GATE ME 2022 Official Paper: Shift 1 (Previous Year Paper)

The correct answer is 'two / too'.

Key Points

• The given words are homophones i.e. similar sounding words.
• Two: equivalent to the sum of one and one; one less than three; 2.
• For eg.- She ate two buns.
• Too: to a higher degree than is desirable, permissible, or possible; excessively.
• For eg.- He was driving too fast.

The complete sentence will be: After playing two hours of tennis, I am feeling too tired to walk back.

• Hence, option 4 is the correct answer.

Additional Information

• In linguistics, homonyms, broadly defined, are words that are homographs or homophones, or both.
• For eg.- Bow-bow, Reed-read, etc.

Explanation:

Average of salaries of M, N, and S = Rs. 4000

Sum of salaries of M, N and S = (4000 × 3) = Rs. 12,000.

Average of salaries of N, S, and P = Rs. 5000

Sum of salaries of N, S and P = (5000 × 3) = Rs. 15,000.

Given salary of P is = Rs. 6000

∴ salary of N and S is = 15000 - 6000 = Rs. 9000.

∴ salary of M will be = (Sum of M, N and S) - (Sum of N and S)

∴ salary of M will be = 12,000 - 9,000 = Rs. 3,000

∴ the monthly salary of M as a percentage of the monthly salary of P is:

\(\frac{M}{P}\times100;%\)

\(\frac{3000}{6000}\times100;%=50;%\)

Explanation:

Let the distance covered be x km with a uniform speed of 10 kmph and (80 - x) km with a uniform speed of 18 kmph.

We know that \(Time =\frac{Distance}{Speed}\)

\(\left(\frac{x}{10}\right)+\left(\frac{80-x}{18}\right)=6\)

\(\frac{9x;+;5(80-x)}{90}=6\)

\(\frac{9x;+;5(80-x)}{90}=6\)

\({9x;+;5(80-x)}{}=540\)

\(4x=540-400 =140\)

x = 35

Therefore the distance of 35 km is covered with a speed of 10 kmph and the distance 45 km is covered with a speed of 18 kmph.

Percentage of the total distance is travelled at a uniform speed of 10 kmph will be:

\(\frac{35}{80}\times 100;%=43.75;%\)

The correct answer is 'Q'.

Key Points

• Based on the above information, Chinese is harder than Japanese, which is harder than Dutch, which is harder than French, which is harder than Spanish.
• Or, Chinese > Japanese > Dutch > French > Spanish
• We can thus conclude that the two most difficult languages are Chinese and Japanese which is the combination of languages studied by 'Q'.
• Hence, option 2 is the correct answer.

Important Points 

• In order to solve such questions, it is advised to arrange the 'given that' portion in ascending or descending order and then refer to the given information to find out the correct answer.

Explanation:

If we view the given figure from the direction of the arrow mentioned in the question we will see a trapezium on the right side placed above the rectangle.

The correct answer is 'Wallets with a key are more likely to be returned because people relate to the suffering of others'.

Key Points

• Let's refer to the passage:
• 'Similarly, wallets that had a key and money are more likely to be returned than wallets with the same amount of money alone.'
• 'This suggests that the primary reason for this behavior is compassion and empathy.'
• From the above-mentioned statements, it is evident that the correct logical inference is that 'wallets with a key are more likely to be returned because people relate to the suffering of others'.
• Hence, option 2 is the correct answer.

Additional Information

• Empathy is the ability to understand and share the feelings of another.
• For eg.- He had empathy with small children.

Explanation:

The rhombus formed by joining the mid-points of square. The given square has sides of 1 unit.

As per the given figure the side of the Rhombus formed will be equal to the diameter inscribed within it.

The diameter of the largest circle will be:

\(D=\sqrt{\left(\frac{a}{2}\right)^2;+;\left(\frac{a}{2}\right)^2}\)

where a = side of the square.

\(D=\sqrt{\left(\frac{1}{2}\right)^2;+;\left(\frac{1}{2}\right)^2}={\frac{1}{\sqrt{2}}}\)

Concept:

Area of an equilateral triangle:

\(A=\frac{\sqrt {3}x^2}{4}\)

where x = side of the triangle.

Area of square = a²

where a = side of the square.

Area of circle = πR²

where R = radius of the circle.

The perimeter of equilateral triangle = 3x

The perimeter of square = 4a

The perimeter of circle = 2πR

Calculation:

Given:

Equal area

\(\frac{\sqrt {3}x^2}{4}=a^2=π R^2\)

\(\frac{ {3^{1/4}}x}{2}=a=\sqrt π R=k\)

Perimeter ratio

\(3x;:;4a;:;2\pi R\)

\(3\times\frac{2k}{3^{1/4}};:;4k;:;2\pi \times\frac{k}{\sqrt \pi}\)

\(3\times\frac{1}{3^{1/4}};:;2;:;\sqrt \pi\)

\(3^{3/4};:;2;:;\sqrt \pi\)

\(\sqrt{3\sqrt{3}};:;2;:;\sqrt \pi;\)

The least possible Venn diagram for the given statement is:

Conclusions:

I. No engineer is a professor → False (There is no direct relation given between Engineer and professor. It is possible but not definite).

II. Some engineers are professors → False (It is possible but not definite).

III. No male is a teacher → True (As all teachers are professors and no professor is male, so no male is a teacher is a definite conclusion).

Hence, the correct answer is "Only conclusion III is correct".

Additional Information

Explanation:

The hands of a clock coincide 11 times in every 12 hours (Since between 11 and 1, they coincide only once, i.e., at 12 o'clock).

AM

12:00,1:05, 2:11, 3:16, 4:22, 5:27, 6:33, 7:38, 8:44, 9:49, 10:55

PM

12:00, 1:05, 2:11, 3:16, 4:22, 5:27, 6:33, 7:38, 8:44, 9:49, 10:55

The hands overlap about every 65 minutes, not every 60 minutes. The hands coincide 22 times in a day.

Explanation:

\(\rm p = \displaystyle\lim_{x \rightarrow π} \left( \frac{x^2 + α x + 2 π^2}{x - π + 2 \sin x} \right)\)

substitue x = π

\(\rm p = \left( \frac{π^2 + α π + 2 π^2}{π - π + 2 \sin π} \right )\), sin π = 0

We can see that denominator becomes zero.

Thus for p to have finite value numerator is also zero.

π² + απ + 2π² = 0

α = - 3π

**Now,  \(\rm p = \left( \frac{\pi ^2 + α π + 2 π^2}{\pi - π + 2 \sin \pi} \right ) = \frac{0}{0} \)**form

Using L'Hospital's rule, we get

\(\rm p = \displaystyle\lim_{x \rightarrow π} \left( \frac{2x\ +\ α }{1\ +\ 2 \cos x} \right ) \)

substituting x = π, α = - 3π

\(\rm p = \left( \frac{2π\ -\ 3π }{1\ +\ 2 \cos π} \right ) \), cosπ = -1

p = \(\frac{-π}{-1}\)

p = π

Concept:

Laplacian function:

div(gradf) = ∇2f = \(\frac{\partial^2f}{\partial x^2} + \frac{\partial^2f}{\partial y^2}+\frac{\partial^2f}{\partial z^2}\)

Calculation:

Given:

∇2T = 0 

Square domain, boundary condition:

T(x, 0) = x; T(0, y) = y; T(x, 1) = 1 + x; T(1, y) = 1 + y

if we will solve it by forming equations it will take a lot of time, hence we will solve it by satisfying boundary conditions to the options.

substituing T(x, 0)= x in all options:

1. T(x, y) = x - xy + y = x - x.0 + 0 = x
2. T(x, y) = x + y = x + 0 = x
3. T(x, y) = -x + y = -x + 0 = -x ⇒ does not satisfy hence eliminated
4. T(x, y) = x + xy + y = x + x.0 + 0 = x

​substituting T(x, 1) = 1 + x in options

1. T(x, y) = x - xy + y = x - x.1 + 1 = 1 ⇒ does not satisfy , thus eliminated
2. T(x, y) = x + y = x + 1 ⇒ It satisfy hence correct option
3. T(x, y) = x + xy + y  = x + x.1 + 1 = 2x + 1 ⇒ does not satisfy , thus eliminated

Hence, option 2 satisfy all the boundary condition.

Important Points Laplace equation in a square domain (0 < x < 1 and 0 < y < 1) means:

∇2T = \(\frac{\partial ^2T}{\partial x^2} + \frac{\partial ^2T}{\partial y^2}\) = 0,  0 < x < 1 and 0 < y < 1

Boundary condition:

f₁(x) = T(x, 0) = x 

f₂(y) = T(0, y) = y

f₃(x) = T(x, 1) = 1 + x

f₄(y) = T(1, y) = 1 + y

Concept:

Gauss divergence theorem:

It states that the surface integral of a vector field over a closed surface is equal to the volume integral of the divergence of that vector field over the volume enclosed by the closed surface.

∯S A.dS =  \(\int \int \int _V \) (∇.A) dV

Calculation:

Given:

\(\rm ϕ = \frac{1}{2} (x^2 + y^2 + z^2) \), ∯S n̂ . ∇ϕ dS = ?

