GATE ME 2015 Official Paper: Shift 2 (Previous Year Paper)

The tone of the sentence is that of regret about something that happened.

In such cases, should have + past participle is the correct choice.

Hence 'should have practiced' is the correct answer.

The connotations of a particular word or name are the ideas or qualities which the word makes you think of.

Hence, the connotation of a road or way can be interpreted as the ideas invoked by a road or way during a journey.

Viaticum means money or supplies for any journey. 

Viaticum and connotation both are associated with a journey.

Hence, viaticum is the most suitable one-word substitute among the given options.

Introvert, atheist and altruist are nouns whereas alternate is a verb.

Thus, the correct statement would be 'Let us alternate'

Hence, alternate is the correct answer.

From the logarithmic property, 

when x > y 

ln x > ln y if x and y both greater than 1.

and x > y 

ln x < ln y, if both x and y are less than 1.

but when x > y 

ex > ey  for any value of x and y.

but when x > y, x² > y²

 also cos x < cos y.

Between is a preposition.

In English, a preposition must be followed by an indirect object pronoun. 

Me is an indirect object pronoun, while I is a subject pronoun. 

Therefore, between has to be followed by me, not I.

Hence, 'It would remain between you and me' is the correct answer.

Statement 2 is about activities enjoyed by the family whereas statement 5 gives information about the activities by the family.

Therefore, statement 2 follows statement 5. i.e., (5 < 2)

Statement 3 is about the activities at night whereas statement 4 talks about activities in the evening.

Therefore, statement 3 follows statement 4. i.e., (4 < 3.) and 4 cannot be the first event (evident from the word 'sunset') and hence option 4 gets eliminated.

Also, statement 3 must be the last part (evident from the word 'night') among the others (2, 3, 4, 5)

Thus, options 2 and 3 get eliminated.

Thus, considering the timeline of events, the best possible logical sequence is 5, 2, 4, 3.

Hence '5, 2, 4, 3'  is the correct answer.

90% of area of circular sheet = surface area of the cone (only the conical part)

⇒ 0.9 × π R² = π.r. l

where R radius of circular sheet, r is radius of cone, l is slant length of cone.

   

Given, R = 30 cm, l = R = 30 cm

⇒ 0.9 × π × 30 × 30 = π × r × 30

⇒ r = 27 cm 

now height of the cone h  \(= \sqrt {{{l}^2} - {{r}^2}}= \sqrt {{{30}^2} - {{27}^2}}\)

= 13.08 cm

Thus r/h = 27 /13.08 = 2.06

As per trigonometric properties

tan θ = cot (90 - θ) and tan θ . cot θ = 1

thus, tan 89° = cot (90 -89° ) =  cot 1°

similarly, tan 88° = cot 2°, tan 87° = cot 3° ans so on...

also from Logarith properties

log m +log n = log (m × n)

thus, log tan 1° + log tan 89° = log (tan1° × tan89°)

= log (tan1° × cot1°)

= log 1

= 0

similarly, log tan 2° + log tan 88° = log (tan1° × cot1°) = log 1 = 0

thus, log tan1° +log tan2° +……..+log tan 89° = 0.

The statement in the first option suggests that Ms X will be in Kochi for one day, only in May but not in other months. This cannot be a definite inference as we do not have sufficient information about her stay in other months.

The statement in the second option suggests that Ms X will not stay in Kochi for more than one day. This is evident in her itinerary. Hence the second inference is valid.

The statement in the third option suggests that Ms X will not be anywhere but Kochi for one day in May. Now, from the given information we can only conclude that she will be in Kochi for one day but not about the other places she could be during that period. Hence, the third option is not valid.

The statement in the fourth option suggests that no one besides Ms X will be in Kochi for one day.

The given information is only about Ms X, therefore, the fourth option is invalid.

Hence, 'Ms. X will be in Kochi for only one day in May' is the logically valid inference.

For a singular matrix [A], determinant of matrix |A| = 0

From eigenvalue properties,

Determinant of A i.e.|A| = product of eigenvalues

So, one of the eigenvalues should be zero for a singular matrix.

Important Point:

The sum of eigenvalues = trace of A i.e sum of diagonal elements.

Explanation:

For negative values of x, f(x) will be positive

For positive values of x, f(x) will be positive

∴ Minimum value of f(x) will occur at x = 0

Concept:

for any vector V(x,y,z) = V_x i + V_y j + V_z k 

Curl of \(V\left( {x,y,z} \right) = \left| {\begin{array}{*{20}{c}} i&j&k\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\ {{V_x}}&{{V_y}}&{{V_z}} \end{array}} \right|\)

Analysis:

Given vector V(x,y,z)  = 2x2i + 3z2j + y3k

curl of \(V\left( {x,y,z} \right) = \left| {\begin{array}{*{20}{c}} i&j&k\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\ {2{x^2}}&{3{z^2}}&{{y^3}} \end{array}} \right|\) 

= i [3y² – 6z] + j [0 – 0] + k [0 – 0]

\(= \left. {\left( {3{y^2} - 6z} \right)i} \right|{_{x = y = z = 1}}\)

= -3i

Concept:

The Laplace transformation

\(L\left( {\sin \omega t} \right) = \frac{\omega }{{{S^2} + {\omega ^2}}}\) and \(L\left( {\cos \omega t} \right) = \frac{S}{{{S^2} + {\omega ^2}}}\)

Calculation:

\(L\left{ {{e^{j5t}}} \right} = L\left( {cos5t + i;sin5t} \right)\)

\(= L\left( {cos5t} \right) + iL\left( {sin5t} \right) \)

=\( \frac{S}{{{S^2} + {5^2}}}\)+\(i\frac{5}{{{S^2} + {5^2}}}\)

= \( \frac{S+5i}{{{S^2} + {25}}}\)

The probability of meeting the design specification of vendor 1 is P₁ = 0.8

The probability of not meeting the design specification of vendor 1 is q₁ = 1 - P₁ =0.2 

The probability of meeting the design specification of vendor 2 is P₂ = 0.7

The probability of  not meeting the design specification of vendor 2 is q₂ = 0.3

The probability of meeting the design specification of vendor 3 is P₃ = 0.5

The probability of not meeting the design specification of vendor 3 is q₃ = 0.5

Probability (at least one will meet specification) = 1 – probability (none will meet specification)

= 1 – (1 – 0.8) × (1 – 0.7) × (1– 0.5) 

= 1 – 0.2 × 0.3 × 0.5

= 1 – 0.03

= 0.97

Concept:

For perfectly elastic collision, both momentum and kinetic energy are conserved and the coefficient of restitution is equal to 1.

