GATE ME 2015 Official Paper: Shift 1 (Previous Year Paper)

The word misogynously is an ‘adverb’ and the remaining words are adjectives.

Use the adjective misogynous to describe a system or a person who shows a strong dislike for women, or who acts in a way that negatively affects women in particular. A government that doesn't allow women to vote, drive, or go to school can be described as misogynous, or being strongly prejudiced against women.

Given that,

Probability that Ram is selected ⇒ \({P_1} = \frac{1}{6}\)

Probability that Ramesh is selected ⇒ \({P_2} = \frac{1}{8}\)

Probability that only one of them is selected = P

\(\begin{array}{l} P = {P_1} \times \overline {{P_2}} + {P_2} \times \overline {{P_1}} \ = \frac{1}{6} \times \left( {1 - \frac{1}{8}} \right) + \frac{1}{8} \times \left( {1 - \frac{1}{6}} \right)\ = \frac{{12}}{{40}} = \frac{1}{4} \end{array}\)

Sometimes the subject is separated from the verb by such words as along with, as well as, besides, not, etc. These words and phrases are not part of the subject. Ignore them and use a singular verb when the subject is singular.

Examples:

The politician, along with the newsmen, is expected shortly.

Excitement, as well as nervousness, is the cause of her shaking.

Dhoni, as well as the other team members of the Indian team, was present on the occasion.

 

SO the electricity consumption per km

For \(M = \frac{{12}}{{20}} = 0.6\)

For \(N = \frac{{13}}{{25}} = 0.555\)

For \(O = \frac{{20}}{{30}} = 0.667\)

For \(P = \frac{{12}}{{25}} = 0.48\)

Awkward: Causing or feeling uneasy embarrassment or inconvenience, causing difficulty; hard to do or deal with

Inept: Having or showing no skill; clumsy

Graceful: Having or showing grace or elegance

Suitable: Right or appropriate for a particular person, purpose, or situation

Dreadful: Causing or involving great suffering, fear, or unhappiness; extremely bad or serious

Inept is a synonym for awkward in a clumsy topics.

Explanation:

There is an error in part P of the sentence. The correct sentence would be: "The student corrected all the errors (P) that the instructor marked (Q) in the answer book (R)."

Symbols F = Film star,  S = Playback singer  D = Film directors

From the Venn-diagram, we can say that both conclusions follow i.e. all film directors are playback singers and some film stars are film directors.

Some of the keywords are:

Lament: express passionate grief about

Anguish: be extremely distressed about something

"Lamenting the gradual sidelining of the arts in school curricula, a group of prominent artists wrote to the Chief Minister last year, asking him to allocate more funds to support arts education in schools. However, no such increase has been announced in this year's Budget."

So these statements below are logically valid and can be inferred from the above statements:

• i. The artists expected funding for the arts to increase this year.
• iv. Schools are giving less importance to arts education nowadays.

We know that

(a + b + c)² ≥ 0

a² + b² + c² + 2(ab + bc + ac) ≥ 0

1 + 2(ab + bc + ac) ≥ 0

(ab + bc + ac) ≥ -1/2

(a - b)² + (b - c)² + (c - a)² ≥ 0

2(a2 + b2 + c2) - 2(ab + bc + ac) ≥ 0

(a2 + b2 + c2) - (ab + bc + ac) ≥ 0

(1) - (ab + bc + ac) ≥ 0

(ab + bc + ac) ≤ 1

⇒ -1/2 ≤ (ab + bc + ac) ≤ 1

(ab + bc + ac) ∈ [-1/2, 1]

1 leap of Tiger = 8 meter

Speed of Tiger = 5 leaps/min = 40 m/min

And

1 leap of Deer = 5 meter

Speed of Deer = 4 leaps/min = 20 m/min

Let at time ‘t’ the tiger catches the deer.

∴ Distance covered by tiger = (distance travelled by deer) + (Initial distance between deer & tiger)

40 × t = 20t + 50 × 8

\(\Rightarrow t = \frac{{400}}{{20}} = 20\ min\)

So, total distance = 400 + 20t = 400 + 20 × 20 = 800 m

Properties of Determinants:

• The value of the determinant remains unchanged if both rows and columns are interchanged.
• If any two rows (or columns) of a determinant are interchanged, then sign of determinant changes.
• If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then the value of the determinant is zero.
• If some or all elements of a row or column of a determinant are expressed as the sum of two (or more) terms, then the determinant can be expressed as the sum of two (or more) determinants.
• det (A^T) = det (A)
• det (kA) = kndet(A)
• det(A.B) = det(A).det(B)

Given:

Velocity of point P = 10m/s (i.e.) V_p = 10 m/s

Radius = r

We know that as body is rotating so angular velocity will be involved.

∴ To find the velocity of point Q in (m/s) we have to find angular velocity at that point

As V_Q = R × ω_Q

Here for point Q; R = 2r.

Concept: Whenever wheel rotates without slipping on horizontal surface then the point at which wheel touches the surface become Instantaneous Centre. Hence we take the distance from that point.

∴ V_Q = 2rω

Now (V = 10 m/s) -----(Given)

 i.e.  (rω = 10 m/s)  ∵ V = rω

∴ V_Q = 2 × 10

⇒ V_Q = 20 m/s

Explanation:

So stress-strain relationship which describes the behaviour of brittle materials is

Explanation:

The interpolator coordinate feed of axes in a CNC machine controller.

In contouring systems, the machining path is usually constructed from a combination of linear and circular segments. It is only necessary to specify in the part program the coordinates of the initial and final points of each segment and the feed rate.

The operation of producing the required shape based on this information is termed interpolation, and the corresponding software algorithm in CNC is the interpolator.

The interpolator coordinates the motion along the machine axes, which are separately driven, to generate the required machining path.

An interpolator coordinates these axis motions in such a way that the programmed path is constantly maintained from the beginning to the end of the movement.

The two most common types of interpolators are linear and circular.

Concept:

Minimum clearance: the difference between the Lower limit of hole and upper limit of shaft is called the minimum clearance.

Maximum clearance: The difference between the upper limit of hole and lower limit of shaft is called the maximum clearance.

Allowance: The difference between the basic dimensions of the mating parts.

When the shaft size is less than the hole size, then the allowance is positive and when the shaft size is greater than the hole size, then the allowance is negative.

i.e. L.L of hole – U.L of shaft > 0, allowance is +ve.

U.L of hole – L. L of shaft < 0, Allowance is – ve.

Allowance is either minimum clearance or maximum interference.

Calculation:

U.L of Hole = 25.040 mm

L.L of hole = 25.020 mm

U.L of shaft =25.005 mm

L.L of shaft = 24.992 mm

Minimum clearance = L.L of Hole – U.L of shaft

= 25.020 – 25.005 = 0.015 mm

Concept:

A) Simpson’s one-third rule:

For applying this rule, the number of subintervals must be a multiple of 2.

