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).

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

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:

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

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:

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:

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:

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

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

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:

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:

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°

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