GATE ME 2014 Official Paper: Shift 4 (Previous Year Paper)

The correct answer is option 1), i.e. dissent.

Key Points 

Let's try to find the meaning of words given in the options-

• Dissent: (noun) the holding or expression of opinions different from those commonly or officially held. (आम तौर पर या आधिकारिक तौर पर आयोजित विचारों से अलग राय की होल्डिंग या अभिव्यक्ति)
• Descent: (noun) an act of moving downwards, dropping, or falling. (नीचे की ओर बढ़ने, गिरने या गिरने की क्रिया)
• Decent: (adjective) of an acceptable standard; satisfactory. (एक स्वीकार्य मानक का; संतोषजनक।)
• Decadent: (adjective) characterized by or reflecting a state of moral or cultural decline. (नैतिक या सांस्कृतिक गिरावट की स्थिति को दर्शाती या दर्शाती है।)

The given sentence mentions the freedom of a citizen to disagree with the government, the only option 1), i.e. dissent would be apt.

The correct answer is option 2), i.e. get back to him. 

Revert: (verb) to reply or respond to someone. It also means to return to a previous state, practice, topic, etc. 

Let's look at the other words given in the options:

• Retract: draw or be drawn back or back in.
• Retreat: withdraw from enemy forces as a result of their superior power or after a defeat.

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

Vindicated: (adjective) justified, cleared of blame. 

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

• Punished: (verb) inflict a penalty or sanction on someone for an offence. It is the past and the past participle form of "punish".
• Substantiated: (verb) proved the truth by giving evidence. It is the past and the past participle form of "substantiate".
• Appreciated: (verb) recognized the full worth of. It is the past and the past participle form of "appreciate".
• Chastened: (verb) to inflict suffering upon for purposes of moral improvement. It is the past and the past participle form of "chasten".

The only word similar in meaning to "vindicated" is "substantiated", i.e. option 2).

Given:

P = xⁿy^m

Now x^’ = 2x, y^’ = y/2

∴ P^’ = (x^’)ⁿ(y^’)^m

P^’ = (2x)ⁿ(y/2)^m

^{\(P' = {\left( {2x} \right)^n}{\left( {\frac{y}{2}} \right)^m} = {2^n}{\left( {\frac{1}{2}} \right)^m}\left( {{x^n}{y^m}} \right) = {2^{n - m}}P\)}

Hence, option 1 is the correct answer.

Concept:

In a consecutive series

\(\frac{{sum,of,numbers}}{{total;numbers}} = middle;number;of;the;series\)

Calculation:

As 425 is the sum of the first 5 consecutive odd numbers

∴ \(\frac{{425}}{5}\) = 85 is the middle of 5 numbers i.e. 3^rd number.

∴ Series is 81, 83, 85, 87,…….101, 103

Sum of last 5 number is 95 + 97 + 99 + 101 + 103 = 495

Explanation:

Sequence is 13 + 4 = 17, 17 + 2 =19, 19 + 4 = 23

M + 4 = Q, Q + 2 = S, S + 4 = W

According to the given code, only consonants are taken and written in reverse order.

KCLFTSB: BST - Best, F - Of, LCK - Luck

SHSWDG: GD - Good, WSHS - Wishes

Similarly

“ace the Exam": C - Ace, T - The, XM - Exam

Concept:

As the increase in annual rate is uniform over time.

∴ it is a case of compound interest.

\(A = P{\left( {1 + \frac{R}{{100}}} \right)^t}\)

where A = amount, P = principle, R = rate, t = time

Calculation:

Given:

A = 2P

\(A = P{\left( {1 + \frac{R}{{100}}} \right)^t}\)

\(2P = P{\left( {1 + \frac{R}{{100}}} \right)^{10}}\)

\({\left( {1 + \frac{R}{{100}}} \right)^{10}} = 2\)

\({\left( {1 + \frac{R}{{100}}} \right)^{10}} = 2\)

R = 7.17 ≃ 7.2 %

 

\(Total;% ;,increase = \frac{{3660000 - 3000000}}{{3000000}} \times 100 = \frac{{660000}}{{3000000}} \times 100 = 22;% \)

Explanation:

The digit in unit place is selected in 4! Ways

The digit in tens place is selected in 4! Ways

The digit in hundreds place is selected in 4! Ways

The digit in the thousand's place is selected in 4! Ways

The digit in ten thousand's place is selected in 4! Ways

Sum of all values for 1

4! × 1 × (10⁰ + 10¹ + 10² + 10³ + 10⁴)

= 4! × 11111 × 1

Similarly for ‘3’ = 4! × (11111) × 3

Similarly for ‘5’ = 4! × (11111) × 5

Similarly for ‘7’ = 4!× (11111) × 7

Similarly for ‘9’ = 4!× (11111) × 9

∴ sum of all such number = 4! × (11111) × (1 + 3 + 5 + 7 + 9) = 24 × (11111) × 25 = 6666600

(P + Q)² = (P + Q).(P + Q)

            = P.P + P.Q + Q.P + Q.Q

            = P² + PQ + QP + Q² 

We cannot write PQ + QP = 2PQ

As matrix multiplication is not commutative.

i.e. PQ ≠ QP

\({\rm{;I}} = \mathop \smallint \limits_0^2 \frac{{{{\left( {{\rm{x}} - 1} \right)}^2}\sin \left( {{\rm{x}} - 1} \right)}}{{{{\left( {{\rm{x}} - 1} \right)}^2} + {\rm{cos}}\left( {{\rm{x}} - 1} \right)}}{\rm{dx}}\)

Let x – 1 = t

then, dx = dt

At x = 0, t = -1

At x = 2, t = 1

\(I = ;\mathop \smallint \limits_{ - 1}^1 \frac{{{t^2}sint}}{{{t^2} + cost}}dt\)

\(f\left( t \right) = ;\frac{{{t^2}sint}}{{{t^2} + cost}}\)

F(-t) = -f(t)

∴ f(t) is an odd function

By the property of the odd function

\(I = \mathop \smallint \limits_{ - a}^a f\left( t \right)dt = 0;\)

Given:

\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = - 2{\rm{xy}},{\rm{y}}\left( 0 \right) = 2\)

\(\Rightarrow \frac{{{\rm{dy}}}}{{{\rm{dx}}}} + 2{\rm{xy}} = 0\)

Comparing with \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {\rm{Py}} = {\rm{Q}}\)

P = 2x, Q = 0

\({\rm{I}}.{\rm{F}} = {{\rm{e}}^{\smallint 2{\rm{xdx}}}} = {{\rm{e}}^{{{\rm{x}}^2}}}\)

Solutions is given by:

\({\rm{y}}\left( {{\rm{I.F}}} \right) = \smallint {\rm{Q}}\left( {{\rm{I.F}}} \right){\rm{dx}} + {\rm{C}}\)

\({\rm{y}}{{\rm{e}}^{{{\rm{x}}^2}}} = 0 + {\rm{C}}\)

at x = 0, y = 2

2e⁰ = C

C = 2

\(∴ {\rm{y}} = 2{{\rm{e}}^{ - {{\rm{x}}^2}}}\)

Concept:

The normal distribution is symmetric with respect to the mean position and the standard normal variate is given by the relation.

