GATE ME 2016 Official Paper: Shift 3 (Previous Year Paper)

When we compare the heights of two people the correct word to be used is ‘taller’. Thus option C is the correct answer.

Let us look at the meanings of the words:

1. Facilitated: Helps improve something
2. Felicitated: Congratulate
3. Fantasize: Dream about something

The other option is not a word. Thus, the correct word to be used in the blank is ‘felicitated’ that will give the meaning of the teacher being congratulated.

The phrase ‘rest is history’ means something that has happened in the past and has become very famous. Thus ‘the rest is well known’ is the correct answer.

Taking squares on both sides,

9 inches = 0.25 yards

M has two times efficiency as that of E.

But E works 12 hrs per day and M works for 6 hrs per day i.e. half of working hrs by E.

So the work done per day by both M and E is same.

M worked half of the days that E worked for.

So working contribution by M to E will be 1:2.

From Venn diagram

n(A) = no of persons reading books = 13 + 44 + 12 + 7 = 76

n(B) = no of persons playing = 15 + 44 + 7 + 17 = 83

n(A ∩ B) = 44 + 7 = 51

n(A ∪ B) = n(A) + n(B) – n(A ∩ B) = 76 + 83 – 51 = 108

The passage does not state the boundaries of discipline for social sciences. Neither can this be inferred. Again, the passage only states that the social sciences were institutionalized because of colonial motives. This is not the same as promoting colonialism. Thus this cannot be inferred either. The knowledge production methods have become obsolete and not the discipline. Thus, option 1 is eliminated.

The passage states that social sciences discipline were in an amorphous form till the colonial period. This means that it existed before the colonial age. Thus option 2 can be inferred.

Two and quarter hours back the time reflected in mirror and the actual time is represented as below,

So now the actual timing will be 10:30 + 2:15 = 12:45.

The given information can be represented as follows,

\(\begin{array}{l} {\rm{Now;distance;between;A;and;C;}} = {\rm{;}}10 + \frac{{10}}{{\sqrt 2 }} - \frac{4}{{\sqrt 2 }} = 14.24\ {\rm{Now;distance;between;B;and;C;}} = {\rm{;}}5 + {\rm{;}}\frac{{10}}{{\sqrt 2 }}; + ;\frac{4}{{\sqrt 2 }} = 14.90\ {\rm{Hence;distance;between;A;and;B;}} = {\rm{;}}\sqrt {{{\left( {14.24} \right)}^2} + ;{{\left( {14.90} \right)}^2}} = 20.61 \end{array}\)

Thus shortest distance between M and N at the end of their travel is 20.61 km

Let’s assume that the piece from which rectangle is made, has length x mm.

Perimeter of rectangle = x

∴ Breadth of rectangle = x/6 and length of rectangle = 2x/6 = x/3

⇒ Area of rectangle \(= \frac{x}{6} \times \frac{x}{3} = \frac{{{x^2}}}{{18}}\)

Perimeter of square = 340 – x

Length of square = (340 – x)/4 = 85 – x/4

⇒ Area of square \(= {\left( {85 - \frac{x}{4}} \right)^2}\)

Total area \(= {\left( {85 - \frac{x}{4}} \right)^2} + \frac{{{x^2}}}{{18}} = f\left( x \right)\)

Now, \(f'\left( x \right) = 2 \times \left( {85 - \frac{x}{4}} \right) \times - 1 + \frac{{2x}}{{18}} = 0\)

Solving, we get: x = 180

Length of square = 85 – x/4 = 85 – 45 = 40 mm

Concept:

A symmetric matrix is a square matrix whose transpose equals to it; that is, it satisfies the condition AT = A.

Skew-symmetric or antisymmetric matrix is a square matrix whose transpose equals its negative; that is, it satisfies the condition AT = −A.

\(B = \left[ {\begin{array}{*{20}{c}} 0&e&f\ { - e}&0&g\ { - f}&{ - g}&0 \end{array}} \right],{B^T} = \left[ {\begin{array}{*{20}{c}} 0&{ - e}&{ - f}\ e&0&{ - g}\ f&g&0 \end{array}} \right] = \left( { - 1} \right)\left[ {\begin{array}{*{20}{c}} 0&e&f\ { - e}&0&g\ { - f}&{ - g}&0 \end{array}} \right] = - B\)

It is a skew-symmetric matrix because of aij = -aji 

To satisfy the above condition, diagonal elements of skew-symmetric matrix are always zero.

Concept:

L' hospital’s rule is used for a function who takes 0/0 or ∞/∞ form.

In this, we differentiate numerator and denominator until they take a finite value.

Calculation:

At, x → 0, the function takes a form 0/0, therefore, we can use L’ hospital’s rule

\(\mathop {{\rm{Lt}}}\limits_{x \to 0} \frac{{{{\log }_e}\left( {1 + 4x} \right)}}{{{e^{3x}} - 1}} = \frac{4}{{\left( {1 + 4x} \right)3{e^{3x}}}}\)

\( = \frac{4}{{\left( {1 + 0} \right)3{e^0}}} = \frac{4}{3}\)

Explanation:

Solutions of Laplace’s equation having continuous second-order partial derivatives is given by

\(\frac{{{d^2}\phi }}{{d{x^2}}} + \frac{{{d^2}\phi }}{{d{y^2}}} = 0\) and it is called as harmonic function (where Φ = any constant).

E.g. Φ = 2xy satisfies \(\frac{{{d^2}\phi }}{{d{x^2}}} + \frac{{{d^2}\phi }}{{d{y^2}}} = 0\) hence it is called as harmonic function.

Conjugate  of Harmonic function:

If f(z) = u + iv is analytic function then imaginary part v is known as conjugate harmonic functions of u (But converse is not true) and u is conjugate harmonic of (-v).

Explanation:

A standard normal distribution having zero mean and unit variance. A standard normal curve has 99.7% area in limits -3 to +3

Note: A standard normal curve has 68% area in limits -1 to +1 and 95% area in limits -2 to +2.

These are the standard results.

Concept:

According to Newton-Raphson Method

\({X_{n + 1}} = {X_n} - \frac{{f\left( {{X_n}} \right)}}{{f'\left( {{X_n}} \right)}}\)

Calculation:

For function f(x) = x³+ x – 1 at x₀ = 1

\({X_1} = {X_0} - \frac{{f\left( {{X_0}} \right)}}{{f'\left( {{X_0}} \right)}} = 1 - \frac{{\left( {1 + 1 - 1} \right)}}{{\left[ {3 \times {{\left( 1 \right)}^2} + 1} \right]}} \)

\(= 1 - \frac{1}{4} = \frac{3}{4}\)

∴ X₁ = 0.75

Explanation:

When we draw a free body diagram we remove all the supports and force applied due to that support are drawn and force or moment will apply in that manner so that it resists forces in any direction as well as any tendency of rotation.

Following is the conclusion:

Option (B) is wrong because the ground support is shown which we never show in FBD.

Option (C) is wrong because the X component is not shown.

Option (D) is wrong because the moment produced due to the eccentric force is not shown.

Hence only option (A) is correct.

