GATE Mechanical Engineering (ME) Official Paper (Held On: 04 Feb 2023) (Previous Year Paper)

Explanation:

• The correct option is "got it fixed."
• This sentence is in the past tense, and the phrase "did not manage to fix the car himself" indicates that the person was unable to fix the car on their own.
• The phrase "got it fixed" means that they had someone else fix the car for them, which is the most logical option based on the context provided.

Explanation:

• The correct answer will be Child.
• The analogy is between planting and raising, both of which involve nurturing something to maturity. The seed is the starting point of a plant's growth, just as a child is the starting point of a human's growth.
• seed needs planting :: Child needs Raising

Calculation:

Given,

Number of colleges, C = 25951

Number of Universities, U = 504

Number of students with grade III in colleges

G₃(C) = 0.23 of 25951= 0.23 × 25951

G₃(C) = 5968.73

Number of students with grade III in universities

G₃(U) = 0.07 of 504 = 0.07 × 504

G₃(U) = 35.28

Thus, the required percentage = \(\rm \frac{Total\ number \ of\ students\ with\ Grade \ III}{Total\ number \ of\ sutdents}\)

 

\(\rm \frac{G_3(c)+G_3(u)}{c+u}=\frac{35.28+5968.73}{25951+504}\)

= 0.2269 or 22.7%

∴ The correct answer is 22.7%.

Calculation:

The minute hand and second hand will cross each other 1 time every minute after 9:15 i.e. 1 time each in 9:16, 9:17, 9.18 and so on up to 9:45. Thus, they will cross each other 30 times.

Additional Information 

 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:

(i) Every row and every column must contain each of 4-symbols.

(ii) Every 2 × 2 also 4 symbols.

→ Observe 3^rd column, the remaining symbol is 'O' (circle)

So 

→ Observe first column * symbol is there. So the first box does not *

→ We have two chances in 1^st row⇒, ☐ So is possible in ? mark.

Explanation:

• Based on the information provided, statement number 4 can be logically inferred with certainty: "Some of Ravi's friends are hardworking and kind."
• This can be inferred from the fact that the teachers mentioned that almost all of Ravi's friends were hardworking and kind, but they did not say that all of Ravi's friends were hardworking and kind. Therefore, it can be concluded that some of Ravi's friends are hardworking and kind.

Explanation:

Given inequalities,

p² - 4q < 4      ...(i)

3p + 2q < 6     ...(ii)

Since p and q are positive integers and satisfy the given inequalities, the only possibility is

p = q = 1

This can be easily verified by putting values of p and q in inequality (ii),

For p = q = 1; 3(1) + 2(1) < 6

For p = 1 and q = 2; 3(1) + 2(2) > 6

And similarly, any other value of p and q will not satisfy the inequality.

Hence p + q = 1 + 1 = 2

Alternatively,

Multiplying inequality (ii) with 2,

6p + 4q < 12

or, 4q < 12 - 6p    ......(iii)

From (i) and (iii)

p² - 4 < 12 - 6p

p² + 6p - 16 < 0

(p + 8) (p - 2) < 0

Using the wavy curve method,

 

∴ p ∈ (-8, 2)

But it is given that p is a positive integer,

∴ p = 1

From (iii), 4q < 12 - 6(1)

4q < 6

⇒ \(\rm q<\frac{3}{2}\)

But again q is a positive integer,

∴ q = 1

Hence, p + q = 1 + 1 = 2.

Explanation:

• The correct sentence sequence that creates a coherent narrative is (3) (ii), (i), (iv), (iii).
• This sequence creates a coherent narrative as it follows a logical order of events. First, the narrator hears unfamiliar voices coming from a shut door in the big drawing room, which prompts them to feel hesitant about knocking. Next, the narrator decides to wander down a passage instead of knocking. Then, the passage suddenly opens up into a bright kitchen. This sequence of events makes sense and flows logically.

(ii) There was a murmur of unfamiliar voices coming from the big drawing room and the door was firmly shut.

(i) I could not bring myself to knock.

(iv) I decided I would rather wander down the passage.

(iii) The passage was dark for a bit, but then it suddenly opened into a bright kitchen.

Explanation:

Let's analyze the problem step by step. The set {1, 2, 3, 4, 5, 6} has 2⁶ = 64 subsets, including the empty set and the set itself.

Now, for each subset S, we need to find the number of subsets T such that S is a subset of T. 

For a given subset S with k elements, there are 2^(6-k) subsets of {1, 2, 3, 4, 5, 6} that contain S. This is because for each element in {1, 2, 3, 4, 5, 6} that is not in S, we can choose to include it or not, giving us 2 options. Since there are 6-k such elements, we have 2^(6-k) possibilities.

Now, summing up these possibilities for all possible sizes of S (from 0 to 6), we get: \(\sum_{k=0}^{6} \binom{6}{k} \cdot 2^{6-k} \)

This sum is equal to \( (1 + 2)^6\) by the binomial theorem, which is 3⁶. Therefore, the total number of pairs (S, T) is 3⁶ = 729.

So, the correct answer is option 1) 729.

Explanation:

Under the given conditions, all the shadows P, R, and S are possible.

Here's why:

Shadow S: This shadow is formed when the pyramid is oriented such that one of its edges is parallel to the direction of the light beam. In this case, the shadow on the screen will be a square.

Shadow R: This shadow is formed when the pyramid is oriented such that one of its edges makes an angle of 45 degrees with the direction of the light beam. In this case, the shadow on the screen will be a trapezoid.

Shadow P: This shadow is formed when the pyramid is oriented such that one of its edges makes an angle of 60 degrees with the direction of the light beam. In this case, the shadow on the screen will be a pentagon.

Therefore, only the shadows P,  R, and S are possible under the given conditions.

Concept:

Poisson distribution is

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

Where m is the mean of the distribution.

For Poisson distribution: Mean = Variance = m

Calculation:

Given:

Probability of defective component, P = 0.015

Total number of components, n = 200

The average number of defective components,

np = 0.015 × 200 = 3

Mean and variance of Poisson distribution are same.

∴ Variance = 3

Explanation:

We know that the graph of y = |x| is

and the graph of y = -|x| is

 

It can be observed that the given graph can be obtained by first shifting the graph of y = -|x| up by 2 units and then taking the modulus of resultant function. Shifting up by 2 units transforms the equation to,

y = 2 - |x|

and, taking modulus gives the resultant equation as

y= |2 - |x||

Explanation:

Curve P₂: Holding cost

Holding cost = \(\rm \frac{Order\ quantity}{2}× \frac{Carrying\ cost}{\frac{unit}{unit\ time}}\)

\(\rm =\frac{Q}{2}× C_c\)

⇒ Holding cos t ∝ Q

Curve P3: Setup cost

Setup cost (or) ordering cost

= No. of setups × setup cost / setup

\(\rm=\frac{Annual\ demand}{Order\ quantity}× \frac{setup\ cos\ t}{setup}\)

\(\rm =\frac{D}{Q}× C_0\)

⇒ Ordering (or setup cost ∝ \(\rm \frac{1}{Q}\)

Curve P₄: Production (or) Material cost

Curve P1: Total cost

Annual production cost = Annual demand × unit cost

= D × C_u

⇒ Production cost is independent of order quantity.

