GATE ME 2022 Official Paper: Shift 2 (Previous Year Paper)

The correct answer is 'board / bored'.

Key Points

• The given words are homophones i.e. similar sounding words.
• Board: a long, thin, flat piece of wood or other hard material, used for floors or other building purposes. (लकड़ी या अन्य कठोर सामग्री का एक लंबा, पतला, सपाट टुकड़ा, जिसका उपयोग फर्श या अन्य भवन उद्देश्यों के लिए किया जाता है।)
• For eg.- Loose boards creaked as I walked on them.
• Bored: feeling weary and impatient because one is unoccupied or lacks interest in one's current activity. (थका हुआ और अधीर महसूस करना क्योंकि कोई खाली नहीं है या किसी की वर्तमान गतिविधि में रुचि नहीं है)
• For eg.- He got bored with staring out of the window.

The complete sentence will be: Writing too many things on the board while teaching could make the students get bored.

• Hence, option 2 is the correct answer.

Additional Information

• Creaked: made a harsh, high-pitched sound when being moved or when pressure or weight is applied. (स्थानांतरित होने पर या दबाव या वजन लागू होने पर एक कठोर, तेज आवाज की।)
• For eg.- The stairs creaked as she went up.

Explanation:

$2-5x≤-\left(\frac{6x-5}{3}\right);$

Multiplying 3 on both sides

$3(2-5x)≤-3\left(\frac{6x-5}{3}\right);$

6 - 15x ≤ 5 - 6x

1 ≤ -6x + 15x

1 ≤ 9x 

x ≥ 1/9

Among the given options (3) matches the given condition.

Explanation:

$y=f(x)=2\ln(\sqrt{e^x})$

The area bounded by the curve y = f(x) and in the interval where x ∈ (0, 2) is given by:

$A=\int_0^22\ln(\sqrt{e^x})dx$

$A=\int_0^22\ln({e^{x/2}})dx$

$A=\int_0^22\times \frac{x}{2}dx$

$A=\left[\frac{x^2}{2}\right]_0^2=[2-0]=2$

Explanation:

A non-leap year is of 28 days and has a maximum of 28 days i.e. 4 weeks.

Given that a person was born on the fifth Monday of February in a particular year means the said year is a leap year.

Also, the fifth Monday or any day will correspond to the 29th of February.

 

Option 1:

The 2nd February of that year is a Tuesday - As 1st February is on Monday, therefore 2nd February will be on Tuesday. (True)

Option 2:

There will be five Sundays in the month of February in that year - Out of all the days, any one day can have 5 counts as the maximum number of days in February is 29. (False)

Option 3:

The 1st February of that year is a Sunday - We have already concluded that 1st February is on Monday. (False)

Option 4:

All Mondays of February in that year have even dates - From the table we can conclude all Mondays do  not have even date as 1st, 15th and 29th is also on Monday. (False)

Explanation:

Option 1:

The above-given image will not be able to produce

 

 as one more triangle is required.

Option 2:

The above-given image will be able to produce

 

 as three triangles will make the bottom part whereas the remaining two will make the above part.

Option 3:

The above-given image will not be able to produce

 

 as there is no scope for trapezium to fit in it.

Option 4:

The above-given image will not be able to produce

 

 as one more triangle is required.

Thus option 2 is the correct  answer here.

The correct answer is 'Having ultra-black skin has both advantages and disadvantages to species S'

Key Points

• Let's refer to the passage:
• 'This helps them not only to avoid predators but also sneakily attack their prey.'
• 'However, having this extra layer of black pigment results in lower collagen on their skin, making their skin more fragile.'
• From the above-mentioned statements, it is evident that the correct logical inference is that 'having ultra-black skin has both advantages and disadvantages to species S'.
• Hence, option 3 is the correct answer.

Additional Information

• Collagen: the main structural protein found in skin and other connective tissues, widely used in purified form for cosmetic surgical treatments.
• For eg.- Vitamin C plays a vital role in the formation of collagen.

Explanation:

$Average = \frac{Sum;of;Numbers}{Total;Numbers}$

Given:

For the past m days, the average daily production at a company was 100 units per day.

