GATE ME 2021 Official Paper: Shift 1 (Previous Year Paper)

Here the correct answer is ii and iv.

Key Points:-

• In the above given options, option 2 and 4 are grammatically correct.
• It is so because 'barely' and 'hardly' here are adverbs which are qualifying the verb 'walk'.
• Barely*(adverb)* - only just; almost not.
• For Example - She nodded, barely able to speak.
• Hardly*(adverb)* - scarcely (used to qualify a statement by saying that it is true to an insignificant degree).
• For Example - The little house in which he lived was hardly bigger than a hut.
• Thus, they need to be used before 'walk' to qualify it.

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Therefore, the correct sentences are ii and iv. 

Additional Information

Whenever a auxiliary verb and a main verb is used in a sentence then the adverb is placed between them to qualify the main verb.

• Note -  Always and never can begin imperative sentences.
• For Examples:-
• Never argue with the referee.
• Always wear your seatbelt.

Concept:

Sunrises in the East and sets in the West.

The shadow always falls in opposite direction of the Sun.

Calculation:

Given:

The morning sun is falling from the East, therefore the shadow of Ms. X is on the West.

Given that the shadow is on the right side of Ms. X, therefore right side is West i.e. Ms X building is facing South.

Concept:

Area of Right-angled triangle = $\frac{1}{2}× base × height$

Area of the traingle = πr²

Calculation:

Given:

ON = OM = Base of triangle = Height of triangle = Radius, Area of triangle = 50 cm².

Area of Right-angled triangle = $\frac{1}{2}× base × height$

$\frac{1}{2}× Radius × Radius =50$

∴ Radius = 10 cm

Area of the circle = πr2 = π × 10 × 10 = 100π cm².

Concept:

Replace the symbol as per their standard sign and solve using BODMAS.

Calculation:

Given:

Δ 2 ⊕ 3 Δ ((4 ⊗ 2) ∇ 4) = + 2 - 3 + ((4 ÷ 2) × 4)

∴ + 2 - 3 + (2 × 4)

∴ + 2 - 3 + 8

∴ 10 - 3 = 7

Here the correct answer is Leading a healthy lifestyle is related to a diet with leafy vegetables.

Key Points:-

• From the above given sentence, it is clear that it is related to increased consumption of leafy vegetables in the state for leading a healthy lifestyle.
• We can follow the elimination method for these type of questions.
• Let's look at option 2 - This is not the correct answer because in the given sentence, it is clearly stated that 'The increased consumption of leafy vegetables in the recent months is a clear indication that the people in the state have begun to lead a healthy lifestyle'. Therefore, it is a clear indicator of a healthy lifestyle.
• Let's look at option 3 - There is no mention of junk food in the sentence. It is only talking about increased consumption of leafy vegetables being a clear indication of healthy lifestyle. Therefore, it is not the correct answer.
• Let's look at option 4 - It is clearly stated that 'The increased consumption of leafy vegetables' which means that there was consumption of leafy vegetables in the past but now it has increased. Therefore, it is not the correct answer.
• Thus, we are left with option 1 which is the correct answer.
  

Therefore, the correct statement is Leading a healthy lifestyle is related to a diet with leafy vegetables.

Here the correct answer is Oxpeckers get a food source, rhinos may be saved from the poachers..

Key Points:-

• In the above given sentences, it is clearly mentioned that 'The oxpeckers warn the rhinos about approaching poachers, thus possibly saving the lives of the rhinos. Oxpeckers also feed on the parasitic ticks found on rhinos'.
• Thus, it is clear from the above lines that the Oxpeckers not only warn the Rhino about approaching poachers but also eat the parasitic ticks found on Rhinos.
• Therefore, Rhinos are saved from poachers and Oxpeckers get a food source in the name of ticks.
• Poachers - A poacher is someone who breaks the law to hunt or fish. Someone who captures wild animals illegally is also a poacher.
• Approaching - coming nearer in distance or time.
  

Hence, the correct answer is Oxpeckers get a food source, rhinos may be saved from the poachers.​

Explanation:

Draw a rectangle and carved out the remaining position.

After the carved-out portion is out, make a mirror image of it.

Explanation:

In Farm P:

Hens = 65, Ducks = 91 and Goats = 169

In Farm Q:

Total number of Hens, Ducks, and Goats = 416

The ratio of Hens : Ducks : Goats = 5 : 14 : 13

Total number of Hens = $\frac{5}{32}\times416=65$

Total number of Ducks = $\frac{14}{32}\times416=182$

Total numbers of Goats = $\frac{13}{32}\times416=169$

From Farm P, all the Hens, Ducks and Goats are sent to Farm Q. 

Total Hens in farm Q = 65 + 65 = 130

Total Ducks in farm Q = 91 + 182 = 273

Total Goats in farm Q = 169 + 169 = 338

New Ratio = 130 : 273 : 338 = 10 : 21 : 26

Concept:

From pie-chart, we can get the total number of executives in each company.

