GATE ME 2013 Official Paper (Previous Year Paper)

The correct answer is Option 1 i.e Specificity

Explanation.

The statement given here is about antonyms. Let's look at the meaning of the first two words given in the sentence. 

• Universalism: The belief that all humankind will eventually be saved.
• Particularism: Exclusive attachment to one's own group, party, or nation.

Here we find that these two words are antonyms as they have opposite meanings. Therefore the next part also follows with antonyms. Hence the correct answer is the word opposite in meaning to Diffuseness.

Let's look at the meaning of the given word and the marked option. 

• Diffuseness​: Widely spread or scattered.
• Specificity: The quality of being unique to one subject.

Here we find that both the words have opposite meanings and therefore Option 1 Specificity is the correct answer. 

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

• Neutrality: The state of not supporting or helping either side.
• Generality: A statement or principle having general rather than specific validity.
• Adaptation: The action or process of adapting or being adapted.

The correct answer is option 1) i.e. would fly

Explanation:

• The given sentence is a conditional sentence. There are 5-6 categories in conditional sentences.

1. Zero condition
2. First Condition
3. Second Condition
4. Third condition
5. Imaginary Condition
6. A few exceptional cases

• If we talk about the Imaginary condition, it describes a condition that is not possible in real life or it has very few possibilities of its occurrence.
• If I were a lion, I would eat you right now.​
• Syntax of Imaginary sentences: If + Sub + were/had + other words, Sub + would + Verb¹  + other words.

Important Point:

• There is one more thing to take care of while dealing with conditional sentences and i.e. Inversion of the verb.
• The inversion of the verb is to use the auxiliary verb before the subject. We use the concept of inversion in many cases and 'conditional sentences' is one of them.
• I am not working on this project.
• Am I not working on this project? (Inversion)
• If You had informed me, I would have come to you.
• Had you informed me, I would have come to you. (Inversion)

Conclusion: 

• The sentence given in the question is an imaginary condition and inversion has been done. So we must use **would + Verb¹ **in the second part.
• Hence, the completed sentence would be: Were you a bird, you would fly in the sky.​

The correct answer is Option 2 i.e Lowest.

Let's look at the meaning of the given word

• Nadir: The lowest or most unsuccessful point in a situation.

From the above word meaning, we find that Nadir is closest in meaning to Lowest which is Option 2

Let's look at the meaning of all the given options

• Highest: Great, or greater than normal, in quantity, size or intensity.
• Lowest: Of less than average height from top to bottom or to the top from the ground.
• Medium: About hallway between two extremes of size or another quality; average.

The correct answer is Option 3 i.e She is an European

Explanation.

A and An are the indefinite articles used in English. As a rule:

• A is used before words starting with a consonant sound
• An is used before words starting with a vowel sound

In the given sentence the word European starts with a consonant sound i.e 'Y' sound. 

Since the word starts with a consonant sound the article before the word European should be A​. Therefore Option 3 is grammatically incorrect and thus the correct answer. 

Important Point.

One should always be aware that:

• The article A or An should be used depending upon the starting sound and not the letter.
• Example: An hour. Here since the 'h' is silent the stating sound is that of a vowel hence we use An.

Concept:

The sum of ‘n’ terms of an AP

\(S_n = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\)

Calculation:

Given:

AP is 44, 42,- - - - - 2.

n = 22, a = 2 and d = 2

The sum of ‘n’ terms of an AP

\(S_n= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\)

S_n = 11[4 + (21 × 2)] = 11 × 46 = 506

Explanation:

The sample space for the event is (S) = 90 (i.e. from 10 to 99)

2-digit integers divisible by 7 = 14, 21, 28,..................91, 98.

i.e. Total 2-digit integers divisible by 7 = 13

∴ Total 2-digit integers not divisible by 7 = 90 - 13 = 77

∴ Required Probability =  \(\frac{Desired~Outcome}{Sample~Space}\) =  \(\frac{77}{90}\)

Concept:

Average speed is given by:

\(Average;speed = \frac{Total~ distance~ covered}{{Total;time;taken}}\)

Calculation:

Given:

Let the total distance covered by ‘D’

Distance covered by train = D/2

Distance covered by bus = D/4

Distance covered by cycle = D/4

Now, \(Average;speed = \frac{D}{{Total;time;taken}}\)

\(V_{avg}= \frac{D}{{\left( {\frac{{\frac{D}{2}}}{60} + \frac{{\frac{D}{4}}}{30} + \frac{{\frac{D}{4}}}{{10}}} \right)}} = \frac{1}{{\frac{1}{{120}} + \frac{1}{{120}} + \frac{1}{{40}}}} = \frac{{120}}{5} = 24~\frac{{km}}{{hr}}\)

 

As the Average speed does not depends on total distance D.

Let the total distance be LCM(60,30,10) = 60 km

time taken by car = \(\frac {30}{60} = 0.5 ~hour\)

time taken by bus = \(\frac{15}{30}=0.5~hour\)

time taken by cycle = \(\frac{15}{10}=1.5~hour\)

Total time taken = 0.5 + 0.5 + 1.5 = 2.5 hour

Average speed is :

\(Average;speed = \frac{Total~ distance~ covered}{{Total;time;taken}}= \frac{60}{2.5}=24 ~km/hour\)

Calculation:

Given:

\(\frac{1}{{\sqrt 1 + \sqrt 2 }} + \frac{1}{{\sqrt 2 + \sqrt 3 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }} + \ldots + \frac{1}{{\sqrt {80} + \sqrt {81} }}\)

We can write as,

\( = ;\frac{1}{{\sqrt 2 + \sqrt 1 }} \times \left( {\frac{{\sqrt 2 - \sqrt 1 }}{{\sqrt 2 - \sqrt 1 }}} \right) + \frac{1}{{\sqrt 3 + \sqrt 2 }} \times \left( {\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}} \right) + \frac{1}{{\sqrt 4 + \sqrt 3 }} \times \left( {\frac{{\sqrt 4 - \sqrt 3 }}{{\sqrt 4 - \sqrt 3 }}} \right) + \ldots + \frac{1}{{\sqrt {81} + \sqrt {80} }} \times \left( {\frac{{\sqrt {81} - \sqrt {80} }}{{\sqrt {81} - \sqrt {80} }}} \right)\)

∵  a² – b² = (a + b) × (a – b)

\( = ;\frac{{\sqrt 2 - \sqrt 1 }}{{{{\left( {\sqrt 2 } \right)}^2} - {{\left( {\sqrt 1 } \right)}^2}}} + \frac{{\sqrt 3 - \sqrt 2 }}{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} + \frac{{\sqrt 4 - \sqrt 3 }}{{{{\left( {\sqrt 4 } \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}}} + \ldots + \frac{{\sqrt {81} - \sqrt {80} }}{{{{\left( {\sqrt {81} } \right)}^2} - {{\left( {\sqrt {80} } \right)}^2}}}\)

\( = ;\frac{{\sqrt 2 - \sqrt 1 }}{{2 - 1}} + \frac{{\sqrt 3 - \sqrt 2 }}{{3 - 2}} + \frac{{\sqrt 4 - \sqrt 3 }}{{4 - 3}} + \ldots + \frac{{\sqrt {81} - \sqrt {80} }}{{81 - 80}}\)

\( = \sqrt 2 - \sqrt 1 + \sqrt 3 - \sqrt 2 + \sqrt 4 - \sqrt 3 + \ldots \sqrt {81} - \sqrt {80} \)

After cancellation of suitable terms,

We get,

\( = \sqrt {81} - \sqrt 1 \Rightarrow 9 - 1 = 8\)

Concept:

Total cost of erection = Rs. (W × T)

where, W = Labour wages in Rs., T = working hours.

Calculation:

Given:

Current erection cost(W × T) = Rs. 13,200, Increase in wages = \(\frac{W}{5}\), Decrease in working hours = \(\frac{T}{24}\)

Now, we know that

Total cost of erection = W × T

Revised wages = W + \(\frac{W}{5}\) = 1.2 W, Revised time = T - \(\frac{T}{24}\) = \(\frac{23T}{24}\)

∴ New cost of erection = 1.2W × \(\frac{23T}{24}\) = 1.2 × \(\frac{23}{24}\) × 13200 

∴ New cost of erection = 1.15 × 13200 = Rs. 15,180.

The correct answer is Option 4 i.e No adversity justifies giving up hope.