$Average = \frac{Sum;of;Numbers}{Total;Numbers}$

$100 = \frac{Sum;of;Numbers}{m}$

Sum of numbers = 100m

New Average after today's production = 110

Todays production = 180 units

$New;Average = \frac{New;Sum;of;Numbers}{Total;New;Numbers}$

$110 = \frac{100m+180}{m+1}$

110(m + 1) = 100m + 180

110m  + 110 = 100m + 180

10m = 70

m = 7

Explanation:

$\rm f(p,q)=\frac{p\times p\times p\times.......\times p}{q\ terms}=p^q;$ and 

$g(p,q)=p^{p^{p^{p^{p^{..^{..^{..^{up\ to\ q\ terms}}}}}}}}$

Option 1:

f(2, 2) = 2² ⇒ 4

g(2, 2) = 22 ⇒ 4

f(2, 2) = g(2, 2) ⇒ True

Option 2:

f(g(2, 2), 2) = f(4, 2) = 42 ⇒ 16

f(2, g(2, 2)) = f(2, 4) = 24 ⇒ 16

f(g(2, 2), 2) < f(2, g(2, 2)) ⇒ Not True

Option 3:

g(2, 1) = 21 ⇒ 2

f(2, 1) = 21 ⇒ 2

g(2, 1) ≠ f(2, 1) ⇒ Not True

Option 4:

f(3, 2) = 32 ⇒ 9

g(3, 2) = 32 ⇒ 9

f(3, 2) > g(3, 2) ⇒ Not True

Explanation:

The person starts from R at 7 AM and visits S and then return to R, if the follows the routes R → Q → P → S → R.

The person starts from R at 7 AM and reaches Q at 8 AM. At Q, if a person starts at 5 AM, then after each 120 mins, the next train will start. 5 AM → 7 AM → 9 AM.

The person starts from Q at 9 AM and reaches P at 10 AM. At P, if a person starts at 8 AM, then after each 90 mins, the next train will start. 8 AM → 9.30 PM → 11 AM.

The person starts from P at 11 AM and reaches S at 12 PM. At S, if a person starts at 8 AM, then after each 45 mins, the next train will start. 8 AM → 8.45 AM → 9.30 AM → 10.15 AM → 11 AM → 11.45 AM → 12.30 PM.

The person starts from S at 12.30 PM and reaches R at 1.30 PM.

∴ the total time taken is (1.30 PM - 7 AM) = 6 hours 30 minutes.

Explanation:

Both square of case M and N are of radius 1 cm.

Case M:

Let the radius be R

∴ 4R = 1 or R = 1/4 cm.

Area of square = 1 cm².

Area of circle = 4 × πR² = π/4 cm²

Area of unshaded region = (1 - π/4) cm2

Case N:

Let the radius be r

∴ 6r = 1 or R = 1/6 cm.

Area of square = 1 cm2.

Area of circle = 9 × πR2 = π/4 cm2

Area of unshaded region = (1 - π/4) cm2

∵ the area of unshaded region in both cases M and N is same, ∴ the ratio is 1 : 1.

Concept:

Fourier Series is defined as

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

where, $a_o = \frac{1}{l}\ \int_{-l}^{l} f(x) dx$

$a_n = \frac{1}{l}\ \int_{-l}^{l} f(x) \ cos\frac{n\pi x}{l} \ dx$

$b_n = \frac{1}{l}\ \int_{-l}^{l} f(x) \ sin\frac{n\pi x}{l} \ dx$

Calculation:

Given:

f(t) is an even periodic function, Since

f(-t) = f(t)

The constant term in the Fourier series is: a_o/2

$a_o = \frac{1}{l}\ \int_{-l}^{l} f(t) dx$

l = 1

$a_o = \frac{1}{1}\ \int_{-1}^{1} f(t) dx$

Since the function is not continious, we need to break the integral according to interval

from -1 to 0, f(t) = 0,

from 0 to 1, f(t) = 4

$a_o = \int_{-1}^{0} 0\ dx + \int_{0}^{1} 4\ dx$ = 4

The constant term is:

$\frac{a_o}{2}$ = 4/2 = 2

Mistake PointsAvoid using the integral property of even function because the given function f(t) is not continuous. Though the function given is an even periodic function but it is not continuous

Concept:

Gauss divergence theorem: $\int_{A}F.dA = \int \int_{V}\int \overrightarrow{\bigtriangledown}.\overrightarrow{F} dV$

where, $\int_A \vec{F}.d\vec{A}$  is called surface integral

$\int \int_{V}\int \overrightarrow{\bigtriangledown}.\overrightarrow{F} dV$  is called volume integral

$\overrightarrow{\bigtriangledown}. \overrightarrow{F}$ is called divergence of F

 $\overrightarrow{\bigtriangledown}. \overrightarrow{F}$ = $\frac{\delta F}{\delta x} + \frac{\delta F}{\delta y} + \frac{\delta F}{\delta z}$

Calculation:

Given:

V = volume of cube = (1)³

$\vec{F}=3x\hat{i}+5y\hat{j}+6z\hat{k}$

$\overrightarrow{\bigtriangledown}. \overrightarrow{F} = \frac{\delta }{\delta x} (3x) + \frac{\delta }{\delta y} (5y) + \frac{\delta }{\delta z}(6z)$ = 3 + 5 + 6 = 14

$\int_{A}F.dA = \int \int_{V}\int \overrightarrow{\bigtriangledown}.\overrightarrow{F} dV$

$\int_{A}F.dA = \int \int_{V}\int 14\ dV$ = 14 × ( V )

$\int_A \vec{F}.d\vec{A}$ = 14

Concept:

Simpson's rule is given by -

**$\int_{x_o}^{x_o\ +\ nh}$**f(x) dx = $\frac{h}{3}$[{y_o + y_n}+ 2 {y₂ + y₄ + ......... + y_n-2} + 4 {y₁+ y3 + ......... + yn-1}]

h = width of interval / step length

yo, y1 ........ yn  - ordinates corresponding to xo, x1 ........ xn

Error = Exact value - approximate value

Calculation:

Given:

Since the given function is second-degree polynomial.

Simpson's 1/3 rd rule also uses a second degree polynomial for approximation.

Hence there will be no error in the result

The value of Is and Ie will be the same and hence 

e = 100 × (Ie - Is)/Ie = 0

Explanation:

$\int^{∞}_{-∞}e^{-a(x+b)^2}dx$

substitute, x + b = t

dx = dt

at x = -∞ , t = -∞ 

 x = ∞ , t = ∞ 

\(\int^{∞}{-∞}e^{-a(x+b)^2}dx\) = \(\int^{∞}{-∞}e^{-at^2}dt\)    ... (1)

Let substitute, at² = y² ⇒ $t = \frac{y}{\sqrt a}$

on differentiating at2 = y2

2 at dt = 2ydy

dt = $\frac{ydy}{at}$ = $\frac{ydy}{a\frac{y}{\sqrt a}}$ = $\frac{dy}{\sqrt a}$

Now substitute dt in (1)

\(\int^{∞}{-∞}e^{-at^2}dt\) = \(\frac{1}{\sqrt a}\int^{∞}{-∞}e^{-y^2}dy\)

It is given that $\int^{∞}_{-∞}e^{-x^2}dx=\sqrt{\pi}$

Hence we get 

$\frac{1}{\sqrt a}\int^{∞}_{-∞}e^{-y^2}dy$ = $\sqrt \frac{\pi}{a}$

Explanation:

Polynomial, ψ(s) = ansn + an-1sn-1 + ......+ a1s + a0 of degree n > 3

It has triple root at s = -σ 

so, at s = -σ 

ψ(s) will be equal to zero, because it is a root of the polynomial

ψ(s) = ψ(-σ) = 0

Since polynomial ψ(s) has triple root at s = -σ 

So, (s + σ)³ is one factor of polynomial

Since the factor has a cube, it will have an inflexion point in the curve as the x³ has.

Hence first and second-order derivatives will be zero at the point of inflexion.

$\frac{d\psi(s)}{ds}=0$, $\frac{d^2\psi(s)}{ds^2}=0$

Explanation:

Ultimate tensile strength is the maximum load that a material can withstand before fracture.

• In the tensile test, UTS is the maximum load attained divided by the original cross-sectional area of the specimen.

Yield stress is the stress value where the stress-strain curve changes from elastic to plastic behaviour.

True stress is the maximum load attained divided by the corresponding instantaneous cross-sectional at that point.

• True stress is the load divided by the actual area at that load in a tensile test
• True strain is the actual strain produced when a load is applied to a material.

Fracture stress is the point where the specimen fractures or ruptures. (point F in figure) 

Tensile test: measures the response of a material to a slowly applied uniaxial force. The yield strength, tensile strength, modulus of elasticity, and ductility are obtained.

Explanation:

The rod will rotate slowly rotate around the z-axis and tend towards the negative z-axis direction.

• It is because the precession(ωp) of the rod is in the clockwise direction when viewed from the top and by using the right-hand thumb rule we can find the axis of precession which is in the downward direction (negative z-axis).

Let's understand from the free body diagram:

If we cut the string AB we will get the gyroscopic effect because the circular disc is rotating about its axis in the x-direction.

• by using the right-hand thumb rule we can see that due to the angular velocity ω there is angular momentum L in the positive x-direction.

Weight (mg) acting at G (centre of gravity) and tension T in string CD will form a couple.