From the ratio table, we can get the numbers of executives with and without a management degree.

Calculation:

Given:

Total executives = 10,000

Percentage of executives in C2 = 5%, C5 = 20%

Ratio of executives with and without management degree C2 = 1 : 4, and C5 = 9 : 1.

Executives in C2 = $\frac{5}{100}\times10000=500$

Executives in C5 = $\frac{20}{100}\times10000=2000$

Executives with Management degree holders in C2 = $\frac{1}{5}\times500=100$

Executives with Management degree holders in C5 = $\frac{9}{10}\times2000=1800$

∴ total executives with management degree holders in C2 and C5 = 100 + 1800 = 1900.

Explanation:

There are two restrictions given:

• Q and R should be separated by one person.
• S should not be seated adjacent to Q.

 

Case I:

When the person 'Q' is in the starting.

Q P R S T

Q P R T S

Q T R P S

Q T R S P.

Case II:

When the person 'Q' is in second from left.

T Q P R S

P Q T R S

Case III:

When the person 'Q' is in the middle.

S T Q P R

S P Q T R

Case IV:

When the person 'R' is in the starting.

R T Q P S

R P Q T S

Case V:

When the person 'R' is in second from left.

S R P Q T

S R T Q P

Case VI:

S T R P Q

S P R T Q

P S R T Q

T S R P Q

∴ the total number of possible arrangements is 16 as per the given restrictions.

Concept:

 $\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right)$

Equation of this type is known as Linear First Order Equation,

The solution is given by:

$y\left( {I.F.} \right) = \smallint \left( {Q × I.F.} \right);dx + c$

where $I.F. = {e^{\smallint P;dx}};$

Calculation:

Given:

$(\sin x) \frac{dy}{dx}+y\cos x = 1$

Dividing both sides by "sin x", the equation is reduced to

$\frac{dy}{dx}+y\cot x = \rm{cosec};x$

The above equation is linear.

By comparing with standard linear equation i.e. $\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right)$

P(x) = cot x and Q(x) = cosec x.

We know that;

$I.F. = {e^{\smallint P;dx}};$

$I.F. = {e^{\smallint \cot x;dx}}={e^{\log \sin x}}=\sin x$

The solution is given by:

$y\left( {I.F.} \right) = \smallint \left( {Q × I.F.} \right);dx + c$

y × sin x = ∫(cosec x × sin x)dx + c

y × sin x = ∫dx + c

y sin x = x + c

Boundary condition: y(π/2) = π/2

$\frac{\pi}{2}\sin \frac{\pi}{2}=\frac{\pi}{2} + c$

∴ c = 0.

y sin x = x 

y(π/6) equals;

$y \sin\left (\frac{\pi}{6}\right)=\frac{\pi}{6}$

$y \left (\frac{1}{2}\right)=\frac{\pi}{6}$

$\therefore y =\frac{\pi}{3}$

Concept: 

L-Hospital Rule: Let f(x) and g(x) be two functions

Suppose that we have one of the following cases,

I.  $\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{0}{0}$

II. $\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{\infty }{\infty }$

Then we can apply L-Hospital Rule as:

$\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}$

Calculation:

Given​

​

$\displaystyle\lim_{x \rightarrow 0} \left(\dfrac{1 - \cos x}{x^2}\right) = \left( {\frac{0}{0}} \right)$

Applying L’ Hospital  rule:

$\displaystyle\lim_{x \rightarrow 0} \left(\dfrac{1 - \cos x}{x^2}\right) =\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{\frac{d}{dx}({1 - \cos x})}{{\frac{d}{dx}(x^2})} =\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{sin;x {\rm{}}}}{{ {2x\rm}}}$

$\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{sin;x {\rm{}}}}{{ {2x\rm}}} = \left( {\frac{0}{0}} \right){\rm{form}}$

Once again by L’ Hospital rule,

${\rm{}}\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{cos; {\rm{x}}}}{{{\rm{2}}}} = \frac{1}{2}$

Explanation:

Dirac Delta Function

\(\delta \left( {t - a} \right) = \mathop {\lim }\limits_{ \epsilon\to 0} \left{ {\begin{array}{*{20}{c}} {0,; - \infty < t < a}\ {\frac{1}{\epsilon},;a \le t \le a + \epsilon }\ {0,;a + \epsilon < t < \epsilon} \end{array}} \right.\)

L[δ(t – a)] = e^-as

∵\(;\mathop \smallint \nolimits_0^\infty f\left( t \right)\delta \left( {t - a} \right)dt = f\left( a \right)\)

\(\therefore L\left[ {\delta \left( {t - a} \right)} \right] = \mathop \smallint \nolimits_0^\infty {e^{ - st}}\delta \left( {t - a} \right)dt = {e^{ - as}}\)

Concept:

CONSISTENCY: A finite difference approximation is considered consistent if by reducing the mesh and time step size, the truncation error terms could be made to approach zero.