S = surface of sphere , V = volume of sphere =\(\frac{4}{3}π r^3\)

r = radius of sphere = 1

∇ϕ = \(\frac{\partial \phi }{\partial x} \hat{i} +\frac{\partial \phi }{\partial y} \hat{j}+\frac{\partial \phi }{\partial z} \hat{k}\)

∇ϕ =  \(\frac{1}{2}(2x\hat{i} +2y\hat{j}+2z\hat{k})\)

∇ϕ = \(x\hat{i} +y\hat{j}+z\hat{k}\)

Using Gauss theorem 

∯S n̂ . ∇ϕ dS = \(\int \int \int _V \) ∇. ∇ϕ dV

=  \(\int \int \int _V \) ∇. (xi +yj +zk) dV

= \(\int \int \int _V \)(1 + 1 + 1)dV

= 3 \(\int \int \int _V \)dV = 3 × \(\frac{4}{3}π r^3 \)

∯S n̂ . ∇ϕ dS  = 4π

Explanation:

f(x) = x³

find f(x) is even or odd

put x = -x

f(-x) = - x³

f(x) = -f(-x) hence it is odd function

for odd function, a_o = a_n = 0 

Fourier Series for odd function has only b_n term

b_n = \(\frac{1}{l} \int_{\alpha}^{\alpha+2l}f(x)sin \frac{n\pi x}{l} dx\)

Hence only sine terms are left in Fourier expansion of x³

Additional Information Fourier Series

f(x) = \(\frac{a_o}{2} + \sum_{n = 1}^{\infty} a_n cos\frac{n\pi x}{l} +\sum_{n = 1}^{\infty} b_n sin\frac{n\pi x}{l} \)

a_o = \(\frac{1}{l} \int_{\alpha}^{\alpha+2l}f(x)dx\)

a_n = \(\frac{1}{l} \int_{\alpha}^{\alpha+2l}f(x)cos \frac{n\pi x}{l} dx\)

bn = \(\frac{1}{l} \int_{\alpha}^{\alpha+2l}f(x)sin \frac{n\pi x}{l} dx\)

Explanation:

A symmetric matrix is one that is equal to its transpose, i.e.

\(A\ =\ \overline{A}\)

\(A=\begin{bmatrix} 10 & 2k + 5 \\ 3k - 3 & k + 5 \end{bmatrix} \)

\(\overline{A} = \begin{bmatrix} 10 & 3k\ -\ 3 \\ 2k + 5 & k + 5 \end{bmatrix} \)

Both the matrix will be equal when the elements of both matrices are the same.

hence, on comparing we get

2k + 5 = 3k - 3

k = 8

Concept:

Euler's formula for buckling is

\(P\ =\ \frac{\pi^2EI}{L^2}\) , where P = buckling load

L = length of the column, EI = section modulus

Calculation:

Given:

W = βπ2EI, L = 2.5

T sin 30^∘  = W

T = 2W

load on beam F = T cos 30^∘ 

F = \(√{3}\) W

To stop buckling of the beam, load on the beam must be equal to Euler's buckling load

\(\frac{\pi^2EI}{L^2}\) = ​√3 βπ2EI

β = \(\frac{1}{\sqrt3}\times\frac{1}{(2.5)^2}\)

β = 0.092

Concept:

Watt Governor Consist of two balls attached to the spindle through four arms

• The upper arm meets at the pivot which is on the spindle axis or offset from the spindle axis.
• The lower arm is connected to the sleeve by a pin joint.

The height of Watt governor is \(h = \frac{g}{ω^2}\)\(\ =\ \frac{895}{N^2} \)

where, ω = angular velocity of arm and ball about the spindle axis in rad/s

N = rpm of arm and ball

Calculation: 

Given:

h₁ = 400 mm , ω₁ = ω 

h₂ = ?, ω₂ = 2ω 

\(h = \frac{g}{ω^2}\) \(\Rightarrow \) \(h \propto \frac{1}{\omega^2}\)

\(\frac{400}{h_2} = \frac{(2\omega)^2}{\omega^2}\)

h₂ = 100 mm

Additional Information If Considering the weight of the arm the height of the governor is:

\(h\ =\ \left (\frac{g}{\omega^2} \right) \left (\ \frac{w\ +\ \frac{W_a}{2}}{w\ +\ \frac{W_a}{3}}\ \right )\)

where, w = weight of the ball, W_a = Weight of arm per unit length

Explanation:

The efficiency of a square threaded screw is the ratio of work output and work input.

\(\eta = \frac{Work\ output}{Work\ input}\)

here,

Work output is equal to Work done by a machine to lift load W per revolution.

Work input is equal to work done by force F to lift load per revolution.

Concept:

Table speed = BLU × frequency

BLU = basic length unit or the distance travel 

Calculation:

Given:

Table speed = 5 m/minute = 5000/60 mm/sec

pitch = 5 mm 

pulse per revolution = 600

So, in 1 revolution distance travel = 5 mm

So in 600 pulses distance travel = 5 mm

i.e. BLU = \(\frac{5}{600}\) mm/pulses

Frequency, f = \(\frac{Table\ speed}{BLU}\) = \(\frac{5000\ \times\ 600}{60\ \times\ 5}\)

f = 10000 Hz = 10 kHz

Explanation:

Transition fit:

• It appears when there is overlap in the tolerance zones.
• A fit with small clearance or interference that allows for accurate location of mating parts.
• It may be a tight fit, push-fit, wringing fit, or press fit.

In transition fit:

Maximum clearance is the difference between the upper limit of hole and the lower limit of shaft.

Maximum interference is the difference between lower limit of the hole and the upper limit of shaft.

Shaft:

Lower limit = 25 - 0.010 = 24.99

Upper limit = 25 + 0.010 = 25.010

Hole:

Lower Limit = 25 - 0.015 = 24.985

Upper Limit = 25 + 0.015 =25.015

Thus we can see that here the fit is transition.

Explanation:

Shadow price:

• The shadow price indicates an additional price to be paid to obtain one additional unit of the resources in order to maximize profit under the resource constraints.
• It is  also defined as the rate of change in the optimal objective value with respect to the unit change in the availability of a resource or RHS of contraint in linear programming problem.

If a resource is not completely utilized, i.e. there is slack, then its marginal return is zero

Additional InformationThe interpretation of rate of change (increase or decrease) in the value of objective function depends on whether Linear programming problem is of maximization or minimization type.

• shadow price for a less than or equal to (≤) type constraint will always be positive. This is because increasing the right-hand side resource value cannot decrease the value of objective function.
• shadow price for a greater than or equal to (≥) type constraint will always be negative because increasing the right-hand side resonance value cannot increase the value of the objective function

Explanation:

The answer is CNC Milling Machine.

Inventory is the stock of any item or resource that is used in manufacturing products in an organization. Inventories are classified as:

Direct Inventory plays a direct role in the manufacturing of products and becomes an integral part of finished goods. These are further classified as -

1. Raw Materials are machined or processed before they are ready to be used in the assembly of finished products.
2. In process Inventories (Work in progress) are semifinished goods at various stages of manufacturing.
3. Purchased Parts these are purchased items from outside suppliers instead of manufacturing in the factory.
4. Finished Goods inventories contain finished goods which are ready for dispatch to the customers.

Indirect Inventory is those stock items that are needed for product manufacturing but are not a direct part of the product. For example fuel, oil, transportation, etc

Important Points

Machines come under the asset category, hence CNC Milling Machine is an asset not inventory.

Explanation:

Clausius inequality is a corollary of the second law of thermodynamics which states that for any thermodynamic cycle the cyclic integral of heat transfer from the surface to the absolute temperature of that surface is less than equal to zero.

 \(\oint \left ( \frac{∂ Q}{T} \right )\leq 0 \) 

where,

∂Q represents the heat transfer at a part of the system boundary during a portion of the cycle

T is the absolute temperature at that part of the boundary.

\(\oint\) is called cyclic integral which means integration is performed over the entire cycle.

Important PointsFor a reversible cycle \(\oint \left ( \frac{∂ Q}{T} \right ) = 0 \)

For an irreversible cycle \(\oint \left ( \frac{∂ Q}{T} \right )< 0 \)

 \(\oint \left ( \frac{∂ Q}{T} \right ) >0 \) : It is not possible

Concept:

The rate of change of temperature with time in vector field is:

\(\frac{DT}{Dt} =\frac{\partial T}{\partial x}. \frac{dx}{dt} + \frac{\partial T}{\partial y}. \frac{dy}{dt} + \frac{\partial T}{\partial z}. \frac{dz}{dt} + \frac{\partial T}{\partial t}. \frac{dt}{dt}\)

where T = temperature, t = time

\(\frac{dx}{dt}\) = u, velocity in x-axis

\(\frac{dy}{dt}\) = v, velocity in y-axis

Calculation:

Given:

\(\vec V = (2x) \hat i + (y + 3t) \hat j,\) T = 2x2 + xy + 4t

at, x =1, y=1, t=1 to find: \(\frac{DT}{Dt}\)

\(\frac{DT}{Dt} =\frac{\partial T}{\partial x}. \frac{dx}{dt} + \frac{\partial T}{\partial y}. \frac{dy}{dt} + \frac{\partial T}{\partial t}. \frac{dt}{dt}\)

\(\frac{dx}{dt}\) = u = 2x, \(\frac{dy}{dt}\) = v = y + 3t

\(\frac{\partial T}{\partial x}\) = 4x + y , \(\frac{\partial T}{\partial y}\) = x, \(\frac{\partial T}{\partial t}\) = 4

\(\frac{DT}{Dt}\) = (4x + y).2x + x.(y + 3t) + 4

put  x =1 , y = 1, t =1

\(\frac{DT}{Dt}\) = (4 +1).2 + 1.(1+3) + 4

\(\frac{DT}{Dt}\) = 10 + 4 + 4 = 18

Explanation:

Natural convection:

• in natural convection, fluid motion occurs by natural forces such as buoyancy.
• viscous forces oppose the fluid motion
• buoyancy forces are expressed in terms of fluid temperature difference through the volume expansion coefficient β =\(\frac{1}{V}\left ( \frac{\partial V}{\partial T} \right) \)
• the term represents buoyancy force per unit mass.