Concept:

Flexural rigidity = EI

Both the solid material bars are made of the same material

∴ E₁ = E₂

Calculation:

\({I_1} = \frac{{{a^4}}}{{12}}\)

\({I_2} = \frac{{\pi {d^4}}}{{64}}\)

Given:

A₁ = A₂

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

d = 2a/√π

Now we are taking the ratio of I₁ and I₂ and substituting the value of d in I₂

\(\frac{{{I_1}}}{{{I_2}}} = \frac{{\frac{{{a^4}}}{{12}}}}{{\pi {d^4}/64}} = \frac{{\frac{{{a^4}}}{{12}}}}{{\frac{\pi }{{64}}{{\left( {\frac{{2a}}{{\sqrt \pi }}} \right)}^4}}}\)

\(\therefore \frac{{{I_1}}}{{{I_2}}} = \frac{\pi }{3}\)

Concept:

Normal stress at a plane  with an ∠θ from plane under the action of σ_x,

\({τ _{xx}} = \left( {\frac{{{σ _x} + {σ _y}}}{2}} \right) + \left( {\frac{{{σ _x} - {σ _y}}}{2}} \right)\cos 2θ + {σ _{xy}}\sin 2θ {{;;;;;}} \ldots \left( 1 \right)\)

The shear stress at a plane with an ∠θ from plane under the action of σ_x,

\({τ _{xy}} = - \left( {\frac{{{σ _x} - {σ _y}}}{2}} \right)\sin 2θ + {σ _{xy}}\cos 2θ {{;;;;;}} \ldots \left( 2 \right)\)

(θ = - 45° as the plane under +P is rotated by 45° anti-clockwise to get the plane under +τxx)

Also,

τxx + τyy = σx + σ_y      ...(3)

Calculation:

Given:

Since the given plane is the principal plane ⇒ σxy = 0**,** σx = P, σy = - P

By using above equation (1),

\({τ _{xx}} = \left( {\frac{{P - P}}{2}} \right) + \left( {\frac{{P - \left( { - P} \right)}}{2}} \right)\cos( -90^\circ) + 0 = 0\)

By using above equation (2),

\({τ _{xy}} = - \left[ {\frac{{P - \left( { - P} \right)}}{2}} \right]\sin \left( {-90^\circ } \right) - 0 = P\)

By using equation (3),

⇒ τyy = (σx + σy) - τxx = 0

∴ τ_xx = τ_yy = 0, τ_xy = P

Alternate method:

The given plane is the principal plane because σ_xy = 0

At 45° from the principal plane, the plane of maximum shear occurs on the plane of maximum shear.

\({τ _{xx}} = {τ _{yy}} = \frac{{{σ _1} + {σ _2}}}{2} = \frac{{P - P}}{2} = 0\)

\({τ _{xy}} = \frac{{{σ _1} - {σ _2}}}{2} = \frac{{P - \left( { - P} \right)}}{2} = P\)

Explanation:

Because PQ is a rigid link so the distance between P and Q can not changes. Therefore, the velocity of Q relative to P is VQP, which has an only component perpendicular to PQ.

∴ V_Pcosβ = V_Qcosθ

∴ V_QP = V_Qsinθ – V_Psinβ

Hence only one component is present and it is perpendicular to PQ.

Explanation:

The general expression for the number of degree of freedom in a plane mechanism having n links, j simple hinge joints, h number of higher pair and F_r redundant degree of freedom is:

N = 3(n – 1) -2j – h - Fr

Concept:

Natural frequency is also calculated by the following formula.

\({\omega _n} = \sqrt {\frac{g}{{\rm{\Delta }}}} \)

Given:

Deflection (Δ) = 1 mm

Acceleration due to gravity (g) = 10 m/s²

Now to calculate the natural frequency we are using the formula as:

\({\omega _n} = \sqrt {\frac{g}{{\rm{\Delta }}}} \)

\(\therefore {\omega _n} = \sqrt {\frac{{10}}{{{{10}^{ - 3}}}}} \)

\(\therefore {\omega _n} = 100;rad/s\)

Explanation:

Bearing: It is a mechanical component used to reduce the friction between two rotating or sliding surfaces.

A plain bearing is divided into two halves, usually associated with a crankcase that can be detached or supports the main bearing cap. The plain bearing is wrapped around the journal and pressurized with oil.

Plain bearings are used in main bearings and connecting rod bearing. Its main application is in the piston and connecting rod in engine.

Roller bearing: It is a type of rolling-element bearing that uses cylindrical rollers to maintain the separation between the bearing races. The load-carrying capacity is more than ball bearing.

It is having 4 types:

1. Cylindrical roller bearing:

• Cylindrical roller bearings can only support radial loads. Axial loads will cause the ends of the rollers to rub against the sides of the races. In addition, because the rollers are fairly wide,
• cylindrical roller bearings cannot accommodate much angular misalignment.
• These bearings have short roller guided in a cage.
• These bearings are relatively rigid against the radial motion and have the lowest coefficient of friction of any form of heavy-duty rolling contact bearings.
• Such types of bearings are used in high-speed service.

2. Spherical roller bearing

• These bearings are self-aligning bearings.
• The self-aligning feature is achieved by grinding one of the races in the form of a sphere.
• These bearings can tolerate angular misalignment in the order of \(\pm1 \frac 12\).
• When used with a double row of rollers, these can carry thrust load in either direction.

3. Needle roller bearing

• These bearings are relatively slender and completely fill the space so that neither a cage nor a retainer is needed.
• These bearings are used when heavy loads are to be carried with an oscillatory motion.
• For example, piston pin bearing in heavy-duty diesel engines where the reversal of motions tends to keep the roller in correct alignment.

4. Taper Roller bearing

• The roller and raceways of these bearings are truncated cones whose elements intersect at a common point.
• Such type of bearing can carry both radial and thrust loads.

Ball-bearing: It is a type of rolling-element bearing that uses balls to maintain the separation between the bearing races.

It is having 6 types:

1. Single row deep groove ball bearing

• During assembly of this bearing, the races are offset and the maximum number of balls are placed between the races.
• The races are then centred and the balls are symmetrically located by the use of a retainer or cage.
• These bearings are used due to their high load-carrying capacity and suitability for high running speeds.

2. Filling notch ball bearing

• These bearings have notches in the inner and outer races which permits more balls to be inserted.
• The notch does not extend to the bottom of the raceway and therefore the balls inserted through the notches must be forced in position.

3. Angular contact bearing

• These bearings have one side of the outer race cut away to permit the insertion of more balls than in a deep groove bearing but without having a notch cut in both races.
• This permits the bearing to carry a relatively large axial load in one direction while carrying a relatively large radial load.
• The angular contact bearing are used in pairs so that thrust load may be carried in either direction.

4. Double row deep groove ball bearing

• These bearings may be made with radial or angular contact between the balls and the races.
• The double row bearing is appreciably narrower than two single-row bearings.
• The load-carrying capacity of such bearing is slightly less than twice that of a single-row bearing.

5. Self-aligned bearing

• These bearings permit shaft deflections with 2-3 degrees.

6. Thrust bearing

• The thrust bearing is used for carrying thrust loads exclusively and at speeds below 2000 rpm.
• It doesn't take any radial load.
• At high speeds, centrifugal force causes the balls to be forced out of the races.

