( = \frac{\partial }{{\partial x}}\left( { - y} \right) + \frac{\partial }{{\partial y}}\left( x \right) = 0 + 0 = 0\)

Concept:

Determinant of a matrix is the product of its eigenvalues.

Determinant of a matrix is same as its transpose.

Determinant of a matrix is reciprocal to its inverse.

Calculation:

Eigenvalues of matrix A are 10 and y.

Determinant of matrix A = 10 × y

50 = 10 × y

∴ y = 5

 the other Eigen value is 5

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:

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 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.

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

Calculation:

σ_x = σ_y = σ_z = τ_xz = τ_zx = τ_yz = 0 and τ_xy = τ_yx = 50 MPa​

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

\(\sigma_{1,2} = 0 \pm \sqrt {0 \pm {{50}^2}} \)

= ± 50 MPa

Alternate Solution: - This is a case of ϕ biaxial pure shear.

Pure shear

Mohr’s circle

\(\sigma_{max}=\sigma_{min}=\tau_{max}=Radius\)

So Maximum normal stress = 50 MPa

Concept:

\(Shear;Force = \frac{{dM}}{{dx}} \)

Calculation:

Given:

M = 5x² + 10x

\(Shear;Force = \frac{{dM}}{{dx}} = 10x + 10\)

Shear force at X = 10 mm

= 10 x 10 + 10 = 110 N

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:

Thermal stress induced in a body is given by:

\({σ _{thermal}} = α {\rm{\Delta }}TE\)

Calculation:

Given:

ΔT = 100°C, α = 11 × 10-6 / 0C and E = 200 GPa

\({σ _{thermal}} = α {\rm{\Delta }}TE\)

σ_thermal = 100 × 11 × 10^-6 × 200 × 10³ = 220 MPa

Concept:

\(\frac{1}{{FOS}} = \frac{{{\sigma _{mean}}}}{{{\sigma _{yield}}}} + \frac{{{\sigma _v}}}{{{\sigma _{endurance}}}}\)

Calculation:

As the machine component made of a ductile material and subjected to completely reversed loading, hence all three fatigue loading theories will give same results

\({\sigma _{mean}} = 0;{\sigma _{variable}} = \frac{50-{\left( { - 50} \right)}}{2} = 50MPa\)

\(\frac{1}{{FOS}} = \frac{{{\sigma _{mean}}}}{{{\sigma _{yield}}}} + \frac{{{\sigma _v}}}{{{\sigma _{endurance}}}}\)

\(\frac{1}{{FOS}} = \frac{0}{{300}} + \frac{{50}}{{100}}\)

FOS = 2

Concept:

From the diagram below,

By using Kennedy's theorem, I24 (Common I center of links 2 and 4) will lie on the line joining I21, I14, and I23, I24. The slider C can be considered as the link of infinite length rotating about point D on the ground. So I24 will coincide with I34. For links 2 and 4 (links of known and unknown variables),

⇒ ω2 × (I12 - I24) = ω4 × (I14 - I24)     

Where,

VS = ω4 (I14 - I24) = Velocity of the slider with respect to the ground  

So, the velocity of the slider,

⇒ Vs = ω2 × (I12 - I24) = ω2 × AC      ...(1) 

Calculation:

Given:

ω₂ = 10 rad/s, AB = 100 mm, BC = 160 mm

So, AC = √ (160² -100²) = 124.89 mm =  0.125 m 

Using equation (1),

⇒ Vs = ω₂ × (I₁₂ - I₂₄) = ω₂ × AB

⇒ V_S = 10 × 0.1

⇒ V_S = 1 m/sec

⇒ V_S = 1 m/sec

Alternate Method 

Calculation:

Given:

ω2 = 10 rad/s, AB = 100 mm, BC = 160 mm

Referring to the diagram below,

Where C is on the slider and D is on the fixed link.