​Volume expansion Coefficient:

β =\(\frac{1}{V}\left ( \frac{\partial V}{\partial T} \right) \) = \(- \frac{1}{ρ} \left ( \frac{\partial ρ }{\partial T} \right )\)

β = \(- \frac{1}{ρ} \left ( \frac{ρ _{\infty } - ρ }{T_{\infty }- T}\right )\)

(ρ_∞ - ρ) = ρβ (T - T_∞) ..... (1)

The Momentum Conservation equation per unit mass is :

\(\rho \left (\rm u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} \right ) = g (\rho_∞ - \rho ) + \nu \frac{\partial^2 u}{\partial y^2}\)

substituting equation 1

\(\rm u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} = g β (T - T_∞) + \nu \frac{\partial^2 u}{\partial y^2} \)

Explanation:

True statement:

A body is subjected to hydrostatic pressure it has no shear stress:

• according to Pascal's body subjected to hydrostatic pressure has equal stress on the body from all directions.
• σ_x = σ_y = σ_z , hence shear stress will be zero in all directions.
• shear stress on the principal plane is  \(\tau = \frac{\sigma_{x}\ -\ \sigma_y}{2}\)
• Mohr circle will become a point, hence every plane will become a principal plane and on all principal planes, shear stress will become zero.

If a long solid steel rod is subjected to tensile load, then its volume increases:

• on applying a tensile load on the rod the tensile strain is produced in the longitudinal direction.
• compressive strain is produced in the lateral direction.
• strain produced in the longitudinal direction is more compared to strain in the lateral direction.
• hence Volume of the rod increases.

Additional InformationMaximum shear stress theory:

• It is suitable for ductile metal not for brittle material.
• for brittle material maximum principal stress theory.

If a portion of a beam has zero shear force, then:

• when a shear force is zero and the bending moment has constant non-zero value then the elastic curve has a circular value.
• when shear force and bending moment both are zero then the elastic curve is a straight line.

Explanation:

Heat treatment are used to impart high mechanical properties. For the surface hardening of steel case hardening methods are used which are:

Carburizing: In this carbon content of the surface layer is increased up to 1% to obtain a hard layer on the workpiece surface after heat treatment.

• it is applied to the low carbon steel up to 0.18% carbon.
• low carbon steel is heated up to 870^∘C in the atmosphere containing carbon.
• Fe + 2CO → FeC + CO₂

 

Cyaniding: A thin layer of high hardness and wear resistance is produced on C-alloy steels.

• the workpiece is immersed in a molten salt bath containing sodium cyanide (NaCN) which is heated to 820 - 860∘C and followed by water quenching.
• cyanide bath contains 20-30% of NaCN, 25-50% of Na₂CO₃.
• the case obtained by cyaniding is high wear-resistant and endurance limit compared to carburizing.

 

Carbonitriding: it is also called gas cyaniding.

• the workpiece is heated to 850∘C in the mixture of ammonia and hydrocarbon for 2 - 10 hours. followed by quenching and tempering at 180∘C
• the alloying elements form carbides and nitrides with carbon and nitrogen on the surface which increases surface hardness.
• this process reduces the endurance limit and ductility of steel due to austenite formation in case.

​ 

Additional InformationAnnealing: is heating of steel to austenite temperature and then cooling slowly in the furnace.

• It reduces the hardness
• It improves machinability
• relieve internal stressess

Explanation:

Rapid prototyping (RP):

• is a family of fabrication methods to make engineering prototypes in minimum possible lead times based on a computer-aided design (CAD) model of the item.
• A number of rapid prototyping techniques are now available that allow a part to be produced in hours or days rather than weeks, given that a computer model of the part has been generated on a CAD system.

The method that use wise as feedstock material are:

1. Fused-deposition modeling (FDM) is an RP process in which a filament of wax or polymer is extruded onto the existing part surface from a workhead to complete each new layer. The filament is basically a wire of ABS material or other polymer.
2. Directed Energy Deposition (DED) allows for the creation of objects by melting the material in powder or as a wire with a focused energy source as it is deposited by a nozzle on a surface.

Additional Information

Stereolithography (STL):

• it is a process for fabricating a solid plastic part out of a photosensitive liquid polymer using a directed laser beam to solidify the polymer. The action of the laser is to harden (cure) the photosensitive polymer where the beam strikes the liquid, forming a solid layer of plastic that adheres to the platform.

Selective laser sintering (SLS):

• it uses a moving laser beam to sinter heat-fusible powders in areas corresponding to the CAD geometric model one layer at a time to build the solid part.
• After each layer is completed, a new layer of loose powders is spread across the surface using a counter-rotating roller.
• The powders are preheated to just below their melting point to facilitate bonding and reduce distortion.

Explanation:

Fatigue strength is the maximum stress level that a material can sustain without failing, for some specified number of cycles.

The process which improves fatigue strength are as follows:

1. Surface finish: scratches present on the surface will increase the stress concentration and reduce the fatigue strength. Therefore enhancing surface finish will reduce the notches or scratches and help in increasing fatigue strength of the shaft.
2. Shot peening: It induces surface compressive residual stress on the surface of components by the cold working process. These induced stresses forbid crack propagation on the surface and thus increase the fatigue strength of the material.
3. Humidity: It can cause corrosion on the surface of mild steel, which will help in crack propagation. Therefore lower relative humidity better the fatigue strength.
4. Cold Rolling: It strengthens the component by strain hardening by changing the properties below the recrystallization temperature. Thus improving fatigue strength as well as the surface finish of the material.

Explanation:

Choked condition: when the back pressure is equal to the critical Pressure the mass flow rate reaches a maximum value and the flow is said to be choked.

• P_back = P_critical

Critical pressure: is the pressure required to increase the fluid velocity to the speed of sound at the exit plane or throat. 

• properties of a fluid at a location where the Mach number is unity are called critical properties.

Now It is given if we increase the pressure while keeping back pressure unchanged the following will happen:

• pressure at the exit plane of the converging nozzle  is equal to critical pressure
• Mach number at the exit plane is unity.
• the mass flow rate is the maximum (or choked) flow rate. which means that the mass flow rate increased from the previous condition.
• because the velocity of the flow is sonic at the throat for the maximum flow rate, a back pressure lower than the critical pressure cannot be sensed in the nozzle upstream flow and does not affect the flow rate

Additional InformationMach number: It is the ratio of the speed of fluid to the speed of sound.

• \(Ma = \frac{V}{C}\)
• flow is Subsonic Ma < 1
• flow is Sonic Ma =1
• flow is supersonic Ma > 1

Concept:

Law of conservation of angular momentum: the angular momentum of a system remains conserved as long as there is no net external torque acting on the system.

angular momentum, \(\overrightarrow{l} = I × ω = \overrightarrow{r} × \overrightarrow{p}\)  = m × V × r

where, m = mass, V = velocity, r = radius

p = linear momentum = m × V

I = moment of inertia = mr²

ω = angular velocity = \(\frac{V}{r} \)

Calculation:

Given:

I = 0.1 kg∙cm2, m = 0.001 kg, v = 200 cm/s, ω = ?

r = 10 cm

Moment of inertia after mass strike rod I_c = 0.1 + 0.001 × 102 

Taking rod and mass as a system, by the law of conservation of angular momentum:

angular momentum before striking of mass = angular momentum after striking mass

m × V × r = I_c × ω

0.001 × 200 × 10 = (0.1 + 0.001 × 10²) × ω

2 = 0.2 ω 

ω = 10 rad/s

Concept:

Time period, T = \(\frac{2π}{ω_n}\)

where, ω_n = natural frequency 

Standard equation form of oscillation: \(\ddot{x} + ω^2_nx = 0 \)

Calculation:

Given:

r2 = R2/2,

applying the Torque method, the disk is disturbed by small-angle θ in anticlockwise direction:

By D'alembert Principal

\(I_A\ddot{θ} + mg× r sinθ\) = 0

since θ is small sinθ = θ 

\(I_A\ddot{θ} + mg× rθ = 0\) .......(1)