Calculation:

Given:

m₁ = 1 kg, u₁ = 12 m/s

m₂ = 2 kg, u₂ = 0 m/s

moment conservation: 

m1u1 + m2u2 = m1v1 + m2v2    

12 = v₁ + 2v₂     ........................ (1)

energy conservation:

\(\frac{1}{2}{m_1}u_1^2 + \frac{1}{2}{m_2}u_2^2 = \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_1}v_2^2\)

\(\frac{1}{2} \times 1 \times 144 + \frac{1}{2} \times 2 \times 0 = \frac{1}{2} \times 1 \times v_1^2 + \frac{1}{2} \times 2 \times v_2^2\)

\(⇒ 144 = v_1^2 + 2v_2^2\)    .................. (2) 

From (1) and (2)

\(144 = 144 + 4v_2^2 - 48{v_2} + 2v_2^2\)

\(6v_2^2 - 48{v_2} = 0\)

\(6{v_2}\left( {{v_2} - 8} \right) = 0\)

⇒v₂ = 8 m/s

Alternative method:

for perfectly elastic collision coefficient of restitution e = 1.

\(e = \frac{{velocity;of;seperation}}{{velocity;of;approach}}\)

for e = 1, velocity of seperation = velocity of approach

⇒ v₂ - v₁ = u₁ - u₂

⇒ v2 - v1 = 12  .................(3)

from (1) and (3)

3v₂ = 24

v₂ = 8 m/s

Concept:

Modulus of elasticity E is defined as the slope of its stress-strain curve in the elastic deformation region.

Slope of σ-ϵ curve = \(E = \frac{{{\rm{Δ }}σ }}{{{\rm{Δϵ }}}}\)

Modulus of rigidity (G), Modulus of Elasticity (E) and Bulk modulus or volumetric modulus of elasticity (K) are related through Poisson's ratio μ. 

\(G = \frac{E}{{2\left( {1 + μ } \right)}}\) and \(K = \frac{E}{{3\left( {1 -2μ } \right)}}\)

Calculation:

Given, change in stress Δσ  = 200 MPa

change in strain Δϵ  = 0.001, μ = 0.3

from, \(E = \frac{{{\rm{Δ }}σ }}{{{\rm{Δϵ }}}}\)

\(\Rightarrow E = \frac{{200}}{{0.001}} = 200 \times {10^3}MPa\)

= 200 GPa

and \(G = \frac{{200}}{{2\left( {1 + 0.3} \right)}} = \frac{{200}}{{2 \times 1.3}} = \frac{{100}}{{1.3}} = \frac{{1000}}{{13}} = 77GPa;\)

Explanation:

Circumferential stress of hoop stress σ_h 

\({σ _1} ={σ _h} = \frac{{pd}}{{2t}} = \frac{{2 \times 14;}}{{2 \times 0.05}} = 280MPa\)

Longitudinal stress σ_L 

\({σ _2} ={σ _L} = \frac{{pd}}{{4t}} = \frac{{2 \times 14;}}{{4 \times 0.05}} = 140MPa\)

As this is the case of a thin cylinder:

Radial stress \({σ _r} =0\)

Maximum shear stress \({τ _{max}} = \max \left{ {\frac{{{\sigma _1} - {\sigma _2}}}{2},\frac{{{\sigma _1}}}{2},\frac{{{\sigma _2}}}{2}} \right}\)

τ_max  = \(\frac{{σ _h}-{σ _r}}{{2}}=\frac{{σ _1} }{2} = \frac{280}{2}=140~ MPa\)

Mistake Points 

Maximum In-Plane shear stress/Surface shear stress:

\(τ_{max,inplane}=\frac{{σ _1}-{σ _2}}{{2}}\)

Maximum wall shear stress/Out plane shear stress/Absolute shear stress:

\(τ_{max,abs}=\frac{{σ _{max}}-{σ _{min}}}{{2}}=\frac{σ_1}{2}\)

Concept:

From Kutzbach's mobility criterion,

F = 3 (n – 1) – 2 j – h

n = Number of links

j = Number of binary joints

h = Number of higher pair

Calculation:

In the gear train system,

no. of links n = 4, no. of joints j = 3 (1 binary joint and 1 ternary joint by link 1, 2 and 3)

and no. of higher pair h = 1.

F = 3 (4 – 1) – 2× 3 – 1 = 2

A planetary gear train has 2 DOF and hence requires two inputs to get the desired output.

Concept:

For a Damped free vibration,

\(ζ = \frac{c}{{{c_c}}} = \frac{c}{{2 \sqrt{km} }}\)

where, ζ is damping factor, C is damping coefficient and C_c is critical damping coefficient.

when ζ =1, c = c_c = 2√(km)   

Calculation:

Given, c_c1 = 0.1 kg/s, m₂ = 2m and k₂ = 8k

\({c_{c1}} = 2\sqrt {{k_1}{m_1}} = 2\sqrt {km}\)

\({c_{c2}} = 2\sqrt {{k_2}{m_2}} = 2\sqrt {8k \times 2m} = 4{c_{c1}} = 4 \times 0.1 = 0.4\) kg/s.

Concept:

If there is uniaxial loading yield stress is σ_y

As per Von Mises failure theory or Distortion energy theory

\(σ _y^2 = σ _1^2 - {σ _1}{σ _2} + σ _2^2\) where, σ₁ and σ₂ are principal stresses  

Under pure shear stress loading

\({σ _1} = - {σ _2} = \tau\) where, \(\tau\) is shear yield stress

then, \(σ _y^2 = {\tau ^2} + {\tau ^2} + {\tau ^2}=3{\tau ^2}\)

\(\Rightarrow \tau = \frac{{{σ _y}}}{{\sqrt 3 }}\)

Calculation:

Given, σ_y = 300MPa 

As shear yield stress  \(;\tau = \frac{{{σ _y}}}{{\sqrt 3 }} = 0.577{σ _y} \) 

thus**, \(;\tau = 0.577{\times 300} = 173.28\)** MPa

Explanation:

From potential function definition,

\(u = - \frac{{\partial ϕ }}{{\partial x}};and;v = - \frac{{\partial ϕ }}{{\partial y}}\)

when, ϕ = ϕ(x,y) 

\(dϕ = \frac{{\partial ϕ }}{{\partial x}}dx + \frac{{\partial ϕ }}{{\partial y}}dy\)

dϕ = - u dx - v dy

Since along potential line, ϕ is constant, so dϕ = 0

0 = - udx - vdy

slope = \(\frac{{dy}}{{dx}} = - \frac{u}{v}\)

Concept:

In a simple Rankine cycle, after the isentropic expansion in the turbine, steam is directly fed into the condenser for the condensation process. But in the reheat system, two turbines (high-pressure turbine and low-pressure turbine) are employed to improve efficiency.

Steam, after expansion from the high-pressure turbine, is sent again to the boiler and heated until it reaches superheated condition. It is then left to expand in the low-pressure turbine to attain condenser pressure. The reheat cycle has been developed to take advantage of the increased efficiency with higher pressures and yet avoid excessive moisture in the low-pressure stages of the turbine.

With reheat, the mean temperature of heat addition TmTm

increases, and so efficiency and work output increase, but the steam rate decreases.