Let variate be x

\({\rm{z}} = \frac{{{\rm{x}} - {\rm{M}}}}{{\rm{σ }}}\)

Where z =  Normal variate, M = Mean, σ = Standard deviation

For P(z ≥ 0) = P(z < 0) = 0.5 or 50%

Calculation:

Given:

M = 500

σ = 50

P(x > 500) = ?

Where X follows the normal distribution

We know that standard normal variable

\({\rm{z}} = \frac{{{\rm{x}} - {\rm{M}}}}{{\rm{σ }}}\)

converting into normalized form

x = 50z + 500

given, x > 500, therefore

50z + 500 > 500 or z > 0

probability, P(z > 0) = 0.5

∴ P(x > 500) = 0.5 or 50%

For x = 500, the standard normal variate is 

\({\rm{z}} = \frac{{500 - 500}}{{50}} = 0\)

∴ P(x > 500) = P(z > 0) = 0.50 or 50%(See figure)

Concept:

If L{f(t)} = F(s), then

• L{e^at f(t)} = F(s – a)
• L{e^-at f(t)} = F(s + a)

Calculation:

\(L\left{ {\cos 4t} \right} = \frac{s}{{{s^2} + {4^2}}}\)

\(\therefore {\rm{L}}\left{ {{{\rm{e}}^{ - 2{\rm{t}}}}\cos 4{\rm{t}}} \right} = \frac{{{\rm{s}} + 2}}{{{{\left( {{\rm{s}} + 2} \right)}^2} + {4^2}}}\)

L{e^-2t cos 4t} \(= \frac{{{\rm{s}} + 2}}{{{{\left( {{\rm{s}} + 2} \right)}^2} + 16}}\)

Explanation:

Truss: 

• A framework composed of members joined at their ends to form a rigid structure.

Planar truss: 

• In these types of trusses, members lie in a single plane.
• The basic element of a planar truss is a triangle i.e. three members are joined at three points.
• For a planar truss, as there are three unknown support reactions (For a node or joint, there will be three unknown but dependent support reactions). so if

1. m = 2j - 3 then system is statically determinate
2. m < 2j - 3 then system is collapsed
3. m > 2j - 3 then system is statically indeterminate

• where, m = number of members, j = total number of joints (including those attached to the supports)

Concept:

Relationship between modulus of elasticity (E), modulus of rigidity (G), and Poisson's ratio (μ) is described by:

E = 2G (1 + μ)

Calculation:

Given:

μ = 0.4

E = 2G (1 + μ)

\(\frac{{\rm{G}}}{{\rm{E}}} = \frac{1}{{2\left( {1 + {\rm{μ }}} \right)}} = \frac{1}{{2\left( {1 + 0.4} \right)}} = 0.357\)

Important Points

Other relationships between various elastic constants are:

• \({\bf{E}} = \frac{{9{\bf{KG}}}}{{3{\bf{K}}; + ,{\bf{G}};}}\)
• E = 3K(1 - 2μ)

​where K is the bulk modulus of elasticity.

Explanation:

Bevel Gears:

The bevel gears are used for transmitting power at a constant velocity ratio between two shafts whose axis intersect at a certain angle.

Helical Gears:

• A helical gear has teeth in the form of a helix around the gear.
• Two such gears can be used to connect two parallel shafts.
• In the case of single helical gears, there is some axial thrust between the teeth which is a disadvantage.
• In order to eliminate this thrust, double helical gears (herringbone gears) are used.
• It is equal to two single helical gears, in which two equal and opposite thrust are provided on each gear and the resulting axial thrust is zero.

​

Worm Gears:

The worm gears are generally used for transmitting power at a high-velocity ratio between non-intersecting shafts that are generally but not necessarily at right angles.

Rack and pinion Gears:

• Rack and pinion gear system are used to transfer rotation motion into linear motion.
• The flat tooth part is the rack and the gear is pinion.
• The rack has teeth cut into it and they mesh with the teeth of the pinion gear.

Explanation:

Isotropic material

• A material is said to be isotropic if it has similar properties in each direction.
• the relationship between various elastic constants is:

E = 2G (1 + μ) = 3K(1 - 2μ)

As seen in the above equations there are a total of 4 elastic constants.

If we know two of them, then we can find other constants also.

∴ there are 2 independent elastic constants required to define the stress-strain relationship for an isotropic material.

Orthotropic material

• A material is said to be orthotropic if it has three different properties in three mutually perpendicular directions.
• there are 9 independent elastic constants required to define the stress-strain relationship for orthotropic material.

Anisotropic material

• A material is said to be anisotropic if it has different properties in each direction.
• there are 21 independent elastic constants required to define the stress-strain relationship for an anisotropic material.

Concept:

Equation of displacement of the simple harmonic motion is given by:

x = Asin ωt or x = Acos ωt 

where A is amplitude, ω is the angular velocity

Calculation:

Given:

A = 10 mm = 0.01 m, frequency f = 4 Hz

Angular velocity (ω)

ω = 2 × π × f = 2 × π × 4 = 8π rad/s

Let x = Acos ωt

Velocity (v)

\({\rm{v}} = \frac{{{\rm{dx}}}}{{{\rm{dt}}}} = - {\rm{Aω }}\sin {\rm{ω t}}\)

Acceleration (a)

\(a = \frac{{dv}}{{dt}} = - Aω \frac{{d(\sin ω t)}}{{dt}} = - A{ω ^2}\cos ω t\)

At t = 0, value of cos ωt is maximum i.e. 1

a_max = -Aω²

∴ |a_max| = Aω² = 0.01 × (8π)² = 6.31 m/s²

Note:

You can also solve by assuming x = A sin ωt, answer would be same.

Concept:

Life of a bearing is given by:

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

where

L₁₀ = basic life rating in million revolutions

C = dynamic load-carrying capacity

W = equivalent dynamic load

n = constant = 3 for ball bearing and \(\frac{{10}}{3}{\rm{;}}\)for roller bearing

Calculation:

Given:

C = 16kN, W = 2kN, n = 3 (ball bearing)

Life of bearing

\({L_{10}} = {\left( {\frac{C}{W}} \right)^n} = {\left( {\frac{{16}}{2}} \right)^3} = 512;million;revolutions = 512 \times {10^6};revolutions\)

∴ p = 512

Explanation:

Gases transfer heat by the collision of molecules.

As the temperature increases, the kinetic energy of molecules of gases also increases and eventually collision between molecules also increases which increases the thermal conductivity of gases.

∴ As temperature increases the thermal conductivity of gases increases.

For liquid and solids, generally as the temperature increases, the thermal conductivity decreases.

