GATE ME 2011 Official Paper (Previous Year Paper)

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

Explanation-

Amalgamate means to unite in or as if in an amalgam, let's have a look at all the given options. 

• Merge: to cause to combine, unite, or coalesce.
• Split: to divide lengthwise usually along a grain or seam or by layers
• Collect: to bring together into one body or place
• Separate:  to set or keep apart: DISCONNECT, SEVER
  

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Amalgamate means to combine or unite to form one organization or structure. So the best option here is split. Separate on the other hand, although a close synonym, it is too general to be the best antonym in the given question while Merge is the synonym; Collect is not related.

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

Explanation-

Inexplicable means incapable of being explained, interpreted or accounted for. 

• Incomprehensible: impossible to comprehend: UNINTELLIGIBLE.
• Indelible: that cannot be removed, washed away, or erased.
• Inextricable: forming a maze or tangle from which it is impossible to get free
• Infallible: incapable of error: UNERRING

Inexplicable means not explicable; that cannot be explained, understood, or accounted for. So the best synonym here is incomprehensible.

Explanation:

Let Log (P) = (1/2) Log (Q) = (1/3) Log (R) = l

∴ P = 10^k, Q = 10^2k, R = 10^3k

Now,

Option 1:

P²=Q³R²

(10^k)² = (10^2k)³ (10^3k)² 

10^2k ≠ 10^12k

Option 2:

Q² = PR

10^4k = (10^3k)(10^k)

10^4k = 10^4k

∴Q² = PR follows the given relation.

Option 3:

Q² = R³P ⇒ 10^2k ≠ 10^7k

Option 4:

R=P²Q² ⇒ 10^3k ≠ 10^6k

∴ Only option(2) is correct.

The correct answer is Option 3 i.e visiting 

Explanation:

Reading the given sentence we find that.

• The word contemplated is a transitive verb.
• A transitive verb always needs to transfer it's action on to something which is an object.
• Hence, here after the transitive verb contemplated we need a noun that will be the object.
• Out of the given options visiting is a gerund. And therefore it grammatically follows the verb contemplated.

Hence the correct answer is Option 3 i.e visiting.

Important Points

• A gerund is an -ing form of verb which acts as a noun.
• In the above case, we need a noun after the verb contemplated.
• And the gerund visiting is the appropriate option as it is a verb acting as a noun.
• Therefore the action of the verb contemplated is transferred to the gerund visiting.

Thus the correct sentence will be: "I contemplated visiting Singapore for my vacation bu decided against it."

The correct answer is Option 2 i.e Restrained

Explanation:

Reading the above statement we find that:

• The tone of the sentence clearly indicates a word similar to understated and tentative.
• The word should also be an antonym of strong.

Let's look at the meaning of the marked option:

• Restrained: Not excessively showy or ornate; understated and timid.

Hence from the above meaning, we find that restrained is similar in meaning to understated and tentative, and fits perfectly in the give sentence.

Thus, the correct answer is option 2 Restrained.

Additional Information 

Let's look at the meaning of the other given options. 

• Hyperbolic: Delibrately; or exaggerated.
• Argumentative: Given to arguing.
• Indifferent: Having no particular interest; unconcerned.

Concept:

${\bf{Volume}};{\bf{of}};{\bf{liquid}};{\bf{left}} = {\bf{x}}{\left( {\frac{{{\bf{x}} - 1}}{{\bf{x}}}} \right)^{\bf{n}}}$

where x = Original amount of liquid, n = Number of times replaced.

Calculation:

Given:

x = 10 litre, n = 3

Now, we know that

${\bf{Volume}};{\bf{of}};{\bf{spirit}};{\bf{left}} = {\bf{x}}{\left( {\frac{{{\bf{x}} - 1}}{{\bf{x}}}} \right)^{\bf{n}}}$

∴  Volume of spirit left = $10{\left( {\frac{{10 - 1}}{{10}}} \right)^3} = \frac{{729}}{{100}} = 7.29$  litres

The correct answer is Option 3 i.e what the leading causes of death are

Explanation:

Reading the above statement we find that.

• The given sentence starts by stating how to deal with bereavement and grief after a tragedy occurs and not about precautions.
• Therefore, irrespective of the causes of death, a school student rarely gets into details of causes
• Hence it is very clear that the leading causes of death are not included in a unit on bereavement.

Hence, the correct answer is option 3 i.e what the leading causes of death are.

Concept:

According to the given information:

Most dangerous microbe ∝ probability that microbe will overcome human immune system.

Most dangerous microbe ∝ area (growth of microbe)

Most dangerous microbe ∝ (quantity required)^-1

$So, the;most;dangerous;microbe;\propto;\frac{Probability;\times;Area}{Quantity;required}$

$So, the;most;dangerous;microbe;=;\frac{K;\times;Probability;\times;Area}{Quantity;required}$

$So, the;most;dangerous;microbe;=;\frac{K;\times;Probability;\times;\pi{d^2}}{4;\times;Quantity;required}$

Calculation:

Given:

For microbe P:

Probability (P) = 0.4, Diameter = 50 mm and Quantity required = 800 milligram of microbe/mass of body in kg.

$The;most;dangerous;microbe;=\frac{\pi{K}}{4}\times\frac{0.4\times50^2}{800}\Rightarrow\frac{\pi{K}}{4}\times\frac{5}{4}=1.25\frac{\pi{K}}{4}$

For microbe Q:

Probability (Q) = 0.5, Diameter = 40 mm and Quantity required = 600 milligram of microbe/mass of body in kg.

$The;most;dangerous;microbe;=\frac{\pi{K}}{4}\times\frac{0.5\times40^2}{600}\Rightarrow\frac{\pi{K}}{4}\times\frac{4}{3}=1.33\frac{\pi{K}}{4}$

For microbe R:

Probability (R) = 0.4, Diameter = 30 mm and Quantity required = 300 milligram of microbe/mass of body in kg.

$The;most;dangerous;microbe;=\frac{\pi{K}}{4}\times\frac{0.4\times30^2}{300}\Rightarrow\frac{\pi{K}}{4}\times\frac{6}{5}=1.2\frac{\pi{K}}{4}$

For microbe S:

Probability (S) = 0.8, Diameter = 20 mm and Quantity required = 200 milligram of microbe/mass of body in kg.