TC = holding cost + setup cost + production cost

Explanation:

Z = 45 X₁ + 60 X₂ 

X₁ ≤ 45 ⇒ \(\rm \frac{X_1}{45}=1\)     .......(i)

X₂ ≤ 50 ⇒ \(\rm \frac{X_2}{50}=1\)     .......(ii)

25 (60 - X₂) + 5X₂ = 750

X₂ = 37.5 (Form (iv))

X₁ = 60 - 37.5 = 22.5

Point B 

X₂ = 50 (From (ii))

10X₁ + 10(50) = 600 (From (iii))

X₁ = 10

Point C

X₂ = 50 (From (ii))

25X₁ + 5 × 50 = 750 (From (iv))

X₁ = 20

Z(A) = 45 × 22.5 + 60 × 37.5 = 3262.5

Z(B) = 45 × 10 + 60 × 50 = 3450

Z(C) = 45 × 20 + 60 × 50 = 3900

Explanation:

3-2-1 principle of location:

• The 3-2-1 principle of location (six-point location principle) is used to constrain the movement of the workpiece along the three axes XX, YY and ZZ.
• This is achieved by providing six locating points, 3-pins in base plate, 2-pins in vertical plane and 1-pin in a plane which is perpendicular to first two planes.
• A three-pin base can restrict five motions and six pins restrict nine motions.
• The “3” in 3-2-1 refers to 3 locators (passive fixture elements) on the primary locating/datum surface.
• The “2” in 3-2-1 refers to 2 locators on the secondary locating/datum surface.
• The “1” in 3-2-1 refers to 1 locator on the tertiary locating/datum surface.

3-2-1 Principle of Location used in Jig & Fixtures:

• It is also known as a six-pin or six-point location principle.
• In the case of three-pin, 5 degrees of freedom are to be eliminated to have a body fixed in space.
• In this, the three adjacent locating surfaces of the blank (workpiece) are resting against 3, 2, and 1 pin respectively, which prevent 9 degrees of freedom.
• The rest three degrees of freedom are arrested by three external forces usually provided directly by clamping.

Methodology of 3-2-1 Principle:

• The workpiece is resting on three pins A, B, and C which are inserted in the base of the fixed body.
• The workpiece cannot rotate about the axes XX and YY and also cannot move downward.
• In this way, the five degrees of freedom 1,2,3,4, and 5 have been arrested.
• Two pins D and E are inserted in the fixed body, in a plane perpendicular to the plane containing pins A, B & C.
• Now the workpiece cannot rotate about the Z-axis and also it cannot move toward the left.
• Hence the addition of pins D and E restrict three more degrees of freedom, namely 6, 7, and 8.
• Another pin F in the second vertical face of the fixed body arrests the degree of freedom 9.

Explanation:

Spring Stiffness (Keq)

• Spring Stiffnessor spring constant represented by k represents how much resistance it offers to displacement when a force is applied to it. The more stiff the spring the lesser it would deflect ( compression or tension).
• \(Maximum~displacement ~(\delta_m);=\frac{{Load ~or~ force}}{{spring ~stiffness}}\)
• \(spring~stiffness~(K_{eq})=\frac{{Load}}{{\delta_m}}\)

The cantilever with a Point Load at its Free End

​Maximum deflection at the free end is given by,

Maximum Deflection \(δ_m=\frac{{Wl^3}}{{3EI}}\)

Spring Stiffness (Keq) of the cantilever with a Point Load at its Free End

\(K_{eq}=\frac {{Load}}{{\delta_m}}=\frac{{W}}{{\frac{{Wl^3}}{{3EI}}}}=\frac{{3EI}}{{l^3}}\)

Additional Information

Explanation:

Nonrigid frames: Relative movement of joints is large then a frame said to be non rigid.

Here, P force causes large deformation, therefore. It is non rigid.

Concept:

The common form of presentation of fatigue data is by using the S-N curve, where the total cyclic stress (S) is plotted against the number of cycles to failure (N) in a logarithmic scale.

The fatigue life reduces with respect to the increase in stress range and at a limiting value of stress, the curve flattens off. The point 4 at which the S-N curve flattens off is called the ‘endurance limit’.

The line between 103 and 106 cycles is taken to represent high cycle fatigue.

From the S-N curve, we can see that the curve becomes asymptotic nearly at 106 cycles for steel.

For non-ferrous materials like Aluminium, Copper, the endurance limit is not definite. Therefore, the S-N curve for non-ferrous materials becomes asymptotic nearly at 108 cycles.

Concept:

For laminar flat plate boundary layer:

\(\frac{\delta }{x}=\frac{5}{\sqrt{R{{e}_{x}}}}\)

\({{C}_{f}}=\frac{{{τ }_{w}}}{\frac{1}{2}ρ {{U}^{2}}}=\frac{0.664}{\sqrt{R{{e}_{x}}}}\)

These results characterize the laminar boundary layer on a flat plate’ they show that laminar boundary layer thickness varies as x1/2 and the wall shear stress varies as 1/x1/2.

The variation of shear stress with distance ‘x’ for the laminar boundary layer is

\(\frac{{{τ }_{1}}}{{{τ }_{2}}}=\sqrt{\frac{{{x}_{2}}}{{{x}_{1}}}}\)

Air, ρ = 1.2 kg/m³, ν = 1.5 × 10^-5 m²/s, U_∞ = 2 m/s

At x₁ = 15 mm τ₁ = τ_w

At x₂ = 30 mm; τ₂ = ?

Reynolds number = \(\rm \frac{U_{\infty}x}{v}=\frac{2\times 30\times 10^{-3}}{1.5\times 10^{-5}}=4000\)

As Re < (Re)_critical

∴ Flow is laminar

For laminar flow over flat plate

we know, \(\rm \tau_w\propto\frac{1}{\sqrt x}\)

\(\rm \frac{\tau_1}{\tau_2}=\sqrt{\frac{x_2}{x_4}}\)

\(\rm \frac{\tau_w}{\tau_2}=\sqrt{\frac{30}{15}}\)

\(\rm \tau_2=\frac{\tau_w}{\sqrt2}\)

Explanation:

Given data:

Isentropic flow or air

γ = 1.4

A = 100 kPa

ρ₁ = 1.2 kg/m³

u₁ = 400 m/s

P = ρRT

P₁ = ρ₁ RT₁

100 = 1.2 × 0.287 × T₁

T₁ = 290.36 (K)

a = \(\sqrt{\gamma RT}\)

\(\rm a=\sqrt{1.4\times 287\times 290.36}\)

a = 341.5649 (m/s)

\(\rm M=\frac{V}{a}=\frac{400}{341.5649}\)

M = 1.1710

Note:

∵ [M > 1]

∴ The flow is supersonic

If M > 1 → velocity increase

M > 1 → M = \(\rm \frac{v}{a}→ v ↑\)

Note:

since velocity (v) ↑ 

pressure must decrease (P) ↓ 

Hence,u₂ > u₁, P₂ < P₁

Explanation:

Bulk mean temperature:

Bulk mean temperature of the fluid at a given cross-section of the pipe is defined as temperature which takes into account the variation of the temperature of the fluid with respect to the radius r at that cross-section of the pipe by averaging it.

Q = ṁ × Cp × Tb

Q is the heat transported by the fluid at that cross-section and ṁ is the mass flow rate and Tb is the bulk mean temperature at that cross-section.

There are two distinct conditions that occurs in the forced convection flow

Constant Heat flux conditions:

In the case of laminar flow condition at a fully developed region with constant heat flux from the pipe wall, bulk means the temperature of fluid and wall temperature both increases in the flow direction. This is explained in the figure below,

Constant wall temperature conditions:

In the case of laminar flow condition at a fully developed region with constant wall temperature, only bulk mean temperature is increasing in the direction of flow.

Note:

For Constant Heat flux condition, Nusselt number = 4.36

For constant wall temperature conditions Nusselt number = 3.66

• For thermal fully developed, in case of pipe Nusselt number is constant (Nu = c), hence Nu is not depends on Reynold's number and Prandtl number.

Concept:

Clausius inequality states that \(\oint \frac{{{\rm{dQ}}}}{{\rm{T}}} \le 0\)

It provides the criteria for the reversibility of a cycle.

If \(\oint \frac{{{\rm{dQ}}}}{{\rm{T}}} = 0\), the cycle is reversible,

\(\oint \frac{{{\rm{dQ}}}}{{\rm{T}}} < 0\), the cycle is irreversible and possible

\(\oint \frac{{{\rm{dQ}}}}{{\rm{T}}} > 0,\) The cycle is impossible

The equality in the Clausius inequality holds for totally or just internally reversible cycles and the inequality for the irreversible ones.

From energy conservation, W = Q_H - Q_L

For option (a)

W = Q_H - Q_L = 2000 - 500

= 1500J ≠ 1000 J

So, this option is incorrect.