Concept:

The principle used here is Conservation of mass

∴ Q₁ = Q₂ + Q₃

Calculation:

Q₁ = Q₂ + Q₃

A₁V₁ = A₂V₂ + A₃V₃

⇒ 0.9 = 300 × 10^-3 × V₂ + 200 × 10^-3 × 3

⇒ 0.9 = 0.3 V₂ + 0.6

⇒ V₂ = 1 m/s

Mistake Points 

The question does not mention that the duct is circular. Here in the place of diameter, width has been taken which implies that the cross-section is of rectangular shape.

The total area we can take here as (width*depth). Since the depth is not given and the inlet and outlet lie in the same plane we can take it as unity.

Concept:

Streamline equation for 2-D flow is:

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

Calculation:

Given,

Velocity as

\(\vec V = 2xy\hat i - {y^2}\hat j\)

u = 2xy ; ν = -y²

\(\frac{{dx}}{{2xy}} = \frac{{dy}}{{ - {y^2}}} \Rightarrow \Rightarrow \)

\(\frac{{dx}}{{2x}} = - \frac{{dy}}{y}\)

Integrating on both the sides we will get:

\(\frac{1}{2}\ln x = - \ln y + \ln c\)

\( \Rightarrow \frac{1}{2}\ln x + \ln y = \ln c \)

\(\Rightarrow\ln x + 2\ln y = 2\ln c\)

ln(xy²) = ln c²

⇒ xy² = C

Concept:

Using Fourier’s law of heat conduction.

Since steady-state one-dimensional flow equating the heat transfer on both the sides to find the temperature T at the interface.

Given:

T₁ = 130°C; k_A = 20 W/m.K; k_B = 100 W/m.K; T₃ = 30°C; b₁ = 0.1m; b₂ = 0.3m

Calculation:

\(\frac{{\left( {{T_1} - {T_2}} \right)}}{{\left( {\frac{{{b_2}}}{{{K_1}{A_1}}}} \right)}} = \frac{{({T_2} - {T_3})}}{{\left( {\frac{{{b_2}}}{{{K_2}{A_2}}}} \right)}}\)

\( \Rightarrow \frac{{\left( {130 - T} \right)}}{{\left( {\frac{{0.1}}{{20}}} \right)}} = \frac{{\left( {T - 30} \right)}}{{\left( {\frac{{0.3}}{{100}}} \right)}}\)

\( \Rightarrow T = 67.5^\circ C\)

Points to remember:

Always take area perpendicular to the heat flow

Hence here A₁ = A₂ i.e. cross-sectional area which is the same for both the slabs.

Explanation:

The dimensionless parameter, which represents the natural convection effects and is called the Grashof number.

Grashof number, Gr, as the ratio between the buoyancy force and the viscous force:

\(Gr = \frac{{g\beta \left( {{T_s} - {T_\infty }} \right)L_c^3}}{{{\nu ^2}}}\)

Nusselt number is a function of the Grashof number and the Prandtl number alone. Nu = f (Gr, Pr)

Important Non-dimensional numbers:

• Biot number  → Ratio of internal thermal resistance to boundary layer thermal resistance
• Grashof number  → Ratio of buoyancy to viscous force
• Prandtl number  → Ratio of momentum to thermal diffusivities
• Reynolds number  → Ratio of inertia force to viscous force

Explanation:

At a critical point, the liquid is directly converted into vapour without having a two-phase transition. So, the enthalpy of vaporization at a critical point is zero. The figure below represents the P-T diagram for a pure substance.

Because at critical point liquid directly convents into the vapour phase. Ice is converted directly into vapour if it is heated at constant pressure which is less than triple point pressure. Hence statement 3 is incorrect.

Concept:

1. When heat capacity rate is minimum the heat will transfer quickly hence for maximum heat transfer we will take Cmin .
2. And more the temperature difference greater will be the value of heat transfer.

C_min = Minimum heat capacity rate

C_max = Maximum heat capacity rate

∴ For maximum heat transfer:

\( \Rightarrow {\dot q_{max}} = {C_{min}}{\rm{\Delta }}{T_{max}}\)

Concept:

Specific work (W) is calculated as

W = [V₁ sin α – V₂ sin β] u

Calculation:

Given:

V₁ = 300 m/s

α = 65°

V₂ = 150 m/s, u = 150 m/s

Given axial velocity is some at entry and exit.∴ V₁ cos α = V₂ cos β

⇒ 300 cos 65° = 150 cos β

⇒ β = 32.30°

∴ Specific work:

W = [V₁ sin α – V₂ sin β] × u

W = [300 sin 65° - (-150 sin 32.30)] × 150

W = 52.0 kJ/kg

Concept:

True strain (ϵ_T) = ln (1 + ϵ)

Calculation:

Given:

Engineering strain \((\epsilon) = 0.100% = \frac{{0.100}}{{100}} \)

True strain (ϵ_T) = ln (1 + ϵ)

\(\epsilon_T=\ln\left(1+\frac{0.100}{100}\right)=\ln \left(\frac{100.1}{100}\right);\)

∈_T = 9.99500 × 10^-4

∈T = 9.99500 × 10-4  × 100 = 0.099 %

Concept:

Solidification rate in the mould in terms of thermal conductively of metal and density of material is

\(\alpha = \frac{k}{{\rho C}}\)

Hence larger the value of α faster will be the solidification rate

Given:

α_steel = 1.77 × 10^-5

α_copper = 1.165 × 10^-4

α_al = 9.72 × 10^-5

Hence from the above value we can see that the copper is having the largest value and steel is having the lowest value

Therefore the order will be:

Copper – aluminium – steel

Explanation:

EDM process is summarized as :

• With the application of voltage, an electric field build-up between the two electrodes at the position of least resistance. The ionization leads to the breakdown of the dielectric which results in the drop of voltage and the beginning of the flow of current.
• Electrons and ions migrate to anode and cathode respectively at very high current density. A column of vapour begins to form and the localized melting of work commences. The discharge channel continues to expand along with the substantial increase of temperature and pressure.
• When the power is switched off, the current drops; no further heat is generated, and the discharge column collapses. A portion of molten metal evaporates explosively and/or is ejected away from the electrode surface. With the sudden drop in temperature, the remaining molten and vaporized metal solidifies. A tiny crater is thus generated at the surface.
• The residual debris is flushed away along with products of decomposition of dielectric fluid. The application of voltage initiates the next pulse and the cycle of events.

Hence for the above procedure, it is necessary that wire and sample are electrically conducting.

Explanation:

• Gear shaping is a generating process. The cutter used is virtually a gear provided with cutting edges. The tool is rotated at the required velocity ratio relative to the gear to be manufactured and anyone manufactured gear tooth space is formed by one complete cutter tooth. This method can be used to produce cluster gears, internal gears, racks, etc with ease, which may not have the possibility to be manufactured in gear hobbing.
• Gear Hobbing is a continuous generating process in which the tooth flanks of the constantly moving workpiece are formed by equally spaced cutting edges of the hob. The main advantage of this process is its versatility to produce a variety of gears including Spur, Helical, Worm Wheels, Serrations, Splines, etc. The main advantage of the method is the higher production rate of the gears due to continuously indexing.
• Gear Milling is one of the initial and best known and metal removal process for making gears. This method requires the usage of a milling machine. This method is right now used only for the manufacture of gears requiring very less dimensional accuracy.
• Gear forming: In gear form cutting, the cutting edge of the cutting tool has a shape identical with the shape of the space between the gear teeth. Two machining operations, milling and broaching can be employed to form cut gear teeth.