Explanation

The passage talks about the following things:

• How a man was on the verge of suicide, after several defeats.
• He observed a spider achieve the near-impossible goal despite countless failures.
• He drew inspiration from the spider that failed a countless number of times but didn't give up hope.

From the above passage and the above points, we find that the passage talks about not giving up despite our multitude of failures. 

The answer that resonates with the meaning of the passage is, therefore, Option 4 and thus the correct answer.

Explanation:

Order of a differential equation

Order is defined by the highest derivative present in the equation.

\(\frac{{{\partial ^2}u}}{{\partial {x^2}}}\) is the highest order derivative here. So, order = 2.

Linear and non-linear differential equation

Linear differential equation

A differential equation is said to be linear when it possesses the following properties:

a) Dependent variable and its derivative should have power ‘1’

b) Dependent variable and its derivatives can have a product with independent variable

c) Dependent variable and its derivatives can’t have product

Non-linear differential equation

 

∴ The given equation is non-linear because the product of 'u' with \(\frac{{\partial u}}{{\partial x}}\) is present.

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

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.

Explanation:

\({\rm{Number;of;intervals}} = \frac{{{\rm{b}} - {\rm{a}}}}{{\rm{h}}}{\rm{;}}\)

where,

b is the upper limit, a is the lower limit, h is the step size

According to the trapezoidal rule

\(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \frac{{\rm{h}}}{2}\left[ {{{\rm{y}}_{\rm{o}}} + {{\rm{y}}_{\rm{n}}} + 2\left( {{{\rm{y}}_1} + {{\rm{y}}_2} + {{\rm{y}}_3}{\rm{;}} \ldots } \right)} \right]\)

It fits for 1-degree polynomial.

According to Simpson's 1/3 rule

\({\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \frac{h}{3}\left[ {\left( {{y_o} + {y_n}} \right) + 2\left( {{y_2} + {y_4} + {y_6} + \ldots } \right)} \right] + 4\left[ {{y_1} + {y_3} + {y_5} + \ldots } \right]}\)

It fits for 2-degree polynomial.

According to Simpson's 3/8 rule

\(\mathop \smallint \limits_a^b ydx = \frac{{3h}}{8}\left[ {({y_0}+y_n) + 3({y_1} + {y_2}+{y_4} +{y_5}..)+ 2({y_3} + {y_6} + {y_9}..)} \right]\)

It fits for 3-degree polynomial.

Concept:

Stress is defined as intensity of internal resisting force developed at a point against the deformation due to applied external load.

Mathematically,

Stress (σ) is:

\(\sigma = \frac{P}{A}\)

where P is applied load and A is the cross-sectional area

Explanation:

As stress depends load applied and cross-sectional area ∴ stress-induced in the above case will be 

\(\sigma = \frac{P}{A}\)

Concept:

Strain energy stored in a bolt of length l, cross-section area A, modulus of elasticity E and is subjected to a load P is given by:

\(U = {\frac{{{P^2}l}}{{2AE}}} \)

Calculation:

Given:

E_A = E_B (∵ material is same), P_A = P_B, L_A = L_B, U_A = 4U_B, d_A = 12 mm

\(\frac{{{U_A}}}{{{U_B}}} = \frac{{{{\left( {\frac{{{P^2}L}}{{2AE}}} \right)}_A}}}{{{{\left( {\frac{{{P^2}L}}{{2AE}}} \right)}_B}}} = \frac{{{A_B}}}{{{A_A}}} \)

\(4 = \frac{{d_B^2}}{{d_A^2}}\)

d_B = 2 × d_A = 24 mm

Concept:

Corriolis  acceleration (α_c) is acted on block is

α_c = 2vω 

Radial acceleration (α_r) towards center is  

α_r = rω²

Calculation:

Given: 

ω = 2 rad/sec, v = 0.75 m/s, r = 1 m

Corriolis  acceleration (α_c) is 

αc = 2 × v × ω

 α_c = 2 × 0.75 × 2 = 3 m/s² 

Radial acceleration (α_r) is:

αr = r × ω²

 α_r = 1 × 2² = 4 m/s²

Resultant acceleration (a) is:

\(a = \sqrt {{{ {\alpha}}_{c} ^2} + {{\alpha}}_{r}^2}=\sqrt {{3^2} + {4^2}} = 5~m/s^2\)

Tangential acceleration = α × r

Concept:

Wall shear stress is given by:

\({\rm{}}{{\rm{\tau }}_{\rm{w}}} = - \frac{{\partial {\rm{P}}}}{{\partial {\rm{x}}}}\times\frac{{\rm{R}}}{2}{\rm{}}\)

where \(\frac{{\partial {\rm{P}}}}{{\partial {\rm{x}}}} = pressure~gradient\), R = radius of pipe = D/2

for length L, the pressure drop is ΔP

∴ Pressure gradient is \( \frac{{\partial {\rm{P}}}}{{\partial {\rm{x}}}} = \frac{{{\rm{Δ P}}}}{{{\rm{L}}}}~\)

Wall shear stress is 

\({\rm{}}{{\rm{\tau }}_{\rm{w}}} = - \frac{{\partial {\rm{P}}}}{{\partial {\rm{x}}}}\times\frac{{\rm{R}}}{2}{\rm{}}= \frac{{{\rm{Δ P}}}}{{\rm{L}}}\times \frac{{\rm{D}}}{{2 \times 2}} = \frac{{{\rm{Δ PD}}}}{{4{\rm{L}}}}\)

Concept:

The specific humidity (ω) is given by:

ω \( = \frac{{0.622 ~{P_v}}}{{P - {P_v}}}\)

Relative humidity (ϕ) is given by:

 \(ϕ = \frac{{{{\rm{P}}_{\rm{v}}}}}{{{{\rm{P}}_{{\rm{vs}}}}}}\)

Calculation:

Given:

P = 1 bar = 100 kPa, P_vs = 4.25 kPa, ϕ = 0.70

Relative humidity (ϕ) is given by:

\(ϕ = \frac{{{{\rm{P}}_{\rm{v}}}}}{{{{\rm{P}}_{{\rm{vs}}}}}}\)

\(0.70 = \frac{{{{\rm{P}}_{\rm{v}}}}}{{4.25}}\) ⟹ P_v = 2.975 kPa

Specific humidity (ω) is

\(\omega = \frac{{0.622 ~{P_v}}}{{P - {P_v}}}\)

\(\omega= \frac{{0.622 ~{P_v}}}{{P - {P_v}}}= \frac{{0.622 \times2.975}}{{\left( {100 - 2.975} \right)}} = 0.019~kg ~water~ vapour/kg ~dry~ air\)

Explanation

The conductivity of the wall, k = k₀ + bT

And T₂ > T₁

Since conductivity is a function of temperature, As we move from left to right temperature increases. Thus thermal conductivity of the wall also increases.

Using Fourier's law of heat conduction.

\(Q = kA\left( {\frac{{dT}}{{dx}}} \right)\)

As thermal conductivity, k is increasing so to keep the heat transfer constant the temperature gradient \(\left( {\frac{{dT}}{{dx}}} \right)\)  must decrease.

Concept:

In a rolling process, during travel of material through the rollers, a vertical force (compression) and a sliding force (shear force) will act on the grains.

Friction Welding : It is comes under Solid welding. It is used to join the cylindrical dissimilar materials 

Gas Metal Arc Welding : It comes under Liquid welding.The electrode is used to weld the metals. The electrode used here is consumable. 

**Tungsten Inert Gas Welding :**It comes under Liquid welding.The electrode is used to weld the metals.The electrode used here is non-consumable.

Electroslag Welding : It also comes under Liquid welding. The two phenomenon is used here i.e. Arc welding and Resistance welding. The thick plates are joined with the help of this welding.

Important Points

Friction Welding

Introduction:

• In friction welding coalescence is produced from the heat generated by rotating one piece against another under controlled axial pressure.
• Although this process is sometimes considered a solid-state welding process, the two surfaces actually reach the melting temperature and the adjacent material becomes plastic.
• At this point the relative motion between the two is stopped and a forging pressure is applied, upsetting the joint slightly. This pressure may equal or exceed the pressure during heating and its value depends on the surface material being welded.
• In the friction welding process, no special preparation of the weld surfaces is necessary except to have them clean and reasonably smooth.
• Equipment must be capable of holding one piece securely while the other rotates, then stopping rotation quickly when the welding temperature is reached.