Concept:

For a reversible Carnot cycle:

\(\oint \frac{{dQ}}{T} = 0\)

Calculation:

Given:

Q₁ = -2.5 kJ/s, T₁ = -5°c = 268 K, T₂ = 35°c = 308 K

As the cycle is reversible

\(\oint \frac{{dQ}}{T} = 0\)

\(\frac{{{Q_1}}}{{{T_1}}} + \frac{{{Q_2}}}{{{T_2}}} = 0\)

\( - \frac{{2.5}}{{268}} + \frac{{{Q_2}}}{{308}} = 0\)

\( \Rightarrow {Q_2} = \frac{{2.5 \times 308}}{{268}} = 2.873\frac{{kJ}}{s}\)

W = Q₁ – Q₂

W = 2.873 – 2.5 = 0.37313 kJ/s = 373.13 J/s = 373.13 W

Explanation:

Total acceleration of a flow is given by:

\(\frac{D\vec V}{Dt}=\frac{\partial\vec V}{\partial t}+u\frac{\partial\vec V}{\partial x}+v\frac{\partial\vec V}{\partial y}+w\frac{\partial\vec V}{\partial z}\)

The total derivative,

\(\frac{D}{Dt}=\frac{\partial}{\partial t}+u\frac{\partial}{\partial x}+v\frac{\partial}{\partial y}+w\frac{\partial}{\partial z}\)

The total differential D/Dt is known as the material or substantial derivative with respect to time.

The first term \(\frac{\partial}{\partial t}\) in the right hand side is known as temporal or local derivative which expresses the rate of change with time, at a fixed position.

The last three terms \(u\frac{\partial}{\partial x}+v\frac{\partial}{\partial y}+w\frac{\partial}{\partial z}\) in the right hand side of  are together known as convective derivative which represents the time rate of change due to change in position in the field.

Explanation:

Biot Number (B_i): 

• It is the ratio of conductive resistance within the body to convective resistance at the surface.

\({B_i} = \frac{{conductive;resistance;within;body}}{{convective;resistance;at;surface}} = \frac{{h{L_c}}}{k}\)

• It is used in conduction mode of heat transfer.

Grashoff Number

• It indicates the relative strength of the buoyant to viscous forces.

\({G_r} = \frac{{gB\left( {{T_s} - {T_\infty }} \right)L_c^3}}{{{\nu ^2}}}\)

• It is used in free convection.
• It has a role in free convection similar to that played by Reynold's number in forced convection.

Prandtl Number

• It is the ratio of molecular diffusivity of momentum to molecular diffusivity of heat.

\({P_r} = \frac{{molecular;diffusivity;of;momentum}}{{molecular;diffusivity;of;heat}} = \frac{{\mu {c_p}}}{k}\)

• It is used in convection mode of heat transfer.

Reynold’s Number

• It is the ratio of inertia force to viscous force.

\({R_e} = \frac{{inertia;force}}{{viscous;force}} = \frac{{\rho vL}}{\mu }\)

• It is used in fluid flow.

Explanation:

Classification of turbines on the various basis is given in the table below:

Concept:

A psychrometric chart is represented as shown in the figure.

Dry Bulb Temperature: Actual temperature of gas or mixture of gases

Wet bulb temperature: Temperature obtained by an accurate thermometer having a wick moistened with distilled water

Dew point temperature: Temperature at which the liquid droplets just appear when the moist air is cooled continuously.

Relative humidity along saturation line is 100%.

Basic Processes in Conditioning of Air:

Sensible heating: Moisture content of air remains constant so specific humidity is constant, temperature increases as it flows over a heating coil

Sensible cooling: Moisture content of air remains constant so specific humidity is constant, but its temperature decreases as it flows over a cooling coil

Cooling and dehumidification: When moist air is cooled below its dew-point by bringing it in contact with a cold surface, some of the water vapour in the air condenses and leaves the air stream as a liquid, as a result, both the temperature and humidity ratio of air decreases.

Heating and Humidification: During winter it is essential to heat and humidify the room air for comfort. It is done by first sensibly heating the air and then adding water vapour to the air stream through steam nozzles, as a result, both the temperature and humidity ratio of air increases.

Cooling & humidification: Air temperature drops and its humidity increases. 

Heating and de-humidification: This process can be achieved by using a hygroscopic material, which absorbs the water vapour from the moisture. If this process is thermally isolated, then the enthalpy of air remains constant, as a result, the temperature of air increases.

Calculation:

Inlet (State 1): 35°C and 100% RH

Outlet (State 2): 25°C and 100% RH

From the above figure, relative humidity is same but as the temperature is decreasing and specific humidity is also decreasing. So the process 1-2 is cooling and dehumidification process.

The name of the device is dehumidifier.

Explanation:

A decision variable is an unknown in an optimization problem.

So in an assignment problem of size n × n, the number of unknowns (decision variables) is n² and the number of constraints is 2n – 1.

Concept:

The expected time during lead time can be calculated by:

\({t_e} = ;\sum DP\left( x \right)\)

where D = Demand, P(x) = Probability

Calculation:

The expected demand during lead time is:

\({t_e} = ;\sum DP\left( x \right)\) = 74.55

Explanation:

The heat-affected zone (HAZ) is the area of the base material, either a metal or a thermoplastic, which is not melted and has had its microstructure and properties altered by welding or heat intensive cutting operations.

The extent and magnitude of property change depend primarily on the base material, the weld filler metal, and the amount and concentration of heat input by the welding process.

Concept:

Maximum uncut chip thickness is given by:

\({t_{max}} = \frac{{2v}}{{NZ}}\sqrt {\frac{d}{D}} \)

where,

d = depth of cut

v = linear speed of the workpiece

Z = number of teeth on the cutter periphery

N = rotational speed of the cutter

D = diameter of cutter

Calculation:

Given:

d₂ = 2d₁

As all other parameters are identical

\({t_{max}} \propto \sqrt d ;\)

\(\therefore \frac{{{t_1}}}{{{t_2}}} = \sqrt {\frac{{{d_1}}}{{{d_2}}}} = \sqrt {\frac{1}{2}} = \frac{1}{{\sqrt 2 }} = 0.707\)

Explanation:

Electrochemical machining is based on electrolysis process.

The electrolysis process is governed by the following two laws proposed by Faraday.

• The amount of chemical change produced by an electric current i.e. the amount of any material dissolved or deposited is proportional to the quantity of electricity passed.
• The amounts of different substances dissolved or deposited by the same quantity of electricity are proportional to their chemical equivalent weights.

Mathematically,

M ∝ Itε

\(M = \frac{{It\varepsilon }}{F}\)

where,

M = weight (in grams) of material dissolved or deposited

I = current (in amperes)

ε = gram equivalent weight of the material

F = Faraday’s constant = 96,500 Coulombs

Concept:

We know that According to merchant analysis:

2ϕ + β - α = π/2

This equation concludes:

• As the rake angle increases or the friction at tool chip interface (rake face) decreases, shear angle increases and the chip becomes thinner.
• Thinner chips mean less energy dissipation because the shear strain is lower.
• As work done during cutting is converted into heat, the temperature rise and cutting force is also lower.