Points to remember:

• Internal gears are manufactured by shaping process with a pinion cutter.
• Hobbing, milling and shaping with rack cutter is mainly used for external gears.

Explanation:

Important G codes

• G00 – Rapid Transverse
• G01 – Linear Interpolation
• G02 – Circular Interpolation (CW)
• G03 – Circular Interpolation (ACW)
• G04 – Dwell​
• G – 97 – Spindle Speed

Important M codes

• M 00 – Program Stop
• M 03 – Spindle (CW)
• M 04 – Spindle (CCW)
• M 05 – Spindle Stop
• M 08 – Coolant on
• M 09 – Coolant off
• M 10 – Clamp-on
• M 11 – Clamp off
• M 02 or M 30 – Program stop, reset to start.

Explanation:

CPM does not directly model uncertainty.

PERT was developed to address the needs of projects which are being done for the first time – a challenge to estimate activity duration.

PERT (Program Evaluation and Review Technique) uses 3 cases:

• Most Optimistic
• Most Pessimistic
• Most likely durations

PERT determines the probability for each duration, whereas CPM considers the most likely duration.

In the standard PERT analysis, the distribution assumed for the activity times is a Beta distribution.

Concept:

If matrix is upper triangular matrix or lower triangular matrix then eigen value of that matrix is diagonal element.

\(A = \left[ {\begin{array}{*{20}{c}} a&d&e\ o&b&f\ o&o&c \end{array}} \right];B = \left[ {\begin{array}{*{20}{c}} a&o&o\ d&b&o\ e&f&c \end{array}} \right]\)

A and B are upper triangular matrix and lower triangular matrix respectively and its eigen value is a, b, c.

Calculation:

\(A = \left[ {\begin{array}{*{20}{c}} 2&1&0\ 0&2&0\ 0&0&3 \end{array}} \right]\)

Above matrix is upper triangular matrix so the eigen value of matrix A is 2, 2, 3.

As the two Eigen value of given matrix are identical, the eigenvectors resulting from these identical eigen values would be identical or linearly dependent.

Therefore, number of linearly independent eigen vectors (since two eigen values are identical)

= n - 1

= 3 - 1

= 2

Concept:

Green’s Theorem: 

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

Calculation:

\(d\vec r = \vec rds;\)

Calculating the given value of the line integral

\(\mathop \oint \nolimits_C \vec F.\overrightarrow {dr} = \oint \left( {y\hat i + 2x\hat j} \right).\overrightarrow {dr};\)

Applying Green’s theorem,

\(\oint \left( {y\hat i + 2x\hat j} \right).\overrightarrow {dr};\)

\(= \oint \left( {y\hat i + 2x\hat j} \right).\left( {dx\hat i + dy\hat j} \right);\)

\(= \oint \left( {ydx + 2xdy} \right);\)

Now,

\(\mathop \oint \nolimits_C \left( {ydx + 2xdy} \right) = \mathop \oint \nolimits_A \left[ {\frac{\partial }{{dx}}\left( {2x} \right) - \frac{\partial }{{dy}}\left( y \right)} \right]dxdy\)

\(\mathop \oint \nolimits_C \left( {ydx + 2xdy} \right) = \mathop \oint \nolimits_A dxdy\)

\(\mathop \oint \nolimits_C \left( {ydx + 2xdy} \right) = \mathop \oint \nolimits_A (Area;of;circle)\)

\(\mathop \oint \nolimits_C \left( {ydx + 2xdy} \right)= \left[ {\pi {{\left( {\frac{4}{{\sqrt \pi }}} \right)}^2}} \right] =16\)

Concept:

Whenever the ∞ - ∞ form occurs then substitute x as (1/t) and then solve it.

Calculation:

\(\mathop {\lim }\limits_{x \to \infty } \sqrt {{x^2} + x - 1} - x\)

\(Let;x = \frac{1}{t},;So,;as;x \to \infty ,;t \to 0\)

\(= \mathop {\lim }\limits_{t \to 0} \sqrt {\frac{1}{{{t^2}}} + \frac{1}{t} - 1} - \frac{1}{t}\)

\(= \mathop {\lim }\limits_{t \to 0} \frac{{\sqrt {1 + t - {t^2}} - 1}}{t}\)

Since the function has (0/0) form, we can apply L’ hospital rule,

\(= \mathop {\lim }\limits_{t \to 0} \frac{{\sqrt {1 + t - {t^2}} - 1}}{t}\)

\(= \mathop {\lim }\limits_{t \to 0} \left{ {\frac{{\left[ {\frac{{\left( {1 - 2t} \right)}}{{2\sqrt {1 + t - {t^2}} }}} \right]}}{1}} \right}\)

\(= \frac{1}{2}\)

Explanation:

Number of ways in which a king can be drawn from the pack of 52 cards is \({}_{}^4{C_1}\) since there are 4 kings in a deck of cards

Similarly, a queen and a jack can be drawn in \({}_{}^4{C_1}\) ways

∴ Number of ways in which a king, a queen and a jack can be drawn is \(= {}_{}^4{C_1} \times {}_{}^4{C_1} \times {}_{}^4{C_1}\)

Number of ways in which three cards can be drawn from the pack of 52 cards is

\(= {}_{}^{52}{C_3}\)

\(\therefore The;required;probability = \frac{{sample;space;}}{{Total;possible;ways}}\)

\(\therefore The;required;probability = \frac{{{}_{}^4{C_1} \times {}_{}^4{C_1} \times {}_{}^4{C_1}}}{{{}_{}^{52}{C_3}}}\)

\(= \frac{{4 \times 4 \times 4}}{{\frac{{52 \times 51 \times 50}}{6}}}\)

\(= \frac{{16}}{{5525}}\)

Concept:

Using Newton’s second law. i.e. F = ma

Where F = Net force on the body, m = mass of body and a = acceleration

Calculation:

Free body diagram of above mass-pulley system

\({m_1}g = 100;N \Rightarrow {m_1} = \frac{{100}}{g} = 10.19kg\)

\({m_2}g = 200;N \Rightarrow {m_2} = \frac{{200}}{g} = 20.38kg\)

Consider block of 100 N:

From Newton’s 2^nd law

T - 100 = m₁a

∴ T = 10.19 a + 100 …i)

Now, Consider block of 200 N:

∴ 200 - T = m₂a

∴ T = 200 - 20.38 a      …ii)

From equation i) & ii)

10.19 a + 100 = 200 - 20.38 a

⇒ a = 3.27 m/s²

Now, substituting the value of a in equation (ii)

T = 200 - 20.38 × (3.27)

⇒ T = 133.32 N

Concept:

Apply the principle of conservation of energy between initial and final position.