$The;most;dangerous;microbe;=\frac{\pi{K}}{4}\times\frac{0.8\times20^2}{200}\Rightarrow\frac{\pi{K}}{4}\times\frac{8}{5}=1.6\frac{\pi{K}}{4}$

Microbe S > Microbe Q > Microbe P > Microbe R.

∴ toxicity is more for microbe S and the company should target it in its first attempt.

Concept:

Total cost = Fixed cost + Variable cost

To find maxima and minima of a function y = f(x), follow these steps.

Step 1

$Find;\frac{{dy}}{{dx}},;and;put\frac{{dy}}{{dx}} = 0.$

Find the value of x and this value is said to be the stationary point, this is a necessary condition to find the extremum value of a function.

Step 2

$Find;\frac{{{d^2}y}}{{d{x^2}}};$

Check the value at the stationary point obtained in Step 1.

A function f(x) has a maxima at x = a if f’(a) = 0 and f”(a) < 0

A function f(x) has a minima at x = a if f’(a) = 0 and f”(a) > 0

A function f(x) has no maxima and minima at x = a if f’(a) = 0 and f”(a) = 0.

Calculation:

Given:

F = 100/q, V = 4q

Total cost (TC) = Fixed cost + Variable cost

$∴;TC=4q;+;\frac{100}{q}$

Step 1:

$\frac{{d(TC)}}{{dq}} = 0$

$\frac{d(TC)}{dq}=4;-;\frac{100}{q^2}$

∴ q = ± 5

$\frac{{d^2(TC)}}{{dq^2}} =+ve;for;minima$

$\frac{{d^2(TC)}}{{dq^2}} =-ve;for;maxima$

$\frac{d^2(TC)}{dq^2}=\frac{200}{q^3}$

At q = 5

$\frac{d^2(TC)}{dq^2}=\frac{200}{5^3}\Rightarrow1.6;;(+ve;,;∴;minima)$

At q = -5

$\frac{d^2(TC)}{dq^2}=\frac{200}{5^3}\Rightarrow-1.6;;(-ve;,;∴;maxima)$

∴ at q = 5, Total cost (TC) will be minimum.

Concept:

The transporter receives the same number of order each day and he currently has pending orders.

Let 'x' be the number of daily orders and 'y' be the number of pending orders.

Calculation:

Given:

Condition I:

Use of 7 trucks each day finishes all orders in 4 days.

∴ total numbers of trucks used = 7 + 7 + 7 + 7 ⇒ 28.

Total orders received in 4 days = 4x.

Pending orders = y.

∴ 4x + y = 28      eq(1).

Condition II:

Use of 3 trucks each day finishes all orders in 10 days.

∴ total numbers of trucks used = 3 × 10 ⇒ 30.

Total orders received in 10 days = 10x.

Pending orders = y.

∴ 10x + y = 30      eq(2).

Solving eq (1) and eq (2).

x = 0.33 ⇒ Daily orders

y = 26.66 ⇒ Pending orders.

Condition III:

Use of 'n' trucks each day finishes all orders in 5 days.

∴ total numbers of trucks used = 5 × n ⇒ 5n.

Total orders received in 5 days = 5x.

Pending orders = y.

∴ 5x + y = 5n   

∴ (5 × 0.33)  + (26.66) = 5n

∴ n = 5.66 ≈ 6

∴ minimum 6 number of trucks required so that there will be no pending order at the end of the 5th day.

Concept:

Streamline: It is an imaginary curve drawn in space such that tangent drawn to it at any point will give the velocity of that fluid particle at a given instant of time. A line along which stream function (ψ) is constant is known as streamline.

Equipotential line: A line along which velocity potential function (ϕ) is constant is known as the equipotential line.

${\left( {\frac{{dy}}{{dx}}} \right)_ϕ } \times {\left( {\frac{{dy}}{{dx}}} \right)_ψ } = - 1$

Slope of equipotential Line × Slope of stream function = -1

They are orthogonal to each line other.

 

For a streamline, $ψ(x,y)=constant$ and the differential of ψ  is zero.

$dψ=\frac{\partialψ}{\partial x}dx+\frac{\partialψ}{\partial y}dy$

$dψ=-vdx+udy$

$(\frac{{\partial y}}{{\partial x}} )_{ψ=const}= \frac{v}{u}$

For an equipotential line, $ϕ(x,y)=constant$ and the differential of ϕ is zero.

$dϕ=\frac{\partialϕ}{\partial x}dx+\frac{\partialϕ}{\partial y}dy$

$dϕ=udx+vdy$

$(\frac{{\partial y}}{{\partial x}} )_{ϕ=const}= -\frac{u}{v}$

\((\frac{{\partial y}}{{\partial x}} ){ψ=const}=- \frac{1}{(\frac{{\partial y}}{{\partial x}} ){ϕ=const}}\)

Concept:

Relative humidity:

• It is defined as the ratio of the mass of water vapour (mv) in a certain volume of moist air to the mass of water vapour (mvs) in the same volume of saturated air (air having the maximum amount of water vapour without condensing) at the same temperature.
• It is denoted by ϕ.

\({\rm{RH}}\left( \phi \right) = \frac{{{{\rm{m}}{\rm{v}}}}}{{{{\rm{m}}{{\rm{vs}}}}}}\)

• If a mass of moist air in an airtight vessel heated to a higher temperature, then the relative humidity of the air decreases.

 

• Heating and Humidifying: A Simultaneous increase in both the dry bulb temperature and humidity ratio of the air.
• Cooling and Dehumidifying: It involves the removal of water from the air as the air temperature falls below the dewpoint temperature.
• Sensible heating: During this process, the moisture content of air remains constant and its temperature increases as it flows over a heating coil.
• Sensible cooling: During this process, the moisture content of air remains constant, but its temperature decreases as it flows over a cooling coil.
• In sensible heating/cooling, specific humidity is constant while relative humidity varies. In heating/cooling with chemical humidification/ dehumidification, both specific humidity and relative humidity varies.

Concept:

Flow configuration in the condenser as shown below.