Concept:

Heat treatment:

• Heat treatment operation can be defined as heating a metal or alloy to various definite temperatures, holding these for the various time durations and cooling at various rates to get the desired grain structure and property in the metal or alloy.

Different heat treatment processes are:

Explanation:

The maximum possible draft is given by:

\({\left( {{t_f} - {t_i}} \right)_{max}} = {\mu ^2}R = \frac{{{\mu ^2}D}}{2}\)

where μ is the coefficient of friction and D is roll diameter.

Concept:

If a complex number is in the form Z = re^iθ

Then ln Z = ln(r) + iθ

where \(r = \sqrt {{x^2} + {y^2}} \)

θ = argument = \({\tan ^{ - 1}}\left( {\frac{y}{x}} \right)\)

Calculation:

\(\left. {\mathop \smallint \limits_5^{3{\rm{i}}} \frac{{{\rm{dz}}}}{{\rm{z}}} = \ln \left( {\rm{z}} \right)} \right|_5^{3{\rm{i}}}\)

ln(3i) – ln(5)

ln(0 + 3i) – ln(5 + 0i)

\(\left{ {\ln \left( {\sqrt {{0^2} + {3^2}} } \right) + i\left( {{{\tan }^{ - 1}}\left( {\frac{3}{0}} \right)} \right)} \right} - \left{ {\ln \left( {\sqrt {{5^2} + {0^2}} } \right) + i\left( {{{\tan }^{ - 1}}\left( {\frac{0}{5}} \right)} \right)} \right}\)

\(\left{ {\ln \left( 3 \right) + i\left( {\frac{\pi }{2}} \right)} \right} - \)(\left{ {\ln \left( 5 \right) + i\left( 0 \right)} \right}\)

\(\ln \left( {\frac{3}{5}} \right) + i\left( {\frac{\pi }{2}} \right)\)

= - 0.511 + i(1.57)

\(\mathop \smallint \limits_0^2 \mathop \smallint \limits_0^{\rm{x}} {{\rm{e}}^{{\rm{x}} + {\rm{y}}}}{\rm{dydx}} = \mathop \smallint \limits_0^2 {{\rm{e}}^{\rm{x}}}\left( {\mathop \smallint \limits_0^{\rm{x}} {{\rm{e}}^{\rm{y}}}{\rm{dy}}} \right){\rm{dx}}\)

\(= \mathop \smallint \limits_0^2 {{\rm{e}}^{\rm{x}}}\left( {{{\rm{e}}^{\rm{y}}}} \right)_0^{\rm{x}}{\rm{dx}} = \mathop \smallint \limits_0^2 {{\rm{e}}^{\rm{x}}}\left( {{{\rm{e}}^{\rm{x}}} - 1} \right){\rm{dx}}\)

\(= \mathop \smallint \limits_0^2 \left( {{{\rm{e}}^{2{\rm{x}}}}-{{\rm{e}}^{\rm{x}}}} \right){\rm{dx}} = \left( {\frac{{{{\rm{e}}^{2{\rm{x}}}}}}{2} - {{\rm{e}}^{\rm{x}}}} \right)_0^2\)

\(= \frac{{{{\rm{e}}^4}}}{2} - {{\rm{e}}^2} - \frac{1}{2} + 1 = \frac{{{{\rm{e}}^4}}}{2} - {{\rm{e}}^2} + \frac{1}{2}\)

\(= \frac{1}{2}\left( {{{\rm{e}}^4} - 2{{\rm{e}}^2} + 1} \right) = \frac{1}{2}{\left( {{{\rm{e}}^2} - 1} \right)^2}\)

Concept:

Poisson distribution

According to the Poisson distribution probability of x = k is given by:

\(P\left( {x = k} \right) = \frac{{{\lambda ^k}{e^{ - \lambda }}}}{{k!}}\)

Calculation:

Given:

λ = 5.2

Let x be random variable which follows Poisson’s distribution

P(x < 2) = P(x = 0) + P(x = 1)

\(= \frac{{{{\rm{e}}^{ - {\rm{\lambda }}}}{{\rm{\lambda }}^0}}}{{0!}} + \frac{{{{\rm{e}}^{ - {\rm{\lambda }}}}}}{{1!}}{{\rm{\lambda }}^1} = {{\rm{e}}^{ - 5.2}}\left( {1+5.2} \right) = 0.0055 \times 6.2 = 0.034\)

Concept:

Runga Kutta method of fourth-order

\(\frac{{dx}}{{dt}} = f\left( {t,x} \right)\)

x(t_o) = x_o

t_n = t_o + nΔt

where Δt is the step size, n = 1,2,3 ..

K₁ = Δt × f(t_o,x_o)

\({K_2} = {\rm{Δ }}t × f\left( {{t_o} + \frac{{{\rm{Δ }}t}}{2},{x_o} + \frac{{{K_1}}}{2}} \right)\)

\({K_3} = {\rm{Δ }}t × f\left( {{t_o} + \frac{{{\rm{Δ }}t}}{2},{x_o} + \frac{{{K_2}}}{2}} \right)\)

K₄ = Δt × f(t_o + Δt, x_o + K₃)

\(K = \frac{1}{6}\left( {{K_1} + 2{K_2} + 2{K_3} + {K_4}} \right)\)

where K is the increment in x.

Calculation:

Given:

\(\frac{{{\rm{dx}}}}{{{\rm{dt}}}} = 4{\rm{t}} + 4\) = f(t,x)

At t = 0,  x = x_o 

Δt = 0.2

f(t,x) does not depend on value of x as it is function of t only.

K₁ =  Δt × f(t_o,x_o) = 0.2 × f(0,x_o) = 0.2 × 4 = 0.8

\({{\rm{K}}_2} = \Delta t×{\rm{f}}\left( {{{\rm{t}}_o} + \frac{{\rm{Δ t}}}{2},{{\rm{x}}_o} + \frac{1}{2}{{\rm{K}}_1}{\rm{}}} \right)\)

K₂ = Δt × f(0 + 0.1,x_o + 0.4) = Δt × f(0.1,x_o + 0.4) = 0.2 × {4(0.1) + 4} = 0.88

\({{\rm{K}}_3} = \Delta t× {\rm{f}}\left( {{{\rm{t}}_o} + \frac{{\rm{Δ t}}}{2},{{\rm{x}}_o} + \frac{{{{\rm{K}}_2}{\rm{}}}}{2}} \right)\)

K₃ = Δt × f{t_o + 0.1,x_o + (0.44)} = Δt × f(0.1,x_o + 0.44) = 0.2 × {4(0.1) + 4} = 0.88

K₄ = Δt × f(t_o + Δt,x_o + K₃)

K₄ = Δt × f(0 + 0.2,x_o + 0.88) = Δt × f(0.2,x_o + 0.88) = 0.2 × {4(0.2) + 4} = 0.96