Explanation:

The Rankine cycle with reheating involves the following processes:

• Isentropic Compression (1-2): Water is pumped from low pressure to high pressure. This process is isentropic, meaning entropy remains constant.
• Isobaric Heat Addition (2-3): Water is heated in the boiler at constant pressure, transforming it into superheated steam.
• Isentropic Expansion in High-Pressure Turbine (3-4): The steam expands isentropically in the high-pressure turbine. Entropy remains constant, and the steam's temperature and pressure drop.
• Isobaric Reheating (4-5): The steam is reheated at constant pressure. This increases the steam's temperature.
• Isentropic Expansion in Low-Pressure Turbine (5-6): The reheated steam expands isentropically in the low-pressure turbine. Entropy remains constant, and the steam's temperature and pressure drop further.
• Isobaric Heat Rejection (6-1): The steam condenses at constant pressure in the condenser, rejecting heat and returning to the liquid phase.

Calculation:

Analyzing the statements:

1. Increase in the average temperature of heat addition:

This statement is true. Reheating increases the average temperature at which heat is added to the cycle, leading to higher thermal efficiency.

2. Reduction in thermal efficiency:

This statement is false. Reheating generally improves the thermal efficiency of the Rankine cycle by increasing the average temperature of heat addition.

3. Drier steam at the turbine exit:

This statement is true. Reheating ensures that the steam remains superheated for a longer portion of the expansion process, resulting in drier steam at the turbine exit.

Conclusion:

Based on the above analysis, the correct statements regarding a Rankine cycle with reheating are:

• (i) Increase in the average temperature of heat addition.
• (iii) Drier steam at the turbine exit.

Therefore, the correct answer is:

3) only (i) and (iii) are correct.

Explanation:

\(\frac{P}{\rho g} + \frac{{{v^2}}}{{2g}} + Z = Constant\)

Bernoulli’s equation:

• It can be derived from the principle of conservation of energy.
• It states that, in a steady flow, the sum of all forms of energy in a fluid along a streamline is the same at all points on that streamline.
• It represented in head form (the total energy per unit weight).

Following assumption are made in deriving Bernoulli’s equation:

Bernoulli’s equation assumptions:

1. The fluid is ideal, i.e. fluid has zero viscosity
2. Flow is steady
3. Flow is continuous
4. Flow is incompressible
5. Flow is irrotational
6. Flow is along a streamline.

Within a boundary layer for a steady incompressible flow, viscosity is present and the viscous forces dominate over inertia forces.

Thus, the Bernoulli equation does not hold within a boundary layer for a steady incompressible flow.

Concept:

The critical radius of insulation r_c is the radius upto which the addition of material increases the rate of heat loss and after critical radius addition of heat results into decrease in heat loss.

for cylinder \({r_c} = \frac{k}{h}\)

for sphere \({r_c} = \frac{2k}{h}\)

where, k is thermal conductivity of insulating material in W/m-K, and h is heat transfer coefficient of surrounding fluid in W/m²-K.

Calculation:

Given, k = 0.1 W/m-K and h_o = 2 W/m²-K

For cylinder,

Critical radius \({r_c} = \frac{k}{{{h_0}}} = \frac{{0.1}}{2} = 0.05m\)

= 5 cm

Concept:

Prandtl number Pr is defined as the ratio of momentum diffusivity to thermal diffusivity.

\(Pr = \frac{{\mu {C_p}}}{K} = \frac{{\left( {\frac{\mu }{\rho }} \right)}}{{\left( {\frac{K}{{\rho {C_p}}}} \right)}}\)

\(Pr = \frac{\nu }{\alpha } = \frac{{momentum;diffusivity}}{{thermal;diffusivity}}\)

In another way, we can define Prandtl number as, the ratio of the rate that viscous forces penetrate the material to the rate that thermal energy penetrates the material.

**\(\frac{δ }{{{δ _T}}} = {\left( {Pr} \right)^{1/3}};\)**where, δ is hydrodynamic boundary layer thickness and δ_T is thermal boundary layer thickness.

Calculation:

Given:

Pr = 0.7 

from, \(\frac{δ }{{{δ _T}}} = {\left( {Pr} \right)^{1/3}};\)=  \({0.7^{\frac{1}{3}}} = 0.88 < 1\)

thus, δ < δ_T .

When              P_r < 1               δ_T > δ 

                        P_r  > 1               δT < δ 

                        P_r  = 1               δ_t = δ

Concept:

For Carnot Heat Pump

\({\left( {{\rm{COP}}} \right)_{HP}} = \frac{{{R.E}}}{{{Win}}} = \frac{{{Q_H}}}{{{Q_H} - {Q_L}}} \)

R.E is Refrigeration Effect = QH for heat pump and QL for refrigerator. 

QL + Win = QH

⇒ Win = QH - QL

Also, \({\left( {{\rm{COP}}} \right)_{HP}} = \frac{{{Q_H}}}{{{Q_H} - {Q_L}}} =\frac{{{T_H}}}{{{T_H} - {T_L}}}\)

Calculation: 

Given, T_H = 37^o C = 37+273 = 310 K  

and T_L = 6^oC = 6 + 273 = 279 K 

\({\left( {{\rm{COP}}} \right)_{HP}} = \frac{{{T_H}}}{{{T_H} - {T_L}}} =\frac{{{310}}}{{{310} - {279}}}= \frac{{310}}{{31}} = 10\)

Explanation:

Vander-Waals equation

• The equation is basically a modified version of the Ideal Gas Law which states that gases consist of point masses that undergo perfectly elastic collisions. Ideal gas equation fails to explain the behaviour of real gases. Therefore, the Van der Waals equation was devised and it helps us define the physical state of a real gas.
• Van der Waals equation is an equation relating the relationship between the pressure, volume, temperature, and amount of real gases. For a real gas containing ‘n’ moles, the equation is written as

\(\left( {p + \frac{a}{{{v^2}}}} \right)\left( {v - b} \right) = RT\)

• The constant "a" provides a correction for the intermolecular forces.
• Constant b adjusts for the volume occupied by the gas particles. It is a correction for finite molecular size and its value is the volume of one mole of the atoms or molecules.

here, p is pressure in N/m², v is the specific volume in m³/kg, R is characteristic gas constant in J/kg-K, \( \frac{a}{{{v^2}}}\) is called the force of cohesion and b is called the co-volume.

(\(p + \frac{a}{{{v^2}}}\)) both terms should give the same unit since they are getting added

i.e. p = \( \frac{a}{{{v^2}}}\) in terms of unit.