By paraller axis theorem

I_A = I_G + mr² .......(2)

To calculate the moment of inertia at G

we will consider an area element of the disc:

where, area of the element, dA =  2πxdx

Mass of differential element dM = \(\frac{M}{A}dA\)

dM = \(\frac{M}{π (R^2 - r^2)} × 2π xdx \)

Moment of inertia of element dI_G = (dM)x²

dIG = \(\frac{M}{π (R^2 - r^2)} × 2π xdx \times x^2 \) 

Integrating both side from limit r to R we get

I_G = \(\frac{m}{2}(R^2 +r^2) \)

Since r2 = R2/2

IG = \(\frac{3}{4}mR^2 \)

substituting in equation (2)

IA = IG + mr2 = \(\frac{3}{4}mR^2 + \frac{1}{2}mR^2 \)

I_A= \(\frac{5}{4}mR^2 \)

substituting in equation (1)

\( \frac{5}{4}mR^2\ddot{θ} + mg\frac{R}{\sqrt{2}}\theta =0\)

\(\ddot{θ} + \frac{4g}{5\sqrt{2}R}\theta =0\)

comparing with the standard equation: \(\ddot{x} + ω^2_nx = 0\)

\(\omega^2_n = \frac{4g}{5\sqrt{2}R}\)

\(\omega_n = \frac{2}{\sqrt{5\sqrt{2}}}\sqrt{\frac{g}{R}} \)

T = \(\frac{2π}{ω_n}\) = \(\frac{2\pi}{ \frac{2}{\sqrt{5\sqrt{2}}}\sqrt{\frac{g}{R}} }\)

\(T = \sqrt{5\sqrt{2}}\pi \sqrt{\frac{R}{g}}\)

comparing with \(T = β π \sqrt{R/g} \)

\(\beta = \sqrt{5\sqrt{2}} \approx 2.66\)

Concept:

ECM: Material removal takes place atom by atom. It is a high current low voltage process

Material removal rate, MRR = \(\frac{eI}{F}\)  in gm/sec

where, e = Chemical equivalent = \(\frac{Atomic\ weight}{Valency}\)

I = current passsed , F = Faraday's constant = 96500

Calculation:

Given:

For Aluminium: amu = 27, valency = 3

For Copper : MRR = 100 mg/s = 100 x 10^-3 gm/s

amu = 63, valency = 2, I = ?

e = 63/2 , F = 96500

MRR = \(\frac{eI}{F}\) = 100 x 10-3 gm/sec

I = \(\frac{100\ \times\ 10^{-3}\ \times\ 96500}{\frac{63}{2}}\) = 306.35 A

Since, Current efficiency is 100% 

For Aluminium:

e = 27/3 = 9

MRR = \(\frac{eI}{F}\)  = \(\frac{9\ \times\ 306.35 }{96500}\)

MRR = 0.02857 gm/sec

MRR = 28.57 mg/sec

Additional Information MRR in volume =\(\frac{eI}{Fρ}\) in cm³/sec

where, ρ = density in gm/cm³

Concept:

Polytropic process is any general process in nature and represented by the equation

PVⁿ = C

Where, P = Pressure of system, V = Volume of system

n = Polytropic index  and 1 < n < \(\gamma\)

Work done in polytropic process is,  \(W = \frac{P_{1}V_{1}^{}\ -\ P_{2}V_{2}^{}}{n\ -\ 1} = \frac{mR (\ T_{1}\ -\ T_{2}\ )}{\ n\ -\ 1}\)

Calculation:

Given:

P₁ = 110 KPa, V₁ = 5 m³ , V₂ = 2.5 m³, P₂ = ?, W = ?

\(P_1V_1^{n}\ =\ P_2V_2^{n}\)

\(110\ \times\ 5^{1.2}\ =\ P_2\ \times\ 2.5^{1.2}\) \(\Rightarrow\) P₂ = 252.7136 KPa

Work done during the process is 

\(W = \frac{P_{1}V_{1}^{}\ -\ P_{2}V_{2}^{}}{n\ -\ 1} = \frac{110\ \times\ 5^{}\ -\ 252.7136\ \times\ 2.5^{}}{1.2\ -\ 1}\)

W = - 408.92 kJ

The magnitude of Work done is 408.92 kJ

Additional Information

The general equation of any thermodynamic process is represented as

PV^k = C

Concept:

Kirchoff's law: the total hemispherical emissivity of a surface at temperature T is equal to its total hemispherical absorptivity for radiation coming from a blackbody at the same temperature.

α(T) = ϵ(T)

α = absorptivity, it is the ratio of absorbed radiation to the incident radiation = \(\frac{G_{abs}}{G}\)

Stefan-Boltzmann law states that the maximum radiation energy emitted from a surface per unit time and per unit surface area is expressed as:

Q = ϵσT⁴

Q = maximum radiation, ϵ = emissivity of surface, σ = Boltzmann constant =  5.669 × 10-8 W/m2-K4

T = absolute temperature of the surface at K

Calculation:

Given:

Solar flux, G_s = 600 W/m²

σ = 5.669 × 10-8 W/m2-K4

ambient Temperature T_a = 25^∘C = 298 K

Surface temperature T_s

absorptivity, α = 0.21

Since the entire solar flux is absorbed by the plate and for steady-state of the slab, by Kirchoff's Law :

Heat absorbed by the plate = net heat transfer between plate and ambient

α = ϵ = 0.21

G_s = q_rad =  ϵσ (T_s⁴ - T_a⁴ )

600 = 0.21 × 5.669 × 10-8 (Ts4 - (298)⁴)

T_s = 491 K = 491 - 273 K

**T_s = 218^∘ **C

Concept:

Residue theorem: if f(z) is an analytic function in a closed curve C except at a finite number of singular points within C then 

• \( \oint_{c} f(z)dz = 2π i \) × (sum of the residues at the singular point within curve C)

Residue for simple pole z = a:

• Res f(a) = \(\displaystyle \lim_{z \to a}[(z-a)f(z)]\)

Calculation:

Given:

\(\rm \oint \left( \frac{6z}{2z^4 - 3z^3 + 7 z^2 - 3z + 5} \right) dz\)

pole z = i

Check for singularity at pole z = i

f(z) = 2z⁴ - 3z³ + 7z² - 3z + 5

f(i) = 2(i)⁴ - 3(i)³ + 7(i)² - 3i + 5 

f(i) = 2 ×1 - 3(-i) - 7 - 3i + 5 = 0

since, f(i) = 0 ⇒ z = i  is a singular point

From Residue theorem:

\(\rm \oint \left( \frac{6z}{2z^4 - 3z^3 + 7 z^2 - 3z + 5} \right) dz\) = 2πi (Residue at z = i )

Residue at z = i :

Res|_{z= i}  = \(\displaystyle \lim_{z \to i} (z-i) \frac{6z}{2z^4 - 3z^3 + 7 z^2 - 3z + 5} \)

at z = i , Res = 0/0 form, applying L'hospital rule 

Res|z= i  = \(\displaystyle \lim_{z \to i} \frac{12z\ -\ 6i}{8z^3 - 9z^2 + 14 z - 3 } \) = \( \frac{12i\ -\ 6i}{8i^3 - 9i^2 + 14 i - 3 } \)

Res|z= i  = \( \frac{6i}{-8i + 9 + 14 i - 3 } \) = \( \frac{i}{ i\ +\ 1} \)

\(\rm \oint \left( \frac{6z}{2z^4 - 3z^3 + 7 z^2 - 3z + 5} \right) dz\) = 2πi × \( \frac{i}{ i\ +\ 1} \)

⇒  \(\frac{2π i^2}{i\ +\ 1} = \frac{-2π}{i\ +\ 1} \times \frac{i\ -\ 1}{i\ -\ 1}\) = \(π (i\ -\ 1)\)

\(\rm \oint \left( \frac{6z}{2z^4 - 3z^3 + 7 z^2 - 3z + 5} \right) dz\) = \(π (i\ -\ 1)\) = = π(-1 + i)

Explanation:

Point C will move in the x- x-direction due to the applied force P. Its deflection will be zero in y- direction.

Free body diagram of the given members is

The deflection of point B in the y direction is zero. Then

(δB)_R = \(\frac{ML^2}{2EI} \) = \(\frac{(Pa)(4a)^2}{2EI}\) = \(\frac{16Pa^3}{2EI}\) and

(δB)_M= \(\frac{PL^3}{3EI}\) = \(\frac{F(4a)^3}{3EI}\)

Total deflection at B is

(δB)_Total  = \(\frac{16Pa^3}{2EI}\) + \(\frac{F(4a)^3}{3EI}\) = 0

⇒ \(R=\frac{-3P}{8}\)

Note that the '- ve' sign indicates downward direction, i.e., R = 3P/8 (↓)

Concept:

Grashof's law for a four-bar linkage states that the sum of the shortest and longest link lengths should be less than or equal to the sum of the other two link lengths:

where:

S is the shortest link length

L is the longest link length

P and Q are the lengths of the other two links

\({S + L \leq P + Q }\)

Calculation:

Given the lengths:

• \(AG = 5 , \text{cm}\)
• \(GF = 3 , \text{cm}\)
• \(FD = 1.5 , \text{cm}\)
• \(AB = 3 , \text{cm}\)
• \(CE \text{ is sufficiently long}\)

\(S = FD = 1.5 , \text{cm}\)

\(L = AG = 5 , \text{cm}\)

\(P = GF = 3 , \text{cm}\)

\(Q = AB = 3 , \text{cm}\)

Apply Grashof's criterion:

\({ S + L \leq P + Q }\)

Substituting the values:

\({ 1.5 + 5 \leq 3 + 3 }\)

Simplifying this:

\({ 6.5 \leq 6 }\)

The inequality does not hold, indicating that the mechanism is non-Grashof.