Since points, A and D are on the fixed link so they have zero relative velocity. So, referring to the velocity triangle

,

Here, cd is the velocity of the slider which is clearly equal to the velocity of point b i.e. ab

cd = ab

⇒ Vs = ab = AB ω2 

⇒ VS = 10 × 0.100

⇒ VS = 1 m/s

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:

\(\eta = \frac{{Actual;heat;transfer}}{{Heat;transfer;keeping;the;entire;fin;at;base;temp}}\)

Calculation:

Heat loss (Q) = 6 W

Effectiveness (ε) = 3

Efficiency (η) = 0.75

\(\eta = \frac{{Actual;heat;transfer}}{{Heat;transfer;keeping;the;entire;fin;at;base;temp}}\)

\(0.75 = \frac{6}{{Heat;transfer;keeping;the;entire;fin;at;base;temp}}\)

Heat transfer keeping the entire fin surface at base temperature = 8 W

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

Concept:

Molar weight of Fe = 56 g and

Molar weight of Carbon(C) = 12 g

∴ Molecular weight of Fe₃C = (3 × 56 + 12 = 180)

Weight of carbon in Fe₃C = 1 × 12 = 12

Weight % of carbon \( = \frac{{12}}{{180}}*100 = 6.67% \)

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.

Concept:

Variance = Sum of square of standard deviation of each dimension

Standard deviation = Square root of Variance of the system

Calculation:

Variance = 9 + 16 = 25

Standard deviation \( = \sqrt {25} = 5\)

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:

Gauss divergence theorem

\(\mathop \int!!!\int \limits_S^; F \cdot nds = \mathop \smallint \limits_S^; div \cdot Fdv\) 

Calculation:

\(\mathop \int!!!\int \limits_S^; F \cdot nds = \mathop \smallint \limits_S^; div \cdot Fdv\)

\( = \mathop \smallint \limits_S^; \nabla \left[ {\left( {x + y} \right)\hat i + \left( {x + z} \right)\hat j + \left( {y + z} \right)\hat k} \right]dv\)

\(= \mathop \smallint \limits_S^; \left( {1 + 0 + 1} \right)dv\)

\( = 2\smallint dv = 2V\)

\(V = \frac{4}{3}\pi {\left( 3 \right)^3} = 36\pi \)

2 × 36 π = 72 π = 226.19

3y”(x) + 27 y(x) = 0

y”(x) + 9y(x) = 0

(D² + 9) y = 0

So characteristic equation is given by:

m² + 9 = 0

m = ± 3i = 0 ± 3i

y = (C₁ cos3x + C₂ sinx)e^0x

y = C₁ cos 3x + C₂ sin 3x

y’ = -3C₁ sin 3x + 3C₂ cos 3x

y’(0) = 2000

\( \Rightarrow 3{C_2} = 2000 \Rightarrow {C_2} = \frac{{2000}}{3}\)

y(0) = 0 ⇒ C₁(1) + C₂(0) = 0

⇒ C₁ = 0

\(y = \frac{{2000}}{3}\sin 3x\)

\(y\left( 1 \right) = \frac{{2000}}{3}\sin (3\times 1 \times\frac{{{180}}}{{{\pi}}}) = 94.08\)

\(A = \left[ {\begin{array}{*{20}{c}} {50}&{70}\ {70}&{80} \end{array}} \right]\)

Eigen values of A are λ₁ and λ₂

According to properties of eigen values,

Sum of Eigen values = Trace of matrix

= sum of diagonal Elements.

So, λ₁ + λ₂ = 80 + 50 = 130

Product of eigen values = Determinant of matrix

λ₁⋅λ₂ = (80 × 50) – (70 × 70) = -900

Now \({X_1} = \left[ {\begin{array}{*{20}{c}} {70}\ {\lambda - 50} \end{array}} \right],;{X_2} = \left[ {\begin{array}{*{20}{c}} {{\lambda _2} - 80}\ {70} \end{array}} \right]\)

\(X_1^T{X_2} = \left[ {\begin{array}{*{20}{c}} {70}&{{\lambda _1} - 50} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{\lambda _2} - 80}\ {70} \end{array}} \right]\)

= 70 λ₂ – 5600 + 70λ₁ – 3500

= 70 (λ₁ + λ₂) – 9100 = 70 × 130 – 9100

= 9100 – 9100 = 0

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:

When a point on one link is sliding along another rotating link, then the Coriolis component of the acceleration is to be calculated.

ac = 2 ωV

Where ω = Angular velocity of the link OA (rad/s) and v = Velocity of slider B with respect to the coincident point C (m/s)

Calculation:

a_centripetal = ω²r = (20)² × 0.1 = 40 m/s²

a_corilios = 2ωv = 2 × 20 × 6 = 240 m/s2

\(a = \sqrt {{{\left( {40} \right)}^2} + {{\left( {240} \right)}^2}} = 243.31~m/s^2\)

Concept:

Mass moment of inertia = mk²

where m = mass and k = Radius of gyration

Calculation:

Radius of gyration (k) = 0.1 m

Mass moment of inertia = m(0.1)² kg-m²

\(I\mathop {\ddot \theta }\limits^{} + mg.l\sin \theta = 0\)

For very small, sin θ ≈ θ

\(\mathop {\ddot \theta }+\frac{{\mathop {}\limits^{} mgl }}{I}\theta = 0 \Rightarrow \mathop {\ddot \theta }\limits^{} + \frac{{m \times 9.81 \times 0.25}}{{m \times {{\left( {0.1} \right)}^2}}} \cdot \theta = 0\)

\({\omega _n} = \sqrt {\frac{{9.81 \times 0.25}}{{{{\left( {0.1} \right)}^2}}}} = 15.66\frac{{rad}}{s}\)

Concept:

​\(\frac{{{N_Q}}}{{{N_P}}} = \frac{{{T_P}}}{{{T_Q}}} = \frac{{24}}{{45}} \Rightarrow \frac{{{N_Q}}}{{400}} = \frac{{24}}{{45}} \Rightarrow {N_Q} = 400 \times \frac{{24}}{{45}} = \frac{{640}}{3}\)

\( \Rightarrow \frac{{{N_Q}}}{{{N_s}}} = \frac{{{T_S}}}{{{T_Q}}} = \frac{{80}}{{45}} \Rightarrow {N_s} = {N_Q} \times \frac{{{T_Q}}}{{{T_s}}} = \frac{{640}}{3} \times \frac{{45}}{{80}} = 120;rpm\)

Concept:

For a helical spring, maximum shear stress developed in the wire, 

\({τ _{shear}} = \frac{{8WD}}{{\pi {d^3}}} {{;;;;;}} \ldots \left( 1 \right)\)

Where, 

D =  Mean coil diameter, d =  Wire diameter, W  = Axial load

Calculation:

Given:

τ_max1 = 24 MPa, d₂ = 2d₁,  D₂ = 2D₁

By using equation (1),

\(\Rightarrow \tau_{shear2} = \frac{{8WD_2}}{{\pi {{\left( {d_2} \right)}^3}}} \)

\(\Rightarrow \frac{τ _{2}}{\tau_1} = \frac{{8WD_2}}{{\pi {d_2^3}}}\times\frac{{\pi {d_1^3}}}{{8WD_1}} \)

\(\Rightarrow \frac{τ _{2}}{\tau_1} = \frac{{D_2}}{{ {D_1}}}\times\left(\frac{{{d_1}}}{{d_2}} \right)^3\)

\(\Rightarrow \frac{τ _{2}}{\tau_1} = \frac{{2D_1}}{{ {D_1}}}\times\left(\frac{{{d_1}}}{{2d_1}} \right)^3=\frac{1}{4}\)

\(\Rightarrow \tau_{shear2} = \frac{\tau_{shear1}}{4} = \frac{24}{4} = 6 \ MPa;\)

Concept:

For static balancing:

∑fX = ∑ mrw2cosθ and ∑fy = = ∑mrw2sinθ = 0

where, m = mass of element, r = distance of element from the fixed point, ω = Angular velocity of masses.