\(\Rightarrow \frac{N}{{{m^2}}} = a{\left( {\frac{{kg}}{{{m^3}}}} \right)^2}\)

\(\Rightarrow a\left( {unit} \right) = \frac{{{m^6}}}{{k{g^2}}}.kg.\frac{m}{{{s^2}{m^2};}} = \frac{{{m^5}kg}}{{k{g^2}{s^2}}} = \frac{{{m^5}}}{{kg{s^2}}}\)

Brake power developed by the engine is given by 

Brake Power = Torque × angular velocity

B.P = T× ω =2πNT/60

Given, B.P = 10 kW =10000 W and ω = 400 rad/s

from  B.P = T× ω

10000 = T × 400

\(T = \frac{{10000}}{{400}} = 25N - m\)

In the crystal structure, atomic packing factor (APF) or packing efficiency or packing fraction is the volume of atoms in a unit cell divided by the volume of the unit cell. It is a dimensionless quantity and always less than unity

The atomic packing factor of different crystal structures is given in the table below:

No of atoms in b.c.c unit cell = 2

\(APF = \frac{{{N_{atoms}}{V_{atom}}}}{{{V_{crystal}}}} = \frac{{2\left( {\frac{4}{3}} \right)\pi {r^3}}}{{{{\left( {a } \right)}^3}}}= \frac{{2\left( {\frac{4}{3}} \right)\pi {r^3}}}{{{{\left( {\frac{{4r}}{{\sqrt 3 }}} \right)}^3}}}\)

for BCC \(a = \frac{{4r}}{{\sqrt 3 }}\)where a is side of the cube and r is atomic radius.

= 0.68

The Atomic packing factor for different crystal structures are given in table.

Explanation:

The unconventional machining process and their characteristics and the application areas are discussed in the table below:

The ‘GO’ guage controls the lower limit of a hole and The ‘NO GO’ gauge controls the upper limit of a hole.

Go size = maximum material limit of component = Lower limit of hole

NO GO gauges are designed for minimum material limit of component = upper limit of hole.

Trick to remember:

In 'GO', the plug gauge should enter into the thus hole lower limit of hole is considered while 'NO GO' plug gauge should not enter into hole so upper limit of hole is considered.

A generative manufacturing process different from a traditional manufacturing process develops a product with the addition of material instead of removal of the excess material as in case of a traditional manufacturing process.

In the generative manufacturing process, plans are generated by means of decision logic, formulas, technology algorithm and geometry-based data to perform uniquely the many processing decision for converting a part from raw material to a finished state.

There are two major components of a generative process planning system.

1. A geometry-based coding scheme.
2. Process knowledge in the form of decision logic and data.

Geometry-based coding scheme:

The objective of a geometric based coding scheme is to define all geometric features for all process related surfaces together with feature dimensions, locations, tolerances and the surface finish desired on the features.

Advantage of Generative Manufacturing Process:

• No tools and fixtures are required.
• No restriction on geometry of the part shape exists.
• Composite parts and assemblies can be produced in one go.
• Small batch production of complex parts is economically viable.

Concept:

In EOQ model,

Annual inventory holding cost = \(\frac{Q}{2} \times {C_h}\)

Annual inventory ordering cost = \( \frac{A}{Q} \times P\)

At optimum total inventory cost (TIC),

Annual inventory holding cost = annual inventory ordering cost

\(\frac{Q}{2} \times {C_h} = \frac{A}{Q} \times P\)

\(Q = \sqrt {\frac{{2.A.P}}{{{C_h}}}} \)

where, Q is economic order quantity or optimum quantity

C_h is holding cost, A is annual demand and P is procurement cost or ordering cost.

Calculation:

Given, A = 50000, P = Rs. 7000 and Q = 10000 

from, \(Q = \sqrt {\frac{{2.A.P}}{{{C_h}}}} \)

\({C_h} = \frac{{2 \times 50000 \times 7000}}{{{{10000}^2}}}\) = Rs. 7.0

Annual inventory holding cost = \(\frac{Q}{2} \times {C_h}\) = \(\frac{{10000}}{2} \times 7\) = Rs. 35000

Alternate method

At optimum total inventory cost (TIC), 

Annual inventory holding cost = annual inventory ordering cost

and annual inventory ordering cost = Number of orders × ordering cost per order

\( = \frac{A}{Q} \times P\)

\(= \frac{{50000}}{{10000}} \times 7000\)

= 5 × 7000

=Rs. 35000

Calculation:

Given:

• When the forecast value shows an increasing trend then the regression will forecast a higher value compared to the moving average.

Let A → student passes the exam

B → student gets above 90% marks

Given the probability of passing the exam P(A) = 0.2

and probability of passing the exam and getting above 90% marks P(A∩B) = 0.05

From conditional probability, 

the probability of happening of B when A has already happened, i.e. \({\rm{P}}\left( {\frac{{\rm{B}}}{{\rm{A}}}} \right) = \frac{{{\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right)}}{{P\left( A \right)}} \)

Required probability of getting above 90% when a student has passed the exam is \({\rm{P}}\left( {\frac{{\rm{B}}}{{\rm{A}}}} \right) = \frac{{{\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right)}}{{P\left( A \right)}} = \frac{{5% }}{{20% }} = \frac{1}{4}\)

By Gauss divergence theorem

\(\mathop \smallint \limits_s \bar F.\bar nds; = \mathop \smallint \limits_v div\bar Fdv\)

Here

div F̅ = 9 – 3 =6

\(\therefore \mathop \int!!!\int \limits_s \frac{1}{\pi }\left( {9x\bar i - 3y\bar j} \right).\bar nds = \frac{1}{\pi }\mathop \smallint \limits_v 6dv = \frac{1}{\pi }6v\)

\(= \frac{1}{\pi }6\left( {\frac{4}{3}\pi {r^3}} \right)\)

= 8(3)³

= 216

Explanation:

\(\frac{{dy}}{{dt}} = -5y\)

\(\Rightarrow \frac{{dy}}{y} = - 5dt\) (variables separable form)

Integrating both side we get,

\(lny = - 5t + c\)         _______________(1)

initial condition: y = 2 at t = 0,

from equation (1), 

\(ln2 = - 5 \times 0 + c\)

\(c = ln2\)

\(lny = - 5t + ln2\)

\({\rm{ln}}\left( {\frac{y}{2}} \right) = - 5t\)

\(y = 2{e^{ - 5t}}\)

now at t = 3, \(y = 2{e^{ - 15}}\)

Concept:

Trapezoidal rule,

\(\mathop \smallint \nolimits_a^b f\left( x \right) = \frac{h}{2}\left[ {\left( {{y_0} + {y_n}} \right) + 2\left( {{y_1} + {y_2} + {y_3} + \ldots + {y_{n - 1}}} \right)} \right];\)

Where,\(y = f\left( x \right);and;{y_n} = f\left( {{x_n}} \right)\) n = 0,1,2,3,…., number of sub-intervals or, the difference between two successive sub-intervals.

\(;h; = ;step;size; = \frac{{b - a}}{n}\) or, the difference between two successive sub-intervals.