Calculation:

∴ loss in potential energy = Gain in kinetic energy

\(mgH = \frac{1}{2}I{\omega ^2} + \frac{1}{2}m{V^2}\) ∵ (Rotational + translational)

∵ V = rω

\(mgH = \frac{1}{2} \times \frac{{m{r^2}}}{2} \times {\left( {\frac{V}{r}} \right)^2} + \frac{1}{2}m{V^2}\)

\(\Rightarrow 10 \times 30 = \frac{{{V^2}}}{4} + \frac{{{V^2}}}{2}\)

⇒ 10 × 30 × 4 = V² + 2V²

⇒ 3V² = 1200

⇒ V² = 400

⇒ V = 20 m/s

Concept:

Use angular velocity ratio theorem to find the velocity since we are using instantaneous centre.

Calculation:

Here we have to do some construction to find the angle ‘θ’.

\(\therefore \tan \theta = \frac{1}{2}\)

\(\therefore \theta = {\tan ^{ - 1}}\left( {0.5} \right)\)

∴ θ = 26.56

Now we will calculate α and β

α = [90° - (45° + θ)]

∴ α = 18.43°

β = 90° - (45 + α)

∴ β = 26.56°

\({V_1} = i + 2\hat j\)      …(Given)

Hence the resultant velocity will be:

\(\left| {{{\vec V}_1}} \right| = \sqrt {1 + 4} = \sqrt 5 m/s\)

Now to calculate the velocity (V₂) we are using the angular velocity ratio theorem:

\(\frac{{{V_2}}}{{IB}} = \frac{{{V_1}}}{{IA}}{\rm{;}}\)

\(\therefore \frac{{{V_2}}}{{{V_1}}} = \frac{{IB}}{{IA}}\)

\(\therefore \frac{{{V_2}}}{{\sqrt 5 }} = \frac{{IB}}{{IA}}\)            ----(1)

Now we have to find the ratio of IB is to IA

Hence in Δ ABI

By sine rule:

\(\frac{{BI}}{{\sin \left( {45 + \beta } \right)}} = \frac{{AI}}{{\sin \left( {45} \right)}}\)

\(\Rightarrow \frac{{BI}}{{AI}} = 1.34\)            ----(2)

Substituting equation (2) in equation (1)

\(\therefore \frac{{{V_2}}}{{\sqrt 5 }} = 1.34\)

∴ V­₂ = 2.99 m/s

Note:

Instantaneous centre is drawn by meeting point of perpendicular on both of the velocity vector.

Concept:

Strain in horizontal direction (X-direction) is calculated as:

As given in the question, assume plane stress condition 

∴ ϵ_z = 0

\({ϵ_x} = \frac{{{\sigma _x}}}{E} - \mu \frac{{{\sigma _y}}}{E}\)

Calculation:

Given:

L = 2 m, δ = 0.001 m, p = 250 MPa

\(- \frac{{2\delta }}{L} = - \frac{p}{E} - \mu \left( { - \frac{p}{E}} \right)\)

\(\Rightarrow - \frac{{2\delta }}{L} = - \frac{p}{E} + \mu \frac{p}{E}\)

\(\Rightarrow - \frac{{2\delta }}{L}\left( E \right) = p \left( { - 1 + \mu } \right)\)

\(\Rightarrow - \frac{{2\delta E}}{{pL}} = - 1 + \mu\)

\(\Rightarrow \mu = 1 - \frac{{2\delta E}}{{p L}}\)

\(\Rightarrow \mu=1 - \frac{{2 \times 0.001 \times 200 \times {{10}^3}}}{{250 \times 2}}\)

∴ μ = 0.2

Concept:

\({\tau _{max}};for;solid;shaft = \frac{T}{J}.r\)

\({\tau _{max}};for;hollow;shaft = \frac{T}{J}.{r_0}\)

where T = Torque, J = Polar Area Moment, r = radius of solid shaft and r₀ = outer radius of hollow shaft

Calculation:

Given:-

L_S = L_H, T_S = T_H = T, J_S = J_H

(Where L = length, T = torque, J = Polar moment of inertia)

∵ J_S = J_H

\(\frac{{\pi {d^4}}}{{32}} = \frac{\pi }{{32}}\left[ {d_0^4 - d_i^4} \right]\)

\(\Rightarrow {d_o} > d \Rightarrow {r_0} > r\)

Now, angle of Twist

\(\theta = \frac{{TL}}{{GJ}}\)

∴ θ_S = θ_H            ----(1) (∵ T, L, G, J all are same for solid and Hollow shaft)

\(\tau = \frac{{TR}}{J}\)

\(\Rightarrow {\tau _{max}};for;solid;shaft = \frac{T}{J}.r\)

\({\tau _{max}};for;hollow;shaft = \frac{T}{J}.{r_0}\)

∵ r₀ > r

∴ (τ_max)_H > (τ_max)_S            ----(2)

Hence from (1) and (2) we can see that the correct option is d.

Concept:

Deflection in cantilever beam due to

\(\to uniformly;distributed;load = \frac{{W{L^4}}}{{8EI}}\)

\(\to point;load;at;end;point = \frac{{W{L^3}}}{{3EI}}\)

Deflection due to uniformly distributed load and deflection due to reaction at simple support will be equal and opposite.

Calculation:

Deflection due to uniform load W over the span of length L will be \(\frac{{W{L^4}}}{{8EI}}\)

Let R be the reaction at simple support at the right end so it is acting as a point load on beam hence deflection due to it will be \(\frac{{R{L^3}}}{{3EI}}\)

Net deflection at the simple support will be zero since from both the sides force is equal and opposite. Hence equating them.

\(\frac{{W{L^4}}}{{8EI}} = \frac{{R{L^3}}}{{3EI}}\)

\(\therefore R = \frac{{3WL}}{8}\)

Concept:

For dynamic equilibrium,

∑ F = 0 & ∑ M = 0 about any is valid.

Calculation:

Given:

m = 1 kg, e = 50 mm, e₁ = 20 mm, b = 0.3 m, a = 2 m, a₁ = 2.5 m.

By symmetry we have,

meω² + m₁e₁ω² = m₁e₁ω² + meω²

Clearly ∑ F = 0

Now for moment,

We will balance all the moment due to forces except reaction, about O.

meω² × (b) + m₁e₁ω² × a₁– meω² × (a + b) = 0

m₁e₁ω² × a₁– meω² × a = 0

\({m_1} = \frac{{mae}}{{{e_1}{a_1}}}\)

\({m_1} = \frac{{1\left( 2 \right)\left( {50} \right)}}{{\left( {2.5} \right)\left( {20} \right)}}\)

∴ m₁ = 2kg

Concept:

The ratio of the amplitude of steady-state response to the static deflection of the spring is nothing but the magnification factor (MF).

Calculation:

Given:

excitation frequency \(= \sqrt {\frac{{3k}}{m}} .\)

To find the ratio of the amplitude of steady-state response to the deflection of the spring:

The ratio of amplitude and static deflection is nothing but a magnification factor.

Also magnification factor = A/x₀

\(\frac{A}{{{x_0}}} = \frac{1}{{\sqrt {{{\left( {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right)}^2} + {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}\)

∵ there is No damper.