ΔT₁ = T_h,i - T_c,i,  ΔT₂ = T_h,o - T_c,o

$LMTD=\frac{{\Delta}T_1~-~{\Delta}T_2}{ln{\frac{\Delta{T_1}}{\Delta{T_2}}{}}}$

 

Calculation:

Given:

T_h,i = T_h,o = 60oC, T_c,i = 30oC, T_c,o = 45oC

 Therefore, ${\rm{Δ }}{T_1} = 30^\circ C,{\rm{Δ }}{T_2} = 15^\circ C,$

Now, we know that $LMTD = \frac{{{\rm{Δ }}{T_1} - {\rm{Δ }}{T_2}}}{{In\left( {\frac{{{\rm{Δ }}{T_1}}}{{{\rm{Δ }}{T_2}}}} \right)}} = \frac{{30 - 15}}{{In\left( {\frac{{30}}{{15}}} \right)}} = 21.6^\circ C$

Concept:

Equilibrium Conditions:

\(\sum {{\rm{F}}{\rm{v}}} = 0{\rm{;and;}}\sum {{\rm{M}}{\rm{P}}} = 0\)

Calculation:

Given:

Now, we know that

Balancing vertical forces by 

$\sum {{\rm{F}}_{\rm{v}}} = 0$

R_P + R_Q = 0       ............ (1)

Now,

Taking moment about point 'P'

$\sum {{\rm{M}}_{\rm{P}}} = 0$

1 = R_Q × 1 ⇒ R_Q = 1 kN (upward)

Then, from equation (1), we get

R_P = 0 - 1 

∴ R_p = -1 kN = 1 kN (downward)

Concept:

Mobility of the mechanism = Degree of Freedom = [3(l - 1) - 2j - h] - F_r

where, l = no. of links, j = no. of the binary joint, h = no. of higher pairs, F_r = redundant link

Here the middle link is acting as a dummy link.

Calculation:

Given:

l = 4, j = 4, h = 0, F_r = 0

DOF = [3(4 - 1) - 2 × 4 - 0] - 0 = 9 - 8 = 1

So degree of freedom of double – parallelogram mechanism is one.

Concept:

Maximum possible draft is given by:

Δh_max = μ²R 

where μ = coefficient of friction and R = radius of roller.

The maximum possible draft can be increased by:

• Increasing the coefficient of friction.
• increasing the radius of roller.

Concept:

Powder Metallurgy (PM): 

• It is the art and science of producing metal powders and making semifinished and finished objects from an individual, mixed, or alloyed powders with or without the addition of nonmetallic constituents.

Impregnation 

• It is the secondary operation that is used to improve the self-lubricating capacity of the sintered part in powder metallurgy.
• In this operation, the inherent pores of the powder metallurgy parts are impregnated with a fluid like oil or grease.
• Such components have a continuous supply of lubricants by capillary action during their service life.

Infiltration,

• The pores of the sintered part with some low melting point metal that results in an improvement in hardness and tensile strength.
• In repressing, the pressure is applied to achieve parts with higher dimensional accuracy.
• Coining is high pressure compacting operation that results in high strength and dimensional accuracy.

Concept:

Fit is a relationship that exists between two mating parts, a hole, and a shaft, with respect to their dimensional difference before assembly.

• Clearance fit: Clearance is the difference between the size of the hole and the size of the shaft which is always positive. Here the tolerance zone of the hole will be above the tolerance zone of the shaft.

Examples: Slide fit, easy sliding fit, running fit, slack running fit, and loose running fit.

The above-given figure represents the clearance fit because the tolerance zone is not meeting.

• Interference fit: Interference is the difference between the size of the hole and the size of the shaft which is always negative i.e. shaft is always larger than the hole size. Here the tolerance zone of the hole will be below the tolerance zone of the shaft.

Examples: Shrink fit, heavy drive fit, and light drive fit.

• Transition fit: It may sometimes provide clearance and sometimes interference. Here the tolerance zones of the hole and shaft will overlap each other.

Examples: Tight fit and push-fit, wringing fit, press fit.

Calculation:

Given:

Hole dimension: \(\phi 9\begin{array}{*{20}{c}} { + 0.015}\ { + 0} \end{array}mm\)

Shaft dimension: \(\phi {9^{\begin{array}{*{20}{c}} { + 0.010}\ { + 0.001} \end{array}}}mm\)

The tolerance zone is overlapping, so it is a transition fit.

Explanation:

Heat and work, both are path functions; Their magnitude depends on the path followed during a process as well as the end state.

• Work and heat are modes of energy transfer.
• Heat is energy transferred due to temperature differences only.
• Work is the energy transfer associated with a force acting through a distance.

Relationship between heat and work:

• Both are recognized at the boundaries of a system as they cross the boundaries; That is, both heat and work are boundary phenomena
• Systems possess energy, but not heat or work
• Both are associated with a process, not a state; Unlike properties, heat or work has no meaning at a state

Properties are classified into two types.

• Intensive property: The properties of the system which are independent of the mass of a system are called Intensive property.
• E.g. pressure, density, temperature, all specific properties, etc.
• Extensive property: The properties of the system which are dependent on the mass of a system are called Extensive property.
• E.g. energy, volume, entropy, etc.

Concept:

Slenderness ratio (λ) is defined as:

$\lambda = \frac{{{L_e}}}{{{k}}}$

The radius of gyration (k) is given by:

$k = \sqrt {\frac{{{I_{min}}}}{A}}$

where Le = Effective length of the column, k= radius of gyration, I_min = Least moment of Inertia, A = Area of the cross-section

Calculation:

Given:

b = 10 mm, d = 20 mm, L = 1 m = 1000 mm

The radius of gyration (k) is:

$k = \sqrt {\frac{{{I_{min}}}}{A}}=\sqrt{\frac{\frac{db^3}{12}}{bd}}=\sqrt{\frac{b^2}{12}}=\frac{b}{\sqrt{12}}$

Slenderness ratio (λ) is:

$λ= \frac{L_e}{k}= \frac{{\sqrt {12} L_e}}{b} = \frac{{\sqrt {12} \times 1000}}{{10}} = 346.41$

Concept:

Taylor expansion series,

$f\left( x \right) = f\left( a \right) + f'\left( a \right).\left( {x - a} \right) + \frac{{f''\left( a \right)}}{{2!}}{\left( {x - a} \right)^2} + \ldots$

At a = 0

f (x) = f(0) + x f'(0) + x²$\frac{f"(0)}{2!}$ + x³$\frac{f'''(0)}{3!}$ + ......