Limitation: 

• The primary limitation of the process is that few configurations can be welded.

Advantages: 

• Advantages claimed for the process are simple equipment, rapid production of sound welds, little preparation required for joints, and low total energy requirements.
• Also, many dissimilar metals can be welded.
• Friction welding finds considerable use in welding plastics.

​Gas metal arc welding (GMAW)

Gas metal arc welding (GMAW) or Metal inert gas arc welding (MIG): In this process the arc is formed between a continuous, automatically fed, metallic consumable electrode and welding job in an atmosphere of inert gas, and hence this is called metal inert gas arc welding (MIG) process.

Tungsten Inert Gas 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.

Electroslag welding

Electroslag welding is a welding process in which the heat is generated by an electric current passing between the consumable electrode (filler metal) and the workpiece through a molten slag covering the weld surface. In this process, the coalescence is formed by molten slag and the molten metal pool remains shielded by molten slag.

Electroslag welding is capable of welding plates with thickness ranging from 50 mm to more than 900 mm and welding is done in one pass.

Concept:

Best wire size (diameter) is given by: 

\(d = \frac{p}{2}{\rm{sec}}\left( {\frac{α }{2}} \right)\)

where p = pitch, α = thread angle

Calculation:

Given:

p = 2 mm, α = 60° 

Best wire size (diameter) is:

\(d = \frac{p}{2}{\rm{sec}}\left( {\frac{α }{2}} \right)\)

 \(d=\frac{2}{2} \times \sec \left( {30} \right) = 1.154 ~{\rm{mm}}\)

Concept:

Utilization factor (ρ) is given by:

ρ = \(\frac{{\rm{\lambda }}}{{\rm{\mu }}}\)

where λ = Arrival rate, μ =  service rate

Waiting time in queue is given by:

W_q = \(\frac{{{{\rm{L}}_{\rm{q}}}}}{{\rm{\lambda }}}\)  where L_q = number of customers in queue

 Number of customer (Lq) in queue isgiven by:

\(L_q=\frac{{{{\rm{\rho }}^2}}}{{1 - {\rm{\rho }}}} \)

Calculation:

Given:

Arrival rate (λ) = 50 cust/hour, service rate (μ) = 1 cust/min = 60 cust/hour

Utilization factor (ρ) is 

\(\rho=\frac{{\rm{\lambda }}}{{\rm{\mu }}}=\frac{{50}}{{60}} = \frac{5}{6} = 0.8333\)

Number of customer (Lq) in queue is

 \(L_q=\frac{{{{\rm{\rho }}^2}}}{{1 - {\rm{\rho }}}} = \frac{{{{0.8333}^2}}}{{1 - 0.8333}} = \frac{{25}}{6}{\rm{customers}}\)

Waiting time in queue is

 \(W_q=\frac{{{{\rm{L}}_{\rm{q}}}}}{{\rm{\lambda }}}=\frac{{\frac{{25}}{6}}}{{50}} = \frac{1}{{12}}~{\rm{hours}}\)

The average waiting time in the queue in minutes = \(\frac{1}{{12}} \times 60 = 5 ~{\rm{min}}\)

The number of customers in the system is given by:

L_s = \(\frac{{\rm{\rho }}}{{1 - {\rm{\rho }}}}\)

Waiting time in the system is given by:

W_s = \(\frac{{{{\rm{L}}_{\rm{s}}}}}{{\rm{\lambda }}}\)

Explanation:

In exponential smoothing method of forecast the forecast for the next period is equal to 

Ft = α Dt-1 + (1 - α) Ft-1 

If we further expand the expression

Ft = α Dt-1 + (1 - α) (α Dt-2 + (1-α) Ft-2

Ft = α Dt-1 + α (1-α ) Dt-2 + (1 - α )2 Ft-2

If we put the value of α = 0.8 (Close to 1) then,

Ft = 0.8 Dt-1 + 0.16 Dt-2 + 0.04 Ft-2 

It can be observed from the above expression that recent data is given more weightage compared to the previous data, therefore, the higher value of exponential smoothing constant (Close to 1) is used for changing pattern of demand or to follow recent demand.

Concept:

The Material removal rate is given by:

MRR = f × d × v

where f = feed, d = depth of cut, v = velocity

Velocity (v) is given by:

v = \(\frac{{{\rm{\pi DN}}}}{{60}}\) 

Calculation:

Given:

D = 200 mm, f = 0.25 mm/rev, d = 4 mm, N = 160 rpm

Velocity (v) is 

 \(v=\frac{{{\rm{\pi DN}}}}{{60}}=\frac{{\pi \times 200 \times 160}}{{60}}= 1675.51 ~mm/s\) 

The material removal rate is:

MRR = f × d × v

MRR \(= \left( {0.25} \right)\times \left( 4 \right) \times1675.51= 1675.51 ~mm^3/s\)

Concept:

The time required for solidification is:

t = \({\rm{C}}{\left( {\frac{{\rm{V}}}{{\rm{A}}}} \right)^2}\)

Calculation:

Given:

t1 = 5 min,

Let weight be 'W'.

W₂ = 8W₁

(ρVg)₂ = 8(ρVg)₁

V₂ = 8V₁

\({\left( {{{\rm{a}}_2}} \right)^3} = 8{\left( {{{\rm{a}}_1}} \right)^3} \Rightarrow {{\rm{a}}_2} = 2{{\rm{a}}_1}\)

For the first cube:

t = \({\rm{C}}{\left( {\frac{{\rm{V}}}{{\rm{A}}}} \right)^2}\)

\(5 = {\rm{C}}{\left( {\frac{{{\rm{a}}_1^3}}{{6{\rm{a}}_1^2}}} \right)^2} = \frac{{{\rm{a}}_1^2}}{{36}}{\rm{C}}\)

For the second cube,

t = \({\rm{C}}{\left( {\frac{{\rm{V}}}{{\rm{A}}}} \right)^2}\)

\({{\rm{t}}_2} = {\rm{C}}{\left( {\frac{{{\rm{a}}_2^3}}{{6{\rm{a}}_2^2}}} \right)^2} = \frac{{{\rm{a}}_2^2}}{{36}}{\rm{C}} = \frac{{{{\left( {2{{\rm{a}}_1}} \right)}^2}}}{{36}}{\rm{C}} = \frac{{4{\rm{a}}_1^2}}{{36}}{\rm{C}}\)

\(∴ \frac{{{{\rm{t}}_2}}}{{{{\rm{t}}_1}}} = \frac{4}{1} \Rightarrow \frac{{{{\rm{t}}_2}}}{5} = \frac{4}{1}\)

∴ t₂ = 4 × t₁ = 4 × 5 = 20 min.

Explanation:

Various properties of a material are:

• Toughness is defined as the ability of the material to absorb energy before fracture takes place. This property is essential for machine components which are required to withstand impact loads. Tough materials have the ability to bend, twist or stretch before failure takes place.
• Hardness is defined as the resistance of a material to penetration or permanent deformation. It usually indicates resistance to abrasion, scratching, indentation, cutting or shaping. It is a surface phenomenon.
• Proof resilience is the quantity of strain energy stored in a body when strained up to the elastic limit (ability to store or absorb energy without permanent deformation).
• Ductility: The property of the material that allows it to be drawn into wires or elongated before failure is known as ductility.
• Malleability: Malleability is the property of a metal by which it can Be extended in any direction by hammering, rolling, etc. without causing a rupture. Lead is an example of a malleable metal.
• Brittleness: The property of a material that produces fracture without any appreciable deformation is known as brittleness. It is opposite of toughness.
• Elasticity: It is the property of a material to regain its original shape after deformation when the external forces are removed.
• Tenacity: The tenacity of a metal is its ability to resist the effect of tensile forces without rupturing. Mild steel, Wrought Iron and copper are some examples of tenacious metals.

Concept:

Hydraulic or blading efficiency of Pelton turbine

\({\eta _h} = \frac{{2u\left( {{V_1} - u} \right)\left( {1 + k\cos{\beta _2}} \right)}}{{{V_1}^2}} = 2\left( {ρ - {ρ ^2}} \right)\left( {1 + k\cos{\beta _2}} \right)\)

where u = bucket speed, V₁ = jet speed

For maximum efficiency: 

\(\frac{d}{{dρ }}{\eta _h} = 0; \to ;ρ = \frac{u}{{{V_1}}} = \frac{1}{2}\)

where ρ is bucket to jet-speed ratio.