ζ = 0

Calculation:

\(\therefore \frac{A}{{{x_0}}} = \frac{1}{{\sqrt {{{\left( {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right)}^2}} }}\)

We know that,

\({\omega _n} = \sqrt {\frac{k}{m}} .\)

⇒ ω/ω_n= √3; which is greater than 1.

So, we will shuffle the equation inside the square root to avoid imaginary values.

\(\frac{A}{{{x_0}}} = \frac{1}{{\sqrt {{{\left( {{{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2} - 1} \right)}^2}} }}\)

\(\frac{A}{{{x_0}}} = \frac{1}{{\left( {3 - 1} \right)}}\)

\(\frac{A}{{{x_0}}} = \frac{1}{2}\)

Hence the magnification factor is 0.5.

Concept:

Since, only torque is applied hence no direct force is acting and all the direct forces on the bolt will be zero.

This is because the applied torque will be resisted by the torque created by secondary force on bolts.

Hence only primary force will be taken into the consideration.

Calculation:

Given:

Diameter = 100 mm; ∴ radius = 50 mm = 0.05 m

Cross-sectional area of each bolt = 25 mm² = 25 × 10^-6 m

Torque (T) = 200 N.m.

Calculation:

\({F_1}{r_1} + {F_2}{r_2} + {F_3}{r_3} + {F_4}{r_4} = T;\)         ----(1)

\(and;also,\frac{F}{r} = C.\)

\(\therefore \frac{{{F_1}}}{{{r_1}}} = \frac{{{F_2}}}{{{r_2}}} = \frac{{{F_3}}}{{{r_3}}} = \frac{{{F_4}}}{{{r_4}}};and;r{;_1} = {r_2} = {r_3} = {r_4}\)

∴ F₁ = F₂ = F₃ = F₄

From the above results we can say that;

4F × r = T

\(F = \frac{{200}}{{4\left( {0.05} \right)}} = 1000N\)

\(Max.;shear;stress = \frac{F}{{Area}} = \frac{{1000}}{{25 \times {{10}^{ - 6}}}}\)

∴ τ_max = 40 MPa.

Concept:

When a flow is fully developed steady laminar flow of an incompressible fluid in a pipe (circular).

Then velocity profile at any radius r:

\(U = - \frac{1}{{4\mu }}\left( {\frac{{dP}}{{dx}}} \right)\left( {{R^2} - {r^2}} \right)\)

Calculation:

Given :

r = R/2; u = U₁

Calculation:

Substituting the above values in the velocity profile we will get:

\({U_1} = - \frac{1}{{4\mu }}\left( {\frac{{dP}}{{dx}}} \right)\left( {{R^2} - \frac{{{R^2}}}{4}} \right)\)

\(\Rightarrow \frac{{ - \left( {4\mu {U_1}} \right) \times 4}}{{3{R^2}}} = \frac{{dP}}{{dx}};\)       ----(1)

We know that;

As the velocity of the incompressible fluid at any point in the fluid is constant, hence the flow is laminar. Shear stress at any distance ‘r’ from the center of the pipe is given by.

Shear stress 

\(\therefore \left( \tau \right) = - \left( {\frac{{dP}}{{dx}}} \right)\frac{R}{2}\)          ----(2)

Substituting equation (1) in equation (2) we get;

\(\Rightarrow \tau = - \left( {\frac{{ - 4\mu {U_1} \times 4}}{{3{R^2}}}} \right)\frac{R}{2}\)

\(\tau = \frac{{8\mu {U_1}}}{{3R}}\)

Comparing it with \(k\frac{{\mu {U_1}}}{R},\) we get k = 8/3

∴ k = 2.667

Concept:

Force = m (V1 – V2) i.e. Change in linear momentum

Calculation:

Given:

Diameter of circular opening (d) = 300 mm,

ρ = 1000 kg/m³.

H = 6.2 m

Calculation:

\({V_{Jet}} = \sqrt {2gH;} \left( {Neglecting;all;losses} \right)\)

∴ V = 11.1355 m/s.

From Newton’s second law

Force = m (V₁ – V₂)

As after striking velocity in horizontal direction is zero.

∴ V₂ = 0

∴ F = mV₁

Also m = ρAV₁

\(F = \rho AV_1^2 = \left( {1000} \right)\left( {\frac{\pi }{4}} \right){\left( {0.3} \right)^2}{\left( {11.1355} \right)^2} = 8765;N\)

F = 8.765 kN

Concept:

i) For incompressible flow, \(\frac{{\partial {u_x}}}{{\partial x}} + \frac{{\partial {u_x}}}{{\partial y}} = 0\)

ii) For steady flow u should not be the function of time i.e. u≠ f(t),

Given:

\(\vec u = \frac{x}{{{x^2} + {y^2}}}\hat i + \frac{y}{{\left( {{x^2} + {y^2}} \right)}}\hat j\)

\(\Rightarrow {u_x} = \frac{x}{{{x^2} + {y^2}}};& ;{u_y} = \frac{y}{{\left( {{x^2} + {y^2}} \right)}}\)

i) For incompressible flow, \(\frac{{\partial {u_x}}}{{\partial x}} + \frac{{\partial {u_x}}}{{\partial y}} = 0\)

\(\frac{\partial }{{\partial x}}\left( {\frac{x}{{{x^2} + {y^2}}}} \right) + \frac{\partial }{{\partial y}}\left( {\frac{y}{{{x^2} + {y^2}}}} \right) = \left{ {\frac{1}{{{x^2} + {y^2}}} - \frac{{x.2x}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}} \right} + \left{ {\frac{1}{{{x^2} + {y^2}}} - \frac{{y.2y}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}} \right}\)

\(\frac{{\partial {u_x}}}{{\partial x}} + \frac{{\partial {u_y}}}{{\partial y}} = \frac{2}{{{x^2} + {y^2}}} - \frac{{2\left( {{x^2} + {y^2}} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} = 0\)

Hence the flow is Incompressible.

ii) For steady flow u should not be the function of time i.e. u≠ f(t), but in the equation it can be clearly seen that u is not the function of time

Hence the flow is Steady.

From here, you can choose the answer by eliminating option.

Since, option 2) is incorrect, therefore (A) & (C) are eliminated.

And out of (B) & (D) only (B) contains option 1.

So correct answer is (B).

Concept:

Heat transfer between two large parallel plates having a small gap is given by:

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

Calculation:

Given:

Two large parallel plates having a gap of 10 mm.

T₁ (Plate 1 temperature) = 1000 K

ε₁ = 0.5

T₂ (Plate 2 temperature) = 400 K.

ε₂ = 0.25

And σ = 5.67 × 10^-8 W/m²K.

To find heat transfer between plates (in kW/m²).

We know that;

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

\(Q = \frac{{\left( {5.67 \times {{10}^{ - 8}}} \right)\left( {{{1000}^4} - {{400}^4}} \right)}}{{\frac{1}{{0.5}} + \frac{1}{{0.25}} - 1}}\)

Q = 11049.7 W/m²

Q = 11.05 kW/m²

Concept:

Using the following equation to calculate the time requied.