Replace x by θ

f (θ) = f(0) + θ f'(0) + θ2$\frac{f"(0)}{2!}$ + θ3$\frac{f'''(0)}{3!}$ + ......

f (θ) = sin θ

f(0) = 0, f'(0) = cos 0 = 1, f''(0) = - sin 0 = 0, f'''(0) = - cos 0 = -1

∴ $\sinθ = θ -~ \frac{{{θ ^3}}}{{3!}}+\frac{{{θ ^5}}}{{5!}} -....$

Important Points

$\cos x = 1 - \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}} - \ldots$

Application:

 $\sin \left( {{x^3}} \right) = {x^3} - \frac{{{x^9}}}{{3!}} + \frac{{{x^{15}}}}{{5!}} - \ldots$

 $\sin \left( {{x^2}} \right) = {x^2} - \frac{{{x^6}}}{{3!}} + \frac{{{x^{10}}}}{{5!}} - \ldots$

 $\cos \left( {{x^3}} \right) = 1 - \frac{{{x^6}}}{{2!}} + \frac{{{x^{12}}}}{{4!}} - \ldots$

 $\cos \left( {{x^2}} \right) = 1 - \frac{{{x^4}}}{{2!}} + \frac{{{x^8}}}{{4!}} - \ldots$

Explanation:

Green sand mould indicates that mould contains moisture.

Sand mould

A mould is an assembly of two or more flasks (metallic) or bonded refractory particles with some cavitites.

Types of sand mould

According to the material used in their construction, the moulds are of following types:

Concept:

$y=\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }}{\theta }$

It gives $\frac{0}{0}$ form

Therefore, by applying L’ Hospital's rule....(differentiate both numerator and denominator)

We will get

$y=\mathop {\lim }\limits_{\theta \to 0} \frac{{\cos \theta }}{1}=\frac{cos~0}{1}=\frac{1}{1}=1$

Explanation:

Eigen values of a real symmetric matrix are always real.

Eigen values and Eigen vector of a square matrix

"λ" is called eigen value and "x" is called eigen vector of a square matrix "A", if

Ax = λx

Important Points

Characteristics of eigen values:

• Tr (A) = Summation of eigen values
• |A| = Product of eigen values
• If A = Upper triangular matrix or lower triangular matrix or diagonal matrix, then its eigen values will be diagonal elements.
• Eigen values of the hermitian matrix and real symmetric matrix are always real.
• Eigen values of skew symmetric and skew hermitian matrix are either zero or purely imaginary.
• Eigen values of the orthogonal matrix and unitary matrix have unit modulus.

Concept:

• Insulation addition on a wire is to increase the heat transfer rate.
• The insulation is added only up to a critical radius.
• Beyond the critical radius, if we add insulation, the heat transfer rate going to decrease.

 

The formula for critical radius for the cylindrical body is, $r_c=\frac{k}{h}$

where, k = thermal conductivity of insulation, h = heat transfer coefficient

Calculation:

Given:

k = 0.05 W/mK. h = 5 W/m²K, r = 12.5 mm

${r_{cr}} = \frac{k}{h} = \frac{0.05}{5}=0.01~m=10\ mm$

r_cr < r

The critical radius is less than the outer radius of the pipe and adding the insulation will reduce the heat loss.

Concept:

Increase in internal energy (ΔU) = rate of heat dissipated by a resistor  

ΔU = I²R

where I is current and R is resistance

According to First law of thermodynamics for any process,

δQ = ΔU + δW

Calculation:

Given:

I = 10 A, R = 23 Ω 

Heat transfer,

Q = 0 (well insulated wall),

ΔU  = I2R = 10² × 23

ΔU = 2.3 kW

According to First law of thermodynamics for any process,

δQ = du + δW

0 = 2.3 + δW

δW = - 2.3 kW

Explanation:

Maximum shear stress theory

(Guest & Tresca’s Theory)

According to this theory, failure of the specimen subjected to any combination of a load when the maximum shearing stress at any point reaches the failure value equal to that developed at the yielding in an axial tensile or compressive test of the same material.

Graphical Representation

\({{\rm{\tau }}{{\rm{max}}}} \le \frac{{{{\rm{\sigma }}{\rm{y}}}}}{2}\) For no failure

\({{\rm{\sigma }}_1} - {{\rm{\sigma }}2} \le \left( {\frac{{{{\rm{\sigma }}{\rm{y}}}}}{{{\rm{FOS}}}}} \right)\) For design

σ1 and σ2 are maximum and minimum principal stress respectively.

Here, τmax = Maximum shear stress

σy = permissible stress

This theory is well justified for ductile materials.

Maximum shear strain energy / Distortion energy theory / Mises – Henky theory.

It states that inelastic action at any point in body, under any combination of stress begging, when the strain energy of distortion per unit volume absorbed at the point is equal to the strain energy of distortion absorbed per unit volume at any point in a bar stressed to the elastic limit under the state of uniaxial stress as occurs in a simple tension/compression test.

 for no failure

 For design

It cannot be applied for material under hydrostatic pressure.

All theories will give the same results if loading is uniaxial.

Maximum principal stress theory (Rankine’s theory)

According to this theory, the permanent set takes place under a state of complex stress, when the value of maximum principal stress is equal to that of yield point stress as found in a simple tensile test.

For the design criterion, the maximum principal stress (σ1) must not exceed the working stress ‘σy’ for the material.

\({{\rm{\sigma }}{1,2}} \le {{\rm{\sigma }}{\rm{y}}}\) for no failure

${{\rm{\sigma }}_{1,2}} \le \frac{{\rm{\sigma }}}{{{\rm{FOS}}}}$ for design

Note: For no shear failure τ ≤ 0.57 σy

Graphical representation

For brittle material, which does not fail by yielding but fail by brittle fracture, this theory gives a satisfactory result.

The graph is always square even for different values of σ1 and σ2.

Important Points

Maximum principal strain theory (ST. Venant’s theory)

According to this theory, a ductile material begins to yield when the maximum principal strain reaches the strain at which yielding occurs in simple tension.

${\epsilon_{1,2}} \le \frac{{{{\rm{\sigma }}_{\rm{y}}}}}{{{{\rm{E}}_1}}}$ For no failure in uniaxial loading.