Despite the non-Grashof nature determined by basic criteria, the specific problem configuration suggests a deeper look:

The given setup involves a slider B and the prismatic joint at D. AB as the crank can fully rotate while oscillating AG and CDEF as the coupler and rocker respectively.

Conclusion:

Given the mechanism setup and the specific constraints that allow the crank AB to fully rotate with AG and CDEF oscillating, the correct answer is:

Answer:

1. a Grashof chain with links AG, AB, and CDEF completely rotatable about the ground link FG.

Concept:

The amplitude in forced damped vibration is:

\(A = \frac{F_o/s}{\sqrt{{\left (1-\frac{ω^2}{ω_n^2} \right )^2}+\left (\frac{2\zetaω}{ω_n}^2 \right )}}\)

where F_o = external force applied

 

Damped frequency: ω_d =   (\(\sqrt{1\ -\ \zeta^2}\) )ωn

Calculation:

Given:

𝑟 = ω/ωn

\(A = \frac{F_o/s}{\sqrt{{\left (1- r^2 \right )^2}+\left (2\zeta r\right )^2}} \)

x = (1 − r2)2 + (2ζr)2

It is given that peak amplitude occurs at a frequency ω =  ωp

it means,  A → maximum, when the denominator (x) is minimum

Minimize x : \(\frac{dx}{dr} = 0\)

dx/dr = 2 (1 − r2). (-2r) + (2ζ)2.(2r) = 0

2 (1 − r2) = (2ζ)2

r2 = 1 - 2ζ2

here r = ω_p/ω_n

\(r = \frac{ω_p}{ω_n}= \sqrt{1\ -\ 2\zeta^2}\)

ω_p = (\(\sqrt{1\ -\ 2\zeta^2}\) )ωn

Now comparing wit with damped frequency:

ωd =   (\(\sqrt{1\ -\ \zeta^2}\) )ωn

We can clearly see that ωd > ωp

We already know that the natural frequency is greater than the damped frequency 

ωn > ω_d

Hence, we can say that:

ωn > ωd> ωp

Concept:

Eccentric loaded riveted joint: when the action of the load does not pass through the center of the rivet system thus all rivets are not equally loaded.

Eccentric loading results in secondary shear caused by the tendency of a force to twist the joint about the center of gravity in addition to direct shear or primary shear.

Primary shear force:

• F_p = F/n where, n = no. of rivet
• The direction of the force will be opposite to F

Secondary shear force:

• F_s = \(\frac{Fe}{\sum_{i\ =\ 1}^{n}r_i^2}\ ×\ r_i\) where,  F = Eccentric load on the joint
• e = eccentricity of the load i.e. the distance between the line of action of the load and the centroid of the rivet system.
• The direction of force is perpendicular to radial distance from the center of gravity and in the opposite direction of the couple (Fe).

Resultant of force R = \(√{F_p^2+F_s^2+2F_pF_scos(\theta)} \)

To avoid failure the permissible shear stress:  \(τ_{per}≥ \frac{Force}{Area} \)

Calculation:

Given:

e = 3√7a, P = 10 kN, 2a = 100 mm

permissible shear stress, τ = 50 MPa 

r = a = 50, e = 3√7 × 50

Primary forces:

F_p = P/2 = 5 kN

Secondary forces:

Pe = F_s (r₁² + r₂²)/ r₁

here r₁ = r₂ and F_s at both the rivet is also equal

F_s = \(\frac{Pe}{2r}\) = (10 × 3√7 × 50)/(2 × 50)

F_s = 39.68 kN

Since the revet is critical at 1 &2

Resultant force R = \(\sqrt{5^2 + 39.68^2} \) = 40 kN

To avoid failure:

\(τ_{per}≥ \frac{Force}{Area} \)

50 ≥ 40 × 10³\ A

A = 800 mm²

Concept:

Eutectoid composition: it is a reaction when a solid phase converts into two different solid phases.

• in Fe-Fe3C diagram gamma(γ) austenite at 725^∘C 0.8%C convert to ferrite(α) and cementite (Fe₃C)

Lever rule: It is used to determine the mass or mole fraction of each phase of binary equilibrium diagram.

• if the weight percent of a component is known then at fix temperature one can determine the composition of phases specifically in two phase region.

​

the mass fraction of Liquid, wt. of L =  \(% wt of = \frac{}{} \)(\frac{C_o -C_α}{C_L - C_α} \)

the mass fraction of solid, wt of α  =  \(% wt of = \frac{}{} \)(\frac{C_L -C_o}{C_L - C_α} \)

Calculation:

Given:

 eutectoid composition, C_L = 0.8 weight % 

α-ferrite, C_α = 0.025 weight % 

alloying element C_o = 0.5 weight % of carbon

Mass fraction of proeutectoid α = \(\frac{C_L -C_o}{C_L - C_α} \times 100\)

= \(\frac{0.8\ -\ 0.5}{0.8\ -\ 0.025} \times 100\) = 0.387

Explanation:

Network diagram:

The latest finish time of activity G is 10.

Steps to make a Network diagram:

1. start from those activities which have no preceding time, mark from a common event 1. here it is A,B,C 

2. Then look for the activity according to the precedence.For example:

• activity A is followed by D ( A → D)
• activity B is followed by E, F.
• activity C  is followed by G.
• similarly, mark other activities

Concept:

Drag force: It is the force experienced by an object submerged in the fluid due to the motion of fluid or object. The drag force acts in the direction of the fluid motion.

F_D = C_DAρV²/2

where, C_D = Drag coefficient, ρV²/2 = dynamic pressure,

A = projected area of the body on a plane perpendicular to the direction of flow

Calculation:

Given:

​Drag force, \(\rm D = (C_D) (0.5 ρ V^2) \left( \frac{\pi d^2}{4} \right), \) ​\(\rm C_D = \frac{24}{Re} \),

Dynamic viscosity, μ = 1.1 × 10-3 kg∙m-1∙s-1

​Density of bead ρ = 11000 kg/m3, d = 0.1 mm

Buoyant force F_b = 0, W = weight of body

The net force on the body sinking is:

W - F_b = D ⇒ W = D

Weight of the sphere bead W = mg = Volume × density × g

Volume of sphere = \(\frac{4}{3}\pi r^3\)= \(\frac{4}{3}\pi \left (\frac{d}{2}\right )^3\)

W = \(\frac{4}{3}\pi \left (\frac{d}{2}\right )^3 ρ g\)

For Drag force calculation we need to calculate:

Reynolds number Re = \(\frac{ρ V d }{μ}\) = \(ρ V\frac{0.1\ ×\ 10^{-3}}{1.1\ × \ 0^{-3}}\)

Re = ρV/11

\(\rm D = (C_D) (0.5 ρ V^2) \left( \frac{\pi d^2}{4} \right)\), \(\rm C_D = \frac{24}{Re} \)

D = \(\frac{24\ ×\ 11}{ρ V} (0.5 ρ V^2) \left( \frac{\pi d^2}{4} \right)\) = 132.V.\(\left( \frac{\pi d^2}{4} \right)\)

Now, W = D

\(\frac{4}{3}\pi \left (\frac{d}{2}\right )^3 ρ g\) =  132 × V \(\left( \frac{\pi d^2}{4} \right)\)

V = \(\frac{ρ gd}{6\ \times\ 33}\) = \(\frac{11000\ \times\ 10\ \times\ 0.1\ \times\ 10^{-3}}{6\ \times\ 33}\)

V = 1/18 m/s

Concept:

Linear momentum equation:

• it states that the vector sum of all external forces acting on a control volume in a fluid flow equals the time rate of change of linear momentum in the control volume.
• external forces are of two kinds, boundary (surface) forces and body forces.
• Fp + Fs + Fb = (Mx)out – (Mx)in , where M = momentum = ρAV, V = Velocity, A = area

Boundary forces consist of

1. Pressure intensities acting normal to a boundary, F_p and
2. Shear stress es acting tangential to a boundary, F_s.

Body forces are those that depend upon the mass of the fluid in the control volume

Calculation:

Given:

diameter, d = 1m, r = 0.5 m, density ρ₁ = 1 kg/m3, V₁ = 100 m/s

Δ P = 10 kPa, V2 = 170 m/s 

A = π r2  = 0.25π m2

mass flow rate \(\dot{m}\) = ρ₁A₁V₁ = 1× 0.25π × 100 = 25π 

F = force exerted by gas, it is a body force.