\(\frac{{T - {T_\infty }}}{{{T_i} - {T_\infty }}} = {e^{ - \frac{t}{\tau }}}\)

Where τ is time constant & \(\tau = \frac{{\rho VC}}{{hA}}\) and t is time in seconds

Calculation:

Given:

Diameter of the rod (D) = 0.01 m

Length of the rod (L) = 0.2 m

T_initial = 750°C;

Immersed fluid temp (T_∞) = 100°C.

h = 250 W/m²K; ρ = 7801 kg/m³; C = 473 J/kg.K

k = 43 W/m-K.

To find the required to reach 300°C in seconds.

We know that, \(\frac{{T - {T_\infty }}}{{{T_i} - {T_\infty }}} = {e^{ - \frac{t}{\tau }}}\)

Where, τ is time constant & \(\tau = \frac{{\rho VC}}{{hA}}\)

\(\tau = \frac{{\left( {7801} \right)|\frac{\pi }{4}{d^2}\left( {0.2} \right)473}}{{\pi d\left( {0.2} \right)250}}\)

τ = 37.037

\(\frac{{300 - 100}}{{750 - 100}} = {e^{ - \frac{6}{{37.037}}}}\)

0.3077 = e^-0.027t

ln (0.3077) = -0.027t

t = 43.654 sec.

Concept:

From steady flow energy equation, we have

\(\dot m\left( {{h_1} + \frac{{c_1^2}}{2} + g{z_1}} \right) + \dot Q = \left( {{h_2} + \frac{{c_2^2}}{2} + g{z_2}} \right)\dot m + w\)

Calculation:

Given:

For an insulated horizontal nozzle as shown

h₁ = 100 kJ/kg, V₁ = 100 m/s, V₂ = 200 m/s.

Now,

From steady flow energy equation,

\(\dot m\left( {{h_1} + \frac{{c_1^2}}{{2000}} + \frac{{g{z_1}}}{{1000}}} \right) + \dot Q = \left( {{h_2} + \frac{{c_2^2}}{2} + \frac{{g{z_2}}}{{1000}}} \right)\dot m + w\)

\({h_1} + \frac{{c_1^2}}{{2000}} = {h_2} + \frac{{c_2^2}}{{2000}}\)

\({h_2} = {h_1} + \frac{{c_1^2}}{{2000}} - \frac{{c_2^2}}{{2000}}\)

\({h_2} = 100 + \frac{{{{\left( {100} \right)}^2}}}{{2000}} - \frac{{{{\left( {200} \right)}^2}}}{{2000}}\)

h₂ = 85 kJ/kg.

Concept:

Humidity ratio \(\left( w \right) = \frac{{0.622;Pv}}{{{P_{total}} - Pv}};kg/k{g_{da}}\)

Calculation:

Given:

P_total = 750 mm of Hg. & P_water = 20 mm of Hg.

To find humidity ratio.

humidity ratio \(\left( w \right) = \frac{{0.622;Pv}}{{{P_{total}} - Pv}};kg/k{g_{da}}\)

\(w = \frac{{\left( {0.622} \right)20}}{{\left( {730} \right)}} = 0.01704;kg/kgda\)

w = 17.04 g/kgda

Concept:

In series of compression the compression ratio will be constant.

i.e. \(\frac{{{p_2}}}{{{p_1}}} = \frac{{{p_3}}}{{{p_2}}} = \frac{{{p_4}}}{{{p_3}}} \)

Calculation:

Given:

p₁ = 100 kPa

\(\begin{array}{l} \frac{{{p_4}}}{{{p_1}}} = 10\ \frac{{{p_2}}}{{{p_1}}} = \frac{{{p_3}}}{{{p_2}}} = \frac{{{p_4}}}{{{p_3}}} \end{array}\)

To calculate p₃ = ?

\(\frac{{{p_4}}}{{{p_3}}} \times \frac{{{p_3}}}{{{p_2}}} \times \frac{{{p_2}}}{{{p_1}}} = 10\)

r_p × r_p × r_p = 10

\(r_p^3 = 10\)

r_p = (10)^1/3

p₂ = r_pp₁

\({p_3} = {r_p}{p_2} = r_p^2{p_1} = {\left( {10} \right)^{\frac{2}{3}}} \times 100\)

= 464.16 kPa

Concept:

Use the following relation to find enthalpy difference:

\({s_1} - {s_0} = \frac{{{h_1} - {h_0}}}{{{T_e}}}\)

Calculation:

Here, we know that for constant temperature process,

\({s_1} - {s_0} = \frac{{{h_1} - {h_0}}}{{{T_e}}}\)

\(\Rightarrow {s_1} - {s_0} = \frac{{{h_1} - {h_0}}}{{260}}\)

Also ∵ Process 1 to 2 is isentropic.

⇒ s₁ = s₂

\(\Rightarrow \frac{{{h_1} - {h_0}}}{{260}} = {s_2} - {s_0}\)     ----(1)

Also \({s_2} - {s_3} = \frac{{{h_2} - {h_1}}}{{{T_c}}} = \frac{{{h_{fg}}}}{{{T_c}}} = \frac{{1054}}{{310}}\)     ----(2)

Also,

\({s_3} - {s_0} = {c_p}\ln \frac{{{T_3}}}{{{T_0}}}\) (∵ process 3 – 0 is in liquid state & c_p is given constant during this state)

\(\therefore {s_3} - {s_0} = 4.8;In;\left( {\frac{{310}}{{260}}} \right)\)     ----(3)

Add (2) & (3)

\(\therefore {s_2} - {s_0} = 4.8;In;\left( {\frac{{310}}{{280}}} \right) + \frac{{1054}}{{310}} = 4.24\)

∴ \({s_2} - {s_0} = \frac{{{h_1} - {h_0}}}{{260}} = 4.24\)

∴ h₁ – h₀ = 1103.51 kJ/kg

Concept:

Calculate the heat supplied for melting and heat required for melting.

And then equating both of them will give you the thickness of nugget.

Calculation:

Given:

thickness of steel sheets = 2 mm.

I = 4 kA, time (t) = 0.2 seconds.

Diameter of Nugget = 5 mm

Assuming cylindrical shape of Nugget.

Volume \(= \frac{\pi }{4}{D^2}{t_n}\), where t_n → Thickness of Nugget

Now,

Heat required to melt the above volume of Nugget = Heat required to melt 1 kg × Total mass (m)

∵ density (ρ) = mass/Volume

∴ m = ρ × volume

∴ Heat required to melt volume of nugget (Q)= Heat required to melt 1 kg × ρ × volume

\(\therefore {\rm{Q}} = \left( {1400} \right) \times \frac{{\rm{\pi }}}{4}{\rm{;}}{{\rm{D}}^2}{{\rm{t}}_{\rm{n}}} \times \left( {8000} \right){\rm{;kJ}}\)

And

Heat supplied = I²Rt= (4× 10³)² × (200 × 10^-6) × 0.2

∴ Heat supplied = 640 J.

Equating the heat calculated by two methods:

\(640 = \left( {1400} \right)\frac{\pi }{4}{\left( {0.005} \right)^2}{t_n}\left( {8000} \right) \times {10^3}\)

640 = (219.911 × 10³) t_n

t_n = 2.91 × 10^-3 m

∴ t_n = 2.91 mm.

Concept:

Shear angle is calculated by the following formula.

