GATE ME 2014 Official Paper: Shift 3 (Previous Year Paper)

India is a homeland to different geographical regions, religions, caste, culture, languages, and dialects for ages along with the world's second-largest population of over 1.1 billion.

• The cultural heritage is the legacy of our cultural attributes in tangible and intangible form that is inherited from past generations which are maintained in the present and bestowed for the benefit of future generations.

Key Points

The rich heritage and cultural diversity of India:

• India is a multilingual country with numerous languages and dialects that reflects the rich cultural diversity and heritage of India.
• As when people speak their native language, it reflects the region or state they belong to along with their culture and heritage i.e., the way they used to greet each other, and the way they pray to their God in the form of prayers, etc.
• There are approximately 1652 languages and thousands of dialects that belong to different language families.
• Around 40 languages are used as a medium of instruction in schools and universities and the English language is one of them.
• Due to the rich cultural heritage and diversity of India, the Indian government has to mention 22 scheduled languages in the 8^th schedule of the Indian constitution.

Hint

 

Hence, it is concluded that the statement "India is home to 22 official languages and thousands of dialects" best supports the above-mentioned claim.

The correct answer is option 2), i.e. depreciated.

Let us try to find out the meaning of the words given in the options- 

• Depressed: (verb) reduced the level or strength of activity in something especially, an economic or biological system. It is the past and the past participle form of "depress".
• Depreciated: (verb) decreased or diminished in value over a period of time. It is the past and the past participle form of "depreciate".
• Appreciated: (verb)  increased in the value or price. It is the past and the past participle form of "appreciate". It is an antonym of "depreciated".
• Stabilized: (verb) made or became fixed or stopped changing; unlikely to change, fail, or decline. It is the past and the past participle form of "stabilize". 

Clearly, the correct option is option 2), i.e. depreciated, as the value of Indian rupee has diminished as compared to the U.S. dollar.

The correct answer is option 2), i.e. a noun. 

Explanation:

• Advice: (noun) guidance or recommendations offered with regard to future action.
• Advise: (verb) to offer suggestions about the best course of action to someone.

​

Important Point:

• Words ending with -ice are nouns most of the time. Ex: sacrifice, prejudice, injustice, cowardice, armistice, etc.
• Words ending with -ise are verbs most of the time. Ex: advertise, advise, apprise, arise, chastise, etc.

Conclusion:

In the question word Advice is ending with -ice. Hence, it will be a noun.

Logic: We can get next term in the sequence by  (\( {{X} \over 3} \)) × 2

(\( {{81} \over 3} \)) × 2= 54

(\( {{54} \over 3} \)) × 2= 36

(\( {{36} \over 3} \)) × 2= 24

(\( {{24} \over 3} \)) × 2= 16

Hence, "16" will come next in the series.

1) M < R > P > S  → False (Relation between P and M cannot be determined)

2) M > S < P < F  → False (Relation between P and M cannot be determined)

3) Q < M < F = P → False ( M < F = P → M < P)

4) P = A < R < M → True ( P = A < R < M → P < M)

Hence, by using option 4) expression P < M be definitely true.

Logic: All are sequence of prime number and alphabet associated with that prime number.

so next prime number in the sequence is "17"

Alphabet with the place value "17" is "Q"

Hence, "17Q" will be the next term in the sequence.

Let's check each option:

1. The sector of butterfly is one of the sectors of bird in the pie chart. Therefore, conclusion (i) is correct.

2. The sector indicating the population of tigers is almost double the size of the sector of the population of red ants. Therefore, conclusion (ii) is correct.

3) There are only two sectors for reptile - snake and crocodile. Therefore, conclusion (iii) is correct.

4. The pie-chart gives information only about the population count of various living beings. It does not suggest anything about their physical characteristics. Therefore, conclusion (iv) is incorrect.

Hence, "(i), (ii) and (iii) only" is the correct answer.

Given:

Man's speed in still water = 8 km per hour

Calculation:

Let the time taken in upstream and downstream be 3x and x respectively

Speed in the upstream and downstream will be x and 3x

Man's speed in still water = (x + 3x)/2

⇒ 16 = 4x

⇒ x = 4

The stream velocity = (Speed in downstream  - Speed in upstream)/2

The stream velocity = (3x - x)/2 = 4 km/h

∴ The stream velocity is 4 km/h

Given:

The expenditure on labour in 2012 = Rs. 4,50,000

Calculation:

Let  the total manufacturing cost of the firm in 2012 be x

⇒ 15% × x = 4,50,000

⇒ x = 30,00,000 

If 20% increased in all expenses then total manufacturing cost in 2013 = 120% × 30,00,000 = Rs. 36,00,000 

So we can find diffecrence between 30% and 20% increased in raw material,

Raw material expenses in 2012 = 20% × 30,00,000 = Rs. 6,00,000 

⇒ Difference = 6,00,000 (30% - 20%) = Rs.60,000

Total manufacturing cost of the firm in 2013 = 36,00,000 + 60,000 = 36,60,000

Total selling price of the firm in 2013 = Cost + Profit = 36,60,000 + 10,00000 = 46,60,000

The selling price of each purifier = 46,60,000/200 = Rs. 23300

∴ The selling price of each purifier in 2013 is Rs. 23,300.

In a batch there are total 100 bulbs of which 5 are defective i.e. 95 are working.

A batch of one hundred bulbs is inspected by testing four randomly chosen bulbs, the probability that the current batch is accepted = 

\(\frac{95}{100}\times \frac{94}{99}\times \frac{93}{98}\times \frac{94}{97}=0.812\)

Concept:

We know that eigenvectors corresponding to distinct eigenvalues of a real symmetric matrix are also orthogonal.

⇒ X₁.X₂^T = 0

Where X₁ and X₂ are the eigenvectors corresponding to distinct eigenvalues of a real symmetric matrix and X₂^T is the transpose of X2

Calculation:

Given:

\(X_{1}=\left[ {\begin{array}{*{20}{c}} {{x_1}}\ {{x_2}}\ {{x_3}} \end{array}} \right]\),\(X_{2}=\left[ {\begin{array}{*{20}{c}} {{y_1}}\ {{y_2}}\ {{y_3}} \end{array}} \right]\)

Using equation (1),

\(⇒ \left[ {\begin{array}{*{20}{c}} {{x_1}}&{{x_2}}&{{x_3}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{y_1}}\ {{y_2}}\ {{y_3}} \end{array}} \right]=0\)

= x₁y₁ + x₂y₂ + x₃y₃ = 0 

Note:

we know that dot product of two vectors is |a||b|cos(α) where |a| and |b| is the magnitude and α is the angle between them, hence its zero when α is 90 degree.

Explanation:

• For a function say f(x), \(\mathop {\lim }\limits_{x \to c} f(x)\) will exist when left and limit and right-hand limit of the function at point 'x = c' will be the same and it will be equal, to the limit of the function at the point 'x = c' where l is a finite value.
• Any function say f(x) is said to be continuous at a point say 'x = c' if and only if the limit exists at point 'x = c' and it will be equal to the value of that function at 'x = c'.

 \(⇒ \mathop {\lim }\limits_{x \to c^{-}} f(x) =\mathop {\lim }\limits_{x \to c^{+}} f(x) = f(c) =l\)

Where,

\(LHL = \mathop {\lim }\limits_{x \to c^{-}} f(x) \) and \(RHL = \mathop {\lim }\limits_{x \to c^{+}} f(x) \)

Referring to the graph below,

If LHL =  RHL ≠ f(c)  ⇒ Limit will exist at 'x = c' but f(x) at 'x= c' will not be continuous. 