So for maximum power:

\(ρ = \frac{u}{V_1} = \frac{1}{2} = 0.5\)

Explanation

To satisfy the steady-state condition while generating heat all the heat generated in the slab must be convected to the fluid both sides of the slab.

• One direction conduction, T = f(x) only
• Uniform conductivity, k = C
• Uniform heat generation throughout, \(\dot q = C\)

Generalised 3D conduction equation is given by, 

\(\frac{{{\partial ^2}T}}{{\partial {x^2}}} + \frac{{{\partial ^2}T}}{{\partial {y^2}}} + \frac{{{\partial ^2}T}}{{\partial {z^2}}} + \frac{{\dot q}}{k} = \frac{1}{\alpha }\left( {\frac{{\partial T}}{{\partial \tau }}} \right)\)

Since there only steady-state conduction in X-direction,

\(\frac{{{\partial ^2}T}}{{\partial {x^2}}} + \frac{{\dot q}}{k} = 0\)

\(\frac{{{\partial ^2}T}}{{\partial {x^2}}} = - \frac{{\dot q}}{k}\)

Integrating

\(\frac{{dT}}{{dx}} = - \frac{{\dot q}}{k}x + {C_1}..........\left( 1 \right)\)

\(T = - \frac{{\dot q}}{k} \times \frac{{{x^2}}}{2} + {C_1}x + {C_{2}} ( {Parabolic~Temperature~distribution}).......( 2 )\)

Applying boundary condition,

At x = 0 , T = 0° C

C₂  = 0

At x = L , T = 100° C

\({C_1} = \frac{{100 + \frac{{\dot q{L^2}}}{{2k}}}}{L}\)

For maximum temperature Equating (1) to zero.

\({C_1} = \frac{{\dot qx}}{k}\)

\(x = \frac{{\left( {100 + \frac{{\dot q{L^2}}}{{2k}}} \right)k}}{{L\dot q}}\)  

\(x = \frac{{100k}}{{L\dot q}} + \frac{L}{2}\)

Hence maximum temperature occurs somewhere at \(x > \frac{L}{2}\)

If we put this value of x in equation (2),

We get T_max > 100° C

Since Maximum temperature in the slab is greater than 100° C, Hence heat can’t flow 100° C to more than 100° C.Thus the statement, direction of heat transfer from 100° C to  0° C is wrong.

Concept:

Work done in a reversible isothermal process is given by 

\(W = C\ln \frac{{{P_1}}}{{{P_2}}} = C\ln \frac{{{V_2}}}{{{V_1}}}\)

Where C = P₁V₁ = P₂V₂

P is the pressure and V is the volume of the gas. 

Calculation:

Given:

P1 = 1 bar = 100 kPa, V1 =  5 m3 and P₂ = 5 bar 

C = P1V1 = 1 bar × 5 m3 = 100 kPa × 5 m3 = 500 kJ

\(W = C\ln \frac{{{P_1}}}{{{P_2}}} = 500\ln \left( {\frac{1}{5}} \right) = -804.718 ;kJ\)

-ve sign shows that work is done on the system.

Concept:

For a long thin cylinder,

Hoop stress is given by:

\({σ _{hoop}} = \frac{{Pd}}{{2t}}\)

Longitudinal stress is given by:

\({σ _{long}} = \frac{{Pd}}{{4t}}\)

Required ratio is:

\(\frac{{{σ _{hoop}}}}{{{σ _{long}}}} = 2\)

For sphere, hoop stress and longitudinal stress is equal

σ_h = σ_long =  \(\frac{{{\bf{Pd}}}}{{4{\bf{t}}}}\)

Concept:

Whirling frequency of shaft is,

f = n² f_c  

where, n = no. of loop (node + 1), f_c = critical frequency

Since it is simply supported critical speed and first frequency = f_n

Calculation:

Given:

N = 1800 rpm (for 2 node), n = 3, Nc = ?

1800 = 3² × (N_c)

∴ N_c = 200 rpm

Concept:

According to Grashof’s Law:

(1) S + L ≤ P + Q (Class I mechanism)

There will be at least one complete revolution between the two links.

• If the shortest link is fixed: Double Crank Mechanism
• If any of the adjacent links of the shortest link is fixed: Crank rocker mechanism
• If the link opposite to shortest link is fixed: Double Rocker Mechanism

(2) S + L > P + Q (Class II mechanism)

• Only double rocker mechanism is possible.

Calculation:

Given:

PQ = 2.0 m, QR = 3.0 m, RS = 2.5 m and SP = 2.7 m

Length of shortest link = PQ = 2 m, Length of largest link = QR = 3 m,

Grashof's Law: The sum of lengths of the shortest and longest link should be less than or equal to the sum of lengths of the other two links.

Here \(s+l\le p+q \ 2+3 \le 2.5+2.7\)

Since the smallest link is the best link for rotation and also governs input and output. So, use the smallest link as a coupler to obtain double rocker mechanism.

Hence, for obtaining Rocker – Rocker mechanism, the link opposite to smallest link (RS) must be fixed.

Concept:

The probability density function of the normal distribution is given by,

\(f\left( x \right) = \frac{1}{{\sigma \sqrt {2\pi } }}{e^{ - \frac{1}{2}{{\left( {\frac{{x - \mu }}{\sigma }} \right)}^2}}}\)

μ = mean

σ² = variance

P (-∞ < X < μ ) = 0.5

P (μ < X < ∞ ) = 0.5

P (μ -σ < X < μ +σ ) = 0.678

P (μ – 2σ < X < μ + 2σ ) = 0.95

P (μ – 3σ < X < μ + 3σ) = 0.997

Calculation:

Given:

Mean, μ = 1

Variance, σ² = 4

σ = 2

From the above probability distribution curve, we can conclude that,

P (X < 0) is greater than zero but less than 0.5

Explanation

We know the following relations

\({\sin ^2}x = 1 - {\cos ^2}x\left( 1 \right)\)

\(cos 2x = 2{\cos ^2}x - 1\left( 2 \right)\)

Putting equation (1) in equation (2)

\(\cos 2x = 1 - 2{\sin ^2}x\)

A linear equation is given by

Y = mx + C

All the three equations are being converted in the form of Y = mx + C, Hence \(\cos 2x,{\sin ^2}x~and~{\cos ^2}x~are~linearly~dependent\)

Concept:

Gauss divergence theorem:

It states that the surface integral of the normal component of a vector function \(\vec F\) taken over a closed surface ‘S’ is equal to the volume integral of the divergence of that vector function \(\vec F\) taken over a volume enclosed by the closed surface ‘S’.

\(\underset{S}{\mathop \iint },\vec{F}.\hat{n}ds=\iiint\limits_{V}{\nabla .\vec{F}dv}\)

Calculation:

Given:

F = xi + yj + zk

\(\nabla .\vec{F}=3\)

\(\frac14 \underset{S}{\mathop \iint },\vec{F}.\hat{n}ds=\frac14\iiint\limits_{V}{\nabla .\vec{F}dv}\)

\(\frac14 \underset{S}{\mathop \iint },\vec{F}.\hat{n}ds=\frac14\iiint\limits_{V}{\nabla .\vec{F}dv}=\frac14\iiint\limits_{V}{3.dv}=\frac34\iiint\limits_{V}{dv}\)

\(\frac14 \underset{S}{\mathop \iint },\vec{F}.\hat{n}ds= \frac{3}{4} \times volume;of;the;sphere\)

\(\frac14 \underset{S}{\mathop \iint },\vec{F}.\hat{n}ds= \frac{3}{4} \times \frac{4}{3} \times\pi\times {\left( 1 \right)^3} = \pi ;(\because;radius = 1)\)

Stokes theorem:

It states that the line integral of a vector field \(\vec F\) around any closed surface C is equal to the surface integral of the normal component of the curl of vector \(\vec F\) over an unclosed surface ‘S’.