\(\frac{{{{\rm{\sigma }}_1}}}{{\rm{E}}} - {\rm{\mu }}\frac{{{{\rm{\sigma }}_2}}}{{\rm{E}}} - {\rm{\mu }}\frac{{{{\rm{\sigma }}3}}}{{\rm{E}}} \le \frac{{{{\rm{\sigma }}{\rm{y}}}}}{{\rm{E}}}\) For no failure in triaxial loading.

\({{\rm{\sigma }}_1} - {\rm{\mu }}{{\rm{\sigma }}_2} - {\rm{\mu }}{{\rm{\sigma }}3} \le \left( {\frac{{{{\rm{\sigma }}{\rm{y}}}}}{{{\rm{FOS}}}}} \right)\) For design, Here, ϵ = Principal strain

σ1, σ2, and σ3 = Principal stresses   

Graphical Representation

This theory overestimates the elastic strength of ductile material.

Maximum strain energy theory (Haigh’s theory)

According to this theory, a body complex stress fails when the total strain energy at the elastic limit in simple tension.

Graphical Representation.

  for no failure

 for design

This theory does not apply to brittle material for which elastic limit stress in tension and in compression are quite different.

Concept:

To multiply two complex numbers, we can use distributive law,

$\left( {a + bi} \right)\left( {c + di} \right) = \left( {ac + bd{i^2}} \right) + \left( {ad + bc} \right)i$

For avoiding binomials, we can apply i2 = -1

Calculation:

Given:

Two complex numbers 1 + i and 2 – 5i

Product of two numbers is:

(1 + i) × (2 – 5i)

2 – 5i + 2i + 5 = 7 – 3i

Concept:

Waiting time in the queue $W_q=\frac{L_q}{λ}$ and $L_q=\frac{ρ^2}{1-ρ}$

$\rho=\frac{\lambda}{μ}$

where, L_q = length of queue, λ = arrival rate, μ = service rate

Calculation:

Given: 

λ = 5 cars per hour, μ = 1 car per 10 minute = 6 cars per hour

⇒ $\rho =\frac{5}{6}$

 $L_q= \frac{(5/6)^2}{1-(5/6)}=\frac{25}{6}$

 $W_q=\frac{25/6}{5}=\frac{5}{6}$ hours = $\frac{5}{6}\times60=50~min$

Explanation:

Kanban

• Kanban is a concept that relates to obtaining materials or required items "just in time" for their introduction into the assembly or process.
• Kanban is a system to signal a need for action.
• This can be done by cards on a board (which is the traditional way) or by other devices that are used as markers, indicating the need to take action.

Economic order quantity

Economic order quantity is that size of the order which helps in minimizing the total annual cost of inventory in the organization.

When the size of the order increases, the ordering costs (cost of purchasing, inspection, etc.) will decrease whereas the inventory carrying costs (costs of storage, insurance, etc.) will increase. Economic Order Quantity (EOQ) is that size of order which minimizes total annual costs of carrying and cost of ordering.

It is evident from above that the minimum total costs occur at a point where the ordering costs and inventory carrying costs are equal.

Capacity Planning - Capacity planning is determining the production capacity needed by an organisation to meet changing demand for its products, the process of determining the production capacity needed by an organisation to meet changing demand for its products, The ability to receive, holds or absorbs the maximum or optimum amount that can be produced.

Product design - Product design is the conceptualization of an idea about a product and transformation of the idea into a reality. To transform the idea into reality a specification about the product is prepared. this specification is prepared by considering different constraints such as production process, customer expectation etc. In the product design stage, every aspect of the product is analyzed. Also, the final decision regarding the product is taken on the basis of the analysis.

Concept:

If f(x) = f(-x) , For all x ϵ Df  Then the function is even function

where Df = Domain of function

 

If f(x) = -f(-x) , For all x ϵ Df  Then the function is odd function.

\(\mathop \smallint \limits_{ - a}^a f\left( x \right)dx = \mathop \smallint \limits_{ - a}^0 f\left( x \right)dx + \mathop \smallint \limits_0^a f\left( x \right)dx\)

\(\mathop \smallint \limits_{ - a}^a f\left( x \right)dx = 2\mathop \smallint \limits_0^a f\left( x \right)dx\)

Therefore,

\(\mathop \smallint \limits_{ - a}^a f\left( x \right)dx = \begin{array}{{20}{c}} {\left{ {2\mathop \smallint \limits_0^a f\left( x \right)dx;} \right.}\ {0;} \end{array}\begin{array}{{20}{c}} {f\left( x \right)is\ even}\ {f\left( x \right)is\ odd} \end{array}\)

f (x) is even function.

Explanation:

Coefficient of restitution (e):

Coefficient of restitution or coefficient of the resilience of a collision is defined as the ratio of the relative velocity of separation after the collision to the relative velocity of the approach before the collision.

Coefficient of restitution (e) =$\frac{Relative;velocity;after;collosion}{Relative;velocity;before;collosion}=\frac{v_B-v_A}{u_A-u_B}$

A perfectly inelastic collision also called perfectly plastic collision is a limiting case of inelastic collision in which the two bodies stick together after impact.

Since both bodies stick together in a perfectly plastic collision. Therefore vA = vB, Thus e = 0

Properties of different types of collision are given in the table below:

Concept:

The hoop stress in a cylinder is the tensile stress exerted tangentially along the circumference of the cylinder and it is given as,

$\sigma_h=\frac{pd}{2t}$

where p = pressure, d = internal diameter, t = thickness of the cylinder.

 

NOTE: Hoop stress brought changes in diameter.

Calculation:

Given:

p = 5 MPa, r_i = 500 mm ⇒ d = 1000 mm, and t = 10 mm

${\sigma _h} = \frac{{pd}}{{2t}} =\frac{{5~\times;1000}}{{2;\times;10}}= 250\ MPa$


Longitudinal stress is the stress which acts along the length and it is also tensile in nature.

Longitudinal stress = $\sigma_l=\frac{\sigma_h}{2}= \frac{pd}{4t}$

NOTE: Longitudinal stress brought changes in length.

Introduction of TIG Welding

Gas Tungsten Arc Welding (GTAW), also known as tungsten inert gas (TIG) welding is a process that produces an electric arc maintained between a non-consumable tungsten electrode and the part to be welded.

Inert Gas in TIG Welding

The heat-affected zone, the molten metal, and the tungsten electrode are all shielded from atmospheric contamination by a blanket of inert gas fed through the GTAW torch.