\(\tan \theta = \frac{{r\cos \alpha }}{{1 - r\sin \alpha }}\)

Where θ = shear angle**; α** = rake angle

Calculation:

Tool material = HSS; Rake angle α = 22°; Chip thickness (t_c) = 0.8 mm; V_c = 48 m/min**;**

Feed = 0.4 mm/rev

Now,

For orthogonal Turning, insert chip thickness = feed

∴ t = 0.4 mm

chip thickness ratio (r) \(= \frac{t}{{{t_c}}} = \frac{{0.4}}{{0.8}} = 0.5\)

Also,

\(\tan \theta = \frac{{r\cos \alpha }}{{1 - r\sin \alpha }}\)

\(\therefore {\rm{tan\theta }} = \frac{{\left( {0.5} \right)\left( {\cos 22} \right)}}{{1 - \left( {0.5} \right)\sin 22}}\)

\(\therefore {\rm{tan\theta }} = 0.5704\)

Where, θ is shear plane angle

∴ θ = tan^-1 (.5704)

∴ θ = 29.70°

Concept:

Force required to punch hole is

F = (τ_u) × πDt

Calculation:

Given:

Thickness of sheet (t) = 2 mm = 2 × 10^-3m

Diameter of hole (D) = 10 mm = 10 × 10^-3m

Yield strength in tension = 100 MPa

Ultimate shear strength (τ_u) = 80 MPa

To find force required to punch hole is

F = (τ_u) × π Dt

Since, the failure is due to shear and area under failure is as shown in fig.

F = (80) π (10)(2) × 10^-6 × 10⁶

F = 5026.55 N.

F = 5.026 kN.

Concept:

From Taylor’s tool life equation, we know that

VTⁿ = constant

Calculation:

Given:

Taylor’s Tool life exponent (n) = 0.25

Cutting speed is reduced by 50%,

To find new life.

Let the initial cutting speed be V₁, Tool life be T₁ and final Tool life be T₂ and cutting speed be V₂

∴ V₂ = 0.5 V₁ (according to the given conditions)

Also from Taylor’s tool life equation, we know that

VTⁿ = constant

\(\therefore {V_1}T_1^n = {V_2}T_2^n\)

\({V_1}T_1^n = 0.5;{V_1}V_2^n\)

\({\left( {\frac{{{T_2}}}{{{T_1}}}} \right)^n} = {\left( {0.5} \right)^{ - 1}} = 2\)

\(\frac{{{T_2}}}{{{T_1}}} = {\left( 2 \right)^{\frac{1}{{0.24}}}} = 16\)

\(\frac{{{T_2}}}{{{T_1}}} = 16\)

∴ T₂ = 16 T₁

∵ Change of Tool life is asked not new tool life in terms of T₁.

Hence the answer is 16.

Concept:

Since λ and θ are changing, and all other parameters are constant.

\(\therefore \frac{{{\theta_1}}}{{{\theta_2}}} = \frac{{{\lambda _1}}}{{{\lambda _2}}}\)

Calculation:

Given:

λ = 450 nm.

Fringe spacing = 1 mm.

To find λ for fringe spacing of 1.5 mm.

We know that: \(\theta = \frac{{D\lambda }}{d}\)

Where, D = Distance between slits and screen

d = Distance between the slits

θ = fringe width.

∵ only λ and θ are changing, other parameters are constant.

\(\Rightarrow \frac{{{\theta_1}}}{{{\theta_2}}} = \frac{{{\lambda _1}}}{{{\lambda _2}}}\)

\({\lambda _2} = \left( {450} \right)\frac{{\left( {1.5} \right)}}{1}\)

∴ λ₂ = 675 nm

Given:

A point P (1, 3, -5).

It is first translated by (2î + 3ĵ - 4k̂) and then rotated counter clockwise by 90°. About z-axis.

To find new position of the point.

Initial point can be written in vector form as.

P^t = (1 + 2)î + (3 + 3)ĵ - (5 + 4)k̂

P^t = 3î + 6ĵ - 9k̂ (p^t = Translated point)

Rotation matrix is

\(R = \left[ {\begin{array}{*{20}{c}} {\cos \theta }&{ - \sin \theta }&0\ {\sin \theta }&{\cos \theta }&0\ 0&0&1 \end{array}} \right]\)

∵ Rotation is counterclockwise:

∴ θ = + 90°

\(R = \left[ {\begin{array}{*{20}{c}} {\cos 90}&{ - \sin 90}&0\ {\sin 90}&{\cos 90}&0\ 0&0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&{ - 1}&0\ 1&0&0\ 0&0&1 \end{array}} \right]\)

\({p_{final}} = \left[ {\begin{array}{*{20}{c}} 0&{ - 1}&0\ 1&0&0\ 0&0&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 3\ 6\ { - 9} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 6}\ 3\ { - 9} \end{array}} \right]\)

∴ option 1 is correct.

Concept:

When smoothing constant is given use the following formula to calculate the forecast.

F_t = α D_t-1 + (1 - α) F_t-1

Calculation:

Given:

Calculation:

To find the forecast for the month of June using exponential smoothening.

We know that

F_t = α D_t-1 + (1 - α) F_t-1

Where f_t is the forecast of current period, F_t-1 is the forecast of previous period, D_t-1 is the Demand for previous period.

So, for calculation of forecast for June, we need to calculate forecast of May.

F_May = α D_A + (1 - α) F_A

F_May = (0.6) (900) + (1 – 0.6) 850 = 880,

F_June = (0.6) D_May + (1 – 0.6) F_May

F_June = (0.6) (1030) + (0.4) (880)

∴ F_June = 970

Concept:

The goal is to maximize profit by determining how many units of parts I and II should be produced, considering the available time constraints for three machines: Turning Center, Milling Center, and Grinding Machine.

Calculation:

Given:

The profit for each unit of Part I is Rs. 40, and for Part II, it is Rs. 100. The machining time required for each part and the total available time on the machines per week are:

• Turning Center: 12 min for Part I, 6 min for Part II, with 6000 min available per week.
• Milling Center: 4 min for Part I, 10 min for Part II, with 4000 min available per week.
• Grinding Machine: 2 min for Part I, 3 min for Part II, with 1800 min available per week.

Objective Function:

We aim to maximize the profit, which is calculated as:

\( P = 40x_1 + 100x_2 \)

Where x_1 represents the number of units of Part I produced, and x_2 represents the number of units of Part II produced.

Constraints:

• \(12x_1 + 6x_2 \leq 6000\) (Turning Center time constraint)
• \(4x_1 + 10x_2 \leq 4000\) (Milling Center time constraint)
• \(2x_1 + 3x_2 \leq 1800\) (Grinding Machine time constraint)
• \(x_1 \geq 0, x_2 \geq 0\) (Non-negativity constraints)

Solving the above equation graphically

By solving the linear programming problem, the optimal solution is:

• x₁ = 375 (Produce 375 units of Part I)
• x₂ = 250 (Produce 250 units of Part II)

The maximum profit is:

\( P = 40(375) + 100(250) = 15000 + 25000 = 40000 , \text{Rs.}\)

Conclusion: The firm should produce 375 units of Part I and 250 units of Part II to achieve a maximum profit of Rs. 40,000 per week, satisfying all the machine time constraints.