For option (b)

W = QH - QL = 2000 - 750

= 1250 J

\(\rm \oint\frac{\delta Q}{T}=\frac{2000}{1000}-\frac{750}{100}=2-7.5 \)

= -5.5 < 0

So, this option is correct.

For option (c)

W = QH - QL = 6000 - 500 = 5500 J

\(\rm \oint\frac{\delta Q}{T}=\frac{6000}{1000}-\frac{500}{100}=6-5 =1>0\)

So, this option is incorrect.

for option (d)

W = QH - QL = 6000 - 600 = 5400 ≠ 5500

So, this option is incorrect.

Explanation:

Brazing

• Brazing is a metal joining process in which parent metal does not melt but only filler metal melts filling the joint with capillary action.

• Brazing is the technique of joining two metals in which the two pieces to be joined are not melted.

• The surfaces to be joined should be made wet by the filler metal during brazing.

• Though brazing is done much below the melting temperature of the metals, the phenomenon of brazing takes place due to diffusion or alloying of the filler metal with the base metal.

• The coalescence in the process is produced by heating to a suitable temperature with a filler metal.

Adhesive bonding

• Adhesive bonding is a material joining process in which an adhesive, placed between the Adherend surfaces, solidifies to produce an adhesive bond.
• Adhesively bonded joints are increasing alternatives to mechanical joints in engineering applications.
• It provides many advantages over conventional mechanical fasteners.
• Among these advantages are lower structural weight, lower fabrication cost, and improved damage tolerance.
• The application of these joints in structural components made of fibre-reinforced composites has increased significantly in recent years.

 

Spot Welding:

• It is also known as resistance spot welding.
• It is a resistance welding process in which two or more metal sheets are welded together by applying pressure and heat from an electric current to the welded area as shown in the given figure below.

Procedure:

There are basically five steps in spot welding which can be seen in the following figure -

1. Parts are inserted between the open electrodes.

2. Electrodes closed and force is applied.

3. Current is switched on i.e. welding starts.

4. Current is turned off but the force is maintained.

5. Electrodes are opened and welded assembly is removed.

Electric arc welding​

• Electric arc welding is the process of joining two metallic pieces.
• In this, the melting of metal is obtained due to the heat developed by an arc struck between an electrode and the metal to be welded or between the two electrodes.
• For the arc welding, the temperature of the arc should be 3500° C.
• At this temperature, mechanical pressure for melting is not required.
• Both AC and DC can be used in arc welding.
• Arc welding usually requires high current (over 80 amperes) and it may need around 12000 amperes in spot welding.
• On the other hand, the voltage requirement is low for arc welding.
• To get the required current and voltage, a welding transformer is required which converts high voltage and low current into a low voltage and high current.

Hence, the correct answer is option 2 and 3.

Concept:

Pure Bending

• Pure bending is a condition of stress where a bending moment is applied to a beam without the simultaneous presence of axial, shear, or torsional forces.
• Pure bending occurs only under a constant bending moment (M) since the shear force (V), which is equal to \(\frac{{dM}}{{dx}}\)= V, has to be equal to zero.
• In reality, this state of pure bending does not practically exist, because such a state needs an absolutely weightless member. The state of pure bending is an approximation made to derive formulas.

Explanation:

(1) When stress is beyond the yield point, plastic regions develop at the outer edges of the cross-sections and progressively move inward. (2) If the applied moment increases further, the elastic core will continue to reduce in size until it disappears completely and finally goes to a fully plastic zone.

Hence, the correct answer is options 3 and 4.

Explanation:

Pattern:

• It is the first step to prepare a model pattern
• The pattern is the replica of the feminine model of final casting to be produced with some allowances.
• Patterns can be used to make moulds.
• Mostly pattern is made of wood, it is very cheap and workability
• For higher durability and strength pattern are made from brass, aluminium alloys.

Flask: 

• The flask is the container box that holds the molding sand. Consisting of two parts, cope (the top half of a flask) and drag (the bottom half of a flask).
• A flask is typically made of aluminum, steel, or wood and can be made in any shape and size.
• To align the cope and drag, an alignment feature is used ensuring a more dimensionally accurate casting.
• Flasks are reusable, some may be used to form a mold and then removed prior to pouring a casting, others contain the mold throughout pouring and are then shaken out.

Cope:

• The top half of a two-part casting flask used in sand casting.

Drag:

• The bottom half of a two-part casting flask used in sand casting.

Parting line:

• The line that divides the cope and drag box is called the parting line.

Mould:

• It is used to create the cavity through which molten metal flows to obtain the product.
• Moulds can be used to make patterns.

Hence,the correct answer is option (1, 2 and 4).

Concept:

Maximum principal stress theory (Rankine’s theory)

 

• According to this theory, permanent set takes place under a state of complex stress, when the value of maximum principal stress is equal to that of yield point stress as found in a simple tensile test.
• For design criterion, the maximum principal stress (σ1) must not exceed the working stress ‘σy’ for the material.
• \({{\rm{σ }}_{1,2}} \le {{\rm{σ }}_{\rm{y}}}\) for no failure
• \({{\rm{σ }}_{1,2}} \le \frac{{\rm{σ }}}{{{\rm{FOS}}}}\) for design

Note:

For no shear failure τ ≤ 0.57 σy

Graphical representation

• For brittle materials, which do not fail by yielding but fail by brittle fracture, this theory gives a satisfactory result.

Maximum shear stress Theory (Tresca theory/ Guest Theory/ Coulomb Theory):

As per this theory, for no failure absolute maximum shear stress should be less than maximum shear stress under uniaxial loading, when the stress is fy.

Maximum shear stress under the uniaxial condition when the stress is fy is given as fy/2

\(\therefore {τ _{abs}}\max < \frac{{{f_y}}}{2}\)

For Design,

\({τ _{abs;max}} < ;\frac{{\left( {\frac{{{f_y}}}{{FOS}}} \right)}}{{;2}}\)

From the maximum shear stress theory, we have

\({τ _{abs;max}} = Max\left{ {\left( {\frac{{{σ _1} - {σ _2}}}{2}} \right),\frac{{{σ _1}}}{2},\frac{{{σ _2}}}{2}} \right} \le \left[ {\frac{{{σ _y}}}{{2 \times \left( {FOS} \right)}}} \right]\)

Explanation:

Gievn, σ₁ = 70, σ₂ = -70 and σ₃​ = 0 and Syt​ = 100

As per maximum principle stress theory,

       (σ1 = 70 MPa) < (S_yt​ = 100)

 so it is safe.

As per maximum shear stress theory,

     (τ_max = 70) > (S_ys = Syt​ /2= 50)

so it is unsafe.

Hence,the correct answer is option 2 and 3.

Explanation:

Mohr Circle:

• It is a two-dimensional graphical representation (σ as x-axis and τ as y-axis) of the state of stress inside a body.
• The abscissa and ordinate of each point on the circle are the magnitudes of the normal stress and shear stress components, respectively.
• In other words, the Mohr circle is the locus of those points which represents the normal and shear stress on various planes passing through a point on a loaded body.

Properties of Mohr Circle:

• Centre of Mohr circle always lies on x-axis (σ-axis) i.e. Mohr circle is always symmetrical about the σ-axis.
• The co-ordinate of centre is \(\left ( \frac{σ_x;+;σ_y}{2} \right );or;\left ( \frac{σ_1;+;σ_2}{2} \right )\) which represents normal stress on the plane of τmax.
• Radius of Mohr circle represents maximum shear stress i.e. \(τ_{max}= \sqrt {{{\left({\frac{{{σ _{x}};-;{σ _{y}}}}{2}} \right)}^2} + τ _{xy}^2};\Rightarrow\frac{σ_1;-;σ_2}{2}\)
• If two-point on the circumference of Mohr circle subtends an angle 2θ at centre, then the angle between those plane will be θ.