Here, LHL = RHL = p but f(c) = q

Referring to the graph below,

If LHL =  RHL = f(c)  ⇒ Limit will exist at 'x = c' and the function f(x) at 'x = c' will be continuous.

Here, LHL = RHL = f(c) = l ⇒ f(x) is continuous at 'x = c'

Concept:

The divergence of the given vector field,

\(div\vec F = \nabla \cdot \vec F{{;;;;;}} \ldots \left( 1 \right)\)

Calculation:

Given:

\(\vec F = {x^2}z\hat i + xy\hat j - y{z^2}\hat k\)

Using equation (1),

\(\Rightarrow \nabla \cdot \vec F= \left( {\hat i\frac{\partial }{{\partial x}} + \hat j\frac{\partial }{{\partial y}} + \hat k\frac{\partial }{{\partial z}}} \right) \cdot \vec F\)

\(\Rightarrow\nabla \cdot \vec F= 2xz + x – 2yz\)

\(\therefore div\vec F;at\left( {1, - 1,1} \right) = 2 + 1 + 2 = 5\)

Concept:

M for men, W for women, E for employed & U for unemployed

Referring to the diagram below,

∴ Total probability when the selected person is unemployed,

P(U) = P(M). P(U/M) + P(W) ⋅ P(U/W)       ...(1)

Total probability when the selected person is emlpoyed,

 ⇒ P(E) = 1 – P(U)      ...(2)

Calculation:

Given:

P(M) = 0.5, P(W) = 0.5, P(U/M) = 0.2 & P(U/W) = 0.5

Using equation (1),

⇒ P(U) = (0.5× 0.2) + (0.5× 0.5)

⇒ P(U) = 0.35

Using equation (1),

⇒ P(E) = 1 - 0.35

⇒ P(E) = 0.65

Concept:

By trapezoidal rule,

\(\mathop \smallint \limits_{{x_0}}^{{x_n}} f\left( x \right)dx = \frac{h}{2}\left[ {\left( {{y_0} + {y_n}} \right) + 2\left( {{y_1} + {y_2} + \ldots } \right)} \right]{{;;;;;}} \ldots \left( 1 \right)\)

Calculation:

Given:

h = 1 &  \(y=\frac{1}{x}\)

Table for calculating y₀, y₁, y₂,  y₃ ......y_n

Using equation (1),

 

\(\Rightarrow \mathop \smallint \limits_1^3 \frac{1}{x}dx = \frac{1}{2}\left[ {\left( {1 + \frac{1}{3}} \right) + 2\left( {0.5} \right)} \right] = 1.167\)

Concept:

A rotating shaft under pure bending will be under reverse fluctuating load so the minimum value of bending stress developed will be have magnitude as maximum bending stress but opposite in sign.

⇒ σ_min = - σmax       ...(1)

\(⇒ {{{σ }}_{{m}}} = \frac{{{{{σ }}_{{{max}}}} + {{{σ }}_{{{min}}}}}}{2} = \frac{{100 - 100}}{{{{}}2}} = 0\)

⇒ σm = 0 

Since σm = 0, so any of the criteria for the design can be considered.

Soderberg equation is given as,

\(\frac{{{{\bf{σ }}_{\bf{m}}}}}{{{{\bf{σ }}_{{\bf{yt}}}}}} + \frac{{{{\bf{σ }}_{\bf{v}}}}}{{{{\bf{σ }}_{\bf{e}}}}} = \frac{1}{{\bf{N}}}\)

\( ⇒ \frac{{{{\bf{σ }}_{\bf{v}}}}}{{{{\bf{σ }}_{\bf{e}}}}} = \frac{1}{{\bf{N}}}{{;;;;;}} \ldots \left( 2 \right)\)

where, σ_m = mean stress, σ_v = variable stress, N = factor of safety  

Calculation:

Given:

σ_max = 100 MPa, S_yt = 300 MPa, S_ut = 500 MPa, S_e = 200 MPa

Using equation (1),

⇒ σmin = - 100 MPa

So, the mean stress and the variable stress,

\(⇒ {{{σ }}_{{m}}} = \frac{{{{{σ }}_{{{max}}}} + {{{σ }}_{{{min}}}}}}{2} = \frac{{100 - 100}}{{{{}}2}} = 0\)

\( ⇒ {{\rm{σ }}_{\rm{v}}} = \frac{{{{\rm{σ }}_{{\rm{max}}}} - {{\rm{σ }}_{{\rm{min}}}}}}{2} = \frac{{100 + 100}}{{{\rm{}}2}} = 100~{\rm{MPa}}\)

Using equation (2),

\(⇒ \frac{100}{200}=\frac{1}{N}\)

⇒ N = 2

Concept: 

The shear stress developed in a circular shaft,

\(\Rightarrow τ = \frac{{16T}}{{\pi {d^3}}}\)

\(\Rightarrow {\rm{τ }} \propto \frac{1}{{{{\rm{d}}^3}}}\)

\(\Rightarrow \frac{{{τ _2}}}{{{τ _1}}} = {\left( {\frac{{{d_1}}}{{{d_2}}}} \right)^3} {{;;;;;}} \ldots \left( 1 \right)\)

Calculation:

Given:

T₁ = T2 , R₁: R₂ = 2 : 1

Using equation (1),

\(\Rightarrow \frac{{{τ _2}}}{{{τ _1}}} = {\left( {\frac{{{R_1}}}{{{R_2}}}} \right)^3} = {2^3} = 8\)

\(\Rightarrow \frac{{{τ _2}}}{{{τ _1}}} = {2^3} = 8\)

Concept:

Because of the excitation force, there will be excitation in the system of damped vibration. So there will be a lag in the response of the damped vibrating system. After reaching the steady-state, there will be the only steady-state response as written by equation (1). The phase difference between the response of the system (x) and excitations (F0sin(ωt)) can be derived from the diagram below,

 

Where,

F = excitation force, KA = Maximum spring force, CAω = Maximum damping force (leading by 90° from spring force response as this proportional to velocity)

MAω2 = Maximum Inertial force (leading by 180° from spring force response as this is proportional to inertial acceleration)

From ΔOab,

\({{tan}}\phi = \frac{{{{CA\omega }}}}{{{{KA}} - {{MA}}{{{\omega }}^2}}}\)

\( ⇒ \phi = ;{\tan ^{ - 1}}(\frac{{{{C\omega }}}}{{{{K}} - {{M}}{{{\omega }}^2}}})\)

\(\Rightarrow\phi ={{\tan }^{-1}}\left( \frac{2\xi \frac{\omega }{{{\omega }_{n}}}}{1-{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}} \right){{;;;;;}} \ldots \left( 1 \right)\)

For ω/ωn > 1 the phase will be 180˚ as the value 1- (ω/ωn)2 will become negative which means out of phase.

• Phase angle varies from zero at low frequencies to 180° at very high frequencies. It changes very rapidly near the resonance and is 90° at resonance irrespective of damping.
• The phase angle between 0 to 90° for all the values of ω/ωn = 0 to 1 and between 90° to 180° for all the values of ω/ωn > 1
• In absence of any damping, the phase angle suddenly changes from zero to 180° at resonance.
• The phase angle φ is zero for all the values of damping ratio at ω/ωn = 0

Calculation:

At resonance, (ω/ωn) = 1

Using equation (1),

\(\Rightarrow\phi ={{\tan }^{-1}}\left( \frac{2\xi \frac{\omega }{{{\omega }_{n}}}}{1-{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}} \right)=\tan^{-1}(\infty)=90^\circ\)

So, the phase angle is 90o.