Increment in x (K)

\(K = \frac{1}{6}\left( {{K_1} + 2{K_2} + 2{K_3} + {K_4}} \right)\)

\(K = \frac{1}{6}\left{ {0.8 + \left( {2 \times 0.88} \right) + \left( {2 \times 0.88} \right) + 0.96} \right}\)

K = 0.88

Concept:

According to Von-Mises theory,

\({\left( {{σ _1} - {σ _2}} \right)^2} + {\left( {{σ _2} - {σ _3}} \right)^2} + {\left( {{σ _1} - {σ _3}} \right)^2} \le \frac{{2{σ _y^2}}}{{FOS}^2}\)

if, σ3 = 0

\(σ _1^2 + σ _2^2 - {σ _1}{σ _2} \le \frac{{σ _y^2}}{{FOS}^2}\)

Calculation:

Given:

σ₁ = 100 MPa, σ₂ = -100 MPa, σ_y = 300 MPa

This is a case of pure shear. And in this case, maximum and minimum principal stresses will be equal and opposite nature and their magnitude will be equal to the shear stress.

According to Von-Mises theory,

\(σ _1^2 + σ _2^2 - {σ _1}{σ _2} \le \frac{{σ _y^2}}{{FOS}^2}\)

\({100^2} + {100^2} +100 \times 100 \le \frac{{{{300}^2}}}{{FOS}^2}\Rightarrow FOS^2=\frac{300^2}{3\times100^2}=\frac{3^2\times100^2}{3\times100^2}=3\)

then, FOS = 1.732

Concept:

In case of thin cylinder,

Hoop Stress, \(\sigma_h = \frac{{Pd}}{{2t}}\)

Longitudinal stress, \(\sigma_l = \frac{{Pd}}{{4t}}\)

The maximum stress in case of the thin cylinder is

\(\sigma = \sigma_h = \frac{{Pd}}{{2t}}\)

Calculation:

Given:

σ = 100 MPa, P = 10 MPa, d = 2r = 200 mm

\(\sigma = \frac{{Pd}}{{2t}}\)

\(100 = \frac{{10 \times 200}}{{2 \times t}}\)

⇒ t = 10 mm

\(\tan \theta = \frac{{1.5}}{2} = \frac{3}{4} ⇒ \theta = {\tan ^{ - 1}}\left( {\frac{3}{4}} \right) = 37^\circ \)

At point P,

 After rearrangement ⇒ 

Using Lami’s theorem

\(\frac{9}{{\sin \left( {90^\circ + 37^\circ } \right)}} = \frac{{{F_{PQ}}}}{{\sin \left( {180^\circ - 37^\circ } \right)}} = \frac{{{F_{PS}}}}{{\sin 90^\circ }}\)  ...(1)

\(\frac{9}{{\cos 37^\circ }} = \frac{{{F_{PQ}}}}{{\sin 37^\circ }}\)

\({F_{PQ}} = \frac{{9 \times \sin 37^\circ }}{{\cos 37^\circ }} = 9 \times \frac{3}{4} = 6.75;kN\)

Now, again from equation (1)

\(\frac{9}{{\sin \left( {90^\circ + 37^\circ } \right)}} = \frac{{{F_{PS}}}}{{\sin 90^\circ }}\)

\(\frac{9}{{\cos 37^\circ }} = \frac{{{F_{PS}}}}{{\sin 90^\circ }}\)

\({F_{PS;}} = \frac{9}{{\cos 37^\circ }} = 11.25;kN;\left( {compressive} \right)\)

At point S,

\(\sum {F_v} = 0\)

F_PS × cos θ = F_SQ × cos θ

F_PS = F_SQ = 11.25 kN

F_SQ = 11.25 kN (tensile)

As both F_PS and F_SQ have opposite directions.

Concept:

As the pinion to gear speed ratio(G) increases, the minimum number of teeth (t) on the pinion decreases.

Minimum Number of Teeth on the Wheel in Order to Avoid Interference:

\(T=\frac{{2{{\rm{A}}_{\rm{w}}}}}{{\sqrt {1 + \frac{1}{G}\left( {\frac{1}{G} + 2} \right){{\sin }^2}\phi } - 1;}}\)

 Minimum Number of Teeth on the Pinion in Order to Avoid Interference:

\(t=\frac{{2{{\rm{A}}_{\rm{p}}}}}{{\sqrt {1 + G\left( {G + 2} \right){{\sin }^2}\phi } - 1;}}\)

therefore if G increases, t decreases.

AP = 1 m, AC = 4 m, PC = 3 m,

and from the given information, we get

m₁ = 3×1/8 = 3/8 kg, m₂ = 3×7/8 = 21/8 kg

moment of inertia of rod about its C.G., 

\({I_C} = \frac{{m{L^2}}}{{12}} = \frac{{3 \times {8^2}}}{{12}} = 16\ kg - {m^2}\)

Then, \({I_P} = {I_C} + m{\left( {PC} \right)^2} = 16 + 3 \times {3^2} = 43\ kg - {m^2}\)

Torque \( = {m_2}g \times \frac{{PB}}{2} - {m_1}g \times \frac{{AP}}{2}\)

\(\begin{array}{l} \Rightarrow T = {m_2} \times 9.81 \times \frac{7}{2} - {m_1} \times 9.81 \times \frac{1}{2}\ \Rightarrow T = 34.335{m_2} - 4.905{m_1} \end{array}\)

\( \Rightarrow T = 34.335 \times \frac{3}{8} - 4.905 \times \frac{{21}}{8} = 88.29\ N - m\)

And we know that, T = I_pα 

\( \Rightarrow \alpha = \frac{T}{{{I_P}}} = \frac{{88.29}}{{43}} = 2.053\ rad/{s^2}\)

Concept:

To determine the effective stiffness of the bolt in the clamped zone, we consider the contributions from both the threaded and unthreaded portions of the bolt.

Calculation:

Given:

Major diameter of the bolt, \( D = 12 , \text{mm} \)

Cross-sectional area of the threaded portion, \( A_t = 84.3 , \text{mm}^2 \)

Length of the threaded portion in grip, \( L_t = 30 , \text{mm} \)

Length of the unthreaded portion in grip, \( L_u = 8 , \text{mm} \)

Young's modulus, \( E = 200 , \text{GPa} = 200 \times 10^3 , \text{N/mm}^2 \)

Cross-sectional area of the unthreaded portion, \( A_u = \frac{\pi (12)^2}{4} \approx 113.1 , \text{mm}^2 \)

The stiffness of the threaded portion, \( k_t \) is:

\( k_t = \frac{A_t E}{L_t} = \frac{84.3 \times 200 \times 10^3}{30} = 562 , \text{MN/m} \)

The stiffness of the unthreaded portion, \( k_u \) is:

\( k_u = \frac{A_u E}{L_u} = \frac{113.1 \times 200 \times 10^3}{8} = 2827.5 , \text{MN/m} \)