Calculation:

Given:

m₁ = 10 kg, m₂ = 5 kg, m₃ = 2.5 kg, r₁ = 0.1 m, r₂ = 0.2 m, r₃ = 0.4 m, ω = 1000 rad$\sec$

Now, we know that

∑f_X = ∑ mrω²cosθ

∑fx = [10 × 0.1 × ω² × cos 0°] – [5 × 0.2 × ω² × cos 60°] – [2.5 × 0.4 × w² × cos 60°] = 0

and

∑f_y  = ∑mrω²sinθ

∑fy = [0.1 × 10 × ω2 sin 0°] + [5 × 0.2 × ω² sin 60°] – [2.5 × 0.4 × ω² sin 60°] = 0

Resultant force is given by

\({F_{resultant}} = \sqrt {{{\left( {\sum {F_X}} \right)}^2} + {{\left( {\sum {F_y}} \right)}^2}} = 0\)

Since this is a balanced system so, net force on bearing is zero and so the magnitude of bearing reaction at point B is zero.

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:

Total frictional force acting in a clutch:

\(W=2\pi\int_{r_i}^{r_o}pr\ dr\)

Total torque carrying capacity in a clutch:

\(T=2\pi μ\int_{r_i}^{r_o}pr^2\ dr\)

Assuming uniform pressure theory (p = constant)

\(W=\pi p(r_o^2;-;r_i^2)\)

In case of a new clutch, the intensity of pressure is approximately uniform, but in an old clutch, the uniform wear theory is more approximate.

Calculation:

Given:

W = p(2 πr)dr

Torque dT = μWr

dT = μP.2πr dr.r

\(T = \mathop \smallint \limits_{{r_i}}^{{r_0}} 0.01r.P.2\pi {r^2}dr\)

\(T = 0.01 \times 2\pi \times P\left[ {\frac{{r_0^4 - r_i^4}}{4}} \right]\)

\(18.85 \times {10^3} = \frac{{0.01 \times 2\pi \times P}}{4}\left( {{{40}^4} - {{20}^4}} \right)\)

P = 0.5 N/mm² = 0.5 MPa

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

Moment of Inertia about \({I_1} = \frac{{M{L^2}}}{{12}} + M{y^2}\)

Now in Δ IOA

\(\frac{y}{{\frac{{\sqrt 2 }}{2}}} = \tan 45^\circ \Rightarrow y = \frac{1}{{\sqrt 2 }}m\)

\({I_1} = \frac{M}{{12}}{\left( {\sqrt 2 } \right)^2} + M{\left( {\frac{1}{{\sqrt 2 }}} \right)^2} = 10\left( {\frac{2}{{12}} + \frac{1}{2}} \right) = 6.67kg - {m^2}\)

Now FBD of rod

Moment balance about point I

mgx = I₁ α

\( \Rightarrow \alpha = \frac{{mgx}}{{{I_1}}}\)

\(\sin 45^\circ = \frac{x}{{\frac{y}{2}}} = \frac{x}{{\frac{{\sqrt 2 }}{2}}}\)

\(\frac{{\sqrt 2 }}{2} \cdot \frac{1}{{\sqrt 2 }} = x \Rightarrow x = 0.5m\)

\(\alpha = \frac{{10 \times 10 \times 0.5}}{{6.67}}\)

= 7.5 rad/s²

Alternate solution

\(L = \sqrt 2 m\)

M = 10 kg

g = 10 m/s²

As both the surfaces are smooth

\(\mathop {\ddot \theta }\limits^{} = \alpha = \frac{{3g}}{{2L}}\sin \theta \)

\(\alpha = \frac{{3 \times 10}}{{2 \times \sqrt 2 }}\sin 45^\circ \)

\( = \frac{{30}}{4} = 7.5\frac{{rad}}{{{s^2}}}\)

\({\rho _m} = 1000\frac{{kg}}{{{m^3}}};;{\rho _g} = 1;kg/{m^3}\)

\(V = \sqrt {2gh\left( {\frac{{{\rho _m}}}{{{\rho _a}}} - 1} \right)} \)

\(20 = \sqrt {2*9.81*h\left( {\frac{{1000}}{1} - 1} \right)} \)

h = 0.0204 m = 20.4 mm of water.