• torque is generated due to this couple in the negative y-axis (by the right-hand thumb rule).

The angular momentum acting on the wheel tends to follow the direction of torque.

• due to this, the rod will start spinning about the z-axis with angular velocity (ω_p) in the clockwise direction and it is called gyroscopic precession.
• Using the right-hand thumb rule we can find the direction of precession which is along the negative z-axis.

 

 Additional InformationRight-hand thumb rule: curl the fingers of your right hand in the direction of angular velocity (motion) and the thumb will point towards the direction of the angular momentum or torque.

Concept:

Maximum distortion energy theory states that the failure or yielding occurs at a point in member when the distortion strain energy per unit volume becomes equal to the limiting distortion strain energy per unit volume at the yielding point.

(σ₁ - σ₂)² + (σ2 - σ3)2 + (σ3 - σ1)2 ≤ $2 \left( \frac{S_y}{N} \right)^2$

S_y = yield stress, N = factor of saftey

Calculation:

Given:

σx = 3P, σy = -2P and τxy = √2 P

Sy = 350 MPa, N = 1

Principal stress are calculated 

σ_1,2 = $\frac{1}{2}[ (σ_x - σ_y)\ \pm\ \sqrt{(σ_x + σ_y)^2 + (2 \tau_{xy})^2} \ ]$

σ1,2 = $\frac{1}{2}[ (3P - 2P)\ \pm\ \sqrt{(3P +2P)^2 + (2 \times \sqrt2 P)^2} \ ]$

σ1,2  = $\frac{1}{2}[ P\ \pm\ \sqrt{33} P\ ]$

σ1 = 3.375 P, σ2 = - 2.375 P

According to Maximum distortion energy theory:

(σ1 - σ2)2 + (σ2 - σ3)2 + (σ3 - σ1)2 ≤ $2 \left( \frac{S_y}{N} \right)^2$

here σ₃ = 0, on simplyfying we get 

σ₁² + σ22 - σ1σ2 = $\frac{S_y}{N}$

(3.375 P)² + ( - 2.375 P)² - (3.375 P)( - 2.375 P) = (350/1)²

11.4 P² + 5.64 P² + 8.015 P² = 350²

25.055 (P)² = (350)²

P = 350 / 5 = 70 MPa

Explanation:

To find the solidification range at 15 weight % of B we need to draw the point at which solidification is completed on the solidus line (AD).

We can see that point C is the solidification point and temperature can be found using similar triangles.

BE = solidification range

By using similar triangles ABC and ADE

$\frac{AB}{BC} = \frac{AE}{ED}$

AE = 700 - 400 = 300

BC = 15 , ED = 30

$\frac{AB}{15} = \frac{300}{30}$

AB = 150^∘C

BE = AE - AB

BE = 300 -150 = 150∘C

Concept:

Allowance is minimum clearance or maximum interference: 

• it is the specified difference in dimensions between mating parts; also called functional dimension.

Clearance is the space between mating parts.

• maximum clearance is the difference between the upper limit of the hole and the lower limit of the shaft.
• minimum clearance is the difference between the lower limit of the hole and the upper limit of the shaft.

 

Calculation:

Given:

shaft of diameter $25^{-0.04}_{-0.07}$ mm 

upper limit of shaft = 25 - 0.04 = 24.96 mm

lower limit of shaft = 25 - 0.07 = 24.93 mm

hole of diameter $25^{+0.02}_{-0.00}$ mm

upper limit  of hole = 25 + 0.02 = 25.02 mm

lower limit of hole = 25 - 0 = 25 mm

Allowance is the minimum clearance in positive allowance:

• allowance = 25 - 24.96 = 0.04 mm

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

• maximum clearance = 25.02 - 24.93 = 0.09 mm

​Maximum material limit for the hole in the lower limit of the hole = 25 mm

Explanation:

Fused deposition modeling:

• Fused deposition modeling is an additive manufacturing process in which the melt extrusion method is used to place filaments of thermoplastic polymer and polyacid materials following a certain pattern to produce a three-dimensional physical model. This process uses polymers as raw materials or filaments

Laminated object manufacturing:

• It is a rapid prototyping system, in it, layers of adhesive-coated paper, plastic or metal laminates are successfully glued together and cut to shape with a knife or laser cutter.
• solid physical model is made by stacking layers of sheet stock, each an outline of the cross-sectional shape of a CAD model that is sliced into layers.
• Starting sheet stock includes paper, thermoplastic, cellulose, green compacted metals, or fibre-reinforced materials

Selective laser sintering:

• It is a type of powder bed fusion wherein a bed of powder polymer, resin or metal is targeted partially (sintering) or fully (melting) by a high power directional heating source such as a laser that results in a solidified layer of fused powder.