Calculation:

Given:

            y₀      y₁    y₂     y₃     y₄

h = x₁ - x₀ = x₂ - x₁ = 0.1

\(\mathop \smallint \limits_0^{0.4} f\left( x \right)dx = \mathop \smallint \limits_0^{0.4} ydx = \frac{h}{2}\left[ {\left( {{y_0} + {y_4}} \right) + 2\left( {{y_1} + {y_2} + {y_3}} \right)} \right]\)

\(=\frac{0.1}{2}\left[ {\left( {0 + 160} \right) + 2\left( {10 + 40 + 90} \right)} \right]\)

= 22

Given,

a = - 0.1V

\(\frac{{dv}}{{dt}} = - 0.1V\)

integration by seperation of variables, we get

lnV = - 0.1t +ln k

V = ke^-0.1t

at t = 0; V = 40, 

∴ k = 40

then, V = 40e^-0.1t

at t = 3 seconds

V = 40 e^-0.1×3 = 29.6327 m/s

Concept:

R_P+R_Q=10

ΣM_P = 0, R_Q × 1 - 10 × 0.5 = 0

R_Q = 5 kN and R_P 5 kN (upward force)

Since this is a case of a propped cantilever, deflection δ at point P will be zero, which will give an opposite 5 kN force acting in the downward direction at point P as shown in the last diagram.

Thus, M_O = R_P × OP = 5 × 2 = 10 kN-m

Concept:

Considering equilibrium in the vertical direction,

R_p + R_Q = 30 kN

ΣM_p =0

R_Q × 3 = 2 × 30

⇒ R_Q = 20 kN

and R_p = 10 kN

for balance at ‘x’ → F_Rx = 20 kN

at ‘RT’ → F_RT cos 45 = 20

\(\Rightarrow {F_{RT}} = \frac{{20}}{{cos45}}\)                (1)

Also F_SR = F_RT cos 45°     (2)

from (1) and (2)

F_SR = 20kN

Concept:

\(\frac{d^2y}{dx^2}=\frac{M}{EI}=-\frac{Px}{EI}\)

\(\frac{dy}{dx}=-\frac{Px^2}{2EI}+c_1\)

At x = L; slope \(\frac{{dy}}{{dx}} = 0\)  where x is distance from free end

\(c_1=\frac{PL^2}{2EI}\)

\(\frac{dy}{dx}=-\frac{Px^2}{2EI}+\frac{PL^2}{2EI}\)

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

Calculation:

given, P = 2 kN, E= 210 GPa, distance from fixed end = 20 mm = 0.2 m.

\(θ = \frac{P}{{2EI}}\left( {{L^2} - {x^2}} \right)\), \(I = \frac{{{a^4}}}{{12}}\) for square section of side a.

\(\theta\) at Q i.e. x = 100 - 20 = 80 mm = 0.08 m from free end.

\(\theta=\frac{2000}{2\times 210\times 10^9\times\frac{(0.006)^4}{12}}[(0.1)^2-(0.08)^2]=0.1587 ;rad\)

Concept:

Maximum and minimum values of normal stresses occur on planes of zero shearing stress. The maximum and minimum normal stresses are called the principal stresses, and the planes on which they act are called the principal plane.

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

But the maximum shear stress planes may or may not contain normal stresses as the case may be.

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

Calculation:

σ_x = 20 MPa, σ_y = 80 MPa and τ_xy = 40MPa

\(\tau_{max}= \sqrt {{{\left( {\frac{{80 - 20}}{2}} \right)}^2} + {{40}^2}} =50;MPa\)

 

\({\tau _{max}} = \max \left{ {\frac{{{\sigma _1} - {\sigma _2}}}{2},\frac{{{\sigma _1}}}{2},\frac{{{\sigma _2}}}{2}} \right}\)

\({\sigma _1}/{\sigma _2} = \frac{{{\sigma _x} + {\sigma _y}}}{2} \pm \sqrt {{{\left( {\frac{{{\sigma _x} - {\sigma _y}}}{2}} \right)}^2} + \tau _{xy}^2} \)

\({\sigma _1}/{\sigma _2} = \frac{{20 + 80}}{2} \pm \sqrt {{{\left( {\frac{{20 - 80}}{2}} \right)}^2} + {{40}^2}} \)

\({\sigma _1}/{\sigma _2} = 50 \pm 50\)

\({\sigma _1} = 100;MPa;and;{\sigma _2} = 0;MPa\)

\({\tau _{max}} = \max \left{ {\frac{{100 - 0}}{2},\frac{{100}}{2},0} \right}\)

\({\tau _{max}} = 50;MPa\)

Concept:

In a slider crank-mechanism, 

The piston displacement: \({x_p} = r\left( {1 - \cos θ } \right)\)

Piston velocity: \({v_p} = \omega r\left( {\sin θ + \frac{{\sin 2θ }}{{2n}}} \right)\)

Piston acceleration: \({a_p} = {\omega ^2}r\left( {\cos θ + \frac{{\cos 2θ }}{n}} \right)\),  where n is equal to l/r 

l length of connecting rod in m and r is crank radius in m.

for maximum acceleration, θ = 0^° \({a_{max}} = {\omega ^2}r\left( {1 + \frac{1}{n}} \right)\)

Calculation:

Given, l = r = 300 mm = 0.3 m, ω = 14 rad/s

\({a_{max}} = {\omega ^2}r\left( {1 + \frac{1}{n}} \right)\)

n = 1, as l = r = 0.3 m

thus, \({a_{max}} = {\omega ^2}r\left( {1 + \frac{1}{1}} \right) = 2{\omega ^2}r\)

\({a_{max}} = 2 \times {14^2} \times 0.3 = 117.6;m/{s^2}\)

Concept:

The amplitude ‘A’ of steady-state vibration in harmonic excitation with damper is,

\(A = {x_{max}} = \frac{{{F_0}}}{{\sqrt {{{\left( {K - m{\omega ^2}} \right)}^2} + {{\left( {C\omega } \right)}^2}} }}\)

Or, \(A = \frac{{{F_0}/K}}{{\sqrt {{{\left( {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right)}^2} + {{\left( {2\xi \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)

also, \(A = \frac{\delta }{{\sqrt {{{\left( {1 - {r^2}} \right)}^2} + {{\left( {2\xi r} \right)}^2}} }}\)

Where δ is static deflection of spring in m, F₀ is sinusoidal force amplitude in N.

, ω is forced frequency and ω_n is natural frequency of spring in rad/s

And ξ is damping factor = C/C_C = actual damping coefficient/critical damping coefficient.

Calculation:

Given, force amplitude F₀ = 10 N, K = 150 N/m, ξ = 0.2 and ω_n = 10 ω

From, \(A = \frac{{{F_0}/K}}{{\sqrt {{{\left( {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right)}^2} + {{\left( {2\xi \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)

\(A = \frac{{10/150}}{{\sqrt {{{\left( {1 - {{\left( {\frac{{\rm{\omega }}}{{10{\rm{\omega }}}}} \right)}^2}} \right)}^2} + {{\left( {2 \times 0.2 \times \frac{{\rm{\omega }}}{{10{\rm{\omega }}}}} \right)}^2}} }}\)

\(A = 0.0673 \approx 0.07 m\)

Concept:

The power transmitted by the shaft is given by,

\(P = T.\omega = \frac{{2\pi NT}}{{60}}\)

Torsion equation,

\(\frac{T}{{{J}}} = \frac{\tau }{{{r}}}=\frac{{Gθ }}{l};\)

Calculation:

Given: 

P = 30 kW, N = 700 rpm, θ = 1^° , G = 80 GPa and d_i = 0.7 d_o 

from  \(P = T.\omega = \frac{{2\pi NT}}{{60}}\)