Special case:

1. Hydrostatic loading / Hydrostatic stress:

In the case of hydrostatic fluid, equal and alike normal stress acts on two mutually perpendicular planes without any shear i.e. σx = σy = σ and τxy = 0.

\(Centre =\left ( \frac{σ_x;+;σ_y}{2} \right );and;Radius=τ_{max}\Rightarrow\sqrt {{{\left({\frac{{{σ _{x}};-;{σ _{y}}}}{2}} \right)}^2} + τ _{xy}^2}\)

∴ centre is at (σ, 0) and radius = 0, which represents a point on x-axis / σ-axis or normal stress axis.

   

2. Pure shear

In pure shear σx and σy = 0, τxy = τ

\(Centre =\left ( \frac{σ_x;+;σ_y}{2} \right );and;Radius=τ_{max}\Rightarrow\sqrt {{{\left({\frac{{{σ _{x}};-;{σ _{y}}}}{2}} \right)}^2} + τ _{xy}^2}\)

∴ centre is at origin (0, 0) and radius = τ.

    

Hence,the correct answer is option 1 and 3.

Explanation:

The temperature distribution for various types of geometry with constant thermal conductivity and no internal heat generation is given in the table below:

So, the correct answer will be option 1 and 2.

Concept:

Module:

It is the ratio of the pitch diameter to the number of teeth.

\(m=\frac{d}{T}\)

where d = pitch circle diameter and T = number of teeth.

When two gears are in mesh, their module is the same.

∴ d = mT

Centre distance between two gears 1, 2 is given by

\(C=\frac{d_1;+;d_2}{2}\Rightarrow\frac{mT_1;+;mT_2}{2}\)

Calculation:

Given:

 

Diametrical pitch, Pd = \(\rm \frac{8 \ teeth}{mm}=\frac{T}{D}\)

⇒ module (m) = \(\rm \frac{D(mm)}{T}=\frac{1}{P_d}\)

∴ \(\rm m=\frac{1}{8}mm\)

\(\rm \frac{\omega_2}{\omega_1}=\frac{1}{4}=\frac{T_1}{T_2}\)

⇒ T2 = 4T1

∵ R + r = 30

⇒ \(\rm \frac{mT_2}{2}+\frac{mT_1}{2}=30\) 

\(\rm \left{\begin{matrix}m=\frac{D}{T}=\frac{2R}{T}\\ \therefore R=\frac{mT}{2}\end{matrix}\right}\)

⇒ \(\rm \frac{m}{2}(T_2+T_1)=30\)

\(⇒\frac{1}{8× 2}(4T_1+T_1)=30\)

⇒ 5T1 = 30 × 16

\(\rm T_1=\frac{30\times 16}{5}=96\)

Concept:

Natural frequency:

\({ω _n} = \sqrt {\frac{k}{m}}\)

\(f = \frac{1}{{2\pi }}\sqrt {\frac{k}{m}}\)

calculation:

Equivalent stiffness k_eq = k + k = 2k = 2 × 1000

k_eq = 2000 N/m

m = 20 kg

Natural frequency ω_n = \(\rm \sqrt{\frac{k_{eq}}{m}}=\sqrt{\frac{2000}{20}}=10 \) rad/s

Time period of oscillation:

T_n = \(\rm \frac{2\pi}{\omega_n}\)

Time taken to complete 10 oscillation = 10 T_n

= \(\rm 10\times \frac{2\pi}{\omega_n}=10\times\frac{2\pi}{10}=2\pi=2\times 3.14=6.28\) sec

CONCEPT:

Gauss divergence theorem:

• It states that the surface integral of the normal component of a vector function \(\vec F\) taken over a closed surface ‘S’ is equal to the volume integral of the divergence of that vector function \(\vec F\) taken over a volume enclosed by the closed surface ‘S’.
• Mathematically, it can be written as:

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

Calculation:

Given,

 \(\rm \vec B(x,y,z)=x\hat i+y\hat j-2z\hat k\)

Since Surface s is a closed surface, Gauss-divergence theorem can be applied

∴ \(\rm\int_s\vec B.\hat nds=\iiint \vec \nabla. \vec B dv\)

where, \(\rm \vec \nabla.\vec B=\frac{\partial }{\partial x}x+\frac{\partial }{\partial y}(y)-\frac{\partial }{\partial z}(2z)=1+1-2=0\)

Thus, \(\rm \int_s\vec B.\hat n\ ds=0\)

Calculation:

Given that, 

x̂ = 3y and ŷ = 2x

∴ \(\rm y=\frac{\hat x}{3}\ and\ x=\frac{\hat y}{2}\)

The given region, 

x² + y² ≤ 1

or \(\rm \left(\frac{\hat y}{2}\right)^2+\rm \left(\frac{\hat x}{3}\right)^2\le1\)

\(\rm \frac{(\hat y)^2}{4}+\frac{(\hat x)^2}{9}\le1\)

This is equation of an ellipse with semi major axis, a = 3 and semi-minor axis, b = 2, thus area of transformed region is

A = πab = 6π ≃ 18.84 units

Concept:

For a function f(z) = u + iv to be analytic, then u and v should obey Cauchy-Riemann equations.

C-R Equations:

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

Calculation:

Given,

f(Z) = u(x, y) + iv(x, y)

f(Z) = e^-kx cos 2y - ie^-kx sin 2y

Here, \(\rm \frac{\partial u}{\partial x}=-ke^{-kx}\cos 2y\)

\(\rm \frac{\partial u}{\partial y}=-2e^{-kx}\sin 2y\)

and, \(\rm \frac{\partial v}{\partial x}=-ke^{-kx} \sin 2y\)

\(\rm \frac{\partial v}{\partial y}=-2e^{-kx}\cos 2y\)

Since f(Z) is analytic, it will satisfy Cauchy-Riemann equation, 

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

⇒ -ke^kx cos 2y = -2e^-kx cos 2y

⇒ k = 2

Concept:

Following are the steps to solve

• Draw the separate FBD (Free Body Diagram)

​

For anticlockwise rotation T1 > T2​

• The relation between tension in the tight and slack sides, coefficient of friction and the angle of wrap,

\(\frac{{{T_1}}}{{{T_2}}} = {e^{\mu \theta }}\)

Using the equation,

\(\rm \frac{T_1}{T_2}=e^{\mu \theta}=e^{0.3\times \frac{3\pi}{2}}\)

⇒ \(\rm \frac{T_1}{T_2}=4.11\)

Concept:

Heat exchanger:

A heat exchanger may be defined as the equipment that transfers the heat from hot fluid to cold fluid with maximum rate, minimum investment, and running cost.

Effectiveness(ϵ): Effectiveness is the ratio of actual heat transfer rate to the maximum possible heat transfer rate. 

Actual heat transfer = mhch (Th1 - Th2 ) = mccc (Tc2 - Tc1)

ϵ = \(\frac{{{{\dot Q}_{actual}}}}{{{{\dot Q}_{max}}}} = \frac{{{C_h}\left( {{T_{h1}} - {T_{h2}}} \right)}}{{{C_{min}}\left( {{T_{h1}} - {T_{c1}}} \right)}}\) = \(\frac{{{C_c}\left( {{T_{c2}}; - ;{T_{c1}}} \right)}}{{{C_{min}}\left( {{T_{h1}}; - ;{T_{c1}}} \right)}}\)

where

Ch = Heat capacity of hot fluid = mhch,

Cc = Heat capacity of cold fluid = mccc,

Cmin = Minimum of Ch and Cc,

Th1 = Inlet temperature of hot fluid, Th2 = Exit temperature of hot fluid, Tc1 = Inlet temperature of cold fluid, Tc2 = Exit temperature of cold fluid

Calculation:

Given:

C₁ = 1000 W/K

C₂ = 400 W/K = C_min

Effectiveness, ∈ = \(\rm \frac{q_{actual}}{q_{max}}\)

∈ = \(\rm \frac{q_{actual}}{C_{min}(T_{h_i}-T_{c_i})}=0.5\)

q_actual = 0.5 × 400 (350 - 300)

= 10000 watt = 10 kW

Concept:

If L-1{F(s)} = f(t)

then

L-1F(s – a) = eat.f(t) and L-1{F(s + a)} = e-at.f(t)

Calculation:

Given, 

\(\rm F(s)=\frac{1}{s^3-s}=\frac{1}{s(s^2-1)}=\frac{1}{s(s-1)(s+1)}\)

On partial fraction decomposition,

\(\rm F(s)=-\frac{1}{s}+\frac{\frac{1}{2}}{s-1}+\frac{\frac{1}{2}}{s+1}\)

Thus, \(\rm L^{-1}(F(s))=L^{-1}\left(-\frac{1}{s}\right)+L^{-1}\left(\frac{\frac{1}{2}}{s-1}\right)+L^{-1}\left(\frac{\frac{1}{2}}{s+1}\right)\)

\(\rm -L^{-1}\left(\frac{1}{s}\right)+\frac{1}{2}L^{-1}\left(\frac{1}{s-1}\right)+\frac{1}{2}L^{-1}\left(\frac{1}{s+1}\right)\)

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

CONCEPT:

• Elastic collision: Elastic collision is a phenomenon where the collision of objects takes place such that the total linear momentum and kinetic energy of the system are conserved.
• Collisions in one dimension:

​Let m1 and m2 be the masses of two objects that undergo elastic collision. 

From the principle of momentum conservation,

⇒ m1v1i + m2v2i = m1v1f + m2v2f      

where m1, m2 are the masses of the colliding bodies, v1i, v2i are the initial velocity and v1f and v2f are their final velocities.

From the principle of kinetic energy conservation,

⇒ m1v1i2 + m2v2i2 = m1v1f2 + m2v2f2      

Calculation:

Velocity of the ball, just before making an impact with the incline

\(\rm u=\sqrt{2gh}=\sqrt{2\times9.8\times4.9}=9.8\) m/s

For perfectly elastic collision, along the line of impact,

velocity of approach = Velocity of separation

9.8 cos 30° = v cos θ    .......(i)

Now, momentum along the inclined plane will remain conserved,

⇒ 2u sin 30° = 2 v sin θ 

⇒ v sin θ = 9.8 sin 30° = 4.9

Using equation (i) and (ii), we get

v = \(\rm 9.8\sqrt{\sin^230+\cos^230^{\circ}}=9.8\) m/s

Therefore, θ = 30°

so, the inclination to the plane for the section = 90° - 30° = 60°

Now, momentum is given by,

\(\rm \vec P_f=2\times 9.8\sin60^{\circ}̂ i+2\times 9.8\cos 60^{\circ}\hat j\)

= 17 î + 9.8 ĵ

Concept:

Kennedy’s theorem states that if three bodies have plane motion relative to one another, then there I-centres i.e. instantaneous centre must lie on a straight line.

Calculation:

Let us assume fixed link as 1, disc as link 2 and rod as link 3.

Applying kennedy theorem

ω₂(I₁₂I₂₃) = ω₃(I₁₃I₂₃)    ........(i)

\(\rm \tanθ =\frac{3}{2}\)

\(\rm \tanϕ=\frac{3}{2}\)

⇒ θ = ϕ 

∴ x cos ϕ = 2

y cos ϕ = 8

From equation (i)

\(\rm ω_1\left(\frac{2}{\cos \phi}\right)=ω_2\left(\frac{8}{\cos \phi}\right)\)

(∵ According to the question, ω₁ is the angular velocity of disc and ω₂ is the angular velocity of rod.)

⇒ ω₁ = 4 ω₂

⇒ ω₂ = 0.25 ω₁

Explanation:

point load on the cantilever beam given linear variation of BM and BM is about positive y-axis

UDL gives parabolic variation of BM about negative z-axis.

Concept:

For a thin cylinder:

Longitudinal stress: \({\sigma _L} = \frac{{pd}}{{4t}}\)

Hoop stress: \({\sigma _h} = \frac{{pd}}{{2t}} = 2{\sigma _L}\)

Circumferential or hoop strain:

\({\epsilon_H} = \frac{1}{E}\left( {{\sigma _H} - \nu {\sigma _L}} \right) = \frac{{{\sigma _L}}}{E}\left( {2 - \nu } \right) = \frac{{pd}}{{4tE}}\left( {2 - \nu } \right)\)

Longitudinal Strain:

\({\epsilon_L} = \frac{1}{E}\left( {{\sigma _L} - \nu {\sigma _H}} \right) = \frac{{{\sigma _L}}}{E}\left( {1 - 2\nu } \right) = \frac{{pd}}{{4tE}}\left( {1 - 2\nu } \right)\)

Calculation:

Given:

σ_L = Iongitudinal stress = σ₁

It is due to pressure at bottom

σ₁ = σ_L = C

\(\rm=\frac{P_{botton}\times d}{4t}=\frac{γ hd}{4t}\) Constant

σ₂ = circumferential stress = \(\rm \frac{pd}{2t}\)

Here 'P' depends on 'y' and it is varying linearly,

P = γy

at y = \(\rm \frac{h}{2}\)

\(\rm σ_2=\frac{\gamma hd}{4t}\)

So, σ₁ = σ₂ ⇒ y = \(\rm \frac{h}{2}\)

Explanation:

For Option (1) For an ideal gas, the enthalpy is independent of pressure

h = u + pv [u = f(T) → For an ideal gas]

h = f(T) + f(T) [pv = RT→ For an ideal gas]

∴ h = f(T) → For an ideal gas

∴ h = f(T) → Enthalpy is independent of pressure

Option (1) is correct

For Option(2) Real gas, reversible adiabatic process,

PV^γ = C

PVγ = C → Derived from ideal gas relations [du = C_v dT] and [dh = C_pdT]

The process PV^γ = C is applicable for an ideal gas going through an adiabatic reversible process.

For Option (3) Ideal gas, reversible polytropic PV^1.5 = C;

\(\rm \frac{P}{R}=\frac{mT}{V}\)

PV = MRT → Ideal gas → correct

PV^1.5 = C → Reversible polytropic → correct

∴ Option (3) is correct

For Option (4) Any real gas behaves as an ideal gas at low pressure and high temperature.

When: \(\rm \left.\begin{matrix}p↓\\ T↑\end{matrix}\right}\) temperature forces are negligible.

when (P↓) & (T↑) → any regal gas behaves as ideal gas.

∴ Option (4) is correct.

Explanation:

In a closed system, the generalized work done equation

\(\rm W=\int_1^2P\ dv\) closed system, reversible process

where: P = P₀ since; reversible process (very slow) But, in the problems, (s₁) is removed instantaneously. It means the process is sudden. It is not reversible.

[P₁ > P₀; P₂ > P₀]

\(\rm \left[W=\int_1^2P\ dv\rightarrow P\ne P_0 \right]\)

∴ W = P₀AdL [∵ dV = A dL]

W = P₀ A(L₂ - L₁)

Since : [P₁ > P₀] & [P₂ > P₀]

The work done by the piston on the atmosphere is

W = P₀ A (L₂ - L₁)

Note:

\(\rm \left[W=\frac{(P_2L_2-P_1L_1}{1-\gamma}\right]\) this equation is for reversible adiabatic.

∴ Option (4) wrong.