Explanation:

Linear motor: 

• it converts the rotary motion of the electric motor to a linear motion and because of it, no lead screw is required to convert rotary motion to linear motion.
• it operates with an AC power supply and servo controller
• the linear motor primary part is connected to the power supply to produce a magnetic field. By changing the current phase in the coils, the polarity of each coil is changed.
• the attractive and repelling forces between the coils in the primary part and the magnets in the secondary part cause the primary to move and generate a linear force.

Ball screw: It is a linear drive.

• A ball screw works in a similar way to a conventional lead screw, but the significant advantage of using a ball screw is that it uses ball bearings running in the helical channel to transmit the load.
• In high precision applications, it is often necessary to translate the rotary motion from a motor to linear motion for the payload. One way of achieving this is by using a ball screw.

Lead screw: 

• It is a mechanical linear actuator that is used to convert the rotary motion to linear motion.
• Its operation relies on the sliding of the screw shaft and the nut threads with no ball bearings between them.
• the screw shaft and the nut are directly moving against each other on a large contact area, so higher energy losses due to friction are produced.

Chain and sprocket:

• sprocket is a toothed wheel that fits onto a shaft. It is prevented from rotating on the shaft by a key that fits into keyways in the sprocket and shaft.
• a chain is used to connect two sprockets and transmit the rotatory motion.
• one sprocket is the driver sprocket and another sprocket is the driven sprocket.
• motion and force can be transmitted from the chain of one sprocket to another, therefore from one shaft to another. Chains that are used to transmit motion and force from one sprocket to another are called power transmission chains.

Explanation:

Intensive property: the properties which are independent of the mass of the system are called intensive properties for example Pressure, Temperature, density, viscosity etc.

Extensive property: the properties which depend on the mass of the system are called extensive properties for example Internal Energy, Energy, Mass, Volume, Entropy etc.

Additional Information

1. All specific properties are intensive properties because specific properties are divided by per unit mass.
2. Thermodynamic property is any property that is measurable and whose value describes the state of the system.
3. Properties are point fiction and these do not depend on the path of the function.

Explanation:

Horizontal divergent channel with the supersonic flow at the inlet.

The condition here formed is called Normal shock, where:

• Pressure at B is greater than the pressure at A, P_B > P_A
• Velocity at B is less than the velocity at A, V_B < V_A

for supersonic flow, Mach number is M > 1

By using the Bernoulli's equation and conservation of momentum the equation obtained for nozzle/diffuser are:

$\frac{dA}{A} = \frac{dP}{\rho V^2} (1 - M^2)$  ..... (1)

$\frac{dA}{A} = \frac{dV}{V} (M^2 - 1)$ ....... (2)

It is given that pressure at B is greater than the pressure at A. So adverse pressure gradient will form and cause flow separation.

At point B

• P_B > P_A so by equation (1) Mach number is less than 1.
• M<1 at point B hence subsonic flow occurs at B.
• the area at B is more than the area at A so from equation (2) we can see the velocity will decrease. i.e VB < VA

At point A

• supersonic flow is there and at B it is subsonic flow so this causes the formation of normal shock.

Additional InformationNormal flow conditions

• Velocity at B is greater than the velocity at A ( V_B > V_A)
• Pressure at B is lesser than the pressure at A ( PB < P_A)

Explanation:

Nusselt number: 

it is the ratio of heat flow rate by convection process to the heat flow rate by conduction process.

it is also equal to the dimensionless temperature gradient at the surface, and it provides a measure of the convection heat transfer occurring at the surface.

Nu = $\frac{Q_{conv}}{Q_{cond}}$

Heat flow in convection, Q_conv = hA ΔT

Heat flow in conduction, Q_cond = $\frac{KA\ \Delta T}{x}$

$Nu = \frac{Q_{conv}}{Q_{cond}} = \frac{hL}{K}$

Additional InformationPrandtl number: It is the ratio of momentum diffusivity to thermal diffusivity.

$Pr = \frac{\nu}{\alpha} = \frac{\mu C_p}{k}$

advection: transfer of heat due to the horizontal movement of fluid in the atmosphere.

Explanation:

To remain in equilibrium all the forces or moments acting on the body should be balanced by the counter-reaction force or moment. 

Here in the given diagram, there is a moment at centroid which must be balanced by the moments produced by the reaction forces at the support to attain equilibrium.