Alternate Solution

Dynamic pressure of gas = (ρgh)_water

1/2 ρ_gas⋅V² = ρ_m × 9.81 × h

1/2 × 1 × 20² = 1000 × 9.81 × h

h = 20.38 mm of water

Concept:

Force = Change in linear momentum

Calculation:

ρ = 1000 kg/m³

Force, F = momentum_inlet - momentum_outlet

\(F = {\left( {\dot mv} \right)_{in}} - {\left( {\dot mv} \right)_{out}}\)

\(F = \rho \left[ {\frac{\pi }{4}\left( {D_1^2 - D_2^2} \right)} \right]{v^2} = 1000 \times \frac{\pi }{4} \times \left[ {{{0.06}^2} - {{0.04}^2}} \right] \times {20^2}\)

= 628.32 N

Concept:

From Buckingham’s π – theorem,

\(\frac{F}{{\rho {U^2}{D^2}}} = \phi \left( {\frac{{\rho UD}}{\mu }} \right) = \phi \left( {Re} \right) = {C_F}\)

For dynamic similarity, Re₁ = Re₂

\(\frac{{{\rho _1}{U_1}{D_1}}}{{{\mu _1}}} = \frac{{{\rho _2}{U_2}{D_2}}}{{{\mu _2}}};;{\rho _1} = {{\rm{\rho }}_2};& ;{\mu _1} = {\mu _2}\)

U₁D₁ = U₂D₂

2 × 100 = U₂ × 200

U₂ = 1 m/s

\(\frac{{{F_1}}}{{{\rho _1}U_1^2D_1^2}} = \frac{{{F_2}}}{{{\rho _2}U_2^2D_2^2}} \Rightarrow \frac{{{F_1}}}{{U_1^2D_1^2}} = \frac{{{F_2}}}{{U_2^2D_2^2}} \Rightarrow \frac{{{F_1}}}{{{2^2} \times {{100}^2}}} = \frac{{{F_2}}}{{{1^2} \times {{200}^2}}}\)

F₁ = F₂

\({F_1} = {C_F} ~\rho _1U_1^2D_1^2 = 0.5 \times 1000 \times {2^2} \times {\left( {0.1} \right)^2}\)

F₁ = 20 N ⇒ F₂ = 20 N

Drag force on second sphere of 200 mm diameter is 20 N

Concept:

Specific Steam Consumption or Steam Rate (SSC) is the mass flow rate of steam required to produce 1 kW net work output.

Specific steam consumption  \(SSC = \frac{1}{{{W_{Net}}}}\frac{{kg}}{{kJ}} = \frac{{3600}}{{{W_{Net}}}};kg/kWh\)

Net Work = Turbine Work - Pump Work

Calculation:

W_T = 2803 – 1800 = 1003 kJ/Kg

W_P = 124 – 121 = 3 kJ/Kg

W_net = W_T – W_P = 1003 – 3 = 1000 kJ/Kg

\(SSC = \frac{{3600}}{{{W_{net}}}} = \frac{{3600}}{{1000}} = 3.6\frac{{Kg}}{{kWh}}\)

Concept:

Steady state energy equation

\(\dot m{h_1} + \dot Q = \dot m{h_2} + W\)

Calculation:

Applying the steady state flow equation

\(\dot m{h_1} + \dot Q = \dot m{h_2} + W\)

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

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

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

\(\dot m = 7\frac{{Kg}}{s};\)

Concept:

The change in entropy of the gas is given by

\({S_2} - {S_1} = {C_p}\ln \frac{{{T_2}}}{{{T_1}}} - R\ln \frac{{{P_2}}}{{{P_1}}}\)

Calculation:

m = 1kg; R = 287 J/Kg K

P₁ = 1 bar; T₁ = 300 K; P₂ = 2 bar; T₂ = 300 K

\({S_2} - {S_1} = - R\ln \frac{{{P_2}}}{{{P_1}}}\left( \because{{T_1} = {T_2}} \right)\)

\({S_2} - {S_1} = - 287\ln \frac{2}{1} = - 198.93\frac{J}{{Kg - K}}\)

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

\(PV = mRT \Rightarrow m = \frac{{PV}}{{RT}} = \frac{{150*2.87}}{{0.287*300}}\)

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

The minimum project completion time is 30 days.