\(\underset{C}{\mathop \oint },\vec{F}.\overrightarrow{dr}=\underset{S}{\mathop \iint },curl~\vec{F}.\hat{n}ds\)

Green's theorem:

\(\oint \left( {{\rm{Mdx}} + {\rm{Ndy}}} \right) = \int!!!\int \left( {\frac{{\partial {\rm{N}}}}{{\partial {\rm{x}}}} - \frac{{\partial {\rm{M}}}}{{\partial {\rm{y}}}}} \right){\rm{dxdy}}\)

Concept:

Given Eq is f”(t) + f(t) = 0

L[f”(t)] + L[f(t)] = L(0)

s²F(s) –sf(0)-f’(0) + F(s) = 0

(s² + 1) F(s) = 4

\(F\left( s \right) = \frac{4}{{{s^2} + 1}}\)

Concept:

Steady flow energy equation

\({h_1} + \frac{{V_1^2}}{2} + \dot Q = {h_2} + \frac{{V_2^2}}{2} + \dot W\)

Given:

h₁ = 3250 kJ/kg, h₂ = 2360 kJ/kg, V₁ = 180 m/s, V₂ = 5 m/s, Q = -5 kW (loss)

Calculation:

\({h_1} + \frac{{V_1^2}}{2} + \dot Q = {h_2} + \frac{{V_2^2}}{2} + \dot W\)

\(\dot W = \left( {h_1 - h_2} \right) + \frac{{\left( {{{V_1}^2} - {V_2^2}} \right)}}{2000} + (\dot Q)\)

\(W = \left( {3250 - 2360} \right) + \frac{{\left( {{{180}^2} - {5^2}} \right)}}{2000} + (-5)\)

W= 901.19 kW

Concept:

Water is coming out and falls on surface shows in the figure.

Point 1 shows datum surface and point 2 shows tap outlet. 

Apply Bernoulli equation at point 1 and 2

\(\frac{{{{\rm{P}}_1}}}{{{\rm{\rho g}}}} + \frac{{{\rm{V}}_1^2}}{{2{\rm{g}}}} + {{\rm{z}}_1} = {\rm{}}\frac{{{{\rm{P}}_2}}}{{{\rm{\rho g}}}} + \frac{{{\rm{V}}_2^2}}{{2{\rm{g}}}} + {{\rm{z}}_2}\)

As pressure is constant everywhere,

\(\frac{{{\rm{V}}_1^2}}{{2{\rm{g}}}} + {{\rm{z}}_1} = {\rm{}}\frac{{{\rm{V}}_2^2}}{{2{\rm{g}}}} + {{\rm{z}}_2}{\rm{}}\)……………….(1)

By using continuity theorem,

A₁V₁ = A₂V₂…………..(2)

Calculation:

Given:

D₂ = 20 mm, V₂ = 2 m/s, z₂ = 0.5 m, z₁ = 0

From eq. (1)

\(\frac{{{\rm{V}}_1^2}}{{2{\rm{g}}}} + {{\rm{z}}_1} = {\rm{}}\frac{{{\rm{V}}_2^2}}{{2{\rm{g}}}} + {{\rm{z}}_2}{\rm{}}\)

\(\frac{{{\rm{V}}_1^2}}{{2 \times 9.81}} + 0 = \frac{{{2^2}}}{{2 \times 9.81}} + 0.5\)

V₁ = 3.716 m/s

From eq. (2)

A₁V₁ = A₂V₂

\(\frac{{\rm{\pi }}}{4}\times{\rm{d}}_1^2 \times 3.716 = \frac{{\rm{\pi }}}{4}\times{20^2} \times 2\)

Therefore, d₁ = 14.67 mm ≈ 15 mm

Concept:

The time required to cool the steel ball in air is

\(\frac{{T - {T_{co}}}}{{{T_i} - {T_{co}}}} = {e^{ - \left( {\frac{{hA }}{{\rho{c}V}}} \right)t}};\)

\(ln(\frac{{T_i - {T_{co}}}}{{{T} - {T_{co}}}}) =\frac{{hA }}{{\rho{c}V}} t\)

Calculation: 

Given:

d = 60 mm = 0.06 m, Ti = 1030°C, T∞ = 30°C, T = 430°C

The time required to cool is:

\(ln(\frac{{T_i - {T_{co}}}}{{{T} - {T_{co}}}}) =\frac{{hA }}{{\rho{c}V}} t\)

\(ln(\frac{{1030 - 30}}{{430 - 30}}) = \frac{{20 × 4 \pi\times(0.03)^2}}{{7800 × 600 ×\frac{4}{3} \times\pi\times(0.03)^3}} t\)

0.9162 = 4.273 × 10^-4 t

∴ t = 2144 sec

Concept:

The difference between the maximum and minimum speeds during a cycle is called the maximum fluctuation of speed. The ratio of the maximum fluctuation of speed to the mean speed is called the coefficient of fluctuation of speed (C_s)

\(C_s=\frac{\omega_{max}-\omega_{min}}{\omega}=\pm2%=0.04\)

As the speed of the flywheel changes from ω_max to ω_min, the maximum fluctuation of energy:

\(\Delta E=I\omega^2 C_s\)

Calculation:

Given:

Cs = 0.04, ΔE = 400 Nm, ω = 20 rad/sec

\(\Delta E=I\omega^2 C_s\)

\(\Rightarrow I = \frac{\Delta E}{\omega^2\times C_s}=\frac{{400}}{{{{20}^2} \times 0.04}} = 25\ kg - {m^2}\)

Concept:

The diameter of a gear is directly proportional to its teeth

d ∝ t 

where, t = no. of teeth, d = diameter of gear.

Calculation:

d_R = m × t_R = 2 × 15 = 30 mm

d_Q = 2 × d_R = 60 mm

\(\frac{{{d_P}}}{{{d_Q}}} = \frac{{{t_P}}}{{{t_Q}}} \Rightarrow {d_P} = \frac{{20}}{{40}} × 60 = 30~mm\)

\(\frac{{{d_R}}}{{{d_S}}} = \frac{{{t_R}}}{{{t_S}}} \Rightarrow {d_S} = dR × \frac{{{t_S}}}{{{t_R}}} = \frac{{30 × 20}}{{15}} = 40~mm\)

\(Centre;distance = \frac{{{d_P}}}{2} + \frac{{{d_Q}}}{2} + \frac{{{d_R}}}{2} + \frac{{{d_S}}}{2}\)

Centre distance = 15 + 30 + 15 + 20 = 80 mm

Explanation:

When the Force F is suddenly removed, then due to W, the rod is in rotating condition with angular acceleration \(\alpha\)

Thus the equation of motion:

\(\Sigma M=I_o\alpha;;;W \times \frac{L}{2} = I\alpha\)

\(As, I = \frac{mL^2}{3}= \frac{1}{3}\times\frac{W}{g}\times{L^2}\)

\(\therefore W \times \frac{L}{2} =\frac{WL^2}{3g}\times\alpha\)

\(\Rightarrow \alpha = \frac{{3g}}{{2L}}\)

\(\therefore Linear;acceleration;at;centre = \alpha \times \frac{L}{2} = \frac{{3g}}{{2L}}\times \frac{L}{2}=\frac{{3g}}{4}\)

Also, the centre of the rod accelerates with linear acceleration a;

\(W-R=F\Rightarrow mg-R=ma\R=mg-ma=mg-\frac{3}{4}mg=\frac{1}{4}mg=\frac{W}{4}\)

Calculation:

Given:

\({V}T_1^{1.6} = 3000,~~and~~,{V}T_2^{0.6} = 200\)

\(\therefore {{\rm{T}}_1} = {\rm{}}{\left( {\frac{{3000}}{{\rm{V}}}} \right)^{\frac{1}{{1.6}}}}{\rm~~{and}}~~{{\rm{T}}_2} = {\rm{}}{\left( {\frac{{200}}{{\rm{V}}}} \right)^{\frac{1}{{0.6}}}}\)

The carbide tool will provide higher tool life means T₁ > T₂

⟹ \(\frac{{{{3000}^{\frac{1}{{1.6}}}}}}{{{{\rm{V}}^{\frac{1}{{1.6}}}}}} > \frac{{{{200}^{\frac{1}{{0.6}}}}}}{{{{\rm{V}}^{\frac{1}{{0.6}}}}}}\)

\(\therefore \frac{{{{\rm{V}}^{\frac{1}{{0.6}}}}}}{{{{\rm{V}}^{\frac{1}{{1.6}}}}}} > \frac{{{{200}^{\frac{1}{{0.6}}}}}}{{{{3000}^{\frac{1}{{1.6}}}}}}\)

V > 39.4 m/s

Concept:

When a mirror point of point P (x₁, y₁) is calculated about a line ax + by + c = 0. It is given by point P' (x₂, y₂).