Concept:

Coefficient of restitution (e)

\(e = \frac{{{\rm{Velocity~of~separation}}}}{{{\rm{Velocity~of~approach}}}}= \frac{{{v_2} - {v_1}}}{{{u_1} - {u_2}}}{{;;;;;}} \ldots \left( 1 \right)\) 

where, v = final velocity of respective masses and u = initial velocity of respective masses

Calculation:

Given:

v₁ = v_{2 = v}, u₁ = 5 m/s, u₂ = 0

Using equation (1),

\(\Rightarrow e= \frac{{v - v}}{{5 - 0}}=0\)

Explanation:

1. The Maximum Shear Stress Theory gives over safe design values among all theories, not the most accurate results.
2. Maximum Distortion Energy Theory gives safe and economic values of stresses in the component.

Maximum Shear Stress Theory and Maximum Distortion Energy Theory are used for Ductile materials.

Concept:

 \(Speed \ of\ gear ∝\frac{1}{{{\rm{No}}.~{\rm{of~teeth}}}}\)

\(\Rightarrow\frac{{{N_5}}}{{{N_2}}} = \frac{{{T_2}}}{{{T_3}}} \times \frac{{{T_4}}}{{{T_5}}} {{;;;;;}} \ldots \left( 1 \right)\)

Calculation:

Given:

T₂ = 20, T₃ = 40, T₄ = 15, T₅ = 30

Using equation (1),

\(\Rightarrow\frac{{{N_5}}}{{{N_2}}} = \frac{{20}}{{40}} \times \frac{{15}}{{30}}\)

⇒ N₅ = 1200 × 0.25 = 300 rpm ccw

Explanation:

The thermal resistance when the heat transfer due to  conduction  takes place radially through the cylinder,

\(R_{cyl}=\frac{r_{2}-r_{1}}{K{A_{m}}}{{;;;;;}} \ldots \left( 1 \right)\)

Where,

A_m = Mean area for radial heat conduction through the cylinder

A_m = 2πr_mL 

\(r_{m}=\frac{r_{2}-r_{1}}{\ln \frac{r_{2}}{r_{1}} }{{;;;;;}} \ldots \left( 2 \right)\)

Calculation:

Using equation (1) and (2),

\(\Rightarrow R_{cyl}=\frac{1}{{2π kL}}\ln \left( {\frac{{{r_o}}}{{{r_i}}}} \right)\)

 

Explanation:

The quantity \(\frac{\delta Q}{T}\) does not depend upon the path. It depends upon the initial and final state of the process. So, the quantity \(\frac{\delta Q}{T}\) must be the property. this property is known as entropy which is represented by 'ds'

From the second law of thermodynamics for any process,

\(\Rightarrow ds≥\frac{{\delta Q}}{T} {{;;;;;}} \ldots \left( 1 \right)\).

For an irreversible process,

\(\Rightarrow ds>\frac{{\delta Q}}{T} {{;;;;;}} \ldots \left( 2 \right)\)

For any cyclic process,

\(\Rightarrow\mathop \oint \nolimits ds = 0\)

Using equation (2),

So, Clausius inequality for the irreversible heat engine is given by, 

\(⇒\mathop \oint \nolimits\frac{{dQ}}{T} < 0\)

For an irreversible heat engine, 

\(⇒ \mathop \oint \nolimits \frac{{dQ}}{T} < 0\)

The heat content of the irreversible heat engine will be positive as heat is given to the system for heat engine

\(\Rightarrow dQ >0\)

Explanation:

• The Reynold number for the flow through the pipe can be defined as,

\({\rm{Re}} = \frac{{{\rm{ρ VD}}}}{{\rm{μ }}}\)

• where, ρ = density of the fluid, V = velocity of the fluid, D = characteristic length (Diameter of the circular pipe). μ = dynamic viscosity
• For Turbulent flow through a pipe ⇒ Re_D > 2300
• The turbulent flow is considered as a well-mixed flow because in it there is a huge order of intermixing of fluid particles and due to intermixing, they mixed with each other very strongly.

Additional Information

Explanation:

About gas turbine plant:

• It can be started up as well as shut down quickly, like a diesel engine.
• It has a low value of weight per kW power output, which is a favorable feature in all vehicles (land, air, and sea).
• Works on the principle of the Brayton cycle.
• Uses any hydrocarbon fuel, from high-octane gasoline to heavy diesel oil and pulverized coal, effectively
• It is light in weight as compared to steam turbine power plant. Therefore, it is used in vehicles.

Concept: 

Carnot cycle is an ideal cycle so it is the most efficient cycle.

Efficiency of Carnot Cycle:

\(\eta=\frac{W_{net}}{Q_{in}}=1-\frac{T_L}{T_H}{{;;;;;}} \ldots \left( 1 \right)\)

Calculation:

Given:

T₁ = 500 K,  T₂ = 27°C = 300 K, Q₁ = Q_in = 1000 kJ

Using equation (1),

\({\eta _{carnot} = 1 - \frac{{300}}{{500}}}=\frac{W_{net}}{1000}\)

⇒ Wnet = 1000 × 0.4 = 400 kJ

⇒ The maximum useful work = 400 kJ

Alternate Method:

Concept:

For reversible Carnot Cycle:

\(⇒\frac{Q_1}{T_1}=\frac{Q_2}{T_2} {{;;;;;}} \ldots \left( 1 \right)\)

The maximum useful work, 

⇒ Wnet = Q1 - Q2       ...(2)

Calculation:

Given:

T1 = 500 K,  T2 = 27°C = 300 K, Q1 = Qin = 1000 kJ

Using equation (1),

\(⇒ \frac{1000}{500}=\frac{Q_2}{300}⇒ Q_2 =600 ;kJ\)

Using equation (2),

⇒W_net = 1000 - 600

⇒ Wnet = 400 kJ

Concept:

Humidity ratio (ω):

It is the mass of water associated with each kilogram of dry air. It is known as specific humidity or absolute humidity.

\(ω = \frac{{{m_v}}}{{{m_a}}} = 0.622\left( {\frac{{{P_v}}}{{{P_t} - {P_v}}}} \right) {{;;;;;}} \ldots \left( 1 \right)\)

Relative humidity (ϕ):

It is defined as the ratio of partial pressure of water vapor (p_v) in a mixture to the saturation pressure (ps) of pure water at the same temperature of the mixture.

\(ϕ=\frac{p_v}{p_{vs}}{{;;;;;}} \ldots \left( 2 \right)\)

Calculation:

Given:

P_t = 85 kPa, P_vs = 4.24 kPa, ϕ = 0.65

Using equation (2),

⇒ P_v  = 0.65 × 4.24 = 2.75 kPa

Using equation (1),

\(⇒ w = 0.622 \times \frac{{{2.75}}}{{{85} - {2.75}}}=0.0208\ kg\ of\ water\ vapour \ per \ kg \ of \ dry\ air\)

\(⇒ w =20.8\ gram\ of\ water\ vapour \ per \ kg \ of \ dry\ air\)

Explanation:

The various processes of the machining are as follows:

Concept:

Moving Average Method:

The moving average method uses the average of the most recent 'n' data values in the time series as the forecast for the next period.

\({F_{t + 1}} = \frac{{{D_t} ;+ ;{D_{t - 1}} ;+ \ldots; +; {D_{t - n + 1}}}}{n}\)

Here, the 'n' past most recent observations are equally weighted.

For the given case when 'n = 4' so the forecast of the sales, using the 4-month moving average method, for the month of February,

\(⇒ {F_{Feb}} = \frac{{{D_{Jan}} ;+ ;{D_{Dec}} ;+;{D_{Nov}} ;+; {D_{Oct}}}}{4}{{;;;;;}} \ldots \left( 1 \right)\)

Calculation:

n = 4,

Using equation (1),

 \(⇒ {F_{Feb}} =\frac{280+250+190+240}{4}=240\)

⇒The forecast of the sales, for the month of February, is 240 units.