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

Maximize, z = 5x₁ + 3x₂

x₁ + 2x₂ ≤ 10

x₁ – x₂ ≤ 8

x₁, x₂ ≥ 0

For simplex

x₁ + 2x₂ + s₁ = 10

x₁ – x₂ + s₂ = 8

and maximize z = 5x₁ + 3x₂ + 0s₁ + 0s₂

where s₁ and s₂ are slack variables

First simplex table

s₂ is leaving variable and x₁ is incoming variable

\({R_1}_{new} = {R_{{1_{old}}}} - {R_{{2_{old}}}}\)

So, from second table value can be read as

s₁ = 2

x₁ = 8

so Max z = 5x₁ + 3x₂ + 0s₁ + 0s₂

Max z = 5 × 8 = 40

Concept:

\(P = 2Tw = 2T × \frac{{2\pi N}}{{60}}\)

T = F × R

where P = Power, F = Rolling Force, R = roll radius, W = angular velocity and N = RPM 

Calculation:

Given:

In this question, the velocity of exit is given. From the velocity of exit, we may assume the velocity of the neutral plane. But the correct answer will be slightly different.

\(Projected;Length,;{L_p} = Rsin;\alpha = \sqrt {R{\rm{\Delta }}h} = \sqrt {150 × \left( {8 - 7.2} \right)} = 10.954mm\)

Projected area, A_p = L_p × b = 10.954 × 120 = 1314.48 mm²

Roll separating force, F = σ₀ × L_p × b = 200 × 1314.48 = 262.9 kN

Arm length = 0.5 L_p for hot rolling = 0.5 × 10.954 mm = 5.477 mm

\(Torque;per;roller;T = F × \frac{a}{{1000}} = 262.9 × 5.477;mm = 1439.9;Nm;\)

Total power for two rollers,

\(P = 2Tw = 2T × \frac{{2\pi N}}{{60}}\)

Now, for finding N, applying the continuity equation

H_ob_oVo = h_fb_fV_f

8 × 120 × V_O = 7.2 × 120 × 30

V_O = 27 m/min

Assuming the velocity of the neutral plane

\(average;velocity,;V = \frac{{{V_o} + {V_f}}}{2} = \frac{{27 + 30}}{2} = 28.5\frac{m}{{min}}\)

V = πDN

28.5 = π × 0.300 × N

⇒ N = 30.24 rpm

\(Total;power;for;two;rollers = 2T × \frac{{2\pi N}}{{60}}\)

\( = 2 × 1439.9 × 2\pi × \frac{{30.24}}{{60}} = 9.12kW\)

Explanation:

In ring gauge, GO ring gauge is used to check upper limit of shaft/pin.

Maximum diameter of pin will be 25.020 mm. On electroplating, thickness increases on both sides so after electroplating maximum diameter

= 25.020 + (2.005) x 2

And since GO ring gauge is used to check upper limit of shaft, so, GO gauge dimension should be 29.030 mm.

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

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

Concept:

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

Calculation:

Gven:

α = 9°

t = 0.2 mm = uncut thickness

\({r_1} = \frac{{0.2}}{{0.25}} = 0.8\)

\(\tan {\phi _1} = \frac{{0.8\cos 9}}{{1 - 0.8\sin 9}} = 0.903\)

\({\phi _1} = {\tan ^{ - 1}}\left( {0.903} \right) = 42.087^\circ \)

\({r_2} = \frac{{0.2}}{{0.4}} = 0.5\)

\(\tan {\phi _2} = \frac{{0.5\cos 9}}{{1 - 0.5\sin 9}} = 0.5357\)

\({\phi _2} = {\tan ^{ - 1}}\left( {0.5357} \right) = 28.18^\circ \)

\(\frac{{{\phi _1}}}{{{\phi _2}}} = \frac{{42.087}}{{28.18}} = 1.49\)