\(\mathop \smallint \nolimits_{{x_0}}^{{x_0} + nh} f\left( x \right)dx = \frac{h}{3}\left[ {\left( {{y_0} + {y_n}} \right) + 4\left( {{y_1} + {y_3} + {y_5} + \ldots + {y_{n - 1}}} \right) + 2\left( {{y_2} + {y_4} + {y_6} + \ldots + {y_{n - 2}}} \right)} \right]\)

B) Simpson’s three-eighths rule:

For applying this rule, the number of subintervals must be a multiple of 3.

\(\mathop \smallint \nolimits_{{x_0}}^{{x_0} + nh} f\left( x \right)dx = \frac{{3h}}{8}\left[ {\left( {{y_0} + {y_n}} \right) + 3\left( {{y_1} + {y_2} + {y_4} + {y_5} + \ldots } \right) + 2\left( {{y_3} + {y_6} + \ldots } \right)} \right]\)

Calculation:

x0 = 0, xn = 1 = x0 + nh

h = 1/n

h = 1/2 = 0.5

 

\(\mathop \smallint \nolimits_{{0}}^{{1} } f\left( x \right)dx = \frac{h}{3}\left[ {\left( {{y_0} + {y_2}} \right) + 4y_1 } \right]\)

 \(\mathop \smallint \limits_0^1 ydx = \frac{{0.5}}{3}\left[ {\left( {\frac{9}{5} + \frac{{12}}{5}} \right) + 4\left( {\frac{{39}}{{20}}} \right)} \right]=2\)

Concept:

Head loss due to friction is given by \({h_f} = \frac{{fL{V^2}}}{{2gD}}\)

\({h_f} = {\frac{{fLv^2}}{{2gD}}}= {\frac{{fLQ^2}}{{2gDA^2}}}\)

Where,

F = friction factor \( = \frac{{64}}{{{R_e}}}\)

Re = Reynold’s number, L = length of pipe, V = velocity of fluid flowing through the pipe, D = Diameter of the pipe

Calculation:

Given: Mass flow rate, density and friction factor to be constant. L' = 2L, d' = 0.5 d

\({h_f}= {\frac{{fLQ^2}}{{2gDA^2}}}\)

Or \(h∝ \frac{L}{{d.A^2 }}∝\frac{L}{{d^5 }}\)

But we know that \(A = \frac{\pi }{d}{d^2}\) or A ∝ d²

\( \frac{{h'}}{h} = \frac{{L'}}{L} \times {\left( {\frac{d}{{d'}}} \right)^5} = 2 \times {\left( 2 \right)^5} = 64\)

h' = 64 h

dh = C_pdT, so specific enthalpy is a function of temperature only.

The ideal gas is defined as a gas which obeys the following equation of state: Pv = RT

The internal energy of an ideal gas is a function of temperature only. That is, u = u(T)

Using the definition of enthalpy and the equation of state of an ideal gas, h = u + Pv = u + RT

Since R is a constant and u = u(T), it follows that the enthalpy of an ideal gas is also a function of temperature only.

h = h(T)

Since u and h depend only on the temperature for an ideal gas, the constant volume and constant pressure specific heats c_v and c_p also depend on the temperature only. c_v = c_v (T), c_P = c_P (T) 

For all ideal gases:

• The specific heat at constant volume (c_v) is a function of T only.
• The specific heat at constant pressure (cp) is a function of T only.
• A relation that connects the specific heats cp, cv , and the gas constant is cp - cv = R
• The specific heat ratio, γ = cp/cv, is a function of T only and is greater than unity.

Efficiency of \(HE - 1,{\eta _1} = 1 - \frac{{500}}{{900}} = 44.4%\)

Or \({\eta _1} = 1 - \frac{{{Q_2}}}{{{Q_1}}} = 0.444\) 

\(\Rightarrow {Q_2} = \left( {1 - 0.444} \right){Q_1}\)

= (1 – 0.444) × 150 = 83.33 MJ

Efficiency of HE – 2,

\({\eta _2} = 1 - \frac{{300}}{{500}} = 0.4\) or 40%

And \({\eta _2} = {1} - \frac{{{Q_3}}}{{{Q_2}}} = 0.4\)

\(\Rightarrow {Q_3} = \left( {1 - 0.4} \right) \times 83.33 = 50MJ\)

Concept:

Mixing of Air Stream:

Consider an adiabatic mixing of two moist air streams during which no condensation of moisture takes place.

when two air streams at state points 1 and 2 ​mixes, the resulting mixture condition 3 can be obtained from mass and energy balance. 

m₁ + m₂ = m₃

m₁ω₁ + m₂ω₂ = m₃ω₃

m₁h₁ + m₂h₂ = m₃h₃

\(\begin{array}{l} \frac{{{w_2} - {w_3}}}{{{w_3} - {w_1}}} = \frac{{{h_2} - {h_3}}}{{{h_3} - {h_1}}} = \frac{{{m_1}}}{{{m_2}}} \end{array}\)

Humidity ratio (w) is the mass of water associated with each kilogram of dry air. It is known as specific humidity or absolute humidity.

\(w =\frac{m_v}{m_a}= 0.622 \times \frac{{{p_v}}}{{{p_a} - {p_v}}}\)

Calculation:

Mass flow rate of moist air = 10.1 kg/s (given)

Humidity ratio \(ω = \frac{{{m_v}}}{{{m_a}}} = 0.01;kg/s\)

Mass of moist air = Mass of dry air + Mass of water vapor

10.1 = m_a+0.01 x m_a

m_a = 10.1/ 1.01 = 10 kg/s

Mass of water vapor  m_v1 = 10.1 - 10 = 0.1 kg/s

m_v2 = 0.1 kg/s (given)

Total mass of water vapour in mixture:

m_v = m_v1 + m_v2 = 0.2 kg/s

Total mass of dry air:

m_a = m_a1 + m_a2 = 10 + 0 = 10 kg/s

Humidity ratio  : \({ω _{final}} = \frac{{{m_v}}}{{{m_a};}} = \frac{{0.2}}{{10}} = 0.020\frac{{kg}}{{kg}}of;dry;air\)

Concept:

The maximum load at which the column tends to have lateral displacement or tends to buckle is known as buckling or crippling load. Load columns can be analyzed with the Euler’s column formulas can be given as

\(P = \frac{{{n^2}{\pi ^2}EI_{min}}}{{{L^2}}}\)

• For both end hinged, n=1
• For one end fixed and other free, n=1/2
• For both end fixed, n=2
• For one end fixed and other hinged, n=√2

Effective Length:

\(L_{eq}=\frac{L}{n}\)

Calculation:

Since both the ends are hinged

\({P _e} = \frac{{{\pi ^2}EI_{min}}}{{L^2}}\)

\(I = {{\mathop{\rm I}\nolimits} _{min}} = \frac{{d{b^3}}}{{12}} = 0.02 \times \frac{{{{0.01}^3}}}{{12}}\)

Then \({P _e} = {\pi ^2} \times 200 \times {10^9} \times \left( {0.02 \times \frac{{{{0.01}^3}}}{{12}}} \right) = 3289.868\ N\)

Concept:

Mean effective pressure is the ratio of the net work done to the displacement volume of the piston.