Free body diagram of given configuration:

Configuration (P):

Since the reaction forces are concurrent at the centre of the plate, which cannot produce a balancing moment about the centre. Hence it will not be in equilibrium.

Configuration (Q):

the reaction forces are acting at the corner and generate a counterclockwise moment at the centre. Hence It will be in equilibrium.

Configuration (R)

the reaction forces acting at two corners along the side of the square plate will produce a counterclockwise moment at the centre to balance the moment and thus will remain in equilibrium.

Configuration (S)

In this configuration, one reaction force is concurrent to the centre and thus will not generate any moment at the centre.

The other reaction force will generate a counterclockwise moment at the centre and will balance the moment at the centre. Hence it will be in equilibrium.

Concept:

Solidification time is the time required for casting to solidify after pouring. It is expressed by Chvorinov's rule: 

$t_s = K \left(\frac{V}{A} \right)^2$ where,

t_s = solidification time, V = volume of casting

A = surface area of casting, K = solidification factor

For casting the solidification time taken by the riser should be more than the solidification time of the casting.

(ts)_Riser ≥ (ts)casting

Calculation:

Given:

ts ∝ V/A

cube of edge length = a

Volume and area of cube:  V = a³, A = 6a²

cylindrical riser with height = h , radius = r

Volume and area of riser: V =  πr²h , A = 2πrh

(ts)Riser ≥ (ts)casting

\(\left(\frac{V}{A}\right){riser} ≥ \left(\frac{V}{A}\right){cube}\)

\(\left(\frac{\pi r^2h}{2\pi rh}\right){riser} ≥ \left(\frac{a^3}{6a^2}\right){cube}\)

$\frac{r}{2}≥ \frac{a}{6}$

r ≥ a/3

r = a/3 

from options r = a/2 also satisfies.

Additional Informationfor the optimum condition the height of the riser is equal to the radius of the riser: h = r

Explanation:

Electric discharge machining:

• material is removed by melting or evaporating from the workpiece by a series of rapidly recurring current discharges between two electrodes.
• the electrodes are separated by dielectric liquid which controls spark discharge.

Laser beam machining:

• the high power laser beam is directed in a programmed manner towards the material required to be cut.
• material is removed by melting/ vaporizing due to the high sped laser beam.

Electron beam machining: 

• the workpiece is placed in the vacuum chamber and a high voltage electron beam is directed towards the workpiece.
• the energy of the electron beam melts or vaporises the selected region

​Electrochemical machining:

• In ECM metal removal is entirely by metallic ion exchange and so there are no cutting forces and the workpiece is left in an undisturbed stress-free state.
• It is never subjected to high temperatures which may cause melting of the workpiece. It doesn’t depend at all upon the physical properties of the workpiece.

Explanation:

$\vec{V}=(4xy)\hat{i}+2(x^2-y^2)\hat{j}$

U = 4xy , V = 2(x² - y²)

δU/δx = 4y , δU/δy = 4x

δV/δx = 4x, δV/δy = -4y

For Incompressible flow: $\overrightarrow{\bigtriangledown } . \overrightarrow{V } = 0$

$\overrightarrow{\bigtriangledown } . \overrightarrow{V } = \frac{δ U}{δ x} + \frac{δ V}{δ y}$ = 4y - 4y = 0

hence the flow is incompressible.

For irrotational flow ω_z = 0

ω_z = $\frac{1}{2} \left( \frac{δ V}{δ x} - \frac{δ U}{δ y}\right)$ =  (4x - 4x) /2 = 0

hence flow is irrotational.

Acceleration at x = 0, y = 0

a_x = $\frac{DU}{Dt} = \left ( U\frac{δ U}{δ x} + V\frac{δ U}{δ y}\right)$ = 0

a_y = $\frac{DV}{Dt} = \left ( U\frac{δ V}{δ x} + V\frac{δ V}{δ y}\right)$ = 0

Hence acceleration in x and y particle is zero.

Explanation:

$\frac{du}{dx}=\frac{-xu^2}{2+x^2u}$

(2 + x²u) du + (xu²) dx = 0

For exact differential, $\frac{dP}{dx} = \frac{dQ}{du}$

Q = (2 + x2u), P = (xu2)

$\frac{dQ}{dx} =\frac{d}{dx} (2 + x^2u)$ = 2ux

$\frac{dP}{du} = \frac{d}{du}(xu^2)$ = 2ux

Since dP/du = dQ/dx, hence it is an exact differential.