Concept:

For single point cutting tool, the uncut chip thickness,

⇒ t = f sin λ  

∵ λ + c_s = 90° 

⇒ t = f cosc_s      ...(1) 

Where,

λ = The approach angle. cs = The side cutting edge angle, f =  feed

Calculation:

Given: 

f = 0.2 mm/rev, Cs = 60°, d = 0.5 mm

Using equation (1),

⇒ t = 0.2 × sin (90-60)

⇒ t = 0.1 mm

Explanation:

A path forms a loop or cycle if it connects a node itself. A spanning tree links all the nodes of the network with no loop allowed.

Rules for Network construction:

• No two or more activities have the same head and tail events.
• The time flows from the left to right direction.
• There should not be looping and dangling on the network diagram.

Explanation:

The various defects of the casting are as follows:

Explanation:

Cutting action involves shear deformation of work material to form a chip, and a chip is removed, a new surface is exposed. Tool wear is a common phenomenon is metal cutting.

Major causes of tool wear are:

• High contact temperatures at the tool-chip and tool–work-piece interfaces that lead to the softening of tool material and promote diffusion and chemical (oxidation) wear
• High contact pressures at these interfaces
• The cyclic nature of the chip formation process, which can cause cracking because of thermal fatigue.
• During machining, the strain also occurs in work material and strain hardening takes place of work and it becomes harder.

Explanation:

The stress-strain curve for steel is shown below the figure.

By comparing with the above curve, we will get,

P → Proportional limit

Q → Elastic limit

R → Upper yield point

S → Lower yield point

T → Ultimate tensile strength

U → Failure

• So it is evident from the graph that the strain is proportional to stress or elongation is proportional to the load giving a straight-line relationship. This law of proportionality is valid up to point A. Point A is known as the limit of proportionality or the proportionality limit.
• For a short period beyond point A, the material may still be elastic in the sense that the deformations are completely recovered when the load is removed. The limiting point B is termed as Elastic Limit.
• Beyond the elastic limit plastic deformation occurs and strains are not totally recoverable. There will be thus permanent deformation or permanent set when the load is removed. These two points are termed as upper and lower yield points respectively. The stress at the yield point is called the yield strength.
• A further increase in the load will cause marked deformation in the whole volume of the metal. The maximum load which the specimen can withstand without failure is called the load at the ultimate strength. The highest point ‘E' of the diagram corresponds to the ultimate strength of a material.
• Beyond point E, the bar begins to form the neck. The load falling from the maximum until a fracture occurs at F.

Explanation:

Hot Tears:

• It is a casting defect, which takes place due to a differential cooling rate.
• The hot tear is a defect that occurs during the solidification of casting. Though steel is poured at high temperatures, this temperature is low relative to the high freezing temperature range of steel.
• Due to non-uniform cooling, internal stresses can be developed. The stresses will be more than the strength of the material and cracks can be expected.
• Discontinuities in the form of cracks (hot tears, also called hot cracks or hot pulls) are caused due to hindrance in the contraction of the casting.
• Hot tears are caused because of the low strength of metal after solidification, causing the metal to fail in coping with the excessively high stress set up by solid shrinkage of the metal.
• The lack of collapsibility of mold or core is another cause. Hot spots are the portions of casting which solidify in the end and are often the cause of the hot tear.
• The use or chills (metallic) helps in the solidification of the hot spot area quicker and thus avoids cracking.

Concept:

The function f(z) is given as,

f(z) = u(x, y) + i v(x, y)

Snce f(z) is analytic so C-R equations are as follows,

\(⇒ \frac{{\partial u}}{{\partial x}} = \frac{{\partial v}}{{\partial y}};and~\frac{{\partial u}}{{\partial y}} = - \frac{{\partial v}}{{\partial x}}{{;;;;;}} \ldots \left( 1 \right)\)

∵ v(x, y) is double valued function,

\(∴ dv = \frac{{\partial v}}{{\partial x}}dx + \frac{{\partial v}}{{\partial y}}dy{{;;;;;}} \ldots \left( 2 \right)\)

Calculation:

Given:

u = x² – y²

\(⇒ \frac{{\partial u}}{{\partial x}} = 2x = \frac{{\partial v}}{{\partial y}}\)

\(⇒\frac{{\partial u}}{{\partial y}} = - 2y = - \frac{{\partial v}}{{\partial x}}\)

Using equation (2),

⇒ dv = 2y dx + 2x dy      ...(3)

This is an exact differential equation of form 'M dx + N dy = 0' as f(z) is analytic and it will be satisfying C-R equations.

Where, 

\(⇒\frac{{\partial M}}{{\partial y}} = \frac{{\partial N}}{{\partial x}}\)

Where M and N are functions of x and y.

Using equation (3),

⇒ ∫ dv = ∫ 2y dx + ∫ 2x dy 

 Keeping y as constant for '∫ 2y dx' and keep all the terms free from x in '∫ 2x dy' 

⇒ v = 2xy + c

Alternate method:

Concept:

We know that,

 \(f^{'}(z)= \frac{{{\rm{du}}}}{{{\rm{dx}}}} + {\rm{i}}\frac{{{\rm{dv}}}}{{{\rm{dx}}}} = \frac{{{\rm{du}}}}{{{\rm{dx}}}} + {\rm{i}}\frac{{ - {\rm{du}}}}{{{\rm{dy}}}} = 2{\rm{x}} + {\rm{i}}\left( {2{\rm{y}}} \right){{;;;;;}} \ldots \left( 2 \right)\)

Since f(z) is analytic,

∴ Put x = z and y = 0

⇒ f’^'(z) = 2z      ...(2)

Calculation:

Given:

u = x2 – y2

Using equation (1),

Integrate both side,

\(⇒ \mathop \smallint \nolimits {\rm{f'}}\left( {\rm{z}} \right) = {\rm{}}\mathop \smallint \nolimits 2{\rm{zdz}}\)

⇒ f(z) = z² + c

For imaginary part,

⇒ f(u + iv) = (x + iy)² + c = (x² – y² + i(2xy)) + c

\∴ u = (x² – y²)  and v = 2xy

Concept:

Given differential equation is

\(\frac{{{d^2}x}}{{d{t^2}}} + x = 0\)

∴ Auxillary equation is

⇒ m2 + 1 = 0      ....(1)

Calculation:

Given:

\(\ x_{1}(0)=1,\ {\left. {\frac{{d{x_1}\left( t \right)}}{{dt}}} \right|_{t = 0}} = 0,;{x_2}\left( 0 \right) = 0,{\left. {\frac{{d{x_2}\left( t \right)}}{{dt}}} \right|_{t = 0}} = 1\)

Using equation (1),

⇒ m = ± i

∴ General solution is

⇒ x = C₁ cost + C₂ sint  

\(⇒ \dot x = - {C_1}\sin t + {C_2}\cos t\)

For,  \({x_1}\left( 0 \right) = 1;& \frac{{d{x_1}\left( 0 \right)}}{{dt}} = 0\)

⇒ x₁ = cos t

For,  \({x_2}\left( 0 \right) = 0;& \frac{{d{x_2}\left( 0 \right)}}{{dt}} = 1\)

⇒ x₂ = sin t

So, the Wronskian  at \(t = \frac{\pi}{2}\),

\(⇒ W = \left| {\begin{array}{*{20}{c}} {{x_1}}&{{x_2}}\ {\frac{{d{x_1}}}{{dt}}}&{\frac{{d{x_2}}}{{dt}}} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {\cos t}&{\sin t}\ { - \sin t}&{\cos t} \end{array}} \right|\)

⇒ W = cos²t + sin²t 

⇒ W = 1

Concept:

The mean value of the number of defective pieces

⇒ μ = E(X) = ∑ x P(X)      ...(1)

The variance of the number of defective pieces

⇒ V(X) = σ² = E(X2) – [(E(X)]2       ...(2)

Calculation:

Given:

Let ‘x’ be the number of defective pieces

Using equation (1),

\(\Rightarrow E(X)= 0 \times \frac{1}{6} + 1 \times \frac{2}{3} + 2 \times \frac{1}{6}\)

⇒ μ = E(x) = 1

Now,

\(\Rightarrow E({X^{2}})= 0 \times \frac{1}{6} + 1 \times \frac{2}{3} + 4 \times \frac{1}{6}=\frac{4}{3}\)

Using equation (2),

\(⇒ V(X)= \frac{4}{3} - 1 = \frac{1}{3}\)

Concept:

By Newton Raphson method

\({x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\)

Calculation:

Let f(x) = 5x – 2cos x – 1

⇒ f’(x) = 5 + 2sin x

f(0) = - 3, f(1) = 2.9

Let x₀ = 1 rad = 57.32° 

Using equation (1),

\(\Rightarrow{{\rm{x}}_1} = 1 - \frac{{\left( {5 \times 1} \right) - \left( {2\cos \left( {57.32^\circ } \right)} \right) - 1}}{{5 + 2\sin \left( {57.32^\circ } \right)}} = 0.5632\)

Put x₁ = .5632 then

\(\Rightarrow{{\rm{x}}_2} = .5632 - \frac{{\left( {5 \times 0.5632} \right) - \left( {2\cos \left( {32.27^\circ } \right)} \right) - 1}}{{5 + 2\sin \left( {32.27^\circ } \right)}} = 0.5425\)

Put x₂ = 0.5425 then

\(\Rightarrow{{\rm{x}}_3} = 0.5425 - \frac{{\left( {5 \times 0.5425} \right) - \left( {2\cos \left( {31.09^\circ } \right)} \right) - 1}}{{5 + 2\sin \left( {31.09^\circ } \right)}} = 0.5424\)

\(\therefore {\rm{real~root}},{\rm{x}} = 0.54\)

⇒ x₁ = 0.5632

⇒ x₂ = 0.5425

⇒ x₃ = 0.5424

So, asnwer is 0.54 (upto two digits of decimal).

Concept:

Referring to the FBD of the lever,

 

Taking moment about pin joint,

⇒ ΣM_O = 0      ...(1) 

The braking torque for the brake shoe,

⇒ T_b = μNR      ...(2)  

Calculation:

Given:

P = 1000 N, μ = 0.2, R = 0.2 m

Using equation (1),

⇒ 1000× 0.8 = N × 0.48 + 0.2 × N × 0.1

⇒ N = 1600 N

Using equation (2),

⇒ T_b = 0.2 × 1600 × 0.2

⇒ T_b = 64 N.m

Concept:

FBD of the block is shown below :

Along the plane,

The net force on the block,

⇒ ΣF = ma      ...(1)

 

Using Newton's 2^nd law of motion,

⇒ v2 – u2 = 2as       ...(2)

where,

v = velocity at end, u = initial velocity, s = distance traveled by block, a = Acceleration

Calculation:

Given:

m = 10 kg, θ = 45°, μ = 0.5, V = 20 m/s  

Using equation (1),

⇒ ma = mg sin 45o - μ mg cos 45o

⇒ a = 3.468 m/s²

Using equation (2),

As block is at top,

u = 0

\(⇒ s = \frac{{{v^2}}}{{2a}} = \frac{{{{20}^2}}}{{2 \times 3.468}} = 57.66 \ m\)

Concept:

Simply supported beam is shown in fig. having point load P at the center.

Shear stress of beam,

 \({\rm{\tau }} = \frac{3}{2} \times \frac{{{{\rm{P}}_{{\rm{max}}}}}}{{\rm{A}}}{{;;;;;}} \ldots \left( 1 \right)\)

Bending stress,

 \({\rm{\sigma }} = \frac{{\rm{M}}}{{{{\rm{Z}}_{{\rm{NA}}}}}}{{;;;;;}} \ldots \left( 2 \right)\) 

Referring to the diagram, the maximum bending moment will be at the center of the simply supported beam

⇒ Pmax = P/2,

⇒ A = 2h2

\(\Rightarrow {{M_{max}}} = \frac{{{P}}}{2} \times 25{{h}}\)

\(\Rightarrow Z_{NA}=\frac{1}{6} \times 2{{\rm{h}}^3}\)

where,

Pmax = maximum load on one end, A = Area, M_max = Maximum bending moment  ZNA = Sectional modulus about neutral axis

Calculation:

Given:

L= 50h, d = h, w = 2h

Using equation (1),

 \(\Rightarrow {{\tau }} = \frac{3}{2} \times \frac{{{P}}}{{2\left( {2{{{h}}^2}} \right)}} = \frac{{3{{P}}}}{{8{{{h}}^2}}}\)

Using equation (2),

\(\Rightarrow {{\sigma }} = \frac{{\left( {\frac{{25{{Ph}}}}{2}} \right)}}{{\left( {\frac{{2{{{h}}^3}}}{6}} \right)}} = \frac{{75{{P}}}}{{2{{{h}}^2}}}\)

The ratio of maximum shear stress to the maximum bending stress in the beam,

\(\Rightarrow \frac{{\rm{\tau }}}{{\rm{\sigma }}} = \frac{{\frac{{3{\rm{P}}}}{{8{{\rm{h}}^2}}}}}{{\frac{{75{\rm{P}}}}{{2{{\rm{h}}^2}}}}} = \frac{1}{{100}} = 0.01\)

Concept:

The damping ratio (ξ) is defined as ratio of viscous damping coefficient to critical viscous damping coefficient.

\(⇒ {{ξ }} = \frac{{{c}}}{{2\sqrt {{{km}}} }}{{;;;;;}} \ldots \left( 1 \right)\)

where, ξ = damping ratio, c = damping coefficient, k = spring constant, 

Calculation: 

Given: 

 m = 1 kg, k = 100 N/m, c = 25 Ns/m

Using equation (1),

\(⇒ {{ξ }} = \frac{{25}}{{2\sqrt {100 \times 1} }}{{}}\)

⇒ ξ = 1.25

Concept:

The disc is rolling without slipping so the point of contact (P) of the annular disc with the ground will be in the state of rest. So about p[oint P there will be pure rotational motion.

 

So, point P can be considered as the instantaneous center of rotation.

The velocity of point P will be the same as the velocity of the center of mass of the disc.

⇒ v = 2Rω 

\(\Rightarrow ω =\frac{V}{2R}{{;;;;;}} \ldots \left( 1 \right)\)

Where 

The kinetic energy of the disc,

\(\Rightarrow KE = \frac{1}{2}{J_p}{ω ^2}{{;;;;;}} \ldots \left( 2 \right)\)

Where,

.J_P = The mass  moment of Inertia of the annular disc about point P, ω = The angular velocity of the annular disc 

Calculation:

The mass  moment of Inertia of the annular disc about point P,

\(\Rightarrow {I_c} = \frac{1}{2}m{\left( {2R} \right)^2} + \frac{1}{2}m{R^2} = \frac{5}{2}m{R^2}\)

Using parallel axis theorem,

\(\Rightarrow {I_p} = {I_c} + m{\left( {2R} \right)^2} = \frac{{13}}{2}m{R^2}{{;;;;;}} \ldots \left( 3 \right)\)

Using equation (1), (2), and (3),

\(\Rightarrow KE= \frac{1}{2} \times \frac{{13}}{2}m{R^2} \times {\left( {\frac{v}{{2R}}} \right)^2}\)

\(\Rightarrow KE= \frac{{13}}{{16}}m{v^2}\)

Concept:

Shift the eccentric force P at the free end 'A' of the cantilever beam. After shifting this eccentric load at point 'A', one moment P(L - x) will be generated and a transverse load P will be acting downward as shown in the figure

Due to point load P, the deflection at the free end of the cantilever beam (downward),

\(⇒{\delta _1} = \frac{{P{L^3}}}{{3EI}}{{;;;;;}} \ldots \left( 1 \right)\)  

Due to moment M = P(L - x) (anticlockwise), the deflection at the free end of the cantilever beam (Upward),

\(⇒{\delta _2} = \frac{{M{L^2}}}{{2EI}}\)

\(⇒ {\delta _2} = \frac{{P\left( {L - x} \right){L^2}}}{{2EI}}{{;;;;;}} \ldots \left( 2 \right)\)

Calculation:

Given:

The displacement at ‘A’ is equal to zero.