\(mep = \frac{{{W_{net}}}}{{{V_s}}}\)

\({V_s} = {V_1} - {V_2} = {V_1}\left( {1 - \frac{{{V_2}}}{{{V_1}}}} \right) = {V_1}\left( {1 - \frac 1r} \right)\)

For the Diesel cycle:

Compression ratio, r = V1/V2 

Calculation:

Given:

r = 21, ρ = 1.0 kg/m3, ṁ = 30 × 10–3 kg/s, W_net = 15 kW

\({V_1} = \frac{{\dot m}}{\rho } = 30 \times {10^{ - 3}};{m^3}/s\)

Now,

\({V_s} = {V_1}\left( {1 - \frac 1r} \right)\)

Mean effective pressure is:

\(mep = \frac{{{W_{net}}}}{{{V_s}}}=\frac{{{W_{net}}}}{{{{V_1}\left( {1 - \frac 1r} \right)}}}\)

\( mep = \frac{{15}}{{30 \times {{10}^{ - 3}}\left[ {1 - \frac{1}{{21}}} \right]}}=525\ kPa\)

Rails – Rolling

Rails are manufactured by the rolling process.

Rolling involves the continuous forcing of metal between two rotating rolls on opposing sides of the workpiece. Structural shapes, such as I - beams, and railroad track rail are produced by rolling. 

The rolling process in the making of rails imparts many beneficial effects to the properties of rail steel. During rolling, the grain structure of steel is refined and the oriented cast structure (harmful to steel) is destroyed. The metal becomes more homogeneous and undesirable brittle films are broken up.

Engine crankshaft – Forging

Crankshaft is main rotating parts of the engine which is installed on the connecting rod, can undertake the (reciprocating) up and down movement into circulation movement (rotation) of the connecting rod.

Crankshafts can be manufactured by either Casting or Forging. The internal grain of forged crankshaft is uniform, eliminates raw material defect and tissue defect and so on.

Aluminum channels – Extrusion

Aluminum channels are formed during the extrusion process, in which round aluminum stock called “billets” or “logs” are pressed by a ram through a die, which is a hollow profile that shapes the aluminum into a specific extruded shape as the billet is squeezed through.

PET water bottles – Blow molding

Blow molding is used extensively for making bottles and other light-weight hollow products.

PET, which stands for polyethylene terephthalate, is a form of polyester (just like the clothing fabric). It is extruded or molded into plastic bottles and containers for packaging foods and beverages, personal care products, and many other consumer products.

Concept:

Taylor’s Tool Life Equation:

VTn = C

Where, V = cutting speed (m/min), T = Time (min) (time taken to develop certain flank wear), n = an exponent that largely depends on tool material, C = constant based on tool and work material and cutting condition.

Calculation:

VTn = C

V₁T₁n = V2T2n

Given that V₂ = 2V₁, T₂ = (1/16)T₁ 

\(\frac{{{V_2}}}{{{V_1}}} = {\left( {\frac{{{T_1}}}{{{T_2}}}} \right)^n} \)

\(n = \frac{{\ln \left( {\frac{{{V_2}}}{{{V_1}}}} \right)}}{{\ln \left( {\frac{{{T_1}}}{{{T_2}}}} \right)}} = \frac{{\ln \left( 2 \right)}}{{\ln \left( {16} \right)}} = 0.25\)

Explanation:

T = Iα (standard Relation)

∵ θ = angular displacement

dθ/dt = Angular velocity = ω

d²θ/dt²= Angular acceleration = α

\(\therefore T = \frac{{I{d^2}\theta }}{{d{t^2}}}\)

\(\therefore \frac{T}{I} = \frac{{{d^2}\theta }}{{d{t^2}}}\)

Integrating the above equation we get;

\(\therefore \frac{{d\theta }}{{dt}} = \frac{T}{I}t + {c_1}\)

Again integrating (∵ we want θ & time relation)

\(\therefore \theta = \frac{T}{I}{t^2} + c_1^t + {c_2}\)

∴ We can see that when we compare the above eqⁿ with standard eqⁿ of a parabola

i.e.  y = aX² + bX + C

The above-obtained equation is also the same hence the graph will be parabolic.

 But in real life due to some function, the graph will look a little distorted.

Hence option (d) is the correct answer.

\(\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \left( {{x^2}} \right)}}{{2{x^4}}}\)

∴ Putting limits we get \(\frac{0}{0}\) form

Hence apply L. Hospital Rule

∴ differentiating numerator & denominator.

\(\therefore \frac{d}{{dx}};\left( {\frac{{1 - \cos \left( {{x^2}} \right)}}{{2{x^4}}}} \right)\)

\(\mathop {\lim }\limits_{x \to 0} \left( {0 - \frac{{( - \sin \left( {{x^2}} \right)2x)}}{{8{x^3}}}} \right)\)

\(\mathop {\lim }\limits_{x \to 0} \frac{{(\cos {x^2})2x \cdot 2x + (\sin {x^2}) \times 2}}{{24{x^2}}}\)

\(\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\cos {x^2}}}{6} + \frac{{\sin {x^2}}}{{12{x^2}}}} \right]\)

\(\mathop {\lim }\limits_{x \to 0} \frac{{\cos {x^2}}}{6} + \mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{\sin {x^2}}}{{12{x^2}}}\)

\( \Rightarrow \frac{1}{6} + \frac{1}{{12}};\mathop {\lim }\limits_{x \to 0} \frac{{\sin {x^2}}}{{{x^2}}}\left( {std;form} \right)\)

\( \Rightarrow \frac{1}{6} + \frac{1}{{12}} \times 1;\)

\( \Rightarrow \frac{1}{4}\)

By vertical force balance, we get

2Fsin 30° = 100

⇒ F = 100 N

Probability distribution function for normal distribution is given by

\(f\left( x \right) = \frac{1}{{\sigma \surd \pi }}{e^ - }\frac{{{{\left( {x - \mu } \right)}^2}}}{{2{\sigma ^2}}}\)

Variance = σ²

For lowest variance, σ should be lowest

And if σ decreased, f (x) will increase

So, curve will have highest peak.