Inert gas is inactive or deficient in active chemical properties. The shielding gas serves to blanket the weld and exclude the active properties in the surrounding air. Inert gases, such as Argon and Helium, do not chemically react or combine with other gases.

Submerged arc welding

Submerged arc welding is an arc welding process in which heat is generated by an arc which is produced between bare consumable electrode wire and the work-piece. The arc and the weld zone are completely covered under a blanket of granular, fusible flux which melts and provides protection to the weld pool from the atmospheric gases.

The molten flux surrounds the arc thus protecting arc from the atmospheric gases. The molten flux flows down continuously and fresh flux melts around the arc. The molten flux reacts with the molten metal forming slag and improves its properties and later floats on the molten/solidifying metal to protect it from atmospheric gas contamination and retards cooling rate. A process of submerged arc welding is illustrated in Figure.

Flux coated arc welding or shielded metal arc welding

Shielded metal arc welding (SMAW) is a fusion welding process that uses a consumable, flux-coated electrode to create an arc between the electrode and the work piece.

This flux coating is used to protect the molten weld metal from the atmosphere and oxidation. This is necessary because no external shielding gas is used for this welding process.

Molten metal travels from the electrode via the electrical arc and is deposited into the work piece.

The flux coating is also melted, and it surfaces on top of the molten weld pool in the form of slag.

Concept:

Austenite is an interstitial solution of Carbon in γ iron (Carbon atoms are accommodated at interstitial spaces). The maximum solubility of carbon in γ iron is 2.11%. In pure form, Austenite has an FCC structure. It has larger interatomic spacings (1.02 A° ) than ferrite. Austenite is soft, tough, and highly ductile. It has high tensile strength also. Austenite is non-magnetic at any temperature.

The liquid molten metal of iron on cooling will become solid at 1539°C (δ-iron) having BCC, on further cooling at 1410°C δ-iron is converted to γ – iron called as austenite having FCC and on further cooling, it counts to α – iron called ferrite at 910°C.

Concept:

For shafts in series, both the shafts carry the same torque T and the total angle of twist at the resisting end is the sum of separate angles of twist of two shafts

• T₁ = T₂
• θ = θ₁ + θ_2​

The angle of twist θ is given by,

$\theta = \frac{{TL}}{{GJ}}$

where, T = Torque applied, L = length of the shaft, G = Modulus of rigidity  J = Polar section modulus

For the solid circular shaft, $J = \frac{\pi }{{32}}{d^4}$

Calculation:

 

Given:

T₁ = T₂ = T, L₁ = L, L₂ = L/2, d₁ = 2d, d₂ = d

θ = θ₁ + θ₂

$\theta = \frac{{{T_1}{L_1}}}{{G{J_1}}} + \frac{{{T_2}{L_2}}}{{G{J_2}}} = \frac{{TL}}{{G \times \frac{\pi }{{32}} \times {{\left( {2d} \right)}^4}}} + \frac{{T \times \frac{L}{2}}}{{G \times \frac{\pi }{{32}} \times {d^4}}} = \frac{{18TL}}{{G\pi {d^4}}}$

$d = {\left( {\frac{{18TL}}{{G\pi \theta }}} \right)^{\frac{1}{4}}}$

Shafts in parallel:

• If two or more shafts are rigidly fixed together such that the applied torque is shared between them then the composite shaft so formed is said to be connected in parallel
• The angle of twist for both the shafts is the same
• θ₁ = θ₂ and T = T₁ + T₂

Concept:

The velocity of flow is given by:

$V=\sqrt{2gh}$ and $h=x(\frac{ρ_m}{ρ}-1)$

where, h = Difference of pressure head,  x = Difference of the manometric fluid level in differential U tube manometer, ρ_m = density of the manometric fluid, ρ = density of fluid flow through the pipe.

Calculation:

Given:

ρ_m = 1000 kg/m³, ρ = 1.2 kg/m³, x = 10 mm = 0.01 m

we know that,

$h=x(\frac{ρ_m}{ρ}-1)$

 $h=0.01(\frac{1000}{1.2}-1)=8.323 ~m$

Velocity of flow is:

$V=\sqrt{2gh}$

 $V=\sqrt{2\times 9.8~\times8.323}=12.77~ m/s$

Concept:

Work done by the turbine = Enthalpy at the inlet - Enthalpy at the outlet of turbine,

W_T = h₃ - h₄

Work input to the pump,

W_p = Enthalpy at outlet – Enthalpy at the inlet of the pump, 

W_p = h₂ – h₁

Net work output, W_net = W_T - W_P

Specific Steam consumption = $\frac{{3600}}{{{W_{net}}}}$ kg/kW-hr

Calculation:

Given:

Enthalpy at the inlet of the turbine, h₃ = 2800 kJ/kg;

Enthalpy at the outlet of turbine, h₄ = 1800 kJ/kg, 

Pump work is negligible, W_p = 0 kJ/kg.

W_net = WT - WP = (h₃ - h₄) - 0 = 2800 - 1800 = 1000 kJ/kg.

Specific steam consumption = $\frac{{3600}}{{{W_{net}}}}$ $= \frac{1}{{1000}} \times 3600 = 3.6$ kg/kW-hr.

The specific steam consumption is 3.6 kg/kW-hour.

Concept:

Simpson’s 1/3 rule is given as:

 \(\mathop \smallint \limits_{{x_0}}^{{x_n}} ydx = \frac{h}{3}\left{ {\left( {{y_0} + {y_n}} \right) + 4\left( {{y_1} + {y_3} + {y_5} + \ldots } \right) + 2\left( {{y_2} + {y_4} + {y_6} + \ldots } \right)} \right}\)

where x₀ = a, x_n = b and $h = \frac{{b - a}}{n}$

Calculation:

Given:

x₀ = a ⇒ 1, x_n = b ⇒ 3, n = 2 and $h = \frac{{b - a}}{n} ⇒\frac{3-1}{2}=1$

<br>

By Simpson’s rule:

\(\mathop \smallint \limits_{{x_0}}^{{x_n}} y~dx = \frac{h}{3}\left{ {\left( {{y_0} + {y_2}} \right) + 4\left( {{y_1}} \right)} \right}\)

\(\Rightarrow \mathop \smallint \limits_{{x_0}}^{{x_n}} y~dx = \frac{1}{3}\left{ {\left( {{1} + {0.33}} \right) + 4\left( {{0.5}} \right)} \right}=1.111\)

Concept:

Load-Life relationship of a Bearing:

${L_{10}} = {\left( {\frac{C}{P}} \right)^n}$

L10 = Rated bearing life (in million revolutions), C = Dynamic load-carrying capacity, P = Load acting on the bearing

n = 3 for ball bearing and n = 10/3 for roller bearing.