STABILITY:  A finite difference approximation is stable if the errors (truncation, round-off etc) decay as the computation proceeds from one marching step to the next.

Calculation:

Given scheme is

$\frac{{y\left( {{t_{n + 1}}} \right) - y\left( {{t_n}} \right)}}{h} = - \pi ;y\left( {{t_n}} \right)$

$\frac{{dy}}{{dt}} = - \pi y$

From the scheme,

⇒ y(t_n+1) – y(t_n) = - πh y(t_n)

⇒ y(t_n+1) = y(t_n) (1 - πh)

$\Rightarrow \frac{{y\left( {{t_{n + 1}}} \right)}}{{y\left( {{t_n}} \right)}} = 1 - \pi h$

The scheme will be stable if$;\left| {\frac{{{y_{n + 1}}}}{{{y_n}}}} \right| < 1$, as the series will converge

⇒ |1 – πh| < 1

⇒ -1 < 1 – πh < 1

⇒ - 2 < - πh < 0

⇒ 0 < πh < 2; 

⇒ 0 < h < 2/π;

Explanation:

Binomial Distribution:

A binomial distribution is a common probability distribution that occurs in practice. It arises in the following situation:

• There are n independent trials.
• Each trial results in a "success" or "failure"
• The probability of success in each and every trial is equal to 'p'.

If the random variable X counts the number of successes in the n trials, then X has a binomial distribution with parameters n and p.

X ~ Bin (n, p).

Properties of Binomial distribution:

If X ~ Bin (n, p), then the probability mass function of the binomial distribution is

f (x) = P (X =x) = nCr p^x(1 - p)^{n - x}

for x = 0, 1, 2, 3,...,n

Mean E (X) = μ = np.

Variance (σ²) = np(1 - p).

Note:

• nCr = $\frac{{n!}}{{x!\left( {n - x} \right)!}}$
• $\sum_{x = 0}^{n}$ nCr px(1 - p)n - x = 1

 

Theorem:

Let X₁, X₂, ..., X_m be independent random variables such that X_i has a BIn (n_i, p) distribution, for i = 1, 2, ..., m. Let

\(Y = ;\mathop \sum \limits_{i = 1}^m {X_i}\)

Then, Y ~ Bin \(\left( {\mathop \sum \limits_{i = 1}^m {n_i},;p} \right)\)

Bernoulli Distribution:

• A Bernoulli experiment/trial has only two possible outcomes, e.g. success/failure, heads/tails, female/male, defective/non-defective, etc.
• The outcomes are typically coded as  0 (failure) or 1 (success).

X ~ Bern (p)

P (X = 1) = 1, P (X = 0) = 1 - p, $0 ≤ p ≤ 1$

Properties:

• The probability mass function is p(x) = p^x (1 - p)^{1 - x} for x = 0, 1.
• The mean is E (X) = μ = (1 × p) + 0 × (1 - p) = p
• Since E (X²) = (1² × p) + 0² × (1 - p) = p,
• σ² = var (X) = E (X²) - μ² = p - p² = p (1 - p).

​Note:

The Bernoulli distribution is a special case of binomial distribution with n = 1.

Exponential Distribution:

The probability density function of the exponential distribution is,

f (x) = λe^-λx x ≥ 0

mean = $\frac{1}{\lambda}$ , variance = $\frac{1}{\lambda^2}$

Normal Distribution:

The probability density function of normal distribution is given by,

 $f\left( x \right) = \frac{{1;}}{{σ \sqrt {2\pi } }}{{\rm{e}}^{ - \frac{1}{2}{{\left( {\frac{{x - μ ;}}{σ }} \right)}^2}}}$ 

Where, - ∞ < x < ∞

mean = μ 

Variance = σ²

Explanation:

Elastic recovery/strain: The strain recovered after the removal of the load is known as elastic strain.

Plastic strain: The permanent changes in dimension after the removal of load is known as plastic strain.

The load is removed when the stress was 200 MPa and the corresponding strain was 0.03

After the removal of load, the body recovered and the final strain found was 0.01.

∴ Elastic strain = 0.03 - 0.01 ⇒ 0.02 and Plastic strain = 0.01 respectively.

Explanation:

Mechanism and shapes formation by cutting:

• The motion responsible for the cutting action is known as the primary motion or cutting motion.
• The motion responsible for gradual feeding the uncut portion is termed as the secondary motion or feed motion.
• The line generated by the cutting motion is called generatrix and the line generated from the feed motion is called the directrix.
• Various geometries can be obtained depending on the shapes of the generatrix and the directrix and their relative directions.

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Generation of various surfaces:

 

Explanation:

The correct sequence of machining operations to be performed to finish a large diameter through-hole is drilling, boring, reaming.

Drilling:

• Drilling is an operation to produce a cylindrical hole in a workpiece.
• The tool used in called “drill bit”.

Boring:

• Boring is the process of enlarging a hole that has already been drilled by means of a single-point cutting tool.
• The single-point cutting tool for boring operation is the Boring cutter.
• The boring process can be executed on various machine tools, including lathes (turning centers) or milling machines.

