GATE ME 2019 Official Paper: Shift 2 (Previous Year Paper)

Once the team of analysts identifies the problem, we _________ in a better position to comment on the issue. 

The 1^st clause is in the present tense, so 'will be', 'are going to be' or 'might be' are appropriate because these are the verbs representing present or future.

'were to be' can't be used because this is the verb of past tense.

• Culmination (चरम बिंदु): It is the point at which an event or series of events ends, having developed until it reaches this point.
• Example: Winning first prize was the culmination of years of practice and hard work.
• Consultation (परामर्श): A meeting to discuss something or to get advice.
• Example: After consultations with our accountants, we've decided on how to cut costs within the company.
• Desperation (निराशा): The feeling of needing or wanting something very much or the feeling of being in such a bad situation that you will take any risk to change it. 
• Example: In desperation, they jumped out of the window to escape the fire.
• Insinuation (कटाक्ष): the action of suggesting, without being direct, that something unpleasant is true.
• Example: He looked insulted by her insinuation.

A final examination is the culmination of a series of evaluations that a student has to go through.

MHO = JNIP; IDK = JEL; SO = TP

In this given code, every letter is being replaced by the letter that is immediately next to it.

So I → J, D → E, C → D

∴ IDC = JED

The product of three integers X, Y and Z is 192.

X.Y.Z = 192

Z is equal to 4.

⇒ X.Y.(4) = 192 ⇒ X.Y = 48

Possibilities are: (1,48), (2,24), (3,16), (4,12), (6,8)

P is equal to the average of X and Y.

P = (X + Y)/2

P_min = (X + Y)_min/2 = (6 + 8)/2 = 7

Here "than" is used which represents the comparison.

Only 'more' is a comparative degree adjective. So 'more' is correct answer.

GDP increased by 10% from 2015 to 2016 ⇒ 

Let GDP in 2015 = X

GDP in 2016 = 1.1 X

Fiscal deficit was 4% of the GDP in 2015 and that increased to 5% in 2016 ⇒

FD in 2015 = 0.04 X

FD in 2016 = 0.05 (1.1 X) = 0.055 X

Percentage increase in the actual fiscal deficit:

\(\frac{{F{D_{2016}} - F{D_{2015}}}}{{F{D_{2015}}}} \times 100 = \frac{{0.055 - 0.04}}{{0.04}} \times 100 = 37.5% \)

Concept:

Basic Concepts and Pipes & Cisterns formula:

• If a pipe can fill a tank in a hrs, then the part filled in 1 hr =1/a.
• If a pipe can empty a tank in b hrs, then the part of the full tank emptied in 1 hr = 1/b.
• If a pipe can fill a tank in a hrs and the another pipe can empty the full tank in b hrs, then the net part filled in 1 hr, when both the pipes are opened =[1/a - 1/b] ∴ Time taken to fill the tank, when both the pipes are opened = ab/(b - a)
• If a pipe can fill a tank in a hrs and another can fill the same tank in b hrs, then the net part filled in 1 hr, when both pipes are opened = [1/a + 1/b] ∴ Time taken to fill the tank = ab/(a + b)
• If a pipe fills a tank in a hrs and another fills the same tank in b hrs, but a third one empties the full tank in c hrs, and all of them are opened together, the net part filled in 1 hr = [1/a+ 1/b-1/c] ∴ Time taken to fill the tank =abc/(bc + ac – ab) hrs.
• A pipe can fill a tank in a hrs. Due to a leak in the bottom it is filled in b hrs. If the tank is full, the time taken by the leak to empty the tank = ab/(b - a) hrs.