For zero displacement at point A,

⇒ δ₁ = δ₂

Using equation (1) and (2),

\(⇒ \frac{{P{L^3}}}{{3EI}} = \frac{{P\left( {L - x} \right){L^2}}}{{2EI}}\)

⇒ x = 0.33 L

Explanation:

Translation of the follower is, d = \(√{{x^2}+{y^2}}\)

Case 1: 

x(t) = cos t, y(t) = sin t

\(d=√{{cos^2t}+{sin^2t}} = 1\)

d_max = 1

Case 2:

x(t) = cos t, y(t) = 2 sin t

\(d=√{{cos^2t}+{\left(2;sin;t\right)^2}} = √{{cos^2t}+{4;sin^2t}} = √{1+3;sin^2t} \)

'd' will be maximum when sin²t will be maximum.

sin t will be maximum at t = π /2 , 3π /2 in the interval of (0, 2π)

\(d_{max}=√{1+3} = 2\)

Case 3:

x(t) = ½ + cos t, y(t) = 2 sin t

\(d=√{{(\frac{1}{2}+cos;t)^2}+{\left(2;sin;t\right)^2}} = √{\frac{1}{4}+cos;t+{cos^2t}+{4;sin^2t}} = √{\frac{5}{4}+cos;t+3;sin^2t} \)

'd' will be maximum when leading term 'sin²t' will be maximum.

sin²t will be maximum at t = π /2 , 3π /2 in the interval of (0, 2π)

\(d_{max}=√{\frac{5}{4}+0+3} = 2.06\)

Case 4:

x(t) = ½ + cos t, y(t) = sin t

\(d=√{{(\frac{1}{2}+cos;t)^2}+{\left(sin;t\right)^2}} = √{\frac{1}{4}+cos;t+{cos^2t}+{sin^2t}} = √{\frac{5}{4}+cos;t} \)

'd' will be maximum when cos t will be maximum.

cos t will be maximum at t = 0 , 2π in the interval of (0, 2π)

\(d_{max}=√{\frac{5}{4}+1} = 1.5\)

Option ‘C’ has maximum net amplitude.

Concept:

From laws of motion,

⇒ ω_F = ω_o – αt      ...(1) 

where,

ωF = final rotation of wheels, ω_o = initial wheel velocity, α = angular acceleration

The torque acting on the wheel of the vehicle,

⇒ ​T = I α      ...(2)

The Inertia I is given as,

⇒ I = mr²

Here, m = mass of vehicle on one single wheel

Using equation (2),

⇒ ​T = mr2α      ...(3)

Calculation:

Given:

4m = 1000 kg ⇒ m = 250 kg , ω_o = 10 rad/s, r = 0.2 m, t = 10 s, ωF = 0 rad/s

Using equation (1),

⇒ 0 = 10 - α × 10

⇒ α = 1 rad/s²

Using equation (3),

\(⇒ T = 250\times {0.2^2} \times 1\)

⇒ T = 10 N.m

Additional Information

For Solid cylinder:

 \(I=\frac{1}{2}mr^2\)

For thin-walled cylinder/hoop/ring:,

 \(I=mr^2\)

Concept:

Let OA be crank, OA = 60 mm = 0.06 m and AB be connecting rod, AB = 240 mm = 0.24 m

Here, I = The instantaneous centre of rotation connecting rod AB for the given configuration.

VA = OA × ωOA       ...(1)

The angular veocity of link AB,

\(⇒{ω _{AB}} = \frac{{{V_A}}}{{AI}} ​​​​{{;;;;;}} \ldots \left( 2 \right)\)

The speed of slider,

⇒ VB = ωAB × IB      ...(3) 

Calculation:

Given:

OA = 60 mm, AB = 240 mm, ω_OA = 10 rad/s²

Using equation (1),

⇒ V_A = 0.06 × 10 = 0.6 m/s

From geometry of above diagram,

\(⇒{\theta _1} = {\tan ^{ - 1}}\left( {\frac{{OA}}{{AB}}} \right) = 14.036^\circ ;\)

⇒ θ₂ = 90 – θ₁ = 75.964°

⇒ AI = AB tan θ₂ = 0.96 m

\(⇒ IB = \frac{{AB}}{{\cos {\theta _2}}} = 0.9895\ m\)

Using equation (2),

\(⇒{ω _{AB}} = \frac{{0.6}}{{0.96}} = 0.625\ rad/s^{2}\)

Using equation (3),

⇒ V_B =0.625 × 0.9895 = 0.618 m/s

⇒ VB = 0.618 m/s

Concept:

Referring to the graphs for given linear equations

The constraints x1 ≥ 0, x2 ≥ 0 and x₁ + 2x₂ ≤ 4 will be having feasible region.

The corner points of the feasible region will give the best feasible solution for the objective function.

Calculation:

Referring to the cornor points of feasible rergion,

The corner point of the feasible region are (0, 0), (0, 2) & (4, 0)

The value of the objective function at corner points,

⇒ Z (0,0) = 0,

⇒ Z (0,2) = 18

⇒ Z (4,0) = 12

∵ The maximum value of Z is at (0, 2)

⇒ Z_max = 18

Concept:

Damping ratio,

\(ξ = \frac{C}{{{C_c}}}=\frac{C}{2\sqrt {km}}{{;;;;;}} \ldots \left( 1 \right)\)

Where,

Critical damping coefficient,

\({C_c} = 2m;{ω _n} =2\sqrt {km}{{;;;;;}} \ldots \left( 2 \right)\)

The amplitude of the steady-state response,

\(A = \frac{{\frac{{{F_0}}}{k}}}{{\sqrt {{{\left[ {1 - {{\left( {\frac{ω }{{{ω _n}}}} \right)}^2}} \right]}^2} + {{\left( {2ξ \frac{ω }{{{ω _n}}}} \right)}^2}} }} = \frac{{{X_s}}}{{\sqrt {{{\left[ {1 - {{\left( {\frac{ω }{{{ω _n}}}} \right)}^2}} \right]}^2} + {{\left( {2ξ \frac{ω }{{{ω _n}}}} \right)}^2}} }}{{;;;;;}} \ldots \left( 3 \right)\)

Calculation:

Given:

m = 10 kg, k = 6250 N/m, F = 10 cos(25 t) N, A = 40 mm, ω = 25 rad/s

Using equation (2),

\(⇒ {C_c} = 2\sqrt {6250 × 10}=500; N.s/m\)

Angular frequency ratio,

\(\because ω_n=\sqrt{\frac{k}{m}}=\sqrt{\frac{6250}{10}}=25\ rad/s\)

\(\frac{ω}{ω_n}=\frac{25}{25}=1\)