Calculation:

Given:

P_P = 30 kN, P_Q = 45 kN, n = 3 (∵ ball bearing) and C_P = C_Q (∵ identical bearing).

${L_{10}} = {\left( {\frac{C}{P}} \right)^n}$

$\therefore \frac{{{L_P}}}{{{L_Q}}} = {\left( {\frac{{{P_Q}}}{{{P_P}}}} \right)^3} \Rightarrow {\left( {\frac{{45}}{{30}}} \right)^3} = {\left( {\frac{3}{2}} \right)^3} \Rightarrow \frac{{27}}{8}$

Concept:

The given linkage is a 4-bar mechanism with 4-revolute joints.

When the input and output link is parallel to each other and perpendicular to AD.

V_AB = V_CD

∴ ω_AB × AB = ω_CD × CD.

Calculation:

Given:

ω_AB = 1 rad/s, CD = 1.5 × AB

V_AB = V_CD

ω_AB × AB = ω_CD × CD

∴ 1 × AB = ω_CD × 1.5 × AB

∴ ω_CD = $\frac{2}{3}$ rad/s.

Concept:

When the force applied is continuous in nature, the work done by the force for small displacement "dx" is given by:

\(W =\mathop \smallint \limits_{x_1}^{x_2} {F_x}dx \)

The energy stored in the rubber band is equal to the work done by the stone.

Calculation:

Given:

F_x = 300x², x₁ = 0 m, x₂ = 0.1 m.

Energy stored in the bar = Work done by the stone

 \(W=\mathop \smallint \limits_0^{0.1} {F_x}dx = \mathop \smallint \limits_0^{0.1} 300{x^2}dx = \left[ {300 \times \frac{{{x^3}}}{3}} \right]_0^{0.1} = 100 \times {0.1^3} = 0.1\ J\)

Concept:

Separation of Variables:

f₁(x)g₁(y)dx + f₂(x)g₂(y)dy = 0

$\int \frac{f_1(x)}{f_2(x)}dx;+;\int \frac{g_2(y)}{g_1(y)}dy;=;C$

Calculation:

Given:

Differential equation:

$\frac{{dy}}{{dx}} = \left( {1 + {y^2}} \right)x$

By separation of variables:

$\frac{1}{1+y^2}dy=xdx$

Integrating on both sides we will get:

$⇒ {\tan ^{ - 1}}y = \frac{{{x^2}}}{2} + C$

$\therefore;y = \tan \left( {\frac{{{x^2}}}{2} + C} \right)$

Concept:

When 'r' is a random variable, then by Binomial distribution

The probability of (r = n) in 'n' observations is given by

${\bf{P}}\left( {{\bf{r}} = {\bf{n}}} \right) = {;^{\bf{n}}}{{\bf{C}}_{\bf{r}}};{{\bf{p}}^{\bf{r}}}{{\bf{q}}^{{\bf{n}} - {\bf{r}}}}$ and p + q = 1

where p and q are the probability of success and failure.

Calculation:

Given:

p = Probability of getting head

q = Probability of not getting head = Probability of getting tail

n = 5, p = q = 0.5

Now, we know that

${\bf{P}}\left( {{\bf{r}} = {\bf{n}}} \right) = {;^{\bf{n}}}{{\bf{C}}_{\bf{r}}};{{\bf{p}}^{\bf{r}}}{{\bf{q}}^{{\bf{n}} - {\bf{r}}}}$

Probability of getting at least head = 1 - Probability of getting zero head

∴  ${\rm{P}}\left( {{\rm{r}} \geq 1} \right) = {\rm{;}}1 - {\rm{P}}\left( {{\rm{r}} = 0} \right) = 1{ - ^5}{{\rm{C}}_0}{\left( {0.5} \right)^0}{\left( {0.5} \right)^5} = 1 - \frac{1}{{32}} = \frac{{31}}{{32}}$

 Alternate Method

The unbiased coin is tossed 5 times.

Sample space = 2⁵ = 32

Probability of getting at least head = 1 - Probability of getting zero head

Now, Probability of getting zero head = Probability of getting tail 5 times

Getting only tail = 1 (T, T, T, T, T)

Probability of getting only tail = $\frac{1}{{32}}$

Probability of getting at least head = $1 - \frac{1}{{32}} = \frac{{31}}{{32}}$

Concept:

Natural frequency of the system

${\omega _n} = \sqrt {\frac{{{k_e}}}{m}}~rad/s$ and $f_n=\frac{\omega_n}{2\pi}~Hz$

where k_e = equivalent stiffness, m = mass attach to the system

Calculation:

Given:

k = 20 kN/m = 20000 N/m, m = 1 kg

As the arrangement of springs are in parallel, 

therefore, k_e = k + k = 2 k = 40000 N/m

${\omega _n} = \sqrt {\frac{{40 \times 1000}}{1}}= 200$ rad/s 

And

 $f_n=\frac{200}{2\pi}=31.83~Hz\approx32 ~Hz$

Concept:

Blanking force is given by:

F = τ × A_s

where, τ = shear strength, A_s = shear area

Calculation:

Given:

τ = 300 MPa, d = 100 mm, t = 1.5 mm

Shear area (A_s) is:

A_s = π × d × t 

Blanking force is:

 $F = τ \times {A_s} = 300 × π \times d\times t$

\(F= 300 × π × 100 × 1.5 = 141.371kN \simeq 141 kN\)

Concept:

The Reynolds number for flow over the plate is 10⁴. Therefore, It is a laminar flow.