Fp = (P₁ - P₂)A it is a pressure force

Using momentum equation

(P1 - P2) A - F = \(\dot{m}\) (V₂ - V₁)

10 × 10³ ×  0.25π - F = 25π  (70)

F = 750π N

Explanation:

Steady-state conduction with internal heat generation:

\(\frac{d^2T}{dx^2} + \frac{\dot{q}}{k} = 0\) ......(i)

It is given that the surface of the cylinder is insulated the heat will transfer only in x-direction i.e. 1-D heat transfer.

The temperature will remain the same even if we change the convective heat transfer coefficient.

Proof:

The rate of heat generation is \(\rm \dot q = \cos \frac{2 π x}{L}\)

To find the temperature profile we will differentiate equation (i) and will use boundary conditions.

\(\frac{d^2T}{dx^2} = - \frac{\dot{q}}{k}\)

Integrating equation we get,

\(\frac{dT}{dx} = \int -\frac{\dot{q}}{k}dx \) = \(-\frac{1}{k}\int\cos \frac{2 π x}{L} dx \)

\(\frac{dT}{dx} = -\frac{1}{k}\times\frac{L}{2π}\ sin\frac{2 π x}{L} + c_1 \) ....... (a)

substitute boundary condition:

at x = 0 , dT/dx = 0, sin (0) = 0

we get c₁ = 0

Integrating again we get

\(T = -\frac{L}{2π k}\times\frac{-L}{2π}\ cos\frac{2 π x}{L} + c_2 \) ........(b)

at x = L 

Heat conducted = Heat convected 

\(-kA\left(\frac{dT}{dx}\right)_{x =L} = hA (T_{L} - T_{∞})\)

\(-k\left(-\frac{1}{k}\times\frac{L}{2π}\ sin\frac{2 π L}{L} \right) = h(T_L - T_{∞})\)

sin 2π = 0 , LHS = 0 hence

T_{x = L} = T_∞

substituting in equation b

\(T_{x\ =\ L}\ =\ T_\infty\ = -\frac{L}{2π k}\times\frac{-L}{2π}\ cos\frac{2 π L}{L} + c_2 \)

c₂ = \(T_\infty - \frac{L^2}{4π^2 k}\)

Hence the equation of temperature is:

\(T = T_\infty+ \frac{L}{2π k}\times\frac{L}{2π}\ cos\frac{2 π x}{L} - \frac{L^2}{4π^2 k} \)

It does not depend on convective heat transfer coefficient h.

Hence on there is no effect on temperature T_A if convective heat transfer is increased to 2h.

Confusion Points

• Generally, the equations of temperature with heat generation that we derive for wall, sphere, cylinder all depend on convective heat transfer.
• But here due to the given value of heat generation, the general equation after integration gave different results.

Concept:

Solution of homogeneous equation AX = 0:

Consistent Infinite many solutions:  if the rank of a matrix is less than the number of variables then it has infinitely many solutions.

rank(r) < no. of variables(n) ⇒ |A| = 0

Calculation:

Given:

\(\rm \begin{pmatrix} \rm x & \rm y \end{pmatrix} \begin{bmatrix} 2 & 5- 2 α \\ α & 1 \end{bmatrix} = \rm \begin{pmatrix} \rm 0 & \rm 0 \end{pmatrix} \)

for non-trivial infinite solution: |A| = 0

A = \(\begin{bmatrix} 2 & 5- 2 α \\ α & 1 \end{bmatrix}\) ⇒ |A| = 2 - α(5 - 2α) = 0

2α² - 5α + 2 = 0

(α - 2) (2α - 1) = 0

α = 2, 1/2

For α = 2 

A = \(\begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix}\)

\(\rm \begin{pmatrix} \rm x & \rm y \end{pmatrix} \begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix} = \rm \begin{pmatrix} \rm 0 & \rm 0 \end{pmatrix} \)

2x + 2 y = 0

x = -y

Substitute option in this equation.

For α = 1/2

A = \(\begin{bmatrix} 2 & 4 \\ \frac{1}{2} & 1 \end{bmatrix} \)

\(\rm \begin{pmatrix} \rm x & \rm y \end{pmatrix} \begin{bmatrix} 2 & 4 \\ \frac{1}{2} & 1 \end{bmatrix} = \rm \begin{pmatrix} \rm 0 & \rm 0 \end{pmatrix} \)

2x + 1/2 y = 0 

4x + y = 0 

from both equations we get,

y = - 4x

Substitute options in equation.

We get, 

option 1. x = 2, y = −2 and

option 2. x = −1, y = 4  satisfies equations.

Concept:

Poisson Distribution: A random variable X, taking on one of the values 0,1,2,..... is said to be a Poisson random variable with parameter λ for λ > 0, Probability is:

P (X = x) = \(\frac{e^{-λ}λ^x}{x!}\)

Calculation:

Given:

Prob(X = 1) = Prob(X = 2)

e = 2.718

\(\frac{e^{-λ}λ^1}{1!} = \frac{e^{-λ}λ^2}{2!}\)

λ = λ²/2

(λ² - 2λ) = 0 ⇒ λ (λ - 2) = 0

Since λ is always greater than zero, hence

λ - 2 = 0 ⇒ λ = 2

Prob(X = 3) = \(\frac{e^{-λ}λ^3}{3!} = \frac{e^{-2}.\ 2^3}{2\ \times\ 3}\)

Prob(X = 3) = \( \frac{ 8}{6. e^2}\) = 0.18

Additional InformationFor Poisson distribution :

Mean = E(x) = λ

Variance = V(x) = λ 

Thus, here λ is an average number of occurrences of events in an observation period Δt.

So, λ = α.Δt

where: α = number of occurrences of event per unit time.

Explanation:

The magnitude of the component of \(\vec a\) orthogonal to |\(\vec b\)|  = | |\(\vec a\)| sinθ|

From the dot product of vector :

\(cos(θ) = \frac{\vec a. \vec b}{|\vec a||\vec b|}\)

\(\rm \vec a = 5 i + 7 j + 2 k \), \(\rm \vec b = 3i - j + 6k\)

a.b = 15 - 7 + 12 = 20

|a| = \(√ {5^2 +7^2 + 2^2} = √ {78} \) 

\(|b| = √ {3^2 +(-1)^2 + 6^2} = √ {46} \)

\(cos(θ) = \frac{\vec a. \vec b}{|\vec a||\vec b|} = \frac{20}{√{78} √{46} }\) = 0.334

θ = 70.49^∘ 

|\(\vec b\)|  =  | |\(\vec a\)| sinθ| = |\(\sqrt{78}\). sin(70.49) |

|\(\vec b\)| = 8.32

Concept:

Method of section:  It is used to determine the forces in members of the truss. The steps involved are:

1. Split the truss into two parts by passing an imaginary section.
2. The imaginary section must not cut more than three members in which the forces are to be determined.
3. Conditions of equilibrium are applied for one part of the truss and unknown force in a member is determined.

Calculation:

Given:

AB = 3, BE = 4, DE = 2, GE = 6

To find force in member CI, split the truss from member CI.

Forces after splitting the section:

applying conditions of equilibrium at E i.e. Summation of the moment at E

∑ M_E = 0

4 × GE + 2 × CD - F_I × CD = 0 .... (i)

To find CD apply similarity in the triangle:

\(\frac{CD}{DE} = \frac{AB}{BE}\)

CD = \(\frac{3}{4} × 2\) = 1.5 m

Substitute in equation (i), we get

4 × 6 + 2 × 1.5 - F_I × 1.5 = 0

1.5. F_I = 24 + 3

F_I = 27/1.5

FI = 18 KN

Concept:

Frictional force, Fs = μsN

When an external force is applied to a body it does not move because frictional force balances the external force applied.

The angle of static friction (ϕ_s):  It is the angle between the resultant force (R_s) and normal reaction (N)

\(tan(ϕ_s) = \frac{F_s}{N} \) = μ_s

where, N = Normal reaction, μs = coefficient of static friction

P = external force, W =weight of body, Rs = Resultant force of N and Fs

Calculation:

Given:

at 𝜃 = 90°, slip occurs simultaneously at points P and Q

\(tan(ϕ) = \frac{12}{5} \), \(tan (ψ) = \frac{5}{12} \)

FBD at P:

Since the angle of static friction is the angle between resultant and normal reaction, thus:

tan(90-ϕ) = μ_p ⇒ cot(ϕ) =  μp

\(μ_p = \frac{1}{tan(ϕ) } = \frac{5}{12} \)

FBD at Q:

tan (90 - ψ ) = μ_Q ⇒ cot(ψ) =  μQ

\(μ_Q = \frac{1}{tan(\psi) } = \frac{12}{5} \)

\(\frac{\mu_Q}{\mu_p} =\frac{12/5}{5/12} = \frac{12}{5}\times \frac{12}{5} \) = 5.76

Important PointsWhile calculating the angle of friction always remember to take the angle between the friction force F_s and the resultant force R_s ( resultant of friction and normal reaction).

The friction force is sometimes called Coulomb friction.