It is given by -

\(\frac{x_2;-;x_1}{a}=\frac{y_2;-;y_1}{b}=\frac{-2(ax_1;+;by_1;+;c)}{a^2;+;b^2}\) 

Calculation:

Given:

P (x₁, y₁) = (5, 10)

A line passing through the origin (0, 0) and making angle 45° is given by y = x ⇒ x - y = 0

Comparing x - y = 0 with ax + by + c = 0

∴ a = 1, b = -1 and c = 0.

\(\frac{x_2;-;x_1}{a}=\frac{-2(ax_1;+;by_1;+;c)}{a^2;+;b^2}\)

\(\Rightarrow \frac{x_2;-;5}{1}=\frac{-2[(1\times5);+;(-1\times 10);+;0]}{1^2;+;(-1)^2}\)

∴ x2 = 10.

\(\frac{y_2;-;y_1}{b}=\frac{-2(ax_1;+;by_1;+;c)}{a^2;+;b^2}\)

\(\Rightarrow \frac{y_2;-;10}{-1}=\frac{-2[(1\times5);+;(-1\times 10);+;0]}{1^2;+;(-1)^2}\)

∴ y2 = 5.

∴ the mirror image of point P (5, 10) on the line y = x, is point P' (10, 5).

Calculation:

From diagram, the two points are in feasible region A and B,

At A (0 , 1.33) and B (2 , 0)

Z_max = 3x + 7y

Therefore, Z_A = 3 × 0 + 7 × 1.33 = 9.33  and

Z_B = 3 × 2 + 7 × 0 = 6

∴ Zmax = Z_A = 9.33

Therefore only one having maximum value. So, it is having exactly one optimal solution.

Concept:

When a shaft is electroplated, the final diameter gets increased.

D_f = D_i + 2t

When a hole is electroplated, the final diameter gets decreased.

D_f = D_i - 2t

where t = plating thickness.

Calculations:

Given:

Basic size = D ⇒  \(25\begin{array}{*{20}{c}} { + 0.020}\ { + 0.010} \end{array}mm\)

Plating = t ⇒  \({30^{ \pm 2}}\ microns\ ={0.03^{ \pm 0.002}} \ mm ;[\because 1;micron;= ;10^{-3};mm]\)

t_max = 0.03 + 0.002 ⇒ 0.032 mm.

⇒ (D_i)_max = 25 + 0.02 ⇒ 25.02 mm.

D_f = D_i + 2t

⇒ (D_f)_max = (D_i)_max + 2t_max

⇒ (D_f)_max = 25.02 + 2 × 0.032

⇒ (D_f)_max = 25.084 mm.

Thus, the size of the GO gauge is 25.084 mm.

Calculation:

The material removal rate in ECM is,

\(MRR = \frac{{IA}}{{FV}} \times η \)

where I = current, A = atomic weight, V = valency, F = faraday's constant = 96500, η = efficiency

Calculation:

Given:

For Titanium: A = 48, V = 3, I = 2000 A, η = 0.9, MRR = ?

The material removal rate is:

\(MRR = \frac{{IA}}{{FV}} \times η \)

\(MRR = \frac{{2000 \times 48}}{{96500 \times 3}} \times 0.90 = 0.30;gm/s\)

Concept:

Impulse (I) = Change in momentum (ΔP)

Calculation

Given

Mass, m = 5 kg

Stiffness of spring, k = 10 kN/m

At, t = 0 , Mass is at rest u = 0

Force, F = 5 kN is applied for time, t = 10^-4 sec

Using impulse momentum theorem

Impulse (I) = Change in momentum (ΔP)

F × t = m × (v – u)

5 × 10³ × 10^-4  = 1 × (v – 0)

\(v = 0.5~\frac{m}{{sec}}\)

Using energy balance

\(\frac{1}{2}m{v^2} = \frac{1}{2}k{X^2}\)

\(1 \times {0.5^2} = 10 \times {10^3} \times {X^2}\)

X = 5 × 10^-3 = 5 mm

Concept:

Mean effective pressure is the ratio of the net work done to the displacement volume of the piston.

\(mep = \frac{{{W_{net}}}}{{{V_s}}}\)

\({V_s} = {V_1} - {V_2} = {V_1}\left( {1 - \frac{{{V_2}}}{{{V_1}}}} \right) = {V_1}\left( {1 - \frac 1r} \right)\)

For the Diesel cycle:

Compression ratio, r = V1/V2 

Calculation:

Given:

r = 21, ρ = 1.0 kg/m3, ṁ = 30 × 10–3 kg/s, W_net = 15 kW

\({V_1} = \frac{{\dot m}}{\rho } = 30 \times {10^{ - 3}};{m^3}/s\)

Now,

\({V_s} = {V_1}\left( {1 - \frac 1r} \right)\)

Mean effective pressure is:

\(mep = \frac{{{W_{net}}}}{{{V_s}}}=\frac{{{W_{net}}}}{{{{V_1}\left( {1 - \frac 1r} \right)}}}\)

\( mep = \frac{{15}}{{30 \times {{10}^{ - 3}}\left[ {1 - \frac{1}{{21}}} \right]}}=525\ kPa\)

Concept:

\(Load = \sin \left( {\frac{{3\pi x}}{L}} \right)\frac{N}{m}\)

\(Total;Load = \mathop \smallint \limits_0^L \sin \left( {\frac{{3\pi x}}{L}} \right)dx\)

\(Total ~load= \frac{{ - L}}{{3\pi }}\left[ {\cos \left( {\frac{{3\pi L}}{L}} \right) - \cos 0} \right] = + \frac{{2L}}{{3\pi }}\)

Since the load is distributed symmetrically at both the sides so it will be same on both the supports

\({R_A} = {R_B} = \frac{{Total;Load}}{2} = \frac{L}{{3\pi }}\)

• Be cautious between uniformly distributed load and varying distributed load.

Concept:

The two plates are kept parallel then the net heat transfer between plates is,

\(\dot Q = \frac{{σ A\left( {T_1^4 - T_2^4} \right)}}{{\frac{1}{{{\varepsilon _1}}} + \frac{1}{{{\varepsilon _2}}} - 1}}\)

Calculation: 

Given:

ϵ₁ = ϵ₂ = 0.8, T₁ = 300 K, T₂ = 400 K, σ = 5.67 × 10^-8 W/m²K⁴

\(\frac{\dot Q}{A} = \frac{{σ \left( {T_1^4 - T_2^4} \right)}}{{\frac{1}{{{\varepsilon _1}}} + \frac{1}{{{\varepsilon _2}}} - 1}}\)

\(\frac{\dot Q}{A} = \frac{{5.67 \times {{10}^{ - 8}}\times\left( {{{400}^4} - {{300}^4}} \right)}}{{\frac{1}{{0.8}} + \frac{1}{{0.8}} - 1}} = 661.5~{\rm{W}}/m^2\)

\(\frac{\dot Q}{A} =0.66~kW/m^2\)

If two cylinders are placed concentrically,

\(\dot Q= \frac{{{\rm{\sigma }}\left( {{\rm{T}}_1^4 - {\rm{T}}_2^4} \right){{\rm{A}}_1}}}{{\frac{1}{{{\epsilon _1}}} + \frac{{{{\rm{A}}_1}}}{{{{\rm{A}}_2}}}\left( {\frac{1}{{{\epsilon _2}}} - 1} \right)}}~{\rm{W}}\)

If a small body is placed in a large room then, 

\(\dot Q = {\rm{\sigma }}\left( {{\rm{T}}_1^4 - {\rm{T}}_2^4} \right){{\rm{A}}_1}{\epsilon_1}~{\rm{W}}\)

Concept:

The pressure force is given by:

F = ρgAh̅ 

Centre of pressure (h)* is given by:

\(h^* = \bar h + \frac{{{I_G}\sin^2θ }}{{A\bar h}}\)

Calculation:

Given:

l = 5 m, θ = 30°, ρ = 1000 kg, width b = 1 m

h̅ = 2.5 × sin 30° = 1.25 m

Area = b × l = 1 × 5 = 5 m²

Moment of inertia (I_G) is:

\({I_G} = \frac{{b{l^3}}}{{12}}=\frac{{1×{(5)^3}}}{{12}}=10.41~m^4\)