Explanation:

The argument of a complex number z = x + iy

\(\arg z = {\tan ^{ - 1}}\left( {\frac{y}{x}} \right)\)

\({Z_1} = 5 + \left( {5\sqrt 3 } \right)i,{Z_2} = \frac{2}{{\surd 3}} + 2i\)

Then \(\arg{Z_1} = {\tan ^{ - 1}}\left( {\frac{{5\sqrt 3 }}{5}} \right) = 60^\circ\)

And \(\arg{Z_2} = {\tan ^{ - 1}}\left( {\frac{2}{{\frac{2}{{\sqrt 3 }}}}} \right) = 60^\circ\)

So, \(\arg \left( {\frac{{{Z_1}}}{{{Z_2}}}} \right) = \arg \left( {{Z_1}} \right) - {\rm{arg}}\left( {{Z_2}} \right)\)

\(\arg \left( {\frac{{{Z_1}}}{{{Z_2}}}} \right) \) = 60° - 60° = 0

Blasius equation for boundary layer theory is

\(\frac{{{d^3}f}}{{d{\eta ^3}}} + \frac{1}{2}f\frac{{{d^2}f}}{{d{\eta ^2}}} = 0\)

An ordinary linear differential equation is

\(\frac{{{d^3}y}}{{d{x^3}}} + {C_1}\frac{{{d^2}y}}{{d{x^2}}} + {C_2}\frac{{dy}}{{dx}} + {C_3} = 0\)

And a non-linear ordinary differential equation is

\(\frac{{{d^3}y}}{{d{x^3}}} + {c_1}y\frac{{{d^2}y}}{{d{x^2}}} + {c_2}\frac{{dy}}{{dx}} + {C_3} = 0\)

From, the basic definition of linear and non-linear differential equations,

Blasius equation,

\(\frac{{{d^3}f}}{{d{\eta ^3}}} + \frac{1}{2}f\frac{{{d^2}f}}{{d{\eta ^3}}} = 0\)  

is a third-order non-linear ordinary differential equation.

Let her speed = x km/hr

She is travelling 10 km in 2 hrs so (along with flow)

\(\therefore Speed = \frac{{distance}}{{time}}\)

\(\therefore Speed = \frac{{10;km}}{{2;hr}}\)

Speed = 5 km/hr (along with flow)    --- (1)

She takes 5 hours for same distance when swimming against the flow

\(\therefore Speed = \frac{{distance}}{{time}}\)

\(Speed = \frac{{10;km}}{{5;hr}}\)

Speed = 2 km/hr (against the flow)      --- (2)

Now from eqⁿ (1) & (2) we get

x km/hr + (flow of river) km/r = 5 km/hr ----------(3)

x km/hr – (flow of river) km/hr = 2 km/hr----------(4)

Adding (3) & (4)

2x km/hr = 7 km/hr

∴ x = 3.5 km/hr

Explanation:

If fluid flows on a surface having a different temperature than the surface, the thermal boundary layer is developed similar to the velocity boundary layer.

For the flow of viscous fluid over a flat plate, if the fluid temperature is the same as the plate temperature, the thermal boundary layer is not formed at all.

Soderberg line is the most conservative fatigue failure criterion. Even yielding is not considered in soderberg criterion. 

The Gerber parabola is the best fit for the failure points of the test data.

The Goodman is safer from design consideration since it lies completely inside the Geber parabola and failure points.

Heat supplied \(= {H_s} = \frac{{VI}}{{{A_b}.v}}{\eta _h}\)

Or \({H_s}\propto \frac{1}{v}\)

The heat input per unit length is inversely proportional to welding speed.

Concept:

θ = Angle of Twist

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

Calculation:

Since Torque is applied at point B

SO, θ = θ_C = θ_B

\(\Rightarrow \theta = \frac{{TL}}{{GJ}} = \frac{{10 \times 0.5 \times 32}}{{77 \times {{10}^9} \times \pi \times (0.02)^4}}\times \frac{180}\pi = 0.236\)

Concept:

The value of Router for which the heat transfer rate is maximum is called the critical radius of insulation.

The thickness up to which heat flow increases and after which heat flow decreases is termed as Critical thickness.

The insulation radius at which resistance to heat flow is minimum and consequently heat flow rate is maximum is called “critical radius”.

Critical radius of insulation for a cylindrical body:

\({r_{cr,cylinder}} = \frac{k}{h}\)

Critical radius of insulation for a spherical shell:

\({r_{cr,sphere}} = \frac{{2k}}{h}\)

The rate of heat transfer from the cylinder increases with the addition of insulation for r2 < rcr, reaches a maximum when r2 = rcr, and starts to decrease for r2 > rcr. Thus, insulating the pipe may increase the rate of heat transfer from the pipe instead of decreasing it when r2 < rcr.

Calculation:

The critical radius of insulation \({r_c} = \frac{k}{h} = \frac{{0.08}}{{10}} = 0.008\)m

Or \({r_c} = 8mm\)

So, if more insulation (greater than 8 mm of outer radius) added, the heat loss increases up to maximum and then decreases.

Given: A 10 mm diameter electrical conductor is covered by the insulation of 2 mm thickness. 

So outer radius is 5 + 2 = 7 mm. 

The rate of heat transfer will increase up to 8 mm (for 7 mm to 8 mm, heat loss will increase continuously to maximum value), and then it will decrease. So, we can say that the addition of further insulation of the same material will increase heat loss to a maximum and then decrease heat loss.

\(\begin{array}{l} {\sigma _{1,2}} = \frac{{\sigma_x + \sigma_y}}{2} + \sqrt {{{\left( {\frac{{\sigma_x - \sigma_y}}{2}} \right)}^2}+{\tau ^2}} \ {\sigma _{1,2}} = \frac{{80 + 20}}{2} + \sqrt {{{\left( {\frac{{80 - 20}}{2}} \right)}^2}+{{\left( {40} \right)}^2}} \end{array}\)

So, σ₁ = 100 MPa,      σ₂ = 0

\(\tau = \frac{{\sigma_1 - \sigma_2}}{2} =\frac{{100 - 0}}{2} = 50 ~MPa\)

Hence \(FOS = \frac{{matrical\ strength}}{{shear\ stress}} = \frac{{100}}{{50}} = 2\)

n = 4, P = 1/6 = Prob. of ‘six’

then prob of failure (other than ‘six’) \(= 1 - \frac{1}{6} = \frac{5}{6}\)

P(x ≥  2) = 1 – P (x < 2)

= 1 – P (0) – P (1)

\(= 1 - {{\rm{4}}_{{C_0}}}{\left( {\frac{1}{6}} \right)^0}{\left( {\frac{5}{6}} \right)^4}{ + ^4}{C_1}{\left( {\frac{1}{6}} \right)^1}{\left( {\frac{5}{6}} \right)^3} = \frac{{19}}{{144}}\)

Calculation:

Given, P = 400N, e = 500mm

The primary shear forces

 \(P_1' = P_2'=P_3' = P_3' = \frac{400}{4} = 100 \ N\)

\({r_1} = {r_2} = {r_3} = {r_4} = \sqrt {{{20}^2} + {{20}^2}} = 20\sqrt 2 ;mm\)

\(C = \frac{{400 \times 500}}{{{{\left( {20\sqrt 2 } \right)}^2} \times 4}} = 62.5\)

The secondary shear Forces

\(P_1^{''} = c{r_1} = 62.5 \times 20\sqrt 2 = 1767.77N\)

Since r₁ = r₂ = r₃ = r_4,

\(so,;P_1^{''} = P_2^{''} = P_3^{''} = P_4^{''} = 1767.77N\)

Here θ = 45°

For the worst loaded riveter

\({P_{net}} = \sqrt {{{\left( {100} \right)}^2} + {{\left( {1767.77} \right)}^2} + 2.\left( {100} \right).\left( {1767.77} \right)\cos 45} \)

= 1839.84 N

Concept:

Gauge pressure:

• Gauge pressure is the pressure relative to atmospheric pressure. Most of the pressure measuring devices are calibrated to read gauge pressure.
• P = ρgh

Atmospheric pressure:

• Atmospheric pressure or barometric pressure is the pressure within the atmosphere of Earth. Atmospheric pressure is closely approximated by the hydrostatic pressure caused by the weight of air above the measurement point.
• Patm = 1.013 bar.