Using equation (3),

\(⇒ A = \frac{{\frac{{{F_0}}}{k}}}{{\sqrt {{{\left[ {1 - {{\left( {\frac{ω }{{{ω _n}}}} \right)}^2}} \right]}^2} + {{\left( {2ξ \frac{ω }{{{ω _n}}}} \right)}^2}} }} ⇒ \frac{F_o}{2 ξ k}\)

\(⇒ A = \frac{F_o}{2 ξ k} ⇒ 0.04=\frac{10}{2 ξ × 6250}⇒ ξ = 0.02\)

Using equation (1),

\(⇒ C=ξ C_c=0.02× 500 \)

⇒ C = 10 N.s/m

Concept:

For a constant pressure process 1 – 2,

P₁ = P₂

Using the ideal gas equation for constant pressure process, 

PV = nRT

⇒ V ∝ T 

\(⇒ \frac{{{V_2}}}{{{V_1}}} = \frac{{{T_2}}}{{{T_1}}}{{;;;;;}} \ldots \left( 1 \right)\)

For a constant volume process 2 – 3,

⇒ V₂ = V₃      ...(2)

Calculation:

Given:

T2 = 3T1/4, T3 = T1/2

Using equation (1),

\(⇒ \frac{{{V_2}}}{{{V_1}}} = \frac{3}{4}{{;;;;;}} \ldots \left( 3 \right)\)

Using equation (2) and (3),

 \(⇒ \frac{{{V_3}}}{{{V_1}}} = \frac{{{V_3}}}{{{V_2}}} \times \frac{{{V_2}}}{{{V_1}}} = 1 \times \frac{3}{4} = \frac{3}{4}\) 

\(⇒ \frac{{{V_3}}}{{{V_1}}} = 0.75\)

Concept:

The general energy equation of the entropy flow,

\({\dot s_{in}} - {\dot s_{out}} + {\dot s_{gen}} = \frac{{ds}}{{dt}}\)

For steady flow. The rate of entropy will be zero.

⇒ \({\dot s_{in}} - {\dot s_{out}} + {\dot s_{gen}} = 0\)

For a plane wall,  

 \(\Rightarrow{\left( {\frac{{\dot Q}}{{{T_1}}}} \right)_{in}} - {\left( {\frac{{\dot Q}}{{{T_2}}}} \right)_{out}} + {\dot s_{gen}} = 0;{{;;;;;}} \ldots \left( 1 \right)\)

Where,

T1 = Inner wall temperature, T2 = Outer wall temperature, Q̇ = Heat transfer rate through the wall,

\({\dot s_{gen}} = The;entropy;generation;in;the;wall\)

Calculation:

Q̇ = 100 kW = 100000 W, T₁ = 127 °C = 400 K, T₂ = 27 °C = 300 K,

Using equation (1),

\(\Rightarrow{\left( {\frac{{100000}}{{400}}} \right)_{in}} - {\left( {\frac{{100000}}{{{3000}}}} \right)_{out}} + {\dot s_{gen}} = 0\)

\(\Rightarrow {\dot s_{gen}}=83.33 \ W/K\)

Concept:

Applying Bernoulli’s equation between the point ‘P’ and ‘R’

\(\frac{{{P_{P}}}}{{\rho g}} + \frac{{V_P^2}}{{2g}} + {Z_P} = \frac{{{P_{R}}}}{{\rho g}} + \frac{{V_R^2}}{{2g}} + {Z_R}{{;;;;;}} \ldots \left( 1 \right)\)

Calculation:

Here, V_P = 0

∵ The points, P, and R are exposed to air so the pressure at these points will be atmospheric pressure.

Using equation (1),

\(\Rightarrow\frac{{{P_{atm}}}}{{\rho g}} + \frac{{V_P^2}}{{2g}} + {Z_P} = \frac{{{P_{atm}}}}{{\rho g}} + \frac{{V_R^2}}{{2g}} + {Z_R}\)

\(\Rightarrow {V_R} = \sqrt {2g\left( {{Z_P} - {Z_R}} \right)} \)

Concept:

When heat flows from a solid wall then total heat transfer from one face to another,

\({\rm{Q}} = \frac{{{{\rm{T}}_1} - {{\rm{T}}_2}}}{{\frac{{{{\rm{b}}_1}}}{{{{\rm{k}}_1}{\rm{A}}}} + \frac{{{{\rm{b}}_2}}}{{{{\rm{k}}_2}{\rm{A}}}}}}\)

For interference temperature calculation,

\(\frac{{{{\rm{T}}_1} - {{\rm{T}}_{\rm{i}}}}}{{\frac{{{{\rm{b}}_1}}}{{{{\rm{k}}_1}{\rm{A}}}}}} = \frac{{{{\rm{T}}_{\rm{i}}} - {{\rm{T}}_2}}}{{\frac{{{{\rm{b}}_2}}}{{{{\rm{k}}_2}{\rm{A}}}}}}{{;;;;;}} \ldots \left( 1 \right)\)

where,

T_i = The interference temperature

Calculation:

Given: 

k1 = k, k2 = 2k, T1 = 600 K, T2 = 300 K, b1 = b2, A1 = A2

From equation (1)

\(⇒\frac{{600 - {{{T}}_{{i}}}{{}}}}{{\frac{1}{{{k}}}}} = \frac{{{{{T}}_{{i}}} - 300}}{{\frac{1}{{2{{k}}}}}}{{}}\)

⇒ T_i = 400 K

Concept:

The pressure drop in the duct is given as, \({\rm{\Delta }}P = \frac{{32\mu L{V_{avg}}}}{{{D^2}}}\)

Where D = 4A/P

Calculation:

Given:

μ = 2 × 10-5 kg/m-s, L = 1 m, Vavg = 1 m/s

\(D = \frac{{4 \times 25 \times 15}}{{2\left( {25 + 15} \right)}} = 18.75~mm = 0.01875~m\),

∴ ΔP =\(\frac{{32 \times \left( {2 \times {{10}^{ - 5}}} \right) \times 1 \times 1}}{{{{\left( {0.01875} \right)}^2}}} = 1.82~{\rm}\) Pa

Concept:

If there is no frictional losses, then the axial thrust is zero and Vf1 = Vf2 and Vr1 = Vr2 

The specific work W = u(V_w1 + V_w2)

Calculation:

Given: 

V₁ = 100 m/s, V_r1 = 78 m/s, u = 25 m/s, α = 25°, V₂ = 58.6 m/s, V_r2 = 78 m/s

\(cos\alpha_1 =\frac{ V_{w1}}{V_1} \Rightarrow cos25 =\frac{ V_{w1}}{100}\)

\({V_{{W_1}}} = {100}\cos {25} = 90.63~\frac{m}{s}\)

\(sin\alpha_1 =\frac{ V_{f1}}{V_1} \Rightarrow sin25 =\frac{ V_{f1}}{100}\)

\({V_{{f_1}}} = {100}\sin {25} = 42.26~\frac{m}{s}\)

\({V_{w2}} = \sqrt {V_2^2 - V_{f2}^2} = \sqrt {{{58.6}^2} - {{42.26}^2}} = 40.59;m/s\)

\(\therefore W = U\left( {{V_{{W_1}}} + {V_{{W_2}}}} \right) = 3280.5~\frac{J}{{kg}}\)

Concept:

F₁₁ + F₁₂ = 1

But F₁₁ = 0

∴ F₁₂ = 1

Now from reciprocating law

A₁ F₁₂ = A₂ F₂₁

Calculation:

\(\Rightarrow 4\pi r_1^2 \times I = 4\pi r_2^2 \times {F_{21}}\)

\(\Rightarrow {F_{21}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2} = {\left( {\frac{2}{3}} \right)^2} = \frac{4}{9}\)

Concept:

In counter-flow heat exchanger, the flowing streams flow in the opposite direction.