Pressure force (F) is:

F = ρgAh̅ = 1000 × 9.81 × 5 × 1.25 = 61312.5 N

Centre of pressure (h)* is:

\(h^* = \bar h + \frac{{{I_G}\sin^2θ }}{{A\bar h}}\)

\({h^*} = 1.25 + \left( {\frac{{10.41 \times {{\sin }^2}30^\circ }}{{5 \times 1.25}}} \right)\)

h = 1.66 m*

Taking moment about hinge,

Moment due to force F = Moment due to W

\(F \times \frac{{{h^*}}}{{\sin 30^\circ }} = W \times 2.5\cos 30^\circ \)

\(613125.5 \times \frac{{1.66}}{{0.5}} = Mg \times 2.165\)

M = 9618.93 kg

∴ Most closely answer is option 4 i.e. 9623 kg

Concept:

Available energy (AE) of system is given as,

AE = (h₁ – h₀) – T₀ (s₁ – s₀) + \(\left( {\frac{{{\rm{V}}_1^2}}{2} - {\rm{}}\frac{{{\rm{V}}_0^2}}{2}} \right)\)

where, 1 denotes present condition and 0 denotes atmospheric condition and

Entropy change is given by:

(s₁ – s₀) = \({{\rm{c}}_{\rm{p}}}\ln \frac{{{{\rm{T}}_1}}}{{{{\rm{T}}_0}}} - {\rm{R}}\ln \frac{{{{\rm{p}}_1}}}{{{{\rm{p}}_0}}}\)

Calculation:

Given:

p₁ = 5 bar, p₀ = 1 bar, T₁ = 500 K, T₀ = 300 K, c_p = 1.005 kJ/kgK, c_v = 0.718 kJ/kgK, v₁ = 50 m/s

Entropy change is:

(s₁ – s₀) = \({{\rm{c}}_{\rm{p}}}\ln \frac{{{{\rm{T}}_1}}}{{{{\rm{T}}_0}}} - {\rm{R}}\ln \frac{{{{\rm{p}}_1}}}{{{{\rm{p}}_0}}}\)

(s₁ – s₀) = \(1.005\ln \frac{{500}}{{300}} - 0.287\ln \frac{5}{1} = 0.05147\frac{{{\rm{kJ}}}}{{{\rm{kgK}}}}\)  (∵ R = c_p - c_v = 1.005 – 0.718 = 0.287 kJ/kgK)

Available energy (AE) of system is:

AE = (h₁ – h₀) – T₀ (s₁ – s₀) + \(\left( {\frac{{{\rm{V}}_1^2}}{2} - {\rm{}}\frac{{{\rm{V}}_0^2}}{2}} \right)\)

AE = 1.005 × (500 - 300) – 300 × (0.05147) + \( \frac{{{{50}^2}}}{2} × {10^{ - 3}}\)     {∵ v_o = 0}

AE = 201 – 15.441 + 1.25 = 186.809 kJ/kg ≈  187 kJ/kg

Concept:

Bayes’ theorem:

Let A₁, A₂, …. , A_n be n mutually exclusive and exhaustive events of the sample space S and A is an event which can occur with any of the events then

\({\rm{P}}\left( {\frac{{{{\rm{A}}_{\rm{i}}}}}{{\rm{A}}}} \right) = {\rm{;}}\frac{{{\rm{P}}\left( {{{\rm{A}}_{\rm{i}}}} \right){\rm{P}}\left( {\frac{{\rm{A}}}{{{{\rm{A}}_{\rm{i}}}}}} \right)}}{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} {\rm{P}}\left( {{{\rm{A}}_{\rm{i}}}} \right){\rm{P}}\left( {\frac{{\rm{A}}}{{{{\rm{A}}_{\rm{i}}}}}} \right)}}\)

Calculation:

Given:

A = The student answer the question correctly

E₁ = Student knows the correct answer

E₂ = Student guesses the answer

\(p\left( {{E_1}} \right) = \frac{2}{3},p\left( {{E_2}} \right) = \frac{1}{3}\)

by using conditional probability,

\(p\left( A \right) = p\left( {{E_1}} \right) \times p\left( {\frac{A}{{{E_1}}}} \right) + p\left( {{E_2}} \right) \times p\left( {\frac{A}{{{E_2}}}} \right)\)

P(A) \(= \frac{2}{3} \times 1 + \frac{1}{3} \times \frac{1}{4} = \frac{3}{4}\)

Using Bayes theorem

\(P\left( {\frac{{{E_1}}}{A}} \right)=\frac{{p\left( {{E_1}} \right) \times p\left( {\frac{A}{{{E_1}}}} \right)}}{{p\left( A \right)}} = \frac{{\frac{2}{3} \times 1}}{{\frac{3}{4}}} = \frac{8}{9}\)

Concept:

If roots are real and different then

y = C₁e^m₁x + C₂e^m₂x

Calculation:

Given:

\(\frac{{{d^2}u}}{{d{x^2}}} - k\frac{{du}}{{dx}} = 0\)

D² – kD = 0    

m(m - k) = 0

m₁ = 0, m₂ = k 

Roots are real and different

u = C₁e^m₁x + C₂e^m₂x

\(u = {C_1}{e^0} + {C_2}{e^{kx}}\)........(1)

u(0) = 0

∴ C₁ + C₂ = 0  .................... (2)

u(L) = U

u = C₁ + C₂e^kL = U  ..................... (3)

solving (2)  and (3), we get

\({C_1} = \frac{U}{{1 - {e^{kL}}}},{C_2}=\frac{{ - U}}{{1 - {e^{kL}}}}\)

Putting values fo C₁ and C₂ in eqn 1

\(u = {C_1}{e^0} + {C_2}{e^{kx}}\)

\(u = \frac{U}{{1 - {e^{kL}}}}+{e^{kx}}\times\frac{{ - U}}{{1 - {e^{kL}}}}\)

\(u = U\left( {\frac{{1 - {e^{kx}}}}{{1 - {e^{kL}}}}} \right)\)

• If Roots are real and equal (m,m) then y = (C₁ + C₂x).e^mx
• If roots are imaginary (α ± iβ) then y = e^αx (C₁ cosβx + C2 sinβx)

Concept:

We know that,

By Parts method

\(\smallint u \cdot v \cdot dx = u \cdot \smallint v \cdot dx - \smallint \left[ {\frac{{du}}{{dx}}\smallint v \cdot dx} \right]dx\)

Where, u, v should follow the ILATE sequence.[I= Inverse, L= Logarithmic, A= Algebraic, T= Trigonometric, E= Exponential terms]

Calculation:

Given:

From the given Equation, \(\mathop \smallint \limits_1^e √ x \ln \left( x \right)dx\)

u = ln(x), v = √x

Now,

\(\smallint u \cdot v \cdot dx = u \cdot \smallint v \cdot dx - \smallint \left[ {\frac{{du}}{{dx}}\smallint v \cdot dx} \right]dx\)

\(\mathop \smallint \limits_1^{\rm{e}} {\rm{lnx}} \cdot √ {\rm{x}} {\rm{dx}} = {\rm{lnx}} \cdot \mathop \smallint \limits_1^{\rm{e}} √ {\rm{x}} {\rm{dx}} - \mathop \smallint \limits_1^{\rm{e}} \left[ {\frac{{{\rm{du}}}}{{{\rm{dx}}}} \cdot {\rm{;}}\smallint √ {\rm{x}} {\rm{dx}}} \right]{\rm{dx}}\)  

\(\mathop \smallint \limits_1^{\rm{e}} {\rm{lnx}} \cdot √ {\rm{x}} {\rm{dx}}= \left[ {\ln \left( x \right) \times \frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_1^e - \smallint \left[ {\frac{1}{x} \times \frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]dx\)

 \(\mathop \smallint \limits_1^{\rm{e}} {\rm{lnx}} \cdot √ {\rm{x}} {\rm{dx}}= \left[ {\ln \left( x \right) \times {x^{\frac{3}{2}}} \times \frac{2}{3} - \frac{4}{9} \times {x^{\frac{3}{2}}}} \right]_1^e\)

∴  \(\mathop \smallint \limits_1^e √ x \ln \left( x \right)dx\)  \(= \frac{2}{9}√ {{e^3}} + \frac{4}{9}\)

Concept:

The maximum load P to avoid crushing failure is,

P_c = σc × (ntdr)

where,  σc = bearing stress, w = width of plate, n = no. of rivets, dr = dia of rivet, t = thickness of plate

Calculation: 

Given:

 σ_c = 150 MPa, w = 200 mm, n = 3, dr = 10 mm, t = 5 mm

The maximum load is:

P_c = σ_c × (ntdr)

P_c  = 150 × 3 × 10 × 5 = 22500 N = 22.5 kN

Concept:

The maximum load P to avoid tearing failure is,

P = σ_t × (w - nd_h) × t

where,  σ_t = tensile stress, w = width of plate, n = no. of rivets, d_h = dia of hole, t = thickness of plate

Calculation: 

Given:

 σ_t = 200 MPa, w = 200 mm, n = 3, d_h = 11 mm, t = 5 mm

The maximum load P to avoid tearing failure is,

P = σ_t × (w - nd_h) × t

P = 200 × (200 – (3 × 11)) × 5 = 167000 N = 167 kN

The maximum load P to avoid shear failure is, P_s = \({{\rm{\tau }}_{\rm{s}}}\left( {\frac{{\rm{\pi }}}{4}{\rm{d}}_{\rm{r}}^2} \right)\)

The maximum load P to avoid bearing/crushing failure is, P_b = \({{\rm{\sigma }}_{\rm{c}}}{\rm{}}\left( {{\rm{nt}}{{\rm{d}}_{\rm{r}}}} \right)\)

Calculation:

Given:

\(q_w'' = 2500x\), ṁ = 0.01 kg/s, T1 = 20°C, d = 50 mm = 0.05 m, l = 3 m, c_p = 4.18 kJ/kgK = 4.18 × 10³ K/kgK

\(x = 0,;q_{w}'' = 0\)

\(x = 3,;q_{w}'' = 2500 × 3 = 7500 ~W/m^2\)

\({q_{avg}} = \frac{{0 + 7500}}{2} = 3750~W/m^2\)

q_avg × Area = ṁ × c_p × ΔT

3750 × π × 0.05 × 3 = 0.01 × 4.18 × 10³ × (T - 20)    {∵ Area = π × d × l}

⇒ T = 42.2 + 20 = 62.2°C

Concept:

Heat transfer rate between the entire pipe is

 \(q_w'' × A = mC_p{\rm{\Delta }}T\)

Given:

q"_w = 5000 = constant, h = 1000 W/m²K, ṁ = 0.01 kg/s, T₁ = 20°C, d = 50 mm = 0.05 m, l = 3 m, c_p = 4.18 kJ/kgK = 4.18 × 103 K/kgK

Calculation:

\(q_w'' × A = mC_p{\rm{\Delta }}T\)

5000 × π × 0.05 × 3 = 0.01 × 4.18 × 10³ × (T₀ - 20)

T₀ – 20 = 56.3 = T₀ = 76.3°C

Heat flux between any two sections is same

Q = hA(T_p – T₀)

But \(\frac{Q}{A} = q_w'' = 5000 = \left( {{T_p} - {T_0}} \right)\)

5000 = 1000(T_p – 76.3)

⇒ T_p = 76.3 + 5 = 81.3°C

Concept:

The above figure represents Merchant's circle diagram.

F_C = Cutting force, F_T = Axial force, F = Friction force, N = Normal force, F_S = Shear force and F_n = Shear normal force.

α = Orthogonal rake angle, β = Friction angle, ϕ_s = Shear plane angle.

Calculation:

Given:

∠ between F_C and F = 90° 

∴ (β - α) + (90 - β) = 90

⇒ α = 0° 

∴ as cutting force F_C and friction force F becomes perpendicular, orthogonal rake angle α become 0°.

Concept:

When cutting force F_C and friction force F becomes perpendicular, orthogonal rake angle α becomes 0°.

Under this condition, the Merchant's circle diagram looks like:

In this situation, forces F, N, F_C and F_T form a rectangle.

where F = F_T and F_C = N.

∴ N = F_C ⇒ 1500 N.

Alternate method:

The relation between Normal force N and Cutting force F_C and Axial force F_T is:

N = F_Ccosα - F_Tsinα

∴ N = 1500 × cos 0

∴ N = 1500 N.

Concept:

The Brayton Cycle is shown in figure, where,

1 – 2 = Ideal work in compressor and 1 – 2’ = Actual work in compressor

For Compressor,  \(\frac{{{{\rm{T}}_2}}}{{{{\rm{T}}_1}}} = {\left( {{{\rm{r}}_{\rm{p}}}} \right)^{\frac{{{\rm{\gamma }} - 1}}{{\rm{\gamma }}}}}~~{\rm{and}}~~{{\rm{\eta }}_{{\rm{th}}}} = \frac{{{{\rm{T}}_2} - {{\rm{T}}_1}}}{{{\rm{T}}_2^{\rm{'}} - {{\rm{T}}_1}}}\)

Calculation:

Given:

r_p = 8, T1 = 300 K, T3 = 1400 K, γ = 1.4

For the compressor,  \(T_2=300 \times {\left( 8 \right)^{\frac{{1.4 - 1}}{{1.4}}}} = 543.434~{\rm{K}}\) 

\(0.8 = \frac{{543.434 - 300}}{{{\rm{T}}_2^{\rm{'}} - 300}} \Rightarrow {\rm{T}}_2^{\rm{'}} = 604.3~{\rm{K}}\)

W_comp = C_p (T^'₂ – T₁)

W_comp = 1(604.3 - 300) = 304.3 kW/kg

Concept:

The Bryton Cycle is shown in figure, where,

1 – 2 = Ideal work in compressor and 1 – 2’ = Actual work in compressor

3 – 4 = Ideal work in turbine and 3 – 4’ = Actual work in turbine

For Compressor,  \(\frac{{{{\rm{T}}_2}}}{{{{\rm{T}}_1}}} = {\left( {{{\rm{r}}_{\rm{p}}}} \right)^{\frac{{{\rm{γ }} - 1}}{{\rm{γ }}}}}~~{\rm{and}}~~{{\rm{\eta }}_{{\rm{th}}}} = \frac{{{{\rm{T}}_2} - {{\rm{T}}_1}}}{{{\rm{T}}_2^{\rm{'}} - {{\rm{T}}_1}}}\)

For Turbine,  \(\frac{{{{\rm{T}}_3}}}{{{{\rm{T}}_4}}} = {\left( {{{\rm{r}}_{\rm{p}}}} \right)^{\frac{{{\rm{γ }} - 1}}{{\rm{γ }}}}}~~{\rm{and}}~~{{\rm{\eta }}_{{\rm{th}}}} = \frac{{{{\rm{T}}_3} - {\rm{T}}_4^{\rm{'}}}}{{{{\rm{T}}_3} - {{\rm{T}}_4}}}\)

Given:

R_p = 8, T₁ = 300 K, T₃ = 1400 K, η_th = 0.8, C_p = 1 kJ/K, γ = 1.4

Calculation:

For compressor, T₂ = \(300 \times {\left( 8 \right)^{\frac{{1.4 - 1}}{{1.4}}}} = 543.434~{\rm{K}}\)

\(0.8 = \frac{{543.434 - 300}}{{{\rm{T}}_2^{\rm{'}} - 300}} \Rightarrow {\rm{T}}_2^{\rm{'}} = 604.3~{\rm{K}}\)

For turbine, T₄ = \(\frac{{1400}}{{{{\left( 8 \right)}^{\frac{{1.4 - 1}}{{1.4}}}}}} = 772.86~{\rm{K}}\)

\(0.8 = \frac{{1400 - {{\rm{T}}_4}{\rm{'}}}}{{1400 - 772.86}} \Rightarrow {\rm{T}}_4^{\rm{'}} = 898.3~{\rm{K}}\)

W_net = W_turb - W_comp

= C_p(T₃ – T’₄) – C_p (T’₂ – T₁)

= 1(1400 - 898.3) – 1 (604.3 - 300)

W_net = 197.4 kW

Heat supply, Q_s = C_p(T₃ – T’₂)

= 1(1400 – 604.3) = 795.7 kW

\({\rm{\eta }} = \frac{{{{\rm{W}}_{{\rm{net}}}}}}{{{{\rm{Q}}_{\rm{s}}}}} = \frac{{197.4}}{{795.7}} = 24.80~{\rm{% }}\)