Absolute pressure:

• Absolute pressure is the sum of gauge pressure and atmospheric pressure.
• Pabs = Pgauge + Patm .

The characteristic gas equation of an ideal gas is given by:

• ⇒ PV= mRT

Calculation:

GIven:

T = 288 K, V = 2 m3, h = 70 cm of mercury, R = 8314 J/kmol-K, P_atm = 1.01325 bar

Pressure p = ρgh = 0.7 × 13600 × 9.81 = 93391.2 Pa

P_actual = p_atm + p = 101325 + 93391.2 = 194716.2 Pa

And pv = mRT \(\Rightarrow m = \frac{{pv}}{{RT}}\)

\(\Rightarrow m = \frac{{194716.2 \times 2}}{{288 \times (\frac{8314}{28})}} = 4.5539\ kg\)

Concept:

The total solidification time is the time required for the casting to solidify after pouring. This time is dependent on the size and shape of the casting by an empirical relationship known as Chvorinov’s rule.

\(T = {C_m}{\left( {\frac{V}{{{A_s}}}} \right)^2}\)

Calculation:

Solidification time \(t\propto {\left( {\frac{V}{A}} \right)^2}\)

For cube:

\(\frac{V}{A} = \frac{{{a^3}}}{{6{a^2}}} = \frac{a}{6}\)

Cube 2: Top face insulated

\(\begin{array}{l} \Rightarrow {t_1}\propto {\left( {\frac{a}{6}} \right)^2}\ {t_2}\propto {\left( {\frac{{{V_2}}}{{{A_2}}}} \right)^2} \Rightarrow {t_2}\propto {\left( {\frac{{a^3}}{{5{a^2}}}} \right)^2} \Rightarrow {t_2}\propto {\left( {\frac{a}{5}} \right)^2} \end{array}\)

Then \(\frac{{{t_2}}}{{{t_1}}} = \frac{{{{\left( {\frac{a}{5}} \right)}^2}}}{{{{\left( {\frac{a}{6}} \right)}^2}}} = \frac{{36}}{{25}}\)

Explanation:

 \(\nabla \times \vec V = 0 - Irrotational;flow;\) 

\(\vec V = \overrightarrow {{V_x}} \widehat {\dot i} + \overrightarrow {{V_y}} \widehat {\dot J} + \overrightarrow {{V_z}} \hat k\)

\(\nabla = \frac{\partial }{{\partial x}}\widehat {\dot i} + \frac{\partial }{{\partial y}}\widehat {\dot J} + \frac{\partial }{{\partial z}}\hat k\)

\(\nabla \times \vec V = Curl\left( {\vec V} \right) = \left| {\begin{array}{*{20}{c}} i&{\dot J}&k\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\ {{V_x}}&{{V_y}}&{{V_z}} \end{array}} \right|;\)

Curl of any vector is zero, which means the flow is not rotating inside the domain or the flow is imitational flow.

\(Curl.\vec V = 0,\) i.e Flow is irrotational

 

\(\nabla .\vec V = 0\) - Incompressible continuity equation

The general continuity equation for any flow field is

\(\frac{{\partial \rho }}{{\partial t}} + \nabla \left( {\rho .\vec V} \right) = 0\)

A fluid is said to be incompressible if the density of the fluid is not varying with time or space i.e. constant.

So,  \(\frac{{\partial \rho }}{{\partial t}} = 0;& ;\nabla \left( {\rho .\vec V} \right) = 0\) 

\(Or,;\rho \left( {\nabla .\vec V} \right) = 0;\)

Or,  \(\nabla \vec{.V} = 0\), incompressible continuity equation.

 

\(\frac{{D\vec V}}{{D\rho }} = 0;\) - Zero acceleration of the fluid particle

D – Total derivative

\(\vec V = {\vec V_x}\widehat {\dot i} + {\vec V_y}\hat J + {\vec V_z}\hat k\)

\(\frac{{D\vec V}}{{Dt}} = \frac{{D{V_x};}}{{Dt}}\widehat {\dot i} + \frac{{D{V_y}}}{{Dt}}\widehat {\dot J} + \frac{{D{V_z}}}{{Dt}}\hat k\)

\(\vec a = {a_x}\widehat {\dot i} + {a_y}\widehat {\dot J} + {a_z}J\hat k\)  –  total acceleration

\({a_x} = \frac{{D{V_x}}}{{Dt}} = \frac{{\partial u}}{{\partial t}} + \frac{{\partial u}}{{\partial x}}.\frac{{\partial x}}{{\partial t}} + \frac{{\partial u}}{{\partial y}};\frac{{\partial y}}{{\partial t}} + \frac{{\partial u}}{{\partial z}}\frac{{\partial z}}{{\partial t}};;\)

\( = \underbrace {\frac{{\partial u}}{{\partial t}}}_{Local,,acceleration} + ,,\underbrace {u\frac{{\partial u}}{{\partial x}} + v\frac{{\partial u}}{{\partial y}} + w\frac{{\partial u}}{{\partial z}}}_{connective,,acceleration}\)

\(\frac{{D\vec V}}{{Dt}}\) – Rate of change of velocity w.r.t. time is acceleration.

As per the given conditions in the question

\(\frac{{D\vec V}}{{Dt}} = 0,\)  So total acceleration i.e. local acceleration + Connective acceleration of fluid-particle is zero.

 

\(\frac{{\partial \vec V}}{{\partial t}} = 0\) – Steady flow.

Steady flow is defined as that type of flow in which the fluid characteristics like velocity, pressure, density, etc. at a point do not change w.r.t. time.

Here velocity is not varying with time i.e. \(\frac{{\partial \vec V}}{{\partial t}} = 0\)

So, the flow is steady flow.