\(30 \times 1000 = T \times \frac{{2π \times 700}}{{60}}\)

∴ T = 409.256 N – m

From Torsion equation,

\(\frac{T}{{{J}}} = \frac{{Gθ }}{l};\)

L = 1 m, θ = 1° = π/180 rad

\(\frac{{409.256}}{{\frac{π }{{32}}\left( {1 - {{0.7}^4}} \right)d_0^4}} = \frac{{80 \times {{10}^9}}}{1} \times \frac{π }{{180}}\)

on Solving, we get

d₀ = 44.52 mm

Given, P = 20 kW, N = 3000 rpm, τ_max= 30 MPa, d_o = 2d_i

The power transmitted by the shaft is given by,

\(P = T.\omega = \frac{{2\pi NT}}{{60}}\)

\(20 \times {10^3} = T \times \frac{{2\pi \times 3000}}{{60}}\)

∴ T = 63.662 N-m

From torsion equation 

\(\frac{T}{{{J}}} = \frac{τ }{r}\Rightarrow τ=\frac{T.r}{J}\)

For hollow shaft: \(J=\frac{\pi}{32}(d_0^4-d_i^4)=\frac{\pi}{32}d_0^4(1-k^4); where ; k=\frac{d_i}{d_0}=\frac{1}{2}\)

\(τ=\frac{T.\frac{d_0}{2}}{\frac{\pi}{32}d_0^4(1-k^4)}=\frac{16T}{\pi d_0^3(1-k^4)}\)

\(d_0^3=\frac{16T}{\pi\timesτ(1-k^4)}=\frac{16\times63.663\times10^3}{\pi \times30\times(1-(\frac{1}{2})^4)}\)

∴ d₀ = 22.59 mm

Concept:

The total emissive power of any real surface is given by, E = σ ϵ A T⁴ in W

and total emissive power per unit area is = σ ϵ T4  in W/m².

where, σ is Stefan-Boltzmann constant, ϵ is the emissivity of the surface and T is temperature of the surface in K.

If temperature is T1, then

Emissive power \(E_1 \propto ϵ T_1^4\)

Calculation:

Given, emissivity of surface is constant i.e.  ϵ1 = ϵ 2

\(500 =\sigma ϵ_1 T_1^4\)                               ____________________(1)

Similarly

\(1200 = \sigma ϵ_2 T_1^4\)                            ____________________(2)

from (1) / (2)

\({\left( {\frac{{{T_1}}}{{{T_2}}}} \right)^4} = \frac{{500}}{{1200}}\) as ϵ₁ = ϵ ₂

\(\frac{{{T_1}}}{{{T_2}}} = {\left( {\frac{{500}}{{1200}}} \right)^{1/4}} = 0.803\)

Concept:

The head loss in laminar flow through a pipe is given by:

\({h_{L}} = \frac{{32{μ ̅ UL}}}{{ \rho gd^2}}\)

where μ = viscosity, L = length of pipe/plate, U̅ = average velocity, d = diameter of the pipe.

Calculation:

Given:

L₂ = L₁, μ₂ = μ₁, U̅₂ = 2U̅₁, d₂ = 0.5d₁

The head loss in laminar flow through a pipe is given by:

\({h_{L}} = \frac{{32{μ ̅ UL}}}{{ \rho gd^2}}\)

\(\frac{h_2}{h_1}= \frac{{{\bar{U_2}}}}{{\bar{U_1}}}\times\frac{{{{d_1^2}}}}{{d_2^2}}\)

\(\frac{h_2}{h_1}= \frac{{{2\bar{U_1}}}}{{\bar{U_1}}}\times\frac{{{{4d_1^2}}}}{{d_1^2}}=8\)

Additional InformationUsing Darcy-Weisbach equation:

\(h_L=\frac{fLV^2}{2gd}\)

where f = friction factor = 64/Re

Reynold's number is given by:

\(Re=\frac{\rho V d}{\mu}\)

Putting the value of friction factor:

\(h_L=\frac{64}{Re}\times\frac{LV^2}{2gd}=\frac{64\mu}{\rho V d}\times\frac{LV^2}{2gd}\)

\(h_L=\frac{64\mu LV}{2 \rho gd^2}\)

When length and viscosity remain same then:

\(h_L\propto\frac{V}{d^2}\) which is the same as above.

Concept:

For a fully developed laminar flow, 

\(u = - \frac{1}{{4\mu }}\frac{{\partial P}}{{\partial x}}\left( {{R^2} - {r^2}} \right)\)

where, R = radius of pipe, r = radial distance from centre (r = 0),  \(\frac{{\partial P}}{{\partial x}}\)= pressure gradient, μ is dynamic viscosity.

\(u = - \frac{1}{{4\mu }}\frac{{\partial P}}{{\partial x}} \times {R^2}\left( {1 - \frac{{;{r^2}}}{{{R^2}}}} \right)\)

At, r = 0, \(;u; = ;{u_{max}} = - \frac{1}{{4\mu }}\frac{{\partial P}}{{\partial x}} \times {R^2}\)

\(u = {u_{max}}\left( {1 - \frac{{{r^2}}}{{{R^2}}}} \right)\)

Calculation:

Given,

 \(\frac{{\partial P}}{{\partial x}} = - 10;Pa/m\) ,radius of pipe R = 5 cm = 0.05 m, μ = 0.001 Pa-s.

From,

  \(u = - \frac{1}{{4\mu }}\frac{{\partial P}}{{\partial x}} \times {R^2}\left( {1 - \frac{{;{r^2}}}{{{R^2}}}} \right)\)

at r = 0.2 cm = 0.002 m,

 \(u = - \frac{1}{{4 \times 0.001}} \times \left( { - 10} \right) \times {0.05^2} \times \left( {1 - \frac{{{{0.002}^2}}}{{{{0.05}^2}}}} \right)\)

u = 6.24 m/s

Concept:

The total heat transfer rate in the heat exchanger is given by

Q = U.A.θ_m

Where, U = overall heat transfer coefficient in W/m²-K,

A = effective surface area of heat exchanger in m²,

θ_m = log mean temperature difference \(\left( {LMTD} \right) = \frac{{{\bf{Δ }}{T_1} - {\bf{Δ }}{T_2}}}{{{\bf{ln}}\left( {\frac{{{\bf{Δ }}{T_1}}}{{{\bf{Δ }}{T_2}}}} \right)}}\)

Also, Q = C_ph (T_hi-T_h0) = C_pc (T_co-T_ci)

i.e. heat lost by hot fluid = heat gain by cold fluid

where, C_ph and C_pc are heat capacity of hot and cold fluid respectively in kJ/K

Since the heat capacity for hot and cold fluids are same and it is a case of counter flow so the LMTD will be the difference of temperature of either side.

Calculation:

Given, A = 20 m², U = 20 W/m²-K, T_ci = 280 K, T_h0 = 300 K

C_ph = C_pc = 0.4× 1000 = 400 J/K.