Thus its solution is given by:

∫Pdx + ∫terms in Q(x,y) not containing x du = C

$∫_C xu^2 dx + ∫ 2du = C$

$\frac{x^2 u^2} {2} + 2u = C$

Explanation:

To topple the block we need an impulse force such that the weight will generate a clockwise moment about point O.

This will happen only when a minimum impulse that can rotate the block such that the Centre of gravity of the body will pass through the point O or beyond.

h = 0.4m, b = 0.3 m, m =1kg

OG^' = $\sqrt{0.2^2 + 0.15^2}$ = 0.25 m

h^' = OG' - h/2 = 0.25 - 0.2 = 0.05 m

Due to force F, the block will topple about A

So By energy balance

(K.E + P.E)initial = (K.E + P.E.)final

$\frac{1}{2}I_Oω ^2$ = 0 + mgh^'

I_o = $\frac{1}{3}(h^2 + b^2)$ = (0.4² + 0.3²)/3 =1/12

$\frac{1}{2}× \frac{1}{12}× ω ^2$ = 1 × 9.81 × 0.05

ω² = 2 × 12 × 9.81 × 0.05

ω = 3.43 rad/s

Now,

Angular impulse = change in angular momentum

IF × d = I_o × (ω_i - ω_f )

IF × 0.3 = 1/12 × ( 3.43 - 0)

IF = 0.953 Ns

Concept:

Normal stress of element under biaxial stress 

$\sigma_n = \frac{σ_{x}+σ_{y}}{2} + \frac{σ_{x}-σ_{y}}{2}\ cos\ (2θ) + τ_{xy}\ sin\ (2θ)$

Calculation:

Given:

σ_x1 = 0, σ_y1 = σ, τ_{x1, y1} = 0, θ = 45^∘ 

 =   

σ_x = $\frac{σ_{x_1}+σ_{x_2}}{2} + \frac{σ_{x_1}-σ_{x_2}}{2}\ cos\ (2θ) + τ_{xy}\ sin\ (2θ)$

σ_x = σ/2

Net normal stress

 + 

 σ_xx = σ + σx  = σ + σ/2

Additional InformationShear stress of element under biaxial stress

$\tau_n = - \frac{\sigma_x - \sigma_y}{2}\ sin\ 2\theta + \tau_{xy}\ cos\ 2\theta$

Explanation:

m_b = √2 kg, r_b = ?

m₁ = m₂ = 1 kg , r₁ = r₂ = 1 cm

Resultant balancing force = m_br_b = √2 rb

Resultant of forces = $\sqrt{(m_1r_1)^2 + (m_2r_2)^2}$ = $\sqrt{(1)^2 +(1)^2}$

√2 rb = $\sqrt{(1)^2 +(1)^2}$

rb = 1 cm

Explanation:

The given system is of forced damped vibration.

(P) c > 0 and $ω=\sqrt{k/m}$

If c > 0 then the damping will be decreasing with time hence the graph will be a decreasing parabola.

c > 0 ⇒ ζ > 1 overdamped condition

(Q) c < 0 and ω ≠ 0

If c < 0 ⇒  ζ < 1 it implies an underdamped condition that means amplitude will be increasing and the curve will be increasing parabola.

R) c = 0 and $\omega=\sqrt{k/m}$

If c = 0, the system will be of natural spring system, with no damping, acting with an external force.

frequency equal to natural frequency ω = ω_n = $\sqrt{k/m}$ then the vibrations will increase suddenly along a straight line with no damping.

(S) c = 0 and $\omega \cong\sqrt{k/m}$

No damping and the amplitude will increase and decrease in a certain period.

Hence , P - i, Q - iv, R - ii, S - iii.

Explanation:

By Johnson's rule of sequencing:

1. Mark the minimum time-consuming operation for each process 

2. It is given that the first parts are machined in the milling machine. Perform the milling of the parts in a sequence from the marked parts that is part which has minimum time consumption has to be machined first. 

here Priority order is: P3 - P4 - P5

3. If minimum time is taken by Polishing then the sequence is done in such a way that the minimum time-consuming part is machined last. 

that is the part that has maximum time must be machined first.

hence Priority order is: P1 - P6 - P2 - P7

4. Final Sequence: P3 - P4 - P5 - P1 - P6 - P2 - P7

5. Now estimate the time consumed corresponding to the optimal sequence. Hence minimum total completion time = 33 hours

Explantion:

2x₁ + 3x₂ ≤ 70   ....... (1)

2x1 + x2 ≤ 50   ......... (2)

2x2 ≤ 40 .................. (3)

For Profit Maximization: 

Z = 150x1 + 100x2 

subtract equation 1, 2  we get:

2x2 = 20 ⇒ x2 = 10

substitute in eq 2 we get: 2x1 = 40 ⇒ x1= 20

A = (20, 10)

From eq 1 and 3 we get:

x2 = 20

2x1 + 3 × 20 = 70 ⇒ x1 = 5

B = (5, 20)

Profit at:

A = (20, 10) ⇒ Z = 150 × 20 + 100 × 10 = Rs. 4000

B = (5, 40) ⇒ Z = 150 × 5 + 100 × 20 = Rs. 2750

C = (25,0) ⇒ Z = 150 × 25+ 100 × 0 = Rs. 3750

D = (0,20) ⇒ Z = 150 ×0 + 100 × 20 = Rs. 2000

So, maximum profit of day is  Rs. 4000

Explanation:

Applying Bernoulli's equation at 1 and 2

 

$\frac{P_1}{{\rho}g}+z_1 = \frac{P_2}{{\rho}g}+\frac{V_2^{2}}{2g}+z_2$

$\frac{P_{atm}+pgh_1}{{\rho}g}+\frac{V_1^2}{2g}+0$ = $\frac{P_{atm}}{{\rho}g}+\frac{V_2^2}{2g}+h_1 + h_2$ [∵ V1 = V] 

$\frac{V_2^2}{2g}=\frac{V_1^2}{2g}+h_1 - (h_1 + h_2)$

∴ $V_2= \sqrt{V^2-2gh_2}$

∴ Q = A₂V₂ = $\frac{\pi d^2}{4}\times \sqrt{V^2-2gh_2}$

Concept:

Navier stoke equation for viscous flow:

$ρ \frac{\Delta V}{\Delta t} = - \frac{δ p}{δ x} + μ(\bigtriangledown ^2u) + ρ g_x$

V = velocity field, p = pressure, μ = viscosity

Calculation:

Given:

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

u = (y² - x²) , v = 2xy

x =1, y = 1, ρ = 1.5 kg/m3

for inviscid flow $\mu(\bigtriangledown^2 u) = 0$, body force (ρg) = 0

So the equation for inviscid flow and negligible body force is:

$ρ \left[u\frac{ δ u}{δ x} + v\frac{δ u}{δ y}\right] = - \frac{δ p}{δ x}$

$\frac{δ p}{δ x} =$ - 1.5 [ (y2 - x2) (-2x) + 2xy (2y)]

at x = 1 y = 1 

δp / δx =  - 1.5 [ 0 + 2 × 2 ] = - 6i

Similarly for y direction: 

$ρ \left[u\frac{ δ v}{δ x} + v\frac{δ v}{δ y}\right] = - \frac{δ p}{δ y}$

δp / δy = - 1.5 [ (y2 - x2) (2y) +  2xy (2x)]

at x = 1 y = 1 

δp / δy = - 1.5 [ 0 +  2 × 2] = - 6j

Hence, Pressure gradient is = - 6i - 6j

Explanation:

h_f = 50 kJ/kg

COP = 3, h₁ = 250 kJ/kg , h₂ = 300 kJ/kg

COP = $\frac{h_1\ -\ h_4}{h_2\ -\ h_1}$ 

3 = $\frac{250\ -\ h_4 }{300\ -\ 250}$ ⇒ h₄ = 100 kJ/kg

h₄ = h_f + x h_fg

100 = 50 + x (h₁ - h_f )

50 = x (250 - 50)

x = 50/200 ⇒ x = 0.25

Explanation:

Matrix = [A]_{3 × 5}

Rank of matrix ρ(A) = 2 

number of unknown variable, n = 5

Since, ρ(A) < n

The given equation will have infinitely many solutions including a zero vector of appropriate value.

Additional InformationIf the rank of a matrix is equal to the rank of an augmented matrix and equal to the number of unknowns then the system is consistent and unique.

if the rank of a matrix is not equal to the rank of the augmented matrix then the system is inconsistent and has no solution.

Explanation:

For member GH,

As point G can move in a slot, hence force provided by member BE will be perpendicular to the member.

Referring to FBD,

ΣM_H = 0 

⇒ R_G = 0

⇒ Force in member GH is zero.

As member AB, CD, and EF are link members, hence they can carry only axial forces.

Referring to FBD,

​

ΣF_x = 0

⇒ F_AB = Q

ΣM_E = 0

⇒ - Q × l/2 + F_CD × l - Q × l/2 + F_AB × l = 0

Putting the value of F_AB = 0.

F_CD = 0

ΣF_y = 0

⇒ F_EF = Q

Hence member CD and GH carry zero force.