Concept:

m = m₁ + m₂

W_T = W_net = W₁ + W₂

= W_1-2 + W_2-3

Process heating is same as feed heating

Heat liberated in process heating, Q_H = m₁(h₂ – h₆)

W_1-2 = m (h₁ – h₂)

W_2-3 = m₂ (h₂ – h₃)

Calculation:

Given, h₁ = 3486 kJ/kg, S₁ = 6.6 kJ /(kg.K)

h₂ = 2776 kJ/kg, m = 100 kg/s

Since, S₁ = S₃ = 6.6 kJ/(kg.K)

S₃ = S_f + x (S_g – S_f)

6.6 = 0.6 + x (8.1 - 0.6)

6 = x × 7.5 

x = 0.8

h₃ = h_f + x (h_g - h_f)

= 174 + 0.8 (2574 - 174) = 2094 kJ/kg

Now,

W_net = W_T = W₁ + W₂

= m(h₁ – h₂) + m₂(h₂ – h₃)

= 100 (3486 - 2776) + 0.8 × 100 × (2776 - 2094)

=125560 kW

= 125.56 MW

Explanation:

Using Bernoulli’s equation: 

  \(\frac{{{P_1}}}{{{ρ _g}}} + \frac{{v_1^2}}{{2g}} + {z_1} = \frac{{{p_2}}}{{ρ g}} + \frac{{v_2^2}}{{2g}} + {z_2}\)

Given that

→ P₁ = 400000 Pa, P₂ = 130000 Pa,   Z₁ = Z₂ and ρ = 1000

_{Apply Continuity Equation:}

 A1V1 = A₂V₂

\(\frac{\pi }{4}{(80)^2}{V_1} = \frac{\pi }{4}{(40)^2}{V_2}\)

_V2 = 4V₁

\({\frac{{{400000} }}{{ρ g}}} + \frac{{V_1^2}}{{2g}} = { {\frac{{130}000}{{ρ g}}}} + {{\frac{{(4{V_1})^2}}{{2g}}}} \)

\(400 + \frac{{V_1^2}}{2} = 130 + \frac{{16V_1^2}}{2}\)

\(\frac{{15V_1^2}}{2} = 270;\)

\(V_1^2 = 36\)

V₁ = 6 m/s

Concept:

𝐹𝑡+1 = 𝛼𝐷𝑡 + (1 − 𝛼) 𝐹𝑡 = 𝐹𝑡 + α (Dt - Ft)

This equation can be used when simple exponential smoothing is used as a forecasting model.

Here Ft represents the forecast for period t, Dt is the known demand for period t and α is the smoothing constant.

Calculation:

Forecaster for February:

F₂ = 400 + 0.75 (500 – 400) = 475

Forecaster for March:

F₃ = 475 + 0.75 (600 – 475) = 568.75

Concept:

The length of the curve in three-dimensional space given in the parametric form is

\(L = \mathop \smallint \limits_a^b \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2} + ;{{\left( {\frac{{dz}}{{dt}}} \right)}^2}} ;.dt,;a \le t \le b\)

And the length of the curve in two- dimensional space given in parametric form is

\(L = \mathop \smallint \limits_a^b \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2};} ;.dt,;a \le t \le b\)

Calculation:

Given, \(x\left( t \right) = cost,;y\left( t \right) = ;sint,;z\left( t \right) = \frac{2}{\pi }t,;0 \le t \le \frac{\pi }{2};;\) 

\(L = \mathop \smallint \limits_0^{\pi /2} \sqrt {{{\left| {\frac{{d\left( {cost} \right)}}{{dt}}} \right|}^2} + {{\left| {\frac{{d\left( {sint} \right)}}{{dt}}} \right|}^2} + {{\left| {\frac{2}{\pi }\frac{{d\left( t \right)}}{{dt}}} \right|}^2};} .dt;\)

\( = \mathop \smallint \limits_0^{\pi /2} \sqrt {{{\sin }^2}t + co{s^2}t + \frac{4}{{{\pi ^2}}}} ;.dt\)

\( = \mathop \smallint \limits_0^{\pi /2} \sqrt {1 + \frac{4}{{{\pi ^2}}}} .dt\)

\( = 1.185\mathop \smallint \limits_0^{\pi /2} dt = 1.185 \times \frac{\pi }{2}\)

= 1.862

≈ 1.86

Concept:

After moving the massless rigid rod to a small angle 'θ', D'Alembert's principle is used to calculate the natural frequency:

\(\Delta T=-I\alpha\)

Calculation:

Given:

k = 400 N/m, m = 1 kg

\(\Delta T=-I\alpha\)

\(m(2r)^2\ddot{θ} \times 2r + k (r)θ (r) = 0\)

\(\therefore \ m (4r^2) \ddot{θ} + k(r^2)\theta=0\)

\(\therefore \ \ddot{θ} + \frac{k(r^2)}{m(4r^2)} θ = 0\)

Comparing with the standard equation:

\(\ddot{θ} + \omega_n^2 θ = 0\)

\(\therefore \omega _n^2 = \frac{k}{{4m}}\)

\(\therefore \omega _n^2 = \frac{{400}}{{4 \times 1}}\)

\(\therefore {\omega _n} = \sqrt {\frac{{400}}{4}} ;rad/s\)

Concept:

Given:

a = 1

 

\(\begin{array}{l} Area = \frac{1}{2} \times base \times height\ = \frac{1}{2} \times a\sqrt 2 \times a\ = \frac{{{a^2}}}{{\surd 2}}\ = \frac{1}{{\sqrt 2 }} = 0.707 \end{array}\)

Concept:

In case of rolling,

Reduction in thickness Δh = 2R (1 - cos α)

where R is the radius of the roller, α = Angle of bite

For maximum reduction

(Δh)max = μ2 R

Calculation:

\(\Delta H=D(1-\cos\theta)\)

\(\cos\theta=1-\frac{\Delta H}{D}=1-\frac{\Delta H}{2R}=1-\frac{\mu^2}{2}=1-\frac{0.1^2}{2}=0.995\)

\(\theta = {\cos ^{ - 1}}\left( {0.995} \right) = 5.73^\circ\)

\(\frac{{{d^2}y}}{{d{x^2}}} = y\left( {{D^2} - 1} \right)y = 0\)

Auxiliary equation is given by

\({m^2} - 1 = 0 \Rightarrow m = \pm 1\)

Then \(y = {c_1}{e^x} + {c_2}{e^{ - x}}\) …………………(i)

Which passes through origine (0,0) and (ln 2, ¾ )

So, \(0 = {c_1}{e^0} + {c_2}{e^{ - 0}} \Rightarrow {c_1} + {c_2} = 0\)………………..(ii)

And \(\frac{3}{4} = {c_1}{e^{ln2}} + {c_2}{e^{ - ln2}}\)

\(\Rightarrow \frac{3}{4} = 2{c_1} + \frac{{{c_2}}}{2}\) …………………….(iii)

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

\({c_1} = \frac{1}{2}\) , \({c_2} = \frac{1}{2}\)

Then \(y = \frac{1}{2}{e^x} - \frac{1}{2}{e^{ - x}} \Rightarrow y = \frac{1}{2}\left( {{e^x} - {e^{ - x}}} \right)\)

Concept:

Since  θ = 45° , so

PQ = QR = 4 m

And \(\mathop \sum \nolimits {M_Q} = 0\)

100 × Cos 60° × 4 = RR × 4

R_R = 50 kN

Now from equilibrium, F_PR cos 45° = 100 cos 60°

\(\Rightarrow {F_{PR}} = 70.71~ kN\)

Concept:

In case of free fall of an object, the sum of kinetic energy + potential energy ie. Total energy at every point of the journey is constant.