Concept:

Law of conservation of angular momentum: the angular momentum of a system remains conserved as long as there is no net external torque acting on the system.

angular momentum, \(\overrightarrow{l} = I × ω = \overrightarrow{r} × \overrightarrow{p}\)  = m × V × r

where, m = mass, V = velocity, r = radius

p = linear momentum = m × V

I = moment of inertia = mr2

ω = angular velocity = \(\frac{V}{r} \)

A is the center of mass(CM), B is the point of contact

Given: 

mass m = 1 kg, r = 0.15 m,  𝜔 = 5 rad/s 

Moment of inertia for the disc I = mr2/2 = 0.01125 mm​

By the law of conservation of angular momentum:

Initial angular momentum = Final angular momentum

\(\therefore I_{o}ω_{o} =I_{f}ω_{f}+ mVr\)  \(\Rightarrow \)  \(\frac{m r^2}{2}\times ω_{o}= \frac{3m r^2}{2}\times ω_{f}\) 

\(\omega_f=\frac{\omega_o}{3}=\frac{5}{5}=1.667 \ rad/sec\)

\(V = \omega_f \times r =1.667 \times 0.15=0.25 \ \frac{m}{sec}\)

Concept:

For thin cylindrical vessel:

Hoop stress or circumferential stress: σ₁ = \(σ_h = \frac{pd}{2t}\)

Longitudinal or axial stress: σ₂ = \(σ_L = \frac{pd}{4t}\)

σ₂ = σ1/2

where, p = pressure, d = diameter, t = thickness

Strain:

Longitudinal strain: \(\epsilon_L = \frac{pd}{4tE} (1 - 2μ)\)

Hoop strain: \(\epsilon_h = \frac{pd}{4tE} (2 - μ)\)

where, μ = Poisson's ratio

Calculation:

Given:

t = wall thickness, r = nominal radius, μ = 1/3,

change in nominal (circumferential) perimeter: hoop strain = 0.1 %

changed wall thickness due to nominal strain(due to radial strain) = t̅

change in the wall thickness of the cylindrical portion \(\frac{δ t}{t}\)= (t̅ − t)/t

\(\epsilon_h = \frac{pd}{4tE} (2 - μ)\) = 0.1/100

\(\frac{pd}{4tE} = \frac{0.001}{2 - \frac{1}{3}}\) = 

Radial strain: \(\frac{δ t}{t} = \frac{1}{E}[σ_3 - \mu(σ_1 + σ_2)]\)

since σ₃ = 0, σ2 = σ1**/2, \(σ_1 = \frac{pd}{2t}\)**

\(\frac{δ t}{t} = - \frac{1}{E}\mu\frac{3σ_1}{2} \) = \(-3\mu\frac{pd}{4tE}\)

δt/t = \(-3×\frac{1}{3}×\frac{0.001}{2\ -\ \frac{1}{3}}\) 

δt/t × 100 = - 0.06 %

Concept:

Velocity ratio: it is the ratio of velocity between two gears

\(\frac{N_1}{N_2} = \frac{T_2}{T_1}\)

where T = number of teeth in gear, N = rotation per minute (rpm)

Modle of gear m = D/T

where D = pitch circle diameter 

Calculation:

Given:

T₁ = ?, T₂ = 20, T₃ = 80

N₃ = 900 rpm (ccw), N₁ = 0 rpm (fixed)

Since the module of all mating gears is always the same, hence radius can be expressed as:

r₃ = r₁ + 2r₂

\(\frac{mT_3}{2} = \frac{mT_1}{2} + 2.\frac{mT_2}{2}\) ⇒ T₃ = T₁ + 2T₂

T₁ = 80 - 40 = 40 teeth

According to the tabular method:

It is given that:

N₁ = 0 ⇒ y + x = 0

N₃ = 900 rpm ⇒  y - (x/2) = 900

on solving we get,

3y/2 = 900 ⇒ y = 600 rpm

x = - 600 rpm

Speed of arm OP = y (from table)

Speed of arm is 600 rpm counter - clockwise.

Concept:

Merchant Force analysis:

Cutting force: F_c = R cos(β - α)

Shear force**: F_s** = R cos(ϕ + β - α)

shear force is also expressed as F_s = \(\tau_s\frac{wt}{sin\ ϕ}\)

where, w = width of cut, t = unct chip thickness, ϕ = shear angle

β = friction angle, α = rake angle

Calculation:

Given:

shear strength, τ_s = 1000 MPa, cutting speed, V = 4 m/s

rake angle α = 5^∘ , uncut chip thickness, t = 0.2 mm

width of cut  w = 3 mm, friction angle β = 45^∘ 

shear angle ϕ = 25^∘

\(\frac{F_c}{F_s} = \frac{R cos(β - α)}{R cos(ϕ + β - α)}\),  Fs = \(\tau_s\frac{wt}{sin\ ϕ}\)

cutting force F_c = \(\tau_s\frac{wt}{sin\ ϕ}\) × \( \frac{cos(β - α)}{cos(ϕ + β - α)}\)

Fc =  \(1000\times \frac{3\ \times\ 0.2}{\sin\ 25^\circ} \times \frac{\cos(40^\circ)}{\cos(65^\circ)}\)

F_c = 2573.40 N

Concept:

Power of cutting is given as

Power = energy of cutting (J/mm³) × Material removal rate (mm³/s)

Calculation:

Given:

specific energy U = 6 J/mm³, axial feed = 1 m/min = 1000/60 mm/sec

tube thickness t = 1mm, diameter = 100 mm

Area of cut = πdt = 3.14 × 100 × 1 = 314 mm²

Material removal rate  = axial feed × area removed

MRR = 314 × 1000/60 mm³/s 

Power = energy × MRR 

Power = 6 × 314 × 1000/60 = 31400 J/s

P = 31.4 kW

HintIn problem like this if we do not remember formulae then we can calculate it by balancing the unit on both side.

unit of Power is Watt(W) = J/s

Power(J/s) = energy(J/mm³) × axial feed ( mm /sec) × area (mm²)

Concept:

True strain 𝜀 = ln (h_o/h_f)

(Δhmax) = μ²R 

Contact length of work piece is \(L = \sqrt{R× \Delta h} \)

average sheet thickness \(\bar h = \frac{h_o\ +\ h_f}{2}\)

average flow stress \(\bar σ\) = \(\frac{1}{ϵ} \int_0^ϵ σ _t dϵ \)

Calculation:

Given:

Initial thickness of sheet, h_o = 4mm, width w = 100 mm,

D = 200 mm ⇒ R = 100 mm

μ = 0.1, flow stress σ = 207 + 414 𝜀

Roll separation force is given: \(F = 1.15 \barσ \left( 1 + \frac{\mu L}{2 \bar h} \right)\)wL

(Δhmax) = ho - hf =  μ2R

hf  = 4 - (0.1)² × 100 = 3 mm

average sheet thickness:

\(\bar h = \frac{h_o\ +\ h_f}{2}\) = (4 + 3)/2 = 3.5 mm 

Contact length of work piece:

\(L = \sqrt{R× \Delta h} \) = √(100 × 1) = 10mm

True strain:

𝜀 = ln (ho/hf) = ln (4/3) = 0.2876

To calculate average flow stress σ_avg = \(\frac{1}{ϵ} \int_0^ϵ σ _t dϵ \)

σ_t = σ_o + kϵⁿ 

comparing σt with given equation σ = 207 + 414 𝜀, we get

σo = 207 , k = 414, n = 1

​\(\bar σ\)= \(\frac{1}{ϵ} \int_0^ϵ (σ_o + kϵ^n)dϵ \) = \(\frac{1}{ϵ} (σ_o \epsilon + \frac{kϵ^{n+1}}{n + 1})\)​

​\(\bar σ\)= \(σ_o + \frac{kϵ^{n}}{n + 1}\)  = \(207 + \frac{414ϵ}{2}\)

\(\bar σ\) = 207 + (207 × 0.2876) = 266.53 MPa

\(F = 1.15 \barσ \left( 1 + \frac{\mu L}{2 \bar h} \right)\)wL = \( 1.15 \times 266.53 \left( 1 + \frac{0.1 \times 10}{2 \times 3.5} \right) \times 100 \times 10\)

F = 351 kN

Concept:

Melting efficiency \(η_m = \frac{H_m}{H_s}\)

H_m = heat required for melting, H_s = heat supplied

Heat supplied H_s = \(\frac{VI}{Av}×\ η_{h}\)

where, V = voltage, I = current, A = cross sectional area, v = velocity

η_h = heat transfer efficiency.

Calculation:

Given:

ηh = 0.7, ηm = 0.6, H_m = 10 J/mm3

V = 20V, I = 150 A, velocity = v = 5 mm/s

feed rate of filler material in joint f = 60% = 0.6

fe = feed rate of electrode, diameter of electrode, d_e = 3 mm

\(η_m = \frac{H_m}{H_s} = \frac{H_m}{\frac{VI}{Av}×\ η_{h}}\)

\(0.6 = \frac{10}{\frac{20\ ×\ 150}{5A}×\ 0.7}\)

Area of bead, A = 25.2 mm²

by balancing the volume flow:

volume of metal from electrode per unit time = volume of metal deposited in joint per unit time

A_e × f_e = A × f × v

π/ 4 (3)² × fe = 25.2 × 0.6 × 5

fe = 10.71 mm/sec

Explanation:

It is given that: 

total processing time of four jobs is 500 minutes

if due to design change, the processing time of job 4 on each machine has increased by 20 minutes.

so, 

The total processing time will also increase by 20 minutes.

revised processing time = 500 + 20 = 520 minutes

Additional InformationHungarian algorithm also known as Munkres assignment algorithm:

It states that if a number is added or subtracted from all the entries of row or column of the matrix then the optimal assignment of resulting matrix is also optimal assignment of original matrix.