LMTD = \(\frac{{{\rm{Δ }}{{\rm{T}}_{\rm{i}}} - {\rm{Δ }}{{\rm{T}}_{\rm{o}}}}}{{\ln \frac{{{\rm{Δ }}{{\rm{T}}_{\rm{i}}}}}{{{\rm{Δ }}{{\rm{T}}_{\rm{o}}}}}}}\) …………….(1)

ΔT_i = T_hi - T_co  and ΔT_o = T_ho - T_ci

Heat loss by hot flow = heat gain by cold flow

Therefore, m_h C (T_hi - T_ho) = m_c C (T_co – T_ci)……….(2)

 

Calculation:

Given:

For Hot stream, m_h = 25 liter/s, C = 4186 J/kgK, T_hi = 46°C, T_ho = ?

For Cold stream, m_c = 19 liter/s, C = 4186 J/kgK, T_ci = 10°C, T_co = 38 °C

From equ. (2)

\(25 \times 4186 \times \left( {46 - {{\rm{T}}_{{\rm{ho}}}}} \right) = 19 \times 4186 \times \left( {38 - 10} \right)\)

\(\therefore {{\rm{T}}_{{\rm{ho}}}} = 24.72{\rm{}}^\circ {\rm{C}}\)

\({\rm{Δ }}{{\rm{T}}_{\rm{i}}} = 46 - 38 = 8{\rm{}}^\circ {\rm{C~and~Δ }}{{\rm{T}}_{\rm{o}}} = 24.72 - 10 = 14.72{\rm{}}^\circ {\rm{C}}\)

From equ. (1)

LMTD = \(\frac{{8 - 14.72}}{{\ln \frac{8}{{14.72}}}} = 11.02{\rm{}}^\circ {\rm{C}}\)

Concept:

The efficiency of diesel engine is,

\({η _{diesel}} = 1 - \frac{1}{{{\gamma r ^{\gamma - 1}}}}\left( {\frac{{{ρ ^\gamma } - 1}}{{ρ - 1}}} \right)\)

where, r = compression ratio, ρ = cut-off ratio

Calculation:

Given:

\(Compression;ratio,;r = \frac{{{V_1}}}{{{V_2}}} = 17\)

\({V_3} - {V_2} = \frac{{10}}{{100}}\left( {{V_1} - {V_2}} \right)\)

\(\Rightarrow \frac{{{V_3}}}{{{V_2}}} - 1 = 0.1\left( {\frac{{{V_1}}}{{{V_2}}} - 1} \right)\)

\(\Rightarrow \frac{{{V_3}}}{{{V_2}}} - 1 = 0.1\left( {17 - 1} \right)\)

\(\Rightarrow \frac{{{V_3}}}{{{V_2}}} = 2.6\)

\(\therefore Cut - off;ratio,;ρ = \frac{{{V_3}}}{{{V_2}}} = 2.6\)

Efficiency is:

\({η _{diesel}} = 1 - \frac{1}{{{\gamma r ^{\gamma - 1}}}}\left( {\frac{{{ρ ^\gamma } - 1}}{{ρ - 1}}} \right)\)

\(η = 1 - \frac{1}{{\left( {1.4} \right){{\left( {17} \right)}^{1.4 - 1}}}}\left( {\frac{{{{2.6}^{1.4}} - 1}}{{2.6 - 1}}} \right)\)

η = 0.596 = 59.6%

Concept:

Duration of project = summation of days on the critical path

Free float (4 – 6) = E S6 – E S4 – D4-6

= 8 – 2 – 4 = 2

The critical path is 1 – 2 – 5 – 6 – 7

∴ Duration of project = summation of days on critical path =13 days.

Critical path: The maximum time-consuming path from the first event to the last event.

Calculation:

Given:

Annual demand in units, r = 8000 × 300 units/year

Production rate of bearings, K = 12000 × 300 unit/years

Set up cost per production run, C_o = Rs. 500/-

The holding cost of a bearing, C_c = Rs. 0.2 × 12 per years

\(EOQ = \sqrt {\frac{{2{C_0}r}}{{{C_c}\left[ {1 - \frac{r}{k}} \right]}}} = 54,772\)

Frequency = EOQ/r  

\(\therefore Frequency;of;production;run = \frac{{54,772}}{{8,000}}\)

= 6.84 days

Concept:

\(V = \frac{\pi }{4}{d^2}h \Rightarrow h = \frac{{4v}}{{\pi {d^2}}}\)

\({A_s} = \pi dh + \frac{\pi }{4}{d^2}\)

\(\Rightarrow {A_s} = \pi d\left( {\frac{{4v}}{{\pi {d^2}}}} \right) + \frac{\pi }{4}{d^2}\)

\(\Rightarrow {A_s} = \frac{{4v}}{d} + \frac{\pi }{4}{d^2}\)

The rate of solidification in riser is minimum when A_s is minimum

\(\therefore \frac{{d{A_s}}}{{dd}} = \frac{{ - 4v}}{{{d^2}}} + \frac{{2\pi }}{4}d = 0\)

\(\Rightarrow \frac{{4v}}{{{d^2}}} = \frac{{2\pi }}{4}d\)

\(\Rightarrow \frac{{4{d^2}h}}{{{d^2}}} = \frac{{2\pi }}{4}d\)

\(\Rightarrow \frac{h}{d} = \frac{1}{2}\)

Calculation:

\(D = \frac{{{d_1}}}{2} + R + \frac{{{d_2}}}{2}\)

R = AB

\(AC = \frac{{{d_1}}}{2} + \frac{{{d_2}}}{2} = 50mm\)

\(BC = \left( {{H_2} + \frac{{{d_2}}}{2}} \right) - \left( {{H_1} + \frac{{{d_1}}}{2}} \right)\)

⇒ BC = 25 mm

\(\therefore R = AB = \sqrt {A{C^2} - B{C^2}} = 43.30mm\)

∴ D = 20 + 43.30 + 30 = 93.30 mm

Explanation:

In orthogonal cutting,

(i) Reduction in friction angle decreases the cutting force

(ii) Reduction in friction angle decreases chip thickness

According to Merchant's theory,

\(ϕ = \frac{{\rm{\pi }}}{4} + \frac{{\rm{α }}}{2} - \frac{{\rm{β }}}{2}\) where, ϕ = shear angle, α = rake angle, β = friction angle

As β decreases. ϕ increases because α is constant.

• If all other factors remain the same, a higher shear angle results in a smaller shear plane area. Since the shear strength is applied across this area, the shear force required to form the chip will decrease when the shear plane area is decreased.

Now \(\frac{t}{{{t_c}}} = \frac{{sinϕ }}{{\cos \left( {ϕ - α } \right)}}\)

\({t_c} = \frac{{\cos \left( {ϕ - α } \right)}}{{sinϕ }} \times t\)

• As ϕ increases, cos (ϕ -α) will decrease and Sinϕ will increase. This chip thickness will decrease.
• Hence from the above options Q and S, both are correct..

Concept:

 Heat supplied to the metal (H_s) = Heat dissipated to the base (H_d) + Heat required for the melting (H_m)

 I²R Ƭ = H_d + H_m\( \times {\rm{V}}\)

Given:

 t = 3 mm, I = 10000 A, Ƭ = 0.2 sec, H_d = 1000 J, H_m = 20 J, V = 1 mm³, R = 0.0002

Calculation:

 \({\left( {10000} \right)^2} \times 0.0002 \times 0.2 = 1000 + 20 \times {\rm{V}}\)

\( \Rightarrow 4000 = 1000 + 20{\rm{V}}\)

\( \Rightarrow {\rm{V}} = 150~{\rm{m}}{{\rm{m}}^3}\)