Explanation:

 

\(\rm A_1=\left(\pi dh+\frac{\pi d^2}{2}\right). A_2=6L^2\)

Summation rule (for surface 1 )

F₁₁ + F₁₂ = 1

F₁₂ = 1 {F₁₁ = 0}

Reciprocity rule (1, 2):

A₁F₁₂ = A₂F₂₁

A₁ = A₂F₂₁ {F₁₂ = 1}

\(\rm F_{21}=\frac{A_1}{A_2}\)

Summation rule (for surface 2)

F₂₁ + F₂₂ = 1

F₂₂ or F_SS = 1 - F₂₁ = \(\rm 1-\frac{A_1}{A_2}\)

F₂₂ = \(\rm 1-\frac{\left(\pi dh+\frac{\pi d^2}{2}\right)}{6L^2}\)

Concept:

\(r = \frac{t}{{{t_c}}} = \frac{{uncut;chip;thickness}}{{chip;thickness;after;cut}}\)

\(\tan ϕ = \frac{{r\cos α }}{{1 - r\sin α }}\)

\(\frac{V}{{\cos \left( {ϕ - α } \right)}} = \frac{{{V_s}}}{{\cos α }}\)

V = cutting speed, Vs = shear speed

\({\rm{Shear;strain;rate}} = \varepsilon = \frac{{{V_s}}}{{{t_s}}}\)

Calculation:

Given:

Cutting velocity, V = 1 m/s,

Rake angle α = 5°,

Shear angle ϕ = 45°,

The thickness of shear plane, δ = 25 μm

Shear velocity \(\rm V_s=\frac{V\cos \alpha}{\cos (\phi-\alpha)}=\frac{1\times \cos 5}{\cos (45-5)}=1.3\) m/s

Shear strain rate = \(\rm \frac{V_s}{\delta}=\frac{1.3}{25\times 10^{-6}}=5.2\times 10^{4}\) s^-1

Explanation:

Given,

Steps per revolution = 50

Total table movement = 100 mm

The pitch of lead screw = 5 mm

So, In 50 steps lead screw moves 5 mm

50 steps = 5 mm

hence 10 steps are required to move 1 mm.

So, number of steps required are = 100 × 10 = 1000

Concept:

The elongation or compression of a bar due to thermal stresses is given as, δ = αLΔT   

where, α = linear coefficient of thermal expansion, ΔT = temperature difference, L = length of a bar

Explanation:

Given:

l1 = l2 = βl, αQ = 2αP

E = young's modulus

Change in temperature = ΔT

\(\rm σ_1=\frac{P}{A}\)

\(\rm σ_2=\frac{P}{A}\)

Where A = cross-sectional area of the bar and is the same for both due to the same radii.

So σ₁ = σ₂

Deformation in P = \(\rm l\alphaΔ T=\frac{Pl}{AE}=Δ_P\)

Deformation in Q = \(\rm 2\alpha lΔ T=\frac{Pl}{2AE}=Δ_Q\)

Δ_Q > Δ_P

Hence, the correct answer will be options 1 and 4.

Concept:

Biot number:

Biot number is defined as the ratio of conductive resistance within the body to convective resistance at the surface.

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

When the Biot number is less than 0.1 then conductive resistance within the body is negligible which is the condition for the applicability of lumped heat analysis.

In the lumped body, there is no temperature gradient exists, the temperature is the function of only and only time.

Values of the Biot number smaller than 0.1 imply that the heat conduction inside the body is much faster than the heat convection away from its surface, and temperature gradients are negligible inside the body.

Calculation:

Bi < 0.1 → Lumped system analysis

\(\rm L_c=\frac{d}{2}\) for plate

\(\rm Bi=\frac{hL_c}{k}\)

Option (1):

h = 10 W/m²K, k = 100 W/mK,

d = 1 mm = 0.001 m

⇒ \(\rm L_c=\frac{0.001}{2}\) m

\(\rm Bi=\frac{hL_c}{k}=\frac{10\times 0.001}{2\times 100}=\frac{0.0001}{2}<0.1\)

Lumped system analysis is applicable

Option (2):

h = 100 W/m²K, K = 100 W/mK, d = 1 m

\(\rm L_c=\frac{1}{2}=0.5\) m

\(\rm Bi=\frac{hL_c}{k}=\frac{100\times 0.5}{100}=0.5\)

Bi > 0.1

The lumped system is not applicable.

Option (3):

h = 100 W/m²K, k = 1000 W/mK, d = 1 mm = 0.001 m

\(\rm L_c=\frac{0.001}{2}\) m

\(\rm Bi=\frac{hL_c}{k}=\frac{100\times 0.001}{1000\times2}=\frac{0.0001}{2}<0.1\)

Lumped system analysis is applicable.

Option (4):

h = 1000 W/m²K, k = 1 W/mK, d = 1 m

\(\rm L_c=\frac{1}{2}=0.5\) m

\(\rm Bi=\frac{hL_c}{k}=\frac{1000\times 05}{1}=500\)

Bi > 0.1

Lumped system analysis is not applicable.

Hence, the correct answer is options 1 and 3.

Explanation:

Let X and y be the dimensions of the rectangle. 

Given, xy = 4

Perimeter, P = 2(x + y)

By equation (i),

\(\rm P = 2\left(x+\frac{4}{x}\right)=2x+\frac{8}{x}\)

For smaller perimeter,

\(\rm \frac{dP}{dx}=0\)

\(\rm \frac{d}{dx}\left(2x+\frac{8}{x}\right)=0\)

\(\rm 2-\frac{8}{x^2}=0\)

⇒ x² - 4 = 0

x = 2 (ignoring negative value)

y = 2

∴ Thus, the smallest perimeter is 

P = 2(2 + 2) = 8 units

Explanation:

It is Euler differential equation

∴ D(D - 1)y + Dy - y = 

[D(D - 1) + (D - 1)]y = 0

Corresponding auxiliary equation,

m² - m + m - 1 = 0

On solving, m² = 1

m = ±1

∴ y = C₁e^t + C₁e^-t

But, t = In x

∴ \(\rm y=C_1x+\frac{C_2}{x}\)

Also, \(\rm \frac{dy}{dx}=C_1-\frac{C_2}{x^2}\)

Now applying boundary conditions,

y (x = 1) = 6; C₁ + C₂ = 6

and, \(\rm \left.\frac{dy}{dx}\right|_{x=1}=2; C_1-C_2=2\)

On solving (i) and (ii),

C₁ = 4; C₂ = 2

Thus, \(\rm y=4x+\frac{2}{x}\)

Substituting, x = 2,

\(\rm y=4(2)+\frac{2}{2}=9\)

Concept:

Explicit Forward Euler Method:

\(\frac{dy}{dx} =f(x,y)\)

yn+1 = yn + hf(xn, yn)

Calculation:

Given:

\(\rm \frac{dy}{dt}+2y=0\) y(0) = 1, h = Δt = positive

t₀ = 1, y₀ = 1

\(\rm \frac{dy}{dt}=-2y\)

f(t, y) = -2y

y_n+1 = y_n + hf(t_n, y_n)

yn+1 = yn + Δt(-2y_n)

yn+1 = yn - Δt2y_n

yn+1 = yn (1 - 2Δt)

\(\rm \frac{y_{n+1}}{y_n}=1-2Δ t\)

\(\rm \frac{|Y_{n+1}|}{|y_n|}≤1\)

⇒ |1 - 2Δt| ≤ 1

⇒ |1 - 2Δt| ≤ 1

-1 ≤ 1 - 2Δt ≤ 1

-2 ≤ -2Δt ≤ 0

0 ≤ 2Δt ≤ 2

0 ≤ Δt ≤ 1

Concept:

Theoretical density of a crystal structure:

\(ρ = \frac{{nA}}{{{N_0}{V_c}}}\)

N: Number of atoms associated with each unit cell

A: Atomic weight

Vc: Volume of unit cell

NA: Avogadro’s number (6.023 × 1023 atoms/mol)

Iridium has FCC unit cell,

n = 4

Vc = a3

Calculation:

Given:

radius of atom, r = \(\left(\frac{\sqrt2}{10}\right)\) nm

Atomic weight of metal, m = 24.092 g/mol

Avogadro's number, A = 6.023 × 10²³ atoms/mol

Volume of unit cell, \(\rm a^3=\left(\frac{4r}{\sqrt2}\right)^3=\left(\frac{4}{\sqrt2}\times \frac{\sqrt2}{10}\right)^3=0.064\) nm³

Density of metal ρ = \(\frac{4\times 24.092\times (10^9)^3}{6.023\times 10^{23}\times 1000\times 0.064}=2500\) kg/m³

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:

Mass fraction of preoeutectiod cementite,

\(\rm m_1=\frac{1.5-0.8}{6.7-0.8}=0.1186\)

Mass fraction of total cementile,

\(\rm m_2=\frac{1.5-0.035}{6.7-0.035}=0.2198\)

\(\rm \frac{m_1}{m_2}=\frac{0.1186}{0.2198}=0.54\)

Concept:

Normal Distribution: 

• Normal distribution, also known as the Gaussian distribution, is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean.
• In graphical form, the normal distribution appears as a "bell curve".