The effective stiffness, \( k_{\text{eff}} \) is calculated as:

\( \frac{1}{k_{\text{eff}}} = \frac{1}{k_t} + \frac{1}{k_u} \)

\( \frac{1}{k_{\text{eff}}} = \frac{1}{562} + \frac{1}{2827.5} \approx 0.002133 \)

\( k_{\text{eff}} \approx \frac{1}{0.002133} = 469 , \text{MN/m} \)

The effective stiffness of the bolt in the clamped zone is:

\( \boxed{469 , \text{MN/m}} \)

Concept:

Strain energy due to bending moment ‘M’ in small length 'dx' is given by:

\({U_{x}} = \frac{{{M^2}dx}}{{2EI}}\)

Total strain energy can be calculated by:

\(U = ;\mathop \smallint \limits_0^L \frac{{{M^2}dx}}{{2EI}}\)

Calculation:

Given:

The horizontal section of the bar has variable moment i.e. M_x = Px

The vertical section of the bar has constant moment i.e. M_y = PL

\({\rm{Strain;energy}}\left( {\rm{U}} \right) = \mathop \smallint \limits_0^{\rm{L}} \frac{{{\rm{M}}_{\rm{x}}^2{\rm{dx}}}}{{2{\rm{EI}}}} + \mathop \smallint \limits_0^{\rm{L}} \frac{{{\rm{M}}_{\rm{y}}^2{\rm{dx}}}}{{2{\rm{EI}}}}\)

\(\frac{1}{2}{\rm{P\delta }} = \mathop \smallint \limits_0^{\rm{L}} \frac{{{{\left( {{\rm{Px}}} \right)}^2}{\rm{dx}}}}{{2{\rm{EI}}}} + \mathop \smallint \limits_0^{\rm{L}} \frac{{{{\left( {{\rm{PL}}} \right)}^2}{\rm{dx}}}}{{2{\rm{EI}}}}\)

\(= \frac{{{{\rm{P}}^2}\left[ {\frac{{{{\rm{x}}^3}}}{3}} \right]_0^{\rm{L}}}}{{{\rm{2EI}}}} + \frac{{{{\rm{P}}^2}{{\rm{L}}^2}}}{{2{\rm{EI}}}}\left[ {\rm{x}} \right]_0^{\rm{L}} = \frac{2}{3}\times\frac{{{{\rm{P}}^2}{{\rm{L}}^3}}}{{{\rm{EI}}}} = \frac{1}{2} \times {\rm{P}} \times \frac{4}{3}\frac{{{\rm{P}}{{\rm{L}}^3}}}{{{\rm{EI}}}}\)

\(\therefore;{\rm{\delta }} = \frac{{{\rm{4P}}{{\rm{L}}^3}}}{{{3\rm{EI}}}}\)

Given:

Mass of wardrobe (m) = 100 kg, height = 4 m, width = 2 m, depth = 1 m

The wardrobe will be on the verge of tip if the normal reaction passes through the point Q as shown in the free body diagram:

Taking moments about Q, ∑M_Q = 0

⇒ W × 1 = P × 2

\(\Rightarrow {\rm{P}} = \frac{{100 \times 9.81}}{2} = 490.5{\rm~{N}}\)

∑H = 0

⇒ F_F = P = 490.5 N

∑V = 0 ⇒ R_N = mg = 981 N

Friction Force = μR_N

\({\rm{\mu }} = \frac{{490.5}}{{981}} = 0.5\)

Concept:

The maximum fluctuation of energy is also given by:

ΔE_max = I × (ω)² × C_s

where,

I = Moment of inertia

ω = Average angular velocity of the flywheel

C_s = Coefficient of fluctuation of energy

Calculation:

From the ω - θ diagram, ω_max = 20 rad/s, ω_min = 10 rad/s

\({\rm{\omega }} = \frac{{{{\rm{\omega }}_{{\rm{max}}}} + {{\rm{\omega }}_{{\rm{min}}}}}}{2} = \frac{{20 + 10}}{2} = 15\)

\({{\rm{C}}_{\rm{s}}} = \frac{{{{\rm{\omega }}_{{\rm{max}}}} - {{\rm{\omega }}_{{\rm{min}}}}}}{{\rm{\omega }}} = \frac{{20 - 10}}{{15}} = 0.6667\)

Since the area above the mean torque line represents the maximum fluctuation of energy , therefore, maximum fluctuation of energy:

\({\rm{\Delta E}} = {\rm{Area;of;T}} - {\rm{\theta\ diagram\ between\ \frac{\pi}{2}\ to \ \pi}} = \frac{{\rm{\pi }}}{2} \times 3000 = 1500{\rm{\pi }}\)

ΔE = Iω²C_s

⇒\(1500\pi = I\times{(15)^2}\times{\frac{2}{3}}\)

⇒ I = 31.416 kg-m²

Concept:

The dynamic magnification factor is given by:

\(MF = \frac{1}{{\sqrt {{{\left[ {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]}^2} + {{\left[ {2\xi \frac{\omega }{{{\omega _n}}}} \right]}^2}} }}\)

Calculation:

Given:

ω = 1.5 rad/s

\(\begin{array}{l} {\omega _n} = \sqrt {\frac{K}{m}} = \sqrt {\frac{8}{2}} = 2\frac{{rad}}{s}\ \xi = 0.02,\frac{\omega }{{{\omega _n}}} = \frac{{1.5}}{2} = 0.75\ \therefore MF = \frac{1}{{\sqrt {{{\left[ {1 - {{\left( {0.75} \right)}^2}} \right]}^2} + {{\left( {2 \times 0.02 \times 0.75} \right)}^2}} }}\ \therefore MF = \frac{1}{{\sqrt {{{\left( {0.4375} \right)}^2} + {{\left( {0.03} \right)}^2}} }} = \frac{1}{{\sqrt {0.1923} }} \end{array}\)

MF = 2.28

For Ladder to be in equilibrium

\(\sum {F_x} = 0\),\(;\sum {F_y} = 0,\) \(\sum {M_{any;point}} = 0\)

From free body diagram of the ladder

W = N_B

P = N_A

\(\sum {M_B} = 0\)

W × 2 – N_A × 3 = 0 

\({N_A} = \frac{{W \times 2}}{3} = \frac{{600 \times 2}}{3} = 400;N\)

∴ P = N_A = 400 N

Given:

m = 10 kg, T = 10°C = 283.15 K, Q = 16.2 MJ = 16.2 × 10³ kJ, S_f = 0.88 kJ/kg.K

Tds = dh - vdp

Tds = dh (constant pressure) = \(\frac{H}{m}\)

283.15 × (s₂ – s₁) = \(\frac{{16.2 × {{10}^3}}}{{10}}\)

\(\therefore {{\rm{s}}_2} = 0.88 + \frac{{16.2 × {{10}^3}}}{{283.15 × 10}} = 6.6013~\frac{{{\rm{kJ}}}}{{{\rm{kg}}}}\)

OR 

Alternate Solution:

Q = mh_fg

16.2 × 10³ kJ = 10h_fg

h_fg = 16.2 × 10² kJ/kg

Now \({s_{fg}} = \frac{{{h_{fg}}}}{T} = \frac{{16.2 × {{10}^2}{\rm{kJ}}/{\rm{kg}}}}{{283}} = 5.7244~kJ/kgK\)

Since \({s_{fg}} = {s_g} - {s_f}\)

\(5.7244 = {s_g} - 0.88\)

\({s_g} = 6.604~kJ/kgK\)

Concept:

Heat transfer through a plane wall is given by:

\(Q = kA\frac{{({T_1} - {T_2})}}{L}\)

where, k = thermal conductivity of wall, A = cross-section area of the wall,

T₁ = higher temperature, T₂ = lower temperature, L = thickness of the wall

Calculation:

Given:

k = 1.15 W/mK, T₁ = 1100 °c, T₂ = 350 °c, \(\frac{Q}{A} = 2500\frac{W}{{{m^2}}}\)

\(\frac{Q}{A} = k\frac{{({T_1} - {T_2})}}{L}\)

\(2500 = \frac{{1.15 \times \left( {1100 - 350} \right)}}{L}\)

\(L = \frac{{862.5}}{{2500}} = 0.345;m\)

Concept:

Streamline:

• It is an imaginary line or curve drawn in space such that the tangent drawn at any point to it will give the direction of velocity.
• As there is no component of velocity in the perpendicular direction, there is no flow across the streamline, i.e., flow occurs along the streamline.
• For steady flow, the path of the particle is streamline

Equation of streamline:

If the velocity vector is given as:

\(\vec V = u\hat i + v\hat j + w\hat k\)

Then the equation of streamline is given by:

\(\frac{{dx}}{u} = \frac{{dy}}{v} = \frac{{dz}}{w}\)

Concept:

If the velocity vector is given as:

\(\vec V = u\hat i + v\hat j + w\hat k\)

Vorticity in the z-direction is defined as:

\({{\rm{\Omega }}_{\rm{z}}} = \frac{{\partial {\rm{v}}}}{{\partial {\rm{x}}}} - \frac{{\partial {\rm{u}}}}{{\partial {\rm{y}}}}\)

Calculation:

Given:

\(\vec V = K\left( {y\hat i + x\hat k} \right)\)

\(\vec u = Ky,;\vec v = 0,;\vec w = Kx\)

\({{\rm{\Omega }}_z} = \frac{{\partial v}}{{\partial x}} - \frac{{\partial u}}{{\partial y}} = - \frac{{K\partial \left( y \right)}}{{\partial y}} = - K\)

Concept:

Reynold’s number for a pipe is given by:

\(Re = \frac{{\rho VD}}{\mu}\)

where, ρ = density of water, V = velocity of water, D = diameter of pipe, μ = dynamic viscosity

Nusselt number for a pipe is given by:

\(Nu = \frac{{hD}}{k}\)

where, h = heat transfer coefficient, D = diameter of the pipe, k = thermal conductivity

Calculation:

Given:

ρ = 1000 kg/m³, μ = 7.25 × 10^-4 N.s/m², k = 0.625 W/mK, Pr = 4.85, V = 1 m/s

D = 25 mm = 0.025 m

Reynold’s number

\(Re = \frac{{\rho VD}}{\mu } = \frac{{1000 \times 1 \times 0.025}}{{7.25 \times {{10}^{ - 4}}}} = 34482.758\)

Now,

 Nu = 0.023 Re^0.8Pr^0.4

Nu = 0.023 × (34482.758)^0.8 × (4.85)^0.4

Nu = 184.5466

\(\frac{{hD}}{k} = 184.5466\)

\(h = \frac{{184.5466 \times 0.625}}{{0.025}} = 4613.66\frac{W}{{{m^2}K}}\)

Concept:

Entropy change of block \(L = {\rm{\Delta }}{S_L} = mc{\log _e}\frac{{{T_2}}}{{{T_1}}}\)

Entropy change of block \(M = {\left( {{\rm{\Delta }}S} \right)_M} = mC{\log _e}\frac{{{T_f}}}{{{T_1}}}\)

Entropy change of Universe

\({\rm{\Delta }}{S_{univ}} = {\rm{\Delta }}{s_L} + {\rm{\Delta }}{s_M} = mclo{g_e}\frac{{{T_2}}}{{{T_1}}} + mclo{g_e}\frac{{{T_f}}}{{{T_1}}};\)

For a reversible refrigerator

\({\left( {{\rm{\Delta }}s} \right)_{univ}} = 0\)

\( \Rightarrow mc{\log _e}\frac{{{T_2}}}{{{T_1}}} + mc{\log _e}\frac{{{T_f}}}{{{T_1}}} = 0\)

\(mc{\log _e}\frac{{{T_2}{T_f}}}{{T_1^2}} = 0\)

or \({\log _e}\frac{{{T_2}{T_f}}}{{T_1^2}} = 0\)

\( \Rightarrow \frac{{{T_2}{T_f}}}{{T_1^2}} = {e^0} = 1\)

Calculation:

Given:

Specific heat (c) = 0.4 kg/kgK

Mass (m) = 5 kg

T₁ = 313 k

Final temperature of Block L = T₂ = 293 K

Let T_f = final temperature of block μ

\( \Rightarrow {T_f} = \frac{{T_1^2}}{{{T_2}}} = \frac{{{{\left( {313} \right)}^2}}}{{293}} = 334.36;K\)

Calculation:

Given:

ṁ₁ = 10 kg/s, h₁ = 3214 kJ/kg, ṁ₂ = 1.5 kg/s, h₂ = 2920 kJ/kg, ṁ₃ = 8.5 kg/s, h₃ = 2374 kJ/kg

As the system is in steady-state

Power output (W)

W = ṁ₁(h₁ – h₂) + ṁ₃(h₂ – h₃)

W = 10(3214 – 2920) + 8.5(2920 – 2374)

W = 2940 + 4641 = 7581 kJ/s = 7581 kW

Concept:

Heat transfer relation when emissivities of plates and shield is same:

\(\frac{{{{\left( Q \right)}_{with;shield}}}}{{{{\left( Q \right)}_{without;shield}}}} = \frac{1}{{N + 1}}\)

where N is the number of shields.