\(i.e.;{\left( {\frac{1}{2}m{V^2} + mgH} \right)_A} = {\left( {\frac{1}{2}m{V^2} + mgH} \right)_B}\)

Calculation:

Given mass m = 0.1 kg, height = 1 m

From,

\(\frac{1}{2}mV_A^2 + mg{H_A} = \frac{1}{2}mV_B^2 + mg{H_B}\)  (Between A & B)

\( \Rightarrow 0 + 0.1 \times 9.81 \times 1.0 = \frac{1}{2} \times 0.1 \times {V^2} + 0\)

\( \Rightarrow V = \sqrt {2 \times 9.81} ;\)

= 4.429 m/s

After impact, the velocity reduced by 20%

i.e. V₂ = 0.8 × V

= 0.8 × 4.429

= 3.543 m/s

Again from,

\(\frac{1}{2}mV_B^2 + mg{H_B} = \frac{1}{2}mV_c^2 + mg{H_c}\left( {between;B& C} \right)\)

\(\frac{1}{2} \times 0.1 \times {3.543^2} + 0 = 0 + 0.1 \times 9.81 \times h\)

\( \Rightarrow h = ;\frac{{{{3.543}^2}}}{{2 \times 9.81}}\)

= 0.639 m

≈ 0.64m

Concept:

The relation among, open-circuit voltage, short circuit current, arc voltage and arc current is given by,

\(V = {V_0} - \frac{{{V_0}}}{{{I_S}}};I\)

where, V0 = open-circuit voltage, Is = short circuit current

Power = V × I

\(P = \left( {{V_0} - \frac{{{V_0}}}{{{I_S}}};I} \right)I\)

\(P = {V_0}I - \frac{{{V_0}}}{{{I_S}}};{I^2}\)

Calculation:

Given:

Maximum power is obtained at, V_o = 80 V , I_s = 300 amp

\(P = {V_0}I - \frac{{{V_0}}}{{{I_S}}};{I^2}\)

For P = P_max , \(\frac{{dP}}{{dI}} = 0\)

\(\frac{{dP}}{{dI}} = {V_0} - 2\frac{{{V_0}}}{{{I_S}}};I = 0\)

∴ \(I = \frac{{{I_S}}}{2}\)

\(I = \frac{{{300}}}{2}=150~A\)

Work done = T.ω 

= 100 × 2π × 1000 

\(\Rightarrow {\rm}W.D. = \frac{{100 \times 2\pi \times 1000}}{{1000}} = 628.3185\ kJ\)

And P₁V₁ = mRT₁

\(\Rightarrow {10^5} \times 1 = 1 \times \left( {1000 - 800} \right){T_1} \Rightarrow {T_1} = 500K\)

Now from the first law, dQ = dU + dW

since, dQ = 0 and dU = C_vdT and work done is on the system i.e. dW  = - Ve

So, W.D. = C_vdT ⇒ \(dT = \frac{628.3185}{0.8}= 785 \ K\)

Then \({T_2} = 500 + 785 = 1285\ K\)

Hence, When the pinion and gear will meet. Let’s assume that they meet at point P. hence at point P according to Newton’s 3^rd law

⋆ F₁ = F₂  ---(1)  (in magnitude)

2. At a point when gear and pinion meet the velocity at the particular point (p) for gear and pinion will remain same. Hence at point (p)

V_pinion = V_gear

∴ r₁θ₁ = r₂θ₂          ---(2)     

Applying newton’s Eqn in pinion

\(I_1 \ddot{\theta}_1 = \tau_1 - F_2r_1 \ \ \ \ \ \ (\because \ F_1 = F_2)\)

Applying Newton’s law in gear

I₂ θ₂ = F₂ r₂

Substituting equation (2) in the above equation.

\(\frac{{{I_2};{r _1};{\theta _1}}}{{\left( {{r _2}} \right)}} = {F_2};{r _2}\)

\(\therefore {F_2} = \frac{{{I_2};{r _1}{\theta _1}}}{{r_2^2}}\)        ---(4)

Now from eq. (3)

\(I_1 \ddot{\theta}_1 = \tau_1 - F_2r_1 \ \ \ \ \ \ (\because \ F_1 = F_2)\) 

\(\therefore \ \tau_1 = I_1 \ddot{\theta}_1 + \left(\frac{I_2r_1 \ddot{\theta}_1}{r_2^2}\right)\cdot r_1\)

\(\therefore \ \tau_1 = I_1 \ddot{\theta}_1 + I_2 \left(\frac{r_1^2}{r_2^2}\right)\ddot{\theta}_1\)   ---(5)

From 1, 4, 5 the correct option is (2)

Concept:

\({δ _A} = \frac{{p{l^3}}}{{3EI}},{θ _A} = \frac{{P{L^2}}}{{2EI}}\)

\(slope;{θ _c} = \frac{{P{l^2}}}{{2EI}},{δ _C} = \frac{{P{l^3}}}{{3EI}}\)

Deflection at A = δ_c + θ_c × (AB - BC)

= δ_c + θ_c × l

\( = \frac{{P{l^3}}}{{3EI}} + \frac{{P{l^2}}}{{2EI}} \times l\)

Calculation:

Given P = 500N; EI = 200 Nm²

l = 50 mm = 0.05 m

2l = 100 mm = 0.10 m

\({\delta _A}\left( {at;tip} \right) = \frac{{500 \times {{0.05}^3}}}{{3 \times 200}} + \frac{{500 \times {{0.05}^2}}}{{2 \times 200}} \times 0.05\)

= 2.604× 10^-4 m

= 0.26 mm

\(\begin{array}{l} \Rightarrow {L_{4max}} = {L_{1max}} - \left( {{L_{2min}} + {L_{3min}}} \right) = 22.01 - \left( {9.995 + 9.995} \right)\ {L_{4max}} = 2.02\ mm \end{array}\)

And \({L_{4min}} = {L_{1min}} - \left( {{L_{2max}} + {L_{3max}}} \right)\)

= 21.99 – (10.005 + 10.005) = 1.98 mm

Hence, \(L_4=2^{±0.02}\ mm\)

Concept:

1. When a rotating body is involved then there comes a force which is known as Centrifugal force (F_c)

It is calculated by using formula

F_c = mr(ω)²

2. Now in the Question, it is mentioned that end Q is just lifted off the ground. So as it is just lifted off the ground there will be no reaction force from that point.

∴ FBD of mobile

M = mass of mobile

m = eccentric mass

r = eccentricity

Hence, Taking a moment about point P = O

∴ Mg × 0.06 - F_c × (0.09) = 0

∴ 90 × 10^-3 × 9.81 × 0.06 = mr(ω)² × (0.09)

90 × 10^-3 × 9.81 × 0.06 = 2 × 10^-3 × 2.19 × 10^-3 (ω)² × 0.09

∴ ω² = 134380.8964

∴ ω = 366.58

\(\therefore \frac{{2\pi N}}{{60}} = 366.58\)

∴ N = 3500 rpm

Transmissibility – It is defined as the ratio of force transmitted (to the foundation) to the force applied.

→ It is the measure of effectiveness of vibration isolating material.