Explanation:

D1 and D2 follow normal distribution.

So, Upper limit = μ + 3σ

Lower limit = μ - 3σ

D̅1 = 20 mm, D̅2 = 35 mm, σ_D2 = 0.4 mm, σ_D1 = 0.3 mm

Mean value of (C̅) = D̅2 - D̅1 = 35 - 20 = 15 mm

The standard deviation of C

\(\rm σ_C^2-σ_{D2}^2-σ^2_{D1}\)

\(σ_C=\sqrt{0.4^2-0.3^2}=0.26457\)

Upper limit of C = C̅ + 6 × σ_C 

= 15 + 6 × 0.26457 = 16.587mm

Explanation:

y-coordinate of points P1, P2, P3 and P4 is zero.

Hence plane fitted between x-z plane.

Distance Q1 from x-z plane = (6 - 0) = 6 mm

Distance Q2 from x-z plane = (6 - 0) = 6 mm

Distance Q3 from x-z plane = (4 - 0) = 4 mm

Distance Q4 from x-z plane = (4 - 0) = 4 mm

Distance between two fitted plane

\(=\frac{6+6+4+4}{4}=5\) mm

Concept:

Time taken to fill mould cavity

\(t = \frac{{Volume}}{{Discharge;Rate}}\)

Discharge rate (Q) = Area of the base of sprue × Velocity at the base of the sprue

Velocity at the base of sprue \(v = \sqrt {2;g{h_t}}\)

Where ht is the length of the sprue

Calculation:

Given:

Nozzle diameter (d) = \(\rm \frac{a}{10}\)

Feed rate (f) = \(\rm \frac{a}{5}\)

Total volume of part = V₁ + V₂ + V₃

V₁ = 9a² × 1.5a = 13.5 a³

V₂ = 4a² × a = 4 a³

V₃ = 0.5 a² × a = 0.5 a³

volume flow rate = \(\frac{\pi}{4}d^2× f=\frac{\pi}{4}× \left(\frac{a}{10}\right)^2× \frac{a}{5}\)

= 1.571 × 10^-3 a³ mm³/mm

Time required = \(\rm\frac{V_1+V_2+V_3}{Volume\ flow\ rate}=\frac{13.5a^3+4a^3+0.5a^3}{1.571\times 10^{-3}a^3}=11459.67\) min = 11460 minutes

Concept:

Optical Flat:

The optical flat is used to check the height of slip gauge blocks.

The slip gauge that is to be checked is kept alongside the reference gauge on a flat table.

An optical flat is then placed on top of both gauges as shown in the figure.

A monochromatic light source is used and the fringe pattern is observed with the help of a magnifying glass.

The fringes seen on both gauges are parallel and the same in number if both the surfaces are perfectly flat, otherwise, the number of fringes form on the two gauges differs.

The number of fringes on the reference block is N over a  width of l mm.  If the distance between the two slip gauges is L and λ is the wavelength of the monochromatic light source, then the difference in height h is given by the following relation.

​\(h = \frac{{\lambda LN}}{{2l}}\)             

Calculation:

Given:

Number of fringes = 12 per 15 mm

Wave length = 0.5 μm

Distance between gauges = 45 mm

Height difference between gauges,

\(\rm h=\frac{h\lambda}{20}=\frac{12}{15}\times \frac{0.5}{2}\times 45=9\) μm

Concept:

Engineering stress-strain is lower than true stress-strain because we consider the initial area only throughout the process while in true stress-strain curve we consider the instantaneous area.

Mathematically,

\(\rm{Engineering~Stress};(σ_o)={Applied~Force\over Initial~Area}\)  

\(\rm{True~Stress};(σ_t)={Applied~Force\over Instantaneous~Area}\)

Mathematically, true stress (σT) and engineering stress (σo) relation is represented as

σt = σo(1 + ϵo)

ϵt = ln (1 + ϵo)

where σo = engineering stress, σt = true stress, and ϵo = engineering strain and ϵt = true strain.

Calculation:

Given:

 \(\rm σ_T=400ε_T^{0.3}\)

At ultimate tensile strength

n = ε_T = 0.3

So, true stress at ultimate tensile strength

σ_T = 400 × 0.3^0.3 = 278.74 MPa

True strain, ε_T = ln(1 + ε)

⇒ 1 + ε = e^0.3 = 1.35

Engineering ultimate tensile strength

\(\rm \sigma=\frac{\sigma_T}{1+\varepsilon}=\frac{278.74}{1.35}=206.55\) MPa

Explanation:

Considering upper half of the section:

Moment of inertia of rectangular section about y-axis:

I_{rectangular sec.} = \(\frac{1}{3}\times 3\times (12)^3=(12)^3\)

Moment of inertia of triangular section about y-axis:

I_{triangular sec.}= \(\frac{1}{12}\times 1.5\times (12)^3=\frac{1}{8}\times {(12)}^3\)

\(\rm I=(12)^3-\frac{(12)^3}{8}=\frac{7}{8}\times (12)^3\)

∴ Moment of inertia for the whole section

\(=2\times \frac{7}{8}\times (12)^3=3024\) m³

Concept:

Strain energy is defined as the energy absorbed by the member when work done by the load deforms that number.

If the load is constant w.r.t length of the bar \(U = \frac{1}{2}P\delta\)

Where, P is the load acting on the bar and δ is the elongation or contraction of the bar by the load P

When the load is varying \(U = \mathop \smallint \limits_0^L \frac{{{P^2}dl}}{{2AE}}\)

For a prismatic bar AE is constant at any cross-section

and axial load due to its own weight = specific weight × volume = \(xAL\)x = specific weight of material, A = cross-sectional area, L = length of bar and E = Modulus of Elasticity

\(U = \mathop \smallint \limits_0^L \frac{{{{\left( {xAl} \right)}^2}dl}}{{2AE}} = \frac{{{x^2}{A^2}{L^3}}}{{6AE}} = ;\frac{{{x^2}A{L^3}}}{{6E}}\)

Calculation:

The strain energy is given by:

U = \(\rm \frac{W^2L}{6AE}=\frac{\gamma^2AL^3}{6E}\)

⇒ \(\rm U=\frac{(2700\times 9.81)^2\times 10\times 5^3}{6\times70\times 10^9}=2.09\) N-m

Concept:

Torsion Equation of a shaft is given by,

\(\frac{T}{J} = \frac{τ }{r} = \frac{{Gθ }}{L}\)

 

The power transmitted by any solid shaft is given by:

\({\rm{P}} = {\rm{Torque;}}\left( {\rm{{\rm T}}} \right) \times {\rm{Angular;velocity;}}({\rm{\omega }}\))

ω = \(\frac{{2{\rm{\Pi }}N}}{{60}}\); where N is in rpm

Where, T = Torque, J = Polar moment of inertia, τ = Shear stress, r = Radius of shaft, G = Shear modulus, θ = Angle of twist and L = Length of shaft

Calculation:

Given:

 N = 500 rpm , τmax = 80 GPa ,d = 100 mm, L =1.5 m  and θ =  0.5° 

Using the equation \(\rm θ=\frac{TL}{GJ}\)

∴ \(\rm T=\frac{GJθ}{L}=\frac{80\times 10^3\times \frac{\pi}{32}\times (100)^4\times0.5\times \frac{\pi}{180}}{1500}\)

⇒ T = 4569261.297 N-mm

⇒ T = 4569.261 N-m

∴ \(\rm P=\frac{2\pi NT}{60}\)

⇒ \(\rm P=\frac{2\pi\times 500\times 4569.261}{60}\)

⇒ P = 239245.96 W

⇒ P = 239.246 kW

Given:

We are given with two gases:

P_t = 100 kPa

V = 10 m³

T = 300 K

X                Y

M_x : 10 kg/k-mol

M_y : 20 kg/k-mol

Using the equation, PV = nR̅T

⇒ 100 × 10 = n × 8.314 × 300

⇒ n = 0.4

Now, n = n_x + n_y

\(\rm n=\frac{m_x}{M_x}+\frac{m_y}{M_y}\)

⇒ \(\rm 0.4=\frac{2}{10}+\frac{m_y}{20}\)

⇒ m_y = 4 kg

Explanation:

Navier-Stokes (N-S) equations represent the momentum conservation equation:

\(\rho \frac{{DV}}{{Dt}} = \rho g - \nabla {\rm{p}} + {\rm{\mu }}{\nabla ^2}V = 0\)

\(\rho \left( {\frac{{\partial u}}{{\partial t}} + u\frac{{\partial u}}{{\partial x}} + v\frac{{\partial u}}{{\partial y}} + w\frac{{\partial u}}{{\partial z}}} \right) = \rho {g_x} - \frac{{\partial \rho }}{{\partial x}} + \mu \left( {\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}} + \frac{{{\partial ^2}u}}{{\partial {z^2}}}} \right)\)

\(\rho \left( {\frac{{\partial v}}{{\partial t}} + u\frac{{\partial v}}{{\partial x}} + v\frac{{\partial v}}{{\partial y}} + w\frac{{\partial v}}{{\partial z}}} \right) = \rho {g_y} - \frac{{\partial \rho }}{{\partial y}} + \mu \left( {\frac{{{\partial ^2}v}}{{\partial {x^2}}} + \frac{{{\partial ^2}v}}{{\partial {y^2}}} + \frac{{{\partial ^2}v}}{{\partial {z^2}}}} \right)\)

\(\rho \left( {\frac{{\partial w}}{{\partial t}} + u\frac{{\partial w}}{{\partial x}} + v\frac{{\partial w}}{{\partial y}} + w\frac{{\partial w}}{{\partial z}}} \right) = \rho {g_z} - \frac{{\partial \rho }}{{\partial z}} + \mu \left( {\frac{{{\partial ^2}w}}{{\partial {x^2}}} + \frac{{{\partial ^2}w}}{{\partial {y^2}}} + \frac{{{\partial ^2}w}}{{\partial {z^2}}}} \right)\)

 

Calculations:

Given:

v(x, y) = k(xî -yĵ)

v_x = kx = u

v_y = -ky = V

Using N-S equation in x-direction

\(\rm -\frac{\partial P}{\partial x}+\mu\rm \frac{\partial ^2u}{\partial x^2}=\rho\left(uf\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}\right)\)

 

\(\rm- \frac{\partial P}{\partial x}=\rho(kx\times k+(-ky)0)\)

\(\rm- \frac{\partial P}{\partial x}=k^2\rho x\)

\(\rm P(x)=\frac{-k^2\rho x^2}{2}+C_1\)

Using N-S equation in y-direction

\(\rm -\frac{\partial P}{\partial y}+\mu\rm \frac{\partial ^2u}{\partial y^2}=\rho \left(u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}\right)\)

\(\rm -\frac{\partial P}{\partial y}=\rho(kx\times 0+(-ky)(-k))\)

\(\rm -\frac{\partial P}{\partial y}=k^2 \rho y\)

P(y) = \(\rm \frac{-k^2\rho y^2}{2}+C_2\)

From equation (i) and (ii), we get

\(\rm P(x,y)=-\frac{k^2 \rho x^2}{2}-\frac{k^2 \rho y^2}{2}+C\)

Now it is given that

P(0, 0) = 100 kPa

100 = 0 + c

c = 100

then, 

\(\rm P(x,y)=\frac{-\rho k^2}{2}(x^2+y^2)+100\)

\(\rm P(2,2)=\frac{-4}{2}(2^2+2^2)+100\)

P(2, 2) = -16 + 100 = 84 kPa

Concept:

Continuity equation:

\(\begin{array}{l} \frac{{\partial \left( {ρ u} \right)}}{{\partial x}} + \frac{{\partial \left( {ρ v} \right)}}{{\partial y}} + \frac{{\partial \left( {ρ w} \right)}}{{\partial z}} + \frac{{\partial ρ }}{{\partial t}} = 0\ \frac{{\partial ρ }}{{\partial t}} + \bar V.\left( {ρ \vec V} \right) = 0 \end{array}\)

Continuity equation:

The continuity equation is the equation of conservation of mass in a fluid flow. 

According to the law of conservation of mass:

Claculation:

Given:

u(0, t) = 1

ρ(t) = 1 + 0.2 e^-t

Using continuity equation

\(\rm \frac{\partial \rho}{\partial t}+\frac{\partial}{\partial x}(\rho u)+\frac{\partial}{\partial y}(\rho v)+\frac{\partial}{\partial z}(\rho \omega)=0\)

\(\rm \frac{\partial \rho}{\partial t}+\frac{\partial}{\partial x}(\rho u)=0\)

\(\rm \frac{\partial \rho}{\partial t}=\frac{\partial (1+0.2e^{-t})}{\partial t}=-0.2e^{-t}\)

\(\rm \frac{\partial}{\partial x}(\rho u)=\frac{\partial}{\partial x}((1+0.2e^{-t})u(x,t))\)

\(=(1+0.2e^{-t}).\frac{\partial u}{\partial x}\)

Now, put in equation (i)

\(\rm -0.2e^{-t}+(1+0.2e^{-t})\frac{\partial u}{\partial x}=0\)

\(\rm \frac{\partial u}{\partial x}=\frac{0.2e^{-t}}{1+0.2e^{-t}}\)

\(\rm u=\left(\frac{0.2e^{-t}}{1+0.2e^{-t}}\right)x+c\)

Using boundary condition,

u(0, t) = 1

c = 1

From equation (ii),

\(\rm u=\frac{0.2e^{-t}}{1+0.2e^{-t}}x+1\)

\(\rm u(2,1)=\frac{0.2e^{-1}}{1+0.2e^{-1}}\times 2+1\)

u(2, 1) = 1.13 m/s

Given:

Making the pressure prism for the system:

P₁ = 10³ × 9.81 × 1

P₁ = 9.81 kPa

P₂ = P₁ + [2000 × 9.81 × 2]

P₂ = 49.05 kPa

The resultant forces F₁ and F₂ due to pressure P₁ and P₂ respectively are given as:

F₁ = Volume of triangular pressure prism

\(\rm =\frac{1}{2}\times P_1\times 1\times 1=4.9 \) kN

F₂ = Volume of trapezoidal pressure prism

\(\rm =\frac{1}{2}\times [P_1+P_2]\times 2\times 1=58.86\) kN

Now, the moment due to forces is given as

\(\rm M=F_1\times \left[2+\frac{1}{3}\right]+F_2\times \left[\frac{2P_1+P_2}{P_1+P_2}\right]\times \frac{2}{3}\)

Putting the values, we get

M = 57.225 kNm

Buckingham’s π theorem

• If there are n variables in a problem and these variables contain m primary dimensions (for example M, L, T) the equation relating all the variables will have (n-m) dimensionless groups.

The repeating variable used in Buckingham’s theorem must have

1. Geometric property, e.g. Length, diameter, height

2. Flow property, e.g. velocity, acceleration

3. Fluid property, e.g. viscosity, density

Quantities must not form a dimensionless group among themselves neither must two variables have the same dimension and the 3 fundamental dimensions must be present in the variable chosen i.e.

Calculation:

Given:

  R = k t^a E^b ρ^c

  M⁰L¹T⁰ = (M⁰L⁰T⁰) (T)^a(ML²T^-2)^b(ML^-3)^c

Comparing powers of M, L, T

b + c = 0

⇒ b = -c

2b - 3c = 1

-2c - 3c = 1

\(\rm c=-\frac{1}{5}\)

⇒ \(\rm b=\frac{1}{5}\)

a - 2b = 0

\(\rm a=2\times\frac{1}{5}=0.4\)