The relation between Nusselt, Reynolds, and Prandtl No. is given as,

$Nu = 0.332{\left( {Re} \right)^{\frac{1}{2}}}{\left( {{\rm{Pr}}} \right)^{\frac{1}{3}}}$ and 

The relation between the ratio of thickness and Prandtl No. is,

$\frac{{{\delta }}}{\delta_t } = P_r^{ \frac{1}{3}}$

Calculation:

Given:

Re = 10⁴

For fluid Q, Pr No. is,

$2 = P_r^{ \frac{1}{3}}\Rightarrow Pr = 8$

$\frac{{35}}{{{{\left( {Nu} \right)}_Q}}} = \frac{{0.332{{\left( {{{10}^4}} \right)}^{\frac{1}{2}}}{{\left( {\frac{1}{8}} \right)}^{\frac{1}{3}}}}}{{0.332{{\left( {{{10}^4}} \right)}^{\frac{1}{2}}}{{\left( 8 \right)}^{\frac{1}{3}}}}}$

∴ (Nu)_Q = 140

Concept:

The swept volume of the cylinder  is given by:

$V_s = \frac \pi4× D^2× L$

where D = diameter of cylinder and L = length of stroke

Length of the stroke is given by:

L = 2r

where r = crank radius

Calculation:

Given:

r = 60 mm, D = 80 mm

Length of the stroke (L) is:

L = 2r = 2 × 60 = 120 mm

Swept volume is

$V_s = \frac \pi4× D^2× L = \frac{\pi }{4} × {80^2} × 120 = 603185.78~ mm^3=603.185\ c{m^3}$

Concept:

Work done by the pump is given as,

$\int_{1}^{2}dW=v\int_{1}^{2}dP$

$\rm{W}=\rm v(P_2-P_1)$

But we know that $ρ=\frac{m}{V}$ and for unit mass, $ρ=\frac{1}{V}$ 

means $V=\frac{1}{ρ}$

$\therefore W=\frac{P_2;-;P_1}{ρ}$ kJ/kg

Calculation:

Given:

ρ = 990 kg/m³, P₂ = 30 bar = 3000 kPa, P₁ = 1 bar = 100 kPa

$W=\frac{P_2;-;P_1}{ρ}$

$\therefore W=\frac{3000;-;100}{990}=2.92~kJ/kg$

Concept:

Biot Number gives an indication of the ratio of internal (conduction) resistance to the surface (convection) resistance. 

$Bi = \frac{{hL_c}}{k} =\frac{R_{th,conduction}}{R_{th,convection}}$

$Bi <<1\Rightarrow \frac{R_{th,conduction}}{R_{th,convection}}<<1\Rightarrow R_{th,conduction} <<R_{th,convection}$

When the value of Bi is small, it indicates that the system has small conduction resistance i.e. relatively small temperature gradient or the existence of practically uniform temperature within the system. 

The convective resistance then predominates and the transient phenomenon is controlled by the convective heat exchange.

Calculation:

Characteristic dimension, $L=\frac{V}{A}=\frac{\frac{4}{3}\pi r^3}{4\pi r^2}=\frac{r}{3}=2~mm$

$Bi = \frac{{hL}}{k} = 5 \times \frac{{0.002}}{{20}} = 0.0005$

For the given condition the Biot number tends to zero, that means conduction resistance is far less than convection resistance.

Concept:

The Ideal Brayton cycle is shown in the figure.

T₁ = Temperature at the start of compression, T₂ = Temperature at the end of compression

T₃ = Temperature at the start of expansion, T₄ =  Temperature at the end of expansion

And $\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{\frac{{γ - 1}}{γ }}}$ and $\frac{{{T_3}}}{{{T_4}}} = {\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{\frac{{γ - 1}}{γ }}}$

Calculation:

Given:

T₁ = 300 K, T₃ = 1500 K, P₂ = 6 bar, P₁ = 1 bar, γ = 1.4

Now, we have

${T_2} = {T_1}{\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{\frac{{γ - 1}}{γ }}} = 300 \times {6^{\frac{{0.4}}{{1.4}}}} = 500 ~K$

And at the end of the expansion, the temperature is, 

${T_4} = \frac{{{T_3}}}{{{{\left( {\frac{{{P_2}}}{{{P_1}}}} \right)}^{\frac{{γ - 1}}{γ }}}}} = \frac{{1500}}{{\frac{{0.4}}{{{6^{1.4}}}}}} = 900~ K$

Calculation:

Taking moments about instantaneous centre ‘A’

IA $\ddot \theta$ + (kx) r = 0

⇒ (I_o + mr²) $\ddot \theta$ + k (θr)r = 0

$\begin{array}{l} \Rightarrow \left( {\frac{1}{2}m{r^2} + m{r^2}} \right)\ddotθ + k\left( {θ {r^2}} \right) = 0 \ \Rightarrow \ddot θ + \frac{{k{r^2}}}{{\frac{3}{2}m{r^2}}}θ = 0 \Rightarrow \ddotθ + \frac{{2k}}{{3m}}θ = 0; \ \therefore {f _n} = \frac{1}{{2\pi }}\sqrt {\frac{{2k}}{{3m}}} \end{array}$

 

Total energy of the system is:

\(\begin{array}{l} E = \frac{1}{2}k{x^2} + \frac{1}{2}m{V^2} + \frac{1}{2}I{\omega ^2}\ E = \frac{1}{2}k{x^2} + \frac{1}{2}m{V^2} + \frac{1}{2}\frac{{M{r^2}}}{2} \times \frac{{{V^2}}}{{{r^2}}}\E = \frac{1}{2}k{x^2} + \frac{3}{2}m{V^2}\ \frac{{dE}}{{dt}} = 0 \Rightarrow {f _n} = \frac{1}{{2\pi }}\sqrt {\frac{k}{{\frac{3}{2}m}}} \ \Rightarrow {f _n} = \frac{1}{{2\pi }}\sqrt {\frac{{2k}}{{3m}}} \end{array}\)

Shortcut Trick

For a cylinder of radius 'r' with surface contact at point 'p',

natural frequency is given by

${{\bf{f}}_{\bf{n}}} = ;\frac{1}{{2{\bf{\pi }}}}\sqrt {\frac{{{\bf{k}} \cdot {{\left( {\bf{r}} \right)}^2}}}{{\frac{3}{2}{\bf{m}}{{\bf{r}}^2}}}} {\rm{;}} = {\rm{;}}\frac{1}{{2{\rm{\pi }}}}{\rm{;}}\sqrt {\frac{{2{\rm{k}}}}{{3{\rm{m}}}}}$