For same heat capacity of hot and cold fluid LMTD = ΔT₁ = ΔT₂ = T_ho-T_ci = 300 – 280 = 20 K.

From, Q = U.A.θ_m = C_ph (T_hi-T_h0) = C_pc (T_co-T_ci)

20× 20× 20 = 400× (T_co - 280)

T_co = 300 K

Alternate Method

Effectiveness NTU Method:

For balanced counter flow heat exchanger, the effectiveness is given by-

\(\epsilon=\frac{T_{hi};-;T_{ho}}{T_{hi};-;T_{ci}}=\frac{T_{co};-;T_{ci}}{T_{hi};-;T_{ci}}\)

Tci = 280 K, Tho = 300 K

\(NTU=\frac{UA}{C_{min}}\)

\(NTU=\frac{20\times 20}{0.4\times 1000}=1\)

Since the heat exchanger is balanced, 

\(\epsilon=\frac{NTU}{1;+;NTU}\)

\(\epsilon=\frac{1}{1;+;1}=0.5\)

\(\epsilon=\frac{T_{hi};-;T_{ho}}{T_{hi};-;T_{ci}}\)

\(0.5=\frac{T_{hi};-;300}{T_{hi};-;280}\)

T_hi - 280 = 2T_hi - 600

T_hi = 320 K

For a balanced heat exchanger, the difference between the inlet and outlet always remains constant.

ΔT₂ = ΔT₁ = T_hi - T_co = T_ho - T_ci

T_ho - T_ci = 20

T_hi - T_co = 320 - T_co

320 - T_co = 20

T_co = 300 K.

Concept: 

For cylinder,

\({T_w} = {T_∞ } + \frac{{{q_g}R}}{{2h}}\)

\({T_c} = {T_w} + \frac{{{q_g}{R^2}}}{{4K}}\)

where T_w is wall or surface temperature in K, T_c is the centre temperature in K, T_∞ is surrounding temperature in K,

q_g is heat generation in W/m³, h is the heat transfer coefficient in W/m²-K, R is the radius of the cylinder in m.

Calculation:

Given:

R = 5 mm, q_g = 4×10⁷ W/m³ and T_∞= 25^° C = 298 K, h = 1000 W/m²K

The surface temp. \(\left( {T_w} \right) = {T_∞ } + \frac{{{q_g}}}{{2h}}R\)

\(= 298 + \frac{{4 × {{10}^7}}}{{2 × 1000}} × 5 × {10^{ - 3}}\)

Tw = 398 K

Concept:

Steady Flow Energy Equation (S.F.E.E) of any system is given by,

\(m\left( {{h_1} + ;\frac{{v_1^2}}{2} + g{z_1}} \right) + Q = m;\left( {{h_2} + \frac{{v_2^2}}{2} + g{z_2}} \right) + W\)

\(m\left( {{u_1+p_1 v_1 } + ;\frac{{v_1^2}}{2} + g{z_1}} \right) + Q = m;\left( {{u_2+p_2 v_2 } + \frac{{v_2^2}}{2} + g{z_2}} \right) + W\)

Calculation:

given, \(v_1 =10 m/s, v_2 =20 m/s, z_2 -z_1 =20 m , \Delta T=T_2 -T_1 =1 K, \) m = 10 kg, C_V = 100 J/kg-K and g = 10 m/s².

Using SFEE ( given that there is no change in any other component of the energy of the system) and for adiabatic system Q = 0.

thus,

\(m\left[ {\left( {\frac{{v_1^2}}{2} - \frac{{v_2^2}}{2}} \right) + \left( {{z_1} - {z_2}} \right)g + \left( {{u_1} - {u_2}} \right)} \right] = W\)

\(10\left[ {\frac{{{{10}^2}}}{2} - \frac{{{{20}^2}}}{2} + \left( { - 20} \right) \times 10 + \left( {100 \times( - 1)} \right)} \right] = W\)

W = - 4500 J = -4.5 KJ

∴ Work done on the system is 4.5 kJ

Concept:

From T-dS relation,

\(Tds=dh-vdp\)

\(Tds=C_p dT-vdp\)

for an isothermal process,

\({s_2} - {s_1} = R\ln \frac{{{V_2}}}{{{V_1}}}\)( = R\ln \frac{{{P_1}}}{{{P_2}}}\)

Calculation:

Given, T₁ = 20^o C = 293 K, V₁ = 0.9 m³

T₂ =  20o C = 293 K, V₂ = 0.6 m³

\({s_2} - {s_1} = R\ln \frac{{{V_2}}}{{{V_1}}}\)

\(s_2 -s_1 = 287\ln \frac{{0.6}}{{0.9}} \)

\(s_2 -s_1 = -116.368 \) J/kg-K

For same values of peak pressure and temperature. Diesel cycle is most efficient and Otto cycle is least. Efficiency of dual cycle lies in between.

η_Diesel > η_Dual > η_Otto

And for same compression ratio and heat rejection or heat addition,

ηOtto > ηDual > ηDiesel

Concept:

The capacity of a steam plant is often expressed in terms of steam rate, which is defined as the rate of steam flow (kg/h) required to produce unit shaft output (1 kW).

Specific steam consumption SSC \(=\frac{{3600}}{{{W_{net}}}}\) \(= \frac{{3600}}{{{W_T} - {W_P}}}\) in kg/kW-h

Calculation:

Given, h₃ = 3159 kJ/kg, h₄ = 2187 kJ/kg and W_P = 2 kJ/kg

W_T = h₃  –  h₄ = 3159 –  2187 kJ/kg

W_T = 972 kJ/kg

W_P = 2 kJ/kg

Thus specific steam consumption \(= \frac{{3600}}{{972 - 2}};\frac{{kg}}{{kW - h}}\)

=3.71 kg/kW.h

Concept:

Solidification Time \(t= k{\left( {\frac{V}{A}} \right)^2}\)

where, V is the volume of casting in m³,

A is surface area of casting in m²

and k is solidification factor in s/m².

Calculation:

given V_C = V_S = 1000 cm³ = 1000× 10^-6 m³ and t_c = 4 s.

for cube, V = a³, A = 6a²

\(t_c = 4 = k.\left( {\frac{{{a^3}}}{{6{a^2}}}} \right) = k.{\left( {\frac{a}{6}} \right)^2}\)

vol. of cube = vol of sphere

\(⇒ \frac{4}{3}\pi {r^3} = {a^3}\)

⇒ \(\frac{\pi }{6}{d^3} = {a^3}\)

⇒\(\frac{d}{a} = {\left( {\frac{6}{\pi }} \right)^{\frac{1}{3}}}\)

Solidification time for sphere, \(V = \frac{4}{3}\pi {r^3} = \frac{\pi }{6}{d^3},;A = 4\pi {r^2} = \pi {d^2}\)

\(t_S=k.{\left( {\frac{{\frac{4}{3}\pi {r^3}}}{{4\pi {r^2}}}} \right)^2} =k. {\left( {\frac{r}{3}} \right)^2}\)( =k. {\left( {\frac{d}{6}} \right)^2}\)

\(\frac{{{t_S}}}{{{t_C}}} = \frac{{k{{\left( {\frac{d}{6}} \right)}^2}}}{{k{{\left( {\frac{a}{6}} \right)}^2}}} = {\left( {\frac{d}{a}} \right)^2}\)( = {\left( {\frac{6}{\pi }} \right)^{\frac{2}{3}}}\)

\({t_S} = 1.539 \times 4\)

= 6.157 sec

Concept:

Fractional reduction = \(\frac{{{\rm{\Delta }}A}}{{{A_i}}} = \frac{{{A_i} - {A_f}}}{{{A_i}}} = 1 - \frac{{{A_f}}}{{{A_i}}}\)

where A_i and A_f are the initial and final cross-sectional area respectively.