Transmissibility \(\left( \epsilon \right) = \frac{{{F_T}}}{{{F_0}}} = \frac{{\sqrt {1 + {{\left( {\frac{{2\zeta \omega }}{{{\omega _n}}}} \right)}^2}} }}{{\sqrt{{{{\left[ {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]}^2} + {{\left( {\frac{{2\zeta \omega }}{{{\omega _n}}}} \right)}^2}}}}}\)

When no damper is used

Then, ζ = 0

\(\therefore \epsilon = \frac{1}{{ \pm \left[ {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]}}\)

Given:

f = 60 Hz, ϵ = 5% = 0.05, m = 1 kg, k = ?

So, ω = 2πf

∴ ω =2π (60)

∴ = 376.99 rad/s.

Now, substitute in the formula given in concept

\(\therefore ;.05 = \frac{1}{{ \pm \left[ {1 - {{\left( {\frac{\omega }{{\omega n}}} \right)}^2}} \right]}}\)

\( - 20 = 1 - {\left( {\frac{\omega }{{\omega n}}} \right)^2}\)

\(\therefore {\left( {\frac{\omega }{{{\omega _n}}}} \right)^2} = 21\)

\(\omega _n^2 = \frac{{{\omega ^2}}}{{21}}\)

\(\therefore {\omega _n} = \sqrt {\frac{{{{\left( {376.99} \right)}^2}}}{{21}}} \)

\(\therefore {\omega _n} = 82.66;rad/s\)

We, know,  \({\omega _n} = \sqrt {\frac{k}{m}} ;;;;;;; \because \ m = 1;kg\)

∴ k = (82.66)² × 1

∴ k = 6767.02 N/m

Concept:

Trick: Whenever activity is mentioned take points on node itself to avoid the confusion.

And take the duration on path.

→ Note: Find the durations of path given in option and then take the path with max duration

Because, Critical path is the minimum time required to completed the task or path with maximum duration

1. 1 – 2 – 3 – 5 = 5 days

2. 1 – 2 – 3 – 4 – 5 (No path is formed) ∵ 3 – 4 path is not a path

3. 1 – 4 – 3 – 5 = 8 days

4. 1 – 4 – 5 → 7 days

Hence from the above calculation we can see that path in the option(c) is the critical path.

Concept:

The method to calculate the mean value of any dependent variable is,

\({U_m} = \frac{1}{R}\mathop \smallint \nolimits_0^R u\left( {r,n} \right)dr\)

\(also,;\bar h = \frac{1}{L}\mathop \smallint \nolimits_0^L {h_x}dx\)

Calculation:

Similarly,

T is dependent on u(r,x) as per the condition given in the question

\(So,;{T_m} = \frac{2}{{{u_m}{R^2}}}\mathop \smallint \nolimits_0^R u\left( {r,x} \right)T\left( {r,x} \right)r;dr\)

\(given;u\left( {r,x} \right) = {c_1},;;T\left( {r,x} \right) = {C_2}\left[ {1 - {{\left( {\frac{r}{R}} \right)}^3}} \right]\)

\({U_m} = \frac{1}{R}\mathop \smallint \nolimits_0^R u\left( {r,x} \right)dr;\)

\(\frac{1}{R}\mathop \smallint \nolimits_0^R {c_1}.dr = \frac{1}{R}.{C_1}.R = {C_1}\)

Now,

\({T_m} = \frac{2}{{{U_m}{R^2}}}\mathop \smallint \nolimits_0^R {C_1}.{C_2}\left[ {1 - {{\left( {\frac{r}{R}} \right)}^3}} \right]r.dr;\)

\( = \frac{{2{C_1}{C_2}}}{{{U_m}{R^2}}}\mathop \smallint \nolimits_0^R \left[ {r - \frac{{{r^4}}}{{{R^3}}}} \right]dr\)

\( = \frac{{2{C_1}{C_2}}}{{{U_m}{R^2}}}\left[ {\frac{{{R^2}}}{2} - \frac{{{R^5}}}{{5{R^3}}}} \right]\)

\( = \frac{{2{C_1}{C_2}}}{{{U_m}{R^2}}} \times \frac{{3{R^2}}}{{10}}\)

\( = \frac{{2{C_1}{C_2}}}{{{C_1}}} \times \frac{3}{{10}} = 0.6{C_2}\)

Distance \(= \frac{1}{4} \times circumference\) 

\(\begin{array}{l} = \frac{1}{4} \times2 \pi(2)= \pi \ \Rightarrow time\ taken = \frac{{distance}}{{speed}} = \frac{\pi }{{1.57}} = 2sec \end{array}\)

\({C_d} = \frac{{1.328}}{{\surd R{e_x}}} = \frac{{1.328}}{{\sqrt {\frac{\rho }{\mu }} }} = \frac{{1.328}}{{\sqrt {\frac{{vd}}{\nu }} }} = \frac{{1.328}}{{\sqrt {\frac{{2 \times 2.5}}{{2 \times {{10}^{ - 5}}}}} }} = 2.656 \times {10^{ - 3}}\)

And A = 2.5 × 1 = 2.5 m²

Then \(F = \frac{1}{2}{C_d}\rho A{V^2} = \frac{1}{2} \times 2.656 \times {10^{ - 3}} \times 1.2\times 2.5 \times {2^2} = 0.015936\)

Concept:

The general form of the continuity equation in cartesian coordinates:

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

For steady flow:

\(\frac{{\partial \rho }}{{\partial t}} = 0\)

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

If the fluid is incompressible, then ρ is constant

\(\frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} + \frac{{\partial w}}{{\partial z}} = 0\)

\(\nabla .\vec V = 0\)

Calculation:

\(\frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} + \frac{{\partial w}}{{\partial z}} = 0\)

\(\frac{\partial }{{\partial x}}({a_1}x + {a_2}y + {a_3}z) + \frac{\partial }{{\partial y}}({b_1}x + {b_2}y + {b_3}z) + \frac{\partial }{{\partial z}}({c_1}x + {c_2}y + {c_3}z) = 0\)

\(\begin{array}{l} \Rightarrow {a_1} + {b_2} + {c_3} = 0 \Rightarrow 2 - 4 + {b_2} = 0 \Rightarrow {b_2} = 2 \end{array}\)

Given that axial feed, fN = 0.4 m/min, N = 400 rpm, λ = 90, t_c = 3 mm

Therefore, feed, \(f=\frac{0.4 × 1000}{400} = 1~mm/rev\)

then uncut chip thickness, t = f sin λ = 1 × sin (90) = 1 mm

Chip thickness ratio  \(r = \frac{t}{{{t_c}}} = \frac{1}{3} = 0.333\)

And \(\tan \phi = \frac{{r\cos \alpha }}{{1 - r\sin\alpha }} = \frac{{0.333\cos5^\circ }}{{1 - 0.333\sin5^\circ }} = 0.342\)

\(\Rightarrow \phi = {\tan ^{ - 1}}\left( {0.342} \right) = 18.86^\circ\)