Calculation:

Given:

N = 1

\(\frac{{{{\left( Q \right)}_{with;shield}}}}{{{{\left( Q \right)}_{without;shield}}}} = \frac{1}{{N + 1}} = \frac{1}{{1 + 1}} = \frac{1}{2}\)

Expansion Ratio: \(r_e=\frac{V_4}{V_3}=\frac{V_1}{V_3}=8\)

Compression Ratio: 

\({r_k} = \frac{{{V_1}}}{{{V_2}}} = {\left( {\frac{{{P_2}}}{{{P_1}}}} \right)^{\frac{1}{\gamma }}} = {\left( {32.42} \right)^{\frac{1}{{1.4}}}} = 11.999 \simeq 12\)

Cut off ratio:

\({r_c} =\frac{V_3}{V_2}=\frac{V_3}{V_1}.\frac{V_1}{V_2}=\frac{\frac{V_1}{V_2}}{\frac{V_1}{V_3}}= \frac{{compression;ratio}}{{expansion;ratio}} = \frac{{12}}{8} = 1.5\)

\(\eta = 1 - \frac{1}{{r_k^{\gamma - 1}}}\left( {\frac{{r_c^\gamma - 1}}{{\gamma \left( {{r_c} - 1} \right)}}} \right)\)

\(= 1 - \frac{1}{{{{12}^{0.4}}}}\left( {\frac{{{{\left( {1.5} \right)}^{1.4}} - 1}}{{1.4\left( {1.5 - 1} \right)}}} \right)\)

η = 59.6%

Concept:

Total float:

Total float represents the amount of time by which an activity can be delayed without delaying the project completion time. Thus, it is the extra time available for an activity.

Total float = L_j – (E_i + t_ij)

Calculation:

Given:

For activity e, L_j = 7, E_i = 4, t_ij = 2

For activity f, L_j = 12, E_i = 4, t_ij = 4

Total float (T.F)

(T.F)_e = L_j – (E_i + t_ij) = 7 – (4 + 2) = 1

(T.F)_f = L_j – (E_i + t_ij) = 12 – (4 + 4) = 4

• Free float (F.F): It is the extra time available for an activity without affecting its successor activity.

F.F = E_j - (E_i + t_ij)

• Independent float (I.F): It is the minimum extra time if preceding activity ends as late as possible and succeeding activity starts as early as possible.

I.F = Ej - (Li + tij)

• For critical activity, T.F = F.F = I.F

Concept:

The number of jobs in the system is given by:

\({N_s} = \frac{\lambda }{{\mu - \lambda }}\)

where, λ = arrival rate, μ = service rate

Calculation:

Given:

λ (arrival rate) = 5 jobs/minute

μ (service rate) = 8 jobs/minute

Number of jobs in the system

\({L_s} = \frac{\lambda }{{\mu - \lambda }} = \frac{5}{{8 - 5}} = 1.67\)

The number of jobs in the queue is given by:

\(​{N_q} = \left( {\frac{\lambda }{\mu }} \right)\left( {\frac{\lambda }{{\mu - \lambda }}} \right)\)

Concept:

• Go gauge is always made for the maximum material limit of the component.
• NoGo gauge is always made for the minimum material limit of the component.
• For hole, the maximum material limit is the lower limit of the hole.
• For the shaft, the maximum material limit is the upper limit.

Calculation:

Given:

The upper limit of hole = 25 + 0.015 = 25.015 mm

The lower limit of hole = 25 – 0.015 = 24.985 mm

Total tolerance = 0.015 + 0.015 = 0.03 mm

Work tolerance = 0.1 × 0.03 = 0.003 mm

As Go gauge is made for the lower limit of the hole

∴ The lower limit of Go gauge is 24.985 – 0.000 (mm)

And Go gauge have unilateral tolerance

∴ The upper limit of Go gauge is 24.985 + 0.003 (mm)

∴ Dimensions of Go gauge is \(24.985_{ - 0.000}^{ + 0.003}\) mm.

Concept:

Solidification time of a body is:

\({T_s} \propto {\left( {\frac{V}{{S.A}}} \right)^2}\)

where, V = volume of the body, S.A = total surface area of the body

Top riser (d = 2h)

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

\(S.A = \frac{\pi }{4}{d^2} + \pi dh\)

\(\frac{V}{{S.A}} = \frac{d}{6}\)

Side riser (d = h)

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

\(S.A = \frac{\pi }{2}{d^2} + \pi dh\)

\(\frac{V}{{S.A}} = \frac{d}{6}\)

Calculation:

Given:

Length of casting (L) = 7 cm, Breadth of casting (B) = 10 cm, Height of Casting (H) = 2 cm

Solidification time for casting (T_s)_casting = 1.36 minutes

For riser, diameter (d) = height (h) = 6 cm

\({\left( {\frac{V}{{S.A}}} \right)_{casting}} = \frac{{LBH}}{{2\left( {LB~ +~ BH ~+ ~HL} \right)}} = \left( {\frac{{7 ~\times~ 10~ \times~ 2}}{{2\left{ {\left( {7~ \times~ 10} \right)~ +~ \left( {10~ \times~ 2} \right) ~+ ~\left( {2~ \times~ 7} \right)} \right}}}} \right)\)

\({\left( {\frac{V}{{S.A}}} \right)_{casting}} = \frac{{140}}{{208}} = 0.673;cm\)

\({\left( {\frac{V}{{S.A}}} \right)_{riser}} = \frac{d}{6} = \frac{6}{6} = 1;cm\)

\(\frac{{{{\left( {{T_s}} \right)}_{riser}}}}{{{{\left( {{T_s}} \right)}_{casting}}}} = {\left( {\frac{{{{\left( {\frac{V}{{S.A}}} \right)}_{riser}}}}{{{{\left( {\frac{V}{{S.A}}} \right)}_{casting}}}}} \right)^2}\)

\(\frac{{{{\left( {{T_s}} \right)}_{riser}}}}{{{{\left( {{T_s}} \right)}_{casting}}}} = {\left( {\frac{1}{{0.673}}} \right)^2} = 2.20;\)

(T_s)_riser = 2.20 × 1.36 = 3 minutes

Concept:

Velocity is given by:

(V) = f × N

where, f = feed (mm/stroke), N = number of cutting strokes per minute

Area of the workpiece (A) = d × L

where, d = depth of cut, L = length of the workpiece

Material removal rate (MRR) is given by:

\(Q = \frac{{f \times N \times d \times L}}{{60}}\)

where Q is in \(\frac{{{\rm{m}}{{\rm{m}}^3}}}{{\rm{s}}}\)

Average power consumption is given by:

P_avg = Q × U

where U is specific cutting energy.

Calculation:

Given:

f = 0.25 mm/stroke, N = 60, d = 4 mm, L = 200 mm and U = \(1.49\frac{J}{{m{m^3}}}\)

Material removal rate (Q)

\(Q = \frac{{f \times N \times d \times L}}{{60}} = \frac{{0.25 \times 60 \times 4 \times 200}}{{60}} = 200\frac{{m{m^3}}}{s}\)

Average power (P_avg)

P_avg = Q × U = 200 × 1.49 = 298 W

Calculation:

Given:

thickness (t) = 8 mm, width (b) = 20 mm, under cut (d) = 3 mm

yield strength (σ_y) = 500 MPa,  ultimate tensile strength (σ_t) = 700 MPa

Area = (t - d)b

Area = (8 - 3) × 20 = 100 mm²

Maximum transverse tensile load,

P_max = σ_t × Area

P_max =  700 N/mm2 × 100 mm² = 70,000 N = 70 kN