Explanation:

From the given product structure diagram, we can see that to produce:

1 unit of P requires 4 units of A and 2 units of B: P(1) = A(4) + B(2)

1 unit of B require 1 unit of C & 3 units of D: B(1) = C(1) + D(3)

1 unit of D requires 2 units of A : D(1) = A(2)

From the above equations, we can say that:

at Level 2: 3 units of D need 6 units of A

at level 1: 2 unit of B needs 12 units of A

Thus,

1 unit of P needs: (4 + 12) unit of A 

P(1) = A(16)

10 unit of P will require: 10 × 16 = 160 unit of A

Inventory given of component A = 50 units

Total unit of A required = 160 - 50 unit = 110 unit

Concept:

Entropy generation: due to internal irreversibility entropy generation takes place.

dS = \(\frac{δ Q}{T}\) + S_gen

Entropy generation is not a property, it is a path function and depends on the path followed by the system. 

Calculation:

Given:

T₁ = 80 °C = 353 K, Q = 4.5 kW/m2 = 4500 W/m² , k = 15 W/m.K , A = 1 m²

x = 0.1 m

by fourier law: Q = KA dT/dx

4500 = 15 × (353 - T₂)/0.1

T₂ = 353 - 30 = 323 K

In control volume i.e slab outer surface, change in entropy is zero.

Entropy in control volume  = S_inlet + S_generation - S_exit = 0

S_gen = S_exit - S_inlet = \(\frac{Q}{T_{exit}} - \frac{Q}{T_{inlet}}\)

S_gen = \(\frac{4500}{323} - \frac{4500}{353}\) = 1.184 W/K. m²

Concept:

In the Reheat Rankine cycle, steam after expansion of high-pressure turbine is sent again to the boiler and heated till it reaches superheated condition. it is then sent to low-pressure turbine to expand till the condenser pressure. 

Efficiency η_cycle = \(\frac{W_{net}}{Q_{in}} = \frac{Q_{in} - Q_{out}}{Q_{in}}\)

W_net = W_HPT + W_LPT - W_P

The T- S diagram: 

Process (1 - 2) = work done by pump(P)

Process (2 - 3) = heating (Q_in ) in Boiler,

Process (3 - 4) = expansion (W_HPT) in High-pressure turbine (HPT),

Process (5 - 6) = expansion (WLPT) in Low-pressure turbine (LPT), 

Process (6 -1) is heat rejection(Q_out) (cooling) in the condenser.

Calculation:

Given:

Work done in HPT, W_T1 = 400 kJ/kg ,

Work done in LPT, W_T2 = 850 kJ/kg

W_p = 15 kJ/kg, η_cycle = 0.32,

Heat rejected in the condenser, Q_out =?

\(\eta_{cycle} = \frac{W_{net}}{Q_{in}} = \frac{W_{T1}\ +\ W_{T2}\ - \ W_{P}}{Q_{in}}\)

0.32 = (400 + 850 - 15) / Q_in

Q_in = 3859.37 kJ/kg

Efficiency is also written as:

\(\eta_{cycle} = \frac{Q_{in} - Q_{out}}{Q_{in}}\)

0.32 = 1 - (Q_out / Q_in )

Q_out = 3859.37 × 0.68

Qout = 2624.37 kJ/kg

Concept:

Otto cycle: 

\(\frac{T_2}{T_1} = \left(\frac{P_2}{P_1} \right)^{\frac{γ -1}{γ}} =\left(\frac{V_1}{V_2} \right)^{{γ -1}} = (r)^ {γ -1} \)

where, r = compression ratio, γ = ratio of specific heat

Calculation:

Given:

V_s = 250 cm3, V₂ = V_c = 35.7 cm3

P₁ = 100 kPa, T₁ = 300 K, γ = C_p/C_v = 1.4

Q_in = 800 kJ/kg, C_v = 0.718 kJ/kg.K, P_max = P₃ = ?

at 2 and 3 ⇒ PV = mRT

Since m, R, V is constant ⇒ P ∝ T

\(\frac{P_3}{P_2} = \frac{T_3}{T_2}\)

To find P₃ we need to find T₂ , T₃ and P₂ :

r = V₁ / V₂ = \(\frac{V_c + V_s}{V_c}\) = 1 + (250/35.7)

r = 8

\(\frac{T_2}{T_1} = (r)^ {γ -1} \)⇒ T2 = 300 (8)0.4

T2 = 689.22 K

Q_in = C_v (T₃ - T₂)

800 = 0.718 (T₃ - 689.22)

T₃ = 1803.426 K

\(\left(\frac{P_2}{P_1} \right)= (r)^ {γ } \) ⇒ P2 = 100 × (8)1.4 

P2 = 1837.92 kPa

\(\frac{P_3}{P_2} = \frac{T_3}{T_2}\)

P₃ = 1837.92 × (1803.426 / 689.22 )

Pmax = P3 = 4809.136 K

Concept:

Stream function: is defined as a scalar function ψ such that its partial derivative with respect to any direction gives velocity component perpendicular to that in the counter-clockwise direction.

\(u = \frac{d\psi}{dy}, v = -\frac{d\psi}{dx}\)

acceleration in x-direction \(a_x = u\frac{du}{dx}+v\frac{du}{dy}+w\frac{du}{dz} +\frac{du}{dt}\)

acceleration in y-direction \(a_y = u\frac{dv}{dx}+v\frac{dv}{dy}+w\frac{dv}{dz} +\frac{dv}{dt}\)

magnitude of acceleration \(|a| = \sqrt{a_x^2 + a_y^2 +a_z^2}\) 

Calculation:

Given:

ψ = kx3y,  k = 1 m-2s-1

at x = 1 m, y = 1 m

\(u =- \frac{d\psi}{dy}, v = \frac{d\psi}{dx}\)

u = - kx³ , v = 3kx²y

u = - x3 , v = 3x2y

Since the flow is steady and two-dimensional:

acceleration in x-direction \(a_x = u\frac{du}{dx}+v\frac{du}{dy}\)

a_x = x3 (-3x²) - 0 = - 3 m/s²

acceleration in y-direction \(a_y = u\frac{dv}{dx}+v\frac{dv}{dy}\)

a_y = - x³ (6xy) + (3x2y) (3x2)

ay = 1 (-6) + (3)(3) = -6 + 9

ay = 3 m/s²

The magnitude of acceleration :

\(|a| = \sqrt{a_x^2 + a_y^2 } = \sqrt{(-3)^2 + 3^2 } \)

|a| = 3√2 = 4.24

Concept:

heat transfer due to conduction is given by Fourier's law: Q = kA \(\left(\frac{δ T}{δ x} \right)\)

heat transfer due to convection is given by Newton's law Q = hA (T₂ - T∞)

heat transfer due to radiation is given by Q = ϵ σ A (T⁴)

Calculation:

Given:

x = 0.2 m, T₁ = ?, 

Apply energy balance equation on surface 2:

heat transfer due to conduction = heat transfer by convection + heat transfer by radiation

\(k\frac{T_1\ -\ T_2}{x} = h ( T_2 - T_∞ ) + σ ϵ (T_2^4 - T_E^4)\)

\(10\frac{T_1\ -\ 300}{0.2} =100 ( 300 - 293 ) + 0.5\times 5.67 \times 10^{-8} (300^4 - 0^4)\)

50 ( T₁ - 300) = 700 + 229.635

T₁ = 300 + 18.59 = 318.59 K

T₁ = 319 K

Concept:

Effectiveness: It is the ratio of actual heat transfer to the maximum heat transfer.

\(\epsilon = \frac{Q_{actual}}{Q_{max}}\)

actual heat transfer, Q = C_h (T_h1 - T_h2) = Cc (T_c2 - Tc1) 

maximum heat transfer, Q_max = C_min (T_h1 - T_c1)

where, subscript 1 = inlet, 2 = exit, h = hot liquid, c = cold liquid

C = heat capacity, T = temperature

Calculation:

Given:

Th2 = 25 °C, Th1 = 37 °C

Tc1 = 4 °C, Tc2 = 18 °C

Using energy balance: to find C_min

Since Cmin is smaller one of Ch and Cc

Ch (Th1 - Th2) = Cc (Tc2 - Tc1)

Ch (37 - 25) = Cc (18 - 4)

Ch (12) = Cc (14)

we can see that Cc is less than Ch, hence

Cmin = C_c

\(\epsilon = \frac{C_{c}\ (T_{c2} - T_{c1})}{C_{min}\ (T_{h1} - T_{c1})}\)

\(\epsilon = \frac{(18^\circ - 4^\circ)}{ (37^\circ- 4^\circ)}\) = 14/33 = 0.42