Fractional reduction = \(1 - \frac{{d_1^2}}{{d_o^2}}\) where, d₁ and d_o are the diameters after and before the 1^st stage of drawing operation.

also, Fractional reduction = \(1 - \frac{{d_2^2}}{{d_1^2}}\) where, d₂ is the diameter of the wire after 2^nd stage of drawing operation

and the overall fractional reduction = \(1 - \frac{{d_2^2}}{{d_o^2}}\) 

Calculation:

Given, Fractional reduction in 1^st stage = 0.4 = \(1 - \frac{{d_1^2}}{{d_o^2}}\) 

which gives, d₁ = √0.6 d_o 

and Fractional reduction in 2^nd stage = 0.3 = \(1 - \frac{{d_2^2}}{{d_1^2}}\)

which gives, d₂ = √0.6 × √0.7 d_o

thus the overall fractional reduction = \(1 - \frac{{d_2^2}}{{d_o^2}}\)

= \(1 - 0.6 \times 0.7 = 1 - 0.42 = 0.58\)

Concept:

True strain \({\epsilon _T}; = ;ln\left( {\frac{{original;area}}{{instantaneous;area}}} \right) = ;ln\left( {\frac{{original;thickness}}{{instantaneous;thickness}}} \right)\), when length is not changing.

Flow stress σ_f = Kϵ_Tⁿ, where K is yield strength in MPa and n is strain hardening component.

Calculation:

True strain ϵ_T \(= ln\frac{{100}}{{95}} = 0.05129\)

σ = 500 × (0.05129)^0.1 ⇒ 371.5147523 = 371.5147523 MPa

Upto elastic limits using hooks law

\(\sigma = E\epsilon = E.\frac{{{\rm{\Delta }}l}}{l}\)

\(371.514 \times {10^6} = 200 \times {10^9} \times \left( {\frac{{{\rm{\Delta }}l}}{{100}}} \right)\)

⇒ Δl = 0.18575 mm = considering this for elastic recovery

∴ This will be added to 95 mm

⇒ Final dimension = 95.18575 mm

Concept:

Total heat input Q = V.I.t =I².R.t  in J (W-s)

where V is arc voltage, I is arc current, R is the resistance and t is time in seconds.

Also heat input per unit length \(q = \frac{{VI}}{v}\)  where v is the welding speed.

Calculation:

given, I₁= 50 A, V₁ = V2**= 60 V, v₁ = 150 mm/min, v₂ =120 mm/min**

 and q₂ = q₁

\(\therefore \frac{{{V_2}{I_2}}}{{{v_2}}} = \frac{{{V_1}{I_1}}}{{{v_1}}}\)

\(\frac{{{I_2}}}{{120}} = \frac{{50}}{{150}}\)

\(\therefore {I_2} = \frac{{120}}{{150}} \times 50\)

I₂ = 40 A

Concept:

The friction force \(F = F_C sinα +F_T cosα \)

The normal friction force \(N=F_C cosα -F_T sinα \)

\(μ = tanβ =\frac{F}{N} = \frac{{{F_C}sin;α + {F_T};cos;α }}{{{F_C}cos;α - {F_T}\sin α }}\)

\(\mu = \frac{{\frac{{{F_T}}}{{{F_C}}} + \tan \alpha }}{{1 - \frac{{{F_T}}}{{{F_C}}}\tan \alpha }}\)

where F_C and F_T are cutting and thrust forces respectively in N.

α is back rake angle

and β is friction angle.

the power consumption P = F_C× V 

where V is cutting speed in m/s.

Calculation:

Given : α  = 0°, V = 180 m/min = 3 m/s F_T = 490 N and μ = 0.7

\(\therefore μ = \frac{{{F_T}}}{{{F_C}}}\)

\(0.7 = \frac{{490}}{{{F_C}}}\)

∴ F_C = 700 N

Power consumption, \(P = {F_C} × {V} = 700 × \frac{{180}}{{60}} × \frac{1}{{1000}}\left( {kW} \right)\)

P = 2.1 kW

Concept:           

In EDM process, the energy E in terms of capacitance C and voltage V is given by the relation

\(E = \frac{1}{2}Q.C = \frac{1}{2}C.{V^2}\) in Joule (J). As Q = C.V

Where, Q is the charge in coulomb (C), C is capacitance in μF, and V is voltage in volt (V).

And Power P is the energy per unit time.

 \(P = \frac{1}{2}\frac{{C{V^2}}}{t}\)  in W(J/s) , where t is spark cycle time in s.

Calculation:

Given, discharge voltage V = 100 V, spark time = 25 μs and average power input = 1kW

From, \(P = \frac{1}{2}\frac{{C{V^2}}}{t}\)

\(1000 = \frac{1}{2} \times \frac{{C \times {{100}^2}}}{{25 \times {{10}^{ - 6}}}}\)

\(C = 50 \times 1000 \times \frac{{{{10}^{ - 6}}}}{{10000}}\)

\(C = 5 \times {10^{ - 6}} = 5;\mu F\)

Concept:

Critical path: Critical path is the minimum time taken for the completion of project in the network.

Critical activities are the activity on the critical path.

Total time taken on any path is the sum of all the activity time in that particular path.

Calculation:

Total time taken,

For path 1-3-5-6 = 12+11+10 = 33 days

For path 1-2-5-6 = 14+ 12+10 = 36 days

For path 1-2-4-5-6 = 14+7+9 +10 =40 days

The longest time taken for path 1-2-4-5-6 is 40 days

or, the minimum time for completion of project is 40 days.

Thus, Path 1-2-4-5-6 is the critical path.

Concept:

Total Cost = Fixed Cost + Variable Cost

Total Cost = F + Q × v

Total Selling Price  S.P = Q × s, where Q is production capacity, v is variable cost per unit and s is selling cost per unit

Total Profit = Selling Price - Total Cost 

Calculation:

Given, F = Rs. 50000, Q = 1500 units, v = Rs. 200, s = Rs. 300

Actual Production capacity = 1500 × 0.8 = 1200 units 

Total Cost = 50000 + 1200 × 200 = Rs. 2,90,000

Total selling price = 1200 × 300 = Rs. 3,60,000

Total profit per month = 3,60,000 - 2,90,000 = Rs. 70,000.