Reamer:

• A reamer is a multipoint cutting tool used for enlarging by finishing previously drilled holes to accurate sizes.
• For reaming with a hand or a machine reamer, the hole drilled should be smaller than the reamer size.

Explanation:

Backlash:

This is any non - movement that occurs during axis reversals. for example, if X-axis is commanded to move 1 inch in the positive direction. Immediately after this movement, if X-axis is commanded to move 1 inch in the negative direction. If any backlash exists in the X-axis, then it will not immediately start moving in the negative direction, and the motion departure will not be precisely 1 inch.

Preloaded Ballscrews:

Ballscrews, also called ball-bearing screws, recirculating ballscrews, etc., consist of a screw spindle and a nut integrated with balls and the balls’ return mechanism, return tubes or return caps. Ballscrews are the most common type of screws used in industrial machinery and precision machines. The primary function of a ballscrew is to convert rotary motion to linear motion or torque to thrust, and vice versa, with the features of high accuracy, reversibility, and efficiency.

There are many benefits in using ballscrews such as high efficiency and reversibility, backlash elimination, high stiffness, high lead accuracy. Compared with the contact thread lead screws as shown in the figure, a ballscrew adds balls between the nut and spindle. The sliding friction of the conventional screw is thus replaced by the rolling motion of the balls.

Features:

(1) Backlash elimination and high stiffness:

• Computer Numerically Controlled (CNC) machine tools require ballscrews with zero axial backlash and minimal elastic deformation (high stiffness).
• Backlash is eliminated by our specially designed Gothic arch form ball track and preload.
• In order to achieve high overall stiffness and repeatable positioning in CNC machines, preloading of the ballscrews is commonly used. However, excessive preload increases friction torque in operation. This induced friction torque will generate heat and reduce life expectancy.

(2) High efficiency and reversibility:

Ballscrews can reach an efficiency as high as 90% because of the rolling contact between the screw and the nut. Therefore, the torque requirement is approximately one-third of that of conventional screws. The mechanical efficiency of ball screws is much higher than conventional lead screws.

(3) Low starting torque and smooth running:

Due to metal to metal contact, conventional contact thread lead screws require high starting force to overcome the starting friction. However, due to rolling ball contact, ballscrews need only a small starting force to overcome their starting friction.

Concept:

Center line average is given by

${R_a} = \frac{{\sum a ;+ ;\sum b}}{n }$

Where ∑a → sum of height in +y direction, ∑b → sum of height in –y direction,

n → number of division

Calculation:

${R_a} = \frac{{{a_1} + {a_2} + {b_1} + {b_2}}}{4}$

${R_a} = \frac{{1 + 1 + 1 + 1}}{4}$

∴ R_a = 1

Explanation:

Carnot cycle:

• The Carnot cycle consists of 4 processes
• 1-2 isothermal heat addition
• 2-3 reversible adiabatic expansion
• 3-4 isothermal heat rejection
• 4-1 reversible adiabatic compression
• A cycle is said to be reversible only when each process in a cycle is reversible.
• Carnot cycle consists of 2 isothermal and 2 adiabatic processes.

Stirling cycle:

• A Stirling cycle consists of two reversible isothermal and two reversible constant volume (isochoric) processes.

Diesel cycle:

• Diesel cycle consists of two isentropic, one constant volume and one constant pressure processes

 

Otto cycle:

• Otto cycle consists of two isentropic, two constant volume processes.

Brayton cycle:

• Brayton cycle consists of two isentropic, two constant pressure processes.

Rankine cycle:

• Rankine cycle consists of two isentropic, two constant pressure processes

Bell-Coleman cycle:

• Bell-Coleman cycle consists of two isentropic, two constant pressure processes
• Bell Coleman cycle is also known as Reversed Brayton cycle or Reversed Joule cycle

Vapour compression refrigeration cycle:

• Vapour compression refrigeration cycle consists of two isentropic, two constant pressure processes

Concept:

The compressibility factor the ratio of actual volume to the volume predicted by the ideal gas law at a given temperature and pressure. It is used to quantify the deviation of real behaviour from the ideal gas behaviour.

$z = \frac{v}{{\left( {\frac{{RT}}{P}} \right)}} = \frac{{Pv}}{{RT}}$

Calculation:

Given:

P = 1500 kPa, ν = 2.75 m³/kmol, Z = 0.95, R = 8.314 J/K/mol

Z = $\frac{Pv}{RT}$

T = $\frac{Pv}{RZ}$ =  $\frac{1500\ \times\ 2.75}{8.314\ \times\ 0.95}$= 522.26 K $\approx$ 522 K

T = 522 - 273 = 249 ^∘C 

Additional Information

• For an ideal gas, Z = 1 at all temperatures and pressures.
• Whereas for real gases Z may be > 1 or < 1.
• The farther away Z is from unity, the more the gas deviates from ideal-gas behaviour.

Don't mark 522 K as an answer as the final answer asked is in degree Celcius 249 ∘C.

Explanation:

• In Unsteady heat conduction, Heisler charts are used to find

1. Centerline body temperature
2. The temperature at any position in the body

• Chart I – It is drawn with help of the reciprocal of Biot number and Fourier number. It is used to find the centerline temperature of the body.
• Chart II – it is drawn with help of the reciprocal of Biot number and dimensionless parameter(x/L,r/R). It is used to find the temperature of the body at any location within the body.

Fourier number (F_o):

It is used in transient heat conduction and equal to the ratio of diffusive or conductive transport rate to the quantity storage rate.

${F_o} = \frac{{Conduction;rate}}{{Thermal;Energy}} = \frac{{\infty t}}{{{L^2}}}$

Where $\infty = \frac{k}{{\rho{C_p}}}$ is the thermal diffusivity

t = characteristic time, L = length through which conduction occur

Biot Number (B_i):

It is defined as the ratio of internal conductive resistance offered by the body to external convective resistance.

${B_i} = \frac{{Internal;conductive;resistance}}{{External;convective;resistance}} = \frac{{hL}}{K}$

Where, L = Characteristic length; h = Convective heat transfer coefficient; K = Thermal conductivity;

Additional Information

Reynolds number: 

$Re = \frac{{\rho \times V \times D}}{\mu }$

Where,

ρ = Density of fluid, V = velocity of fluid, D = Diameter of pipe, μ = Dynamic viscosity of the fluid

Prandtl number 

$Pr = \frac{{\mu {C_p}}}{K} = \frac{{\left( {\frac{\mu }{\rho }} \right)}}{{\left( {\frac{K}{{\rho {C_p}}}} \right)}}$

$Pr = \frac{\nu }{\alpha } = \frac{{momentum;diffusivity}}{{thermal;diffusivity}}$

$\frac{δ }{{{δ _T}}} = {\left( {Pr} \right)^{1/3}};$

where, δ = Hydrodynamic boundary layer thickness; δT = Thermal boundary layer thickness

Grashof number

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

Concept:

For long fins, the rate of heat loss from fin is given by:

$Q=\sqrt{hpkA};θ_o$

Where k = thermal conductivity, p = perimeter of the fin, h = heat transfer coefficient, A = cross-sectional area,  and θ₀ is the temperature difference.

$Q\propto\sqrt{k};$ if other parameter remains constant.

Calculation:

Given:

Let the amount of heat loss be Q̇₁ when thermal conductivity is (k₁) = k

Let the amount of heat loss be **Q̇₂ **when thermal conductivity is (k₂) = 2k

$\frac{{{Q_{2}}}}{{{Q_{1}}}} = \sqrt {\frac{{k_2{}}}{{k_1}}}$

$\frac{{{Q_{2}}}}{{{Q_{1}}}} = \sqrt {\frac{{2k{}}}{{k}}}$

$\frac{{{Q_{2}}}}{{{Q_{1}}}} = \sqrt {{{2{}}}{{}}}$

Concept:

Relative humidity (ϕ):

If is defined as the ratio of the mass of vapour when air is not saturator to the mass of vapour when air is saturated. It represents the water vapour absorbing capacity.

$\phi = \frac{{{m_v}}}{{{m_{vs}}}}$

Also,

$\frac{{{m_v}}}{{{m_{vs}}}} = \frac{{{P_v}}}{{{P_{vs}}}}$

Therefore, $\phi = \frac{{{P_v}}}{{{P_{vs}}}}$

Where, P_v = Partial pressure of vapour when air is not saturated; P_vs = Partial pressure of vapour, when air is saturated(P_sat).

Calculation:

Given:

ϕ = 0.5,

$\phi = \frac{{{P_v}}}{{{P_{vs}}}}$

∴ P_v = 0.5 P_vs

Or

Pv = 0.5 Psat

Explanation:

The unbalanced force due to reciprocating masses varies in magnitude but constant in direction while due to the revolving masses, the unbalanced force is constant in magnitude but varies in direction.

Unbalanced force:

${F_U}= m{ω ^2}R\left( {cosθ + \frac{{cos2θ }}{n}} \right) = m{ω ^2}R\cosθ + m{ω ^2}R \times \frac{{\cos2θ }}{n}$

$F_P=m{ω ^2}R\cosθ$ = Primary unbalance force.

$F_S=m{ω ^2}R \times \frac{{\cos2θ }}{n}$ = Secondary unbalance force.

The expression of the secondary unbalance force can also be written in the following way:

$F_S=m{ω ^2}R \times \frac{{\cos2θ }}{n}=m{(2ω )^2}R \times \frac{{\cos2θ }}{4n}$

Given that the primary and secondary unbalance force is equal:

$m{ω_{eq} ^2}R_{eq}\cosθ=m{(2ω^{} )^2} \times \frac{{R}}{4n}\times\cos 2θ$

Thus, when the actual crank turned through an angle θ = ωt, the imaginary crank would have turned an angle of 2θ = 2ωt.

$R_{eq}=\frac{R}{4n}=\frac{R^2}{4L}$

where $n=\frac{L}{R}$ is known as the obliquity ratio.    

i.e. the effect of the secondary unbalance force is equivalent to an imaginary crank of length $\frac{R^2}{4L}$ rotating at double angular velocity i.e. twice the engine speed.

Concept:

From the Double integration method, we know,

$EI \times \frac{{{d^2}y}}{{d{x^2}}} = {M_x}$    -- (i)

On integrating (i) we get slope (dy/dx) at a given point

$EI \times \frac{{dy}}{{dx}} = {M_x} \times x + {C_1}$ -- (ii)

On integrating (ii), we get deflection (y) at a given point

$EI \times \left( y \right) = {M_x} \times \frac{{{x^2}}}{2} + {C_1}x + {C_2}$

Where EI is flexural rigidity

Mx is the moment at section x – x

Calculation

Given,

For cantilever beam at \(x = 0,\frac{{dy}}{{dx}} = 0;& ;x = 0,~y = 0\) (x taken from fixed-end).

From this we get C1 = C2 = 0,  Hence

$y = \frac{{M{x^2}}}{{2EI}}$  

Now at x = L/2, the deflection will be 

$y = \frac{{M{L^2}}}{{8EI}}$

Concept:

Stress: Force (F) applied per unit area (A) is called stress.

$σ = {F \over A}$

Unit of stress: N/m2.

Since the given beam is prismatic i.e. constant cross-sectional area, the stress produced is directly proportional to force acted on the section.

Calculation:

Given:

Draw the FBD of the given beam to analyze the force acting on each separate section.

 

Therefore we can see that the force acting on the section PQ, QR, RS, and ST are tensile whose magnitude is 10 kN, 20 kN, 5 kN, and 25 kN respectively.

∴ the maximum force is 25 kN on ST and the minimum force is 5 kN on RS respectively.

∵ σ ∝ F

∴ σ_max is on ST and σ_min is on RS respectively.

Explanation:

Fig. (a) Pure torsional stress.

Fig. (b) Direct shear stress.

Fig. (c) Combined direct shear and torsional shear stress in the coil wire cross-section.

Torsional shear stress in the bar:

$\Rightarrow {\tau _1} = \frac{{8PD}}{{\pi {d^3}}}$

Direct shear stress in the bar:

${\tau _2} = \frac{P}{{\frac{\pi }{4}{d^2};}} = \frac{{4P}}{{\pi {d^2}}}$

Combining both equations

∴ τ = τ1 + τ2

$\Rightarrow \frac{{8PD}}{{\pi {d^3}}} + \frac{{4P}}{{\pi {d^2}}}$

$\Rightarrow \frac{{8PD}}{{\pi {d^3}}}\left( {1 + \frac{d}{{2D}}} \right)$

$\Rightarrow \frac{{8PD}}{{\pi {d^3}}}\left( {1 + \frac{1}{{2C}}} \right);,;{\rm{where;C}} = \frac{{\rm{D}}}{{\rm{d}}} = {\rm{Spring;index}}$

$\Rightarrow \frac{{8PD}}{{\pi {d^3}}}\left( {\frac{{2C + 1}}{{2C}}} \right)$

$\Rightarrow {K_s}\left( {\frac{{8PD}}{{\pi {d^3}}}} \right)$

${{\rm{K}}_{\rm{s}}} = {\rm{shear;stress;correction;factor}} \Rightarrow \frac{{2C + 1}}{{2C}}$

Concept:

Eigen values and Eigen vector:

Is a linear transformation of a matrix A multiplied by a vector X to form:

AX = Y Where, Y is new vector.

AX = λX

Where λ = Eigen value, X → Eigen vector

The vector (λX) is in same direction of X but different in magnitude.

Calculation:

Given:

A is 2 × 2 matrix and p is Eigen vector of matrix A with Eigen value λ

p’ = Ap

By definition of Eigen vector & values:

Ap = λp

p’ = λp

||p’|| = ||λp||

||p’|| = λ||p||

Direction of vector p’ will be same as vector p = θ

Concept:

Cauchy's Integral Formula:

For Simple Pole:

If f(z) is analytic within and on a closed curve c and if a (simple pole) is any point within c, then

$\oint_{c}^{}\frac{f(z)}{z-a}dz=2\pi i.f(a)$

$\cosh z=\frac{e^z+e^{-z}}{2}$

Calculation:

Given:

$\mathop \oint \nolimits_C^{} \frac{{\cosh 3z}}{{2z}}dz$ where C represents unit circle i.e. radius is unity.

The above equation can be written in standard form i.e. $\mathop \oint \nolimits_C^{} \frac{\frac{\cosh 3z}{2}}{{z-0}}dz$

Therefore $f(z)=\frac{\cosh3z}{2}$ and a = 0.

The pole of the given function is at z = 0, and lie inside the circle.

$f(z)=\frac{\cosh3z}{2}=\frac{e^{3z}+e^{-3z}}{2\times2}=\frac{e^{3z}+e^{-3z}}{4}$

At z = 0,

$f(0)=\frac{e^{0}+e^{-0}}{4}=\frac{2}{4}=\frac{1}{2}$

Cauchy's Integral Formula:

$\oint_{c}^{}\frac{f(z)}{z-a}dz=2\pi i.f(a)$

$\mathop \oint \nolimits_C^{} \frac{\frac{\cosh 3z}{2}}{{z-0}}dz=2\pi i\times f(0)$

$\mathop \oint \nolimits_C^{} \frac{\frac{\cosh 3z}{2}}{{z-0}}dz=2\pi i\times \frac{1}{2}=\pi i$

Explanation:

By Johnson’s rule of sequencing

1. Mark the minimum time consuming operation for each process

2. It is mentioned in problem to first process turning operation.

Perform that job in machine x which has minimum time consumption.

It is A – E

3. If minimum time is force y then perform that activity from last

Here B is minimum & performed at last.

Final sequence is:

A – E – D – F – H – C – G – B

Explanation:

If there is a relation between x, y & z, then z may be expressed as a function of x & y.

If an equation of type dZ = Mdx + Ndy  is an exact differential, then it can be written as ${\left( {\frac{{\partial M}}{{\partial y}}} \right)_x} = {\left( {\frac{{\partial N}}{{\partial x}}} \right)_y}$

Now,

Comparing this thermodynamic relation  dU = TdS + τdL with ${\left( {\frac{{\partial M}}{{\partial y}}} \right)_x} = {\left( {\frac{{\partial N}}{{\partial x}}} \right)_y}$ we get,

Z = U, M = T, X = S, N = τ, y = L

The thermodynamic relation can be written as, ${\left( {\frac{{\partial T}}{{\partial L}}} \right)_S} = {\left( {\frac{{\partial \tau}}{{\partial S}}} \right)_L}$

Concept:

Degree of freedom:

• The number of independent coordinates required to describe a vibratory system is known as the degree of freedom.
• A simple spring-mass system or a simple pendulum oscillating in one plane are examples of a single-degree of freedom.
• A two-mass, two-spring system, constrained to move in one direction, or a double pendulum belongs to two degrees of freedom.

 

Lagrangian (L): It is the difference between the kinetic energy (T) and potential energy (V) of a dynamic system constrained to move in one direction.

Calculation:

Given:

 

Here the two-mass and one spring is our system and constrained to move along X-direction.

Bigger mass M:

The kinetic energy in the X-direction is T_M = $\frac{1}{2}M\dot{x}^2$

Rod:

The elemental mass of the rod is $dm = \frac{m}{b}dy$

Rod element displacement in X-direction = $x+y\sin \theta$

Rod element velocity in X-direction = $\dot {x}+y\cos\theta \dot{\theta}$

$dT=\frac{1}{2}\times element;mass \times (element;velocity)^2$

$dT=\frac{1}{2}\times \left(\frac{m}{b}dy \right)\times (\dot {x}+y\cos\theta \dot{\theta})^2$

\(T=\frac{1}{2}\times\frac{m}{b}\times\left{\int_{0}^{b}(\dot{x}^2+y^2\dot{\theta}^2+2\dot{x}y\dot{\theta}\cos \theta)dy\right}\)

$T=\frac{m}{2b}\times\left(\dot{x}^2b+\frac{b^3}{3}\dot{\theta}^2+\dot{x}\dot{\theta}b^2\cos \theta\right)$

$T_{rod}=\frac{1}{2}m\dot{x}^2+\frac{mb^2}{6}\dot{\theta}^2+\frac{mb}{2}\dot{x}\dot{\theta}\cos \theta$

Potential energy of rod:

$V_{rod}=-mg\frac{b}{2}\cos \theta$

Potential energy of spring:

$V_{spring} =\frac{1}{2}kx^2$

Combining all these equations, the Lagrangian is given by:

L = T - V

L = T_M + T_rod - V_spring - V_rod

$L=\frac{1}{2}M\dot{x}^2+\frac{1}{2}m\dot{x}^2+\frac{mb^2}{6}\dot{\theta}^2+\frac{mb}{2}\dot{x}\dot{\theta}\cos \theta-\frac{1}{2}kx^2-(-mg\frac{b}{2}\cos \theta)$

$L=\frac{1}{2}(M+m)\dot{x}^2+\frac{mb^2}{6}\dot{\theta}^2+\frac{mb}{2}\dot{x}\dot{\theta}\cos \theta-\frac{1}{2}kx^2+mg\frac{b}{2}\cos \theta$

Concept:

The deflection due to a load is generally given as

$\delta L = \frac{{PL}}{{AE}}$

Here,

The vertical deflection at any point of the cone due to self-weight will be the contraction in the height due to the weight of the below part from that point. 

Because the weight is opposed by the ground (normal reaction).

Calculation:

Given:

A right solid circular cone standing on its base:

Height H and base radius R, specific weight W, and elastic modulus E.

Let’s assume a small section of height dx at a distance of x from the top as shown in the below figure.

The radius of the section be r,

From the similar triangles,

$\frac{r}{x} = \frac{R}{H} \Rightarrow r = \frac{{Rx}}{H}$

Now the weight of the upper part will be

$P = W \times {V_x} = w \times \frac{{\pi {r^2}x}}{3} = W \times \pi \times \frac{{{R^2}{x^2}}}{{3{H^2}}} \times x$

$\Rightarrow P = \frac{{W\pi {R^2}{x^3}}}{{3{H^2}}}$

The deflection will be

${\delta _x} = \frac{{{P_x}dx}}{{{A_x}E}}$   

$\Rightarrow {\delta _x} = \frac{{\left( {\frac{{W\pi {R^2}{x^3}}}{{3{H^2}}}} \right)}}{{\left( {\frac{{\pi {R^2}{x^2}}}{{{H^2}}}} \right)E}}dx$  

$\Rightarrow {\delta _x} = \frac{{Wx;dx}}{{3E}}$  

At midpoint, we have to integrate from x = H/2 to x = H;

\( \Rightarrow \delta = \mathop \smallint \nolimits_{\frac{H}{2}}^H \frac{{Wx}}{{3E}}dx\)

$\Rightarrow \delta = \left| {\frac{{W{x^2}}}{{6E}}} \right|_{\frac{H}{2}}^H$

$\Rightarrow \delta = \frac{{W{H^2}}}{{8E}}$

Explanation:

Given:

(I_rad)_A = 10^-4 kg/m², S₁ = 15 N/mm, S₂ = 10 N/mm, a = 3.5 cm, b = 2.5 cm

Now,

Applying Torque method

I_A θ̈ + (S₁ a² + S₂ b²) θ = 0

$\mathop \theta \limits^ . \frac{{\left( {{S_1}{a^2} + {S_2}{b^2}} \right)\theta }}{{{I_A}}} = 0$

Now,

By comparing the above equation with the equation of motion,

$\mathop \theta \limits^ . \omega _n^2;\theta = 0$

${\omega _n} = \sqrt {\frac{{{S_1}{a^2} + {S_2}{b^2}}}{{{I_A}}}}$

${\omega _n} = \sqrt {\frac{{1.5 \times 1000 \times {{\left( {0.035} \right)}^2} + 10 \times 1000 \times {{\left( {0.025} \right)}^2}}}{{{{10}^{ - 4}}}}}$

ω_n = 496.236 rad/s

Now,

$N = \frac{{60 \times {\omega _n}}}{{2\pi }} = \frac{{60 \times 496.236}}{{2\pi }} = 4738.7;rpm$

∴ N ≈ 4739 rpm

Concept:

For the Poisson process of rate λ, and for any t > 0, the Probability mass function for N(t) (i.e., the number of arrivals in (0,t]) is given by the Poisson PMF

${P_{N\left( t \right)}}\left( n \right) = \frac{{{{\left( {\lambda t} \right)}^n}\exp \left( { - \lambda t} \right)}}{{n!}}$

If the arrivals of a Poisson process are split into two new arrival processes, each new process is independent.

Calculation:

Given:

Customers arrive at a shop according to the Poisson distribution with a mean of 10 customers/hour.

⇒ λ = 10 customers/60 min = 1 customer/6 min = 1/6 customers per minute;

The manager notes that no customer arrives for the first 3 minutes after the shop opens,

⇒ t = 3 min and n = 0;

From the Poisson’s PMF,

The probability of zero customers in 3 minutes will be

${P_{N\left( 3 \right)}}\left( 0 \right) = \frac{{{{\left( {\frac{1}{6} \cdot 3} \right)}^0}\exp \left( { - \frac{1}{6} \cdot 3} \right)}}{{0!}} = {e^{ - 0.5}} = 0.606$

Now,

The probability that the customer arrives in the next 3 minutes = 1 - P(0)

∴ The probability that the customer arrives in the next 3 minutes = 1 - 0.606

∴ The probability that the customer arrives in the next 3 minutes = 0.39

Concept:

By lagrangian mean value theorem,

$f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}};,{\rm{;where;c \epsilon ;}}{\rm{;}}\left( {{\rm{a}},{\rm{;b}}} \right);$

Calculation:

Given:

f(x) = x² – 2x + 2, x ϵ [1, 3]

By lagrangian mean value theorem

$f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$

$f'\left( c \right) = \frac{{f\left( 3 \right) - f\left( 1 \right)}}{{3 - 1}}$

$2c - 2 = \frac{{\left( {9 - 6 + 2} \right) - \left( {1 - 2 + 2} \right)}}{{3 - 1}}$

$2c - 2 = \frac{{5 - 1}}{2}$

c = 2