Calculation:

P → 6 hrs (Fill), Q → 9 hrs (Fill), R → 12 hrs (Empty)

In 1 hr, (P + R) can fill: [1/6 - 1/12]

In 4 hrs, (P + R) can fill: \(4\left( {\frac{1}{6} - \frac{1}{{12}}} \right) = \frac{1}{3}\)

After 4 hrs, P is closed and Q is opened for 6 hours:

In 6 hrs, (R + Q) can fill: \(6\left( {-\frac{1}{12} + \frac{1}{{9}}} \right) = \frac{1}{6}\)

Total tank filled in 10 hours (4 + 6 hrs): \(\left( {\frac{1}{3} + \frac{1}{6}} \right) = \frac{1}{2}\)

After 6 more hours, R is closed. So the remaining 1/2 part of the tank will be filled by Q only.

To fill 1/2 part, Q will take: 9/2 = 4.5 hrs

Total time taken to fill the tank = 10 + 4.5 = 14.5 hrs

"While teaching a creative writing class in India, I was surprised at receiving stories from the students that were all set in distant places"

So the writer of this passage is surprised by the creative writing assignments of his students because none of the students had written stories set in India.

 

The percentage share of revenue contributed by Prime to the total revenues of Mola:

\(\frac{{{\rm{Prime}}}}{{{\rm{Pool + Mini + Prime}}}} = \frac{{465}}{{895 + 580 + 465}} = 0.2397 = 23.97% \)

X is an online media provider. By offering unlimited and exclusive online content at attractive prices for a loyalty membership, X is almost forcing its customers towards its loyalty membership. If its loyalty membership continues to grow at its current rate, within the next eight years more households will be watching X than cable television. 

That means X is offering good deals to its customers and its loyalty membership is growing. Most of the cable television operator is opting to subscribe to X’s loyalty membership.

Most households that subscribe to X’s loyalty membership discontinue watching cable television.

Concept:

• The roots of characteristic equation |A - λI| = 0 are known as Eigen values of matrix A.
• To each Eigen value of λ if there exists a non-zero vector X such that AX = λX then X is called Eigen vector of matrix A corresponding to the Eigen value λ.

Calculation:

Given:

 \(\left[ A \right] = \left[ {\begin{array}{*{20}{c}} 4&8&4\ 8&{16}&{ - 4}\ 4&{ - 4}&{15} \end{array}} \right],;\left{ x \right} = \left{ {\begin{array}{*{20}{c}} 2\ 1\ 4 \end{array}} \right};and;\left{ R \right} = \left{ {\begin{array}{*{20}{c}} {32}\ {16}\ {64} \end{array}} \right}\)

\(\left[ A \right]\left{ x \right} = \left{ R \right} = \left[ {\begin{array}{*{20}{c}} 4&8&4\ 8&{16}&{ - 4}\ 4&{ - 4}&{15} \end{array}} \right]\left{ {\begin{array}{*{20}{c}} 2\ 1\ 4 \end{array}} \right} = \left{ {\begin{array}{*{20}{c}} {32}\ {16}\ {64} \end{array}} \right} = 16\left{ {\begin{array}{*{20}{c}} 2\ 1\ 4 \end{array}} \right}\)

This is in the form of AX = λX, where λ is Eigenvalue.

∴ One of the Eigenvalue of matrix [A] is 16

Concept:

Directional derivative of a function f along the vector \(\hat u \) is given by

\(DD = \nabla f.\frac{{\vec u}}{{\left| u \right|}}\)

grad f or ∇ f is defined by the equation,

\(grad;f = \nabla f = i\frac{{\partial f}}{{\partial x}} + j\frac{{\partial f}}{{\partial y}} + k\frac{{\partial f}}{{\partial z}}\)

Calculation:

f (x, y) = x² + y²

Line vector from (0,0) to (1,1) is

(1 – 0) î + (1 – 0) ĵ = î + ĵ

\(\nabla f = i\frac{\partial }{{\partial x}}\left( {{x^2} + {y^2}} \right) + \hat j\frac{\partial }{{\partial y}}\left( {{x^2} + {y^2}} \right)\)

\(\nabla f = \hat i\left( {2x} \right) + \hat j\left( {2y} \right) \Rightarrow {\left( {\nabla f} \right)_{1,1}} = 2\hat i + 2\hat j\)

∴ Directional derivative of f (x, y) at (1, 1) in the direction of î + ĵ is

\({\left( {\nabla f} \right)_{1,1}}.\frac{{\left( {\hat i + \hat j} \right)}}{{\left| {\hat i + \hat j} \right|}}\)

\( \Rightarrow \left( {2\hat i + 2\hat j} \right).\frac{{\left( {i + \hat j} \right)}}{{\left| {\hat i + \hat j} \right|}} = \frac{{2 + 2}}{{\sqrt {{1^2} + {1^2}} ;}} = \frac{4}{{\sqrt 2 }} = 2\sqrt 2 \)

Concept:

If the differential equation is in the form of:

\(\frac{{dy}}{{dx}} + Py = Q\)

Integrating factor: 

\(IF = {e^{\smallint Pdx}}\)

Solution for equation:

\({\rm{y}}\left( {{\rm{IF}}} \right) = \smallint \left( {{\rm{IF}}} \right){{\rm{Q}}{\rm{}}}{\rm{dx}} + {\rm{C}}\)

Calculation:

\(\frac{{dy}}{{dx}} + 4y = 5\)

where 0 ≤ x ≤ 1 and y (0) = 2.25

This differential equation is in the linear form,

\(\frac{{dy}}{{dx}} + Py = Q\)

Where P = 4, Q = 5

\(\therefore IF = {e^{\smallint Pdx}} = {e^{\smallint 4dx}} = {e^{4x}}\)

Solution of differential equation

\(y\left( {{e^{4x}}} \right) = \smallint 5.{e^{4x}}.dx + c\)

\(y{e^{4x}} = 5\frac{{{e^{4x}}}}{4} + c\)

Using y (0) = 2.25

\(2.25 \times {e^0} = 5\frac{{{e^0}}}{4} + c = 1.25 + c\)

c = 1

\(\therefore y{e^{4x}} = \frac{5}{4}{e^{4x}} + 1 \Rightarrow y = \frac{5}{4} + {e^{ - 4x}} \Rightarrow y = {e^{ - 4x}} + 1.25\)

Concept:

Analytic function f(z) = u (x, y) + iv (x, y)

If f(z) = u + iv is analytic function, then u and v must satisfy

Cauchy-Riemann (C-R) equation

\(\frac{{\partial u}}{{dx}} = \frac{{dv}}{{dy}};and~\frac{{du}}{{dy}} = - \frac{{\partial v}}{{dx}}\)

Concept:

If V_A and V_B are velocity at point A and B respectively then the I-centre of this system will be finding by the meeting point of perpendicular drawn on the both velocity vector.

If w is the angular velocity of system, then 

\(\frac{{{{\vec V}_A}}}{{A{I_C}}} = \frac{{{{\vec V}_B}}}{{B{I_C}}} = \omega \)

Calculation:

∵ AI = AC sin 60° = AC × √3/2 = √3/2

CI = AC cos 60° = AC × 1/2 = 1/2 

\(\therefore {V_p} = \left( {AI} \right)w \Rightarrow w = \frac{{{V_P}}}{{AI}} = \frac{2}{{\sqrt 3 }}{V_p}\)

V_R = (CI)w ⇒ w = V_R/CI = 2V_R

Concept:

\({\epsilon_x} = \frac{{{\sigma _x}}}{E} - \mu \left( {\frac{{{\sigma _y}}}{E} + \frac{{{\sigma _z}}}{E}} \right)\)

Calculation:

σ₁ = 100 MPa, μ = 0.3 and ϵ = 200 MPa

∵ It is uniaxial tensile stress so, σ₂ = σ3 = 0

\({\epsilon_1} = \frac{{{\sigma _1}}}{E} - \mu \left( {\frac{{{\sigma _2}}}{E} + \frac{{{\sigma _3}}}{E}} \right)\)

\( \Rightarrow {\epsilon_1} = \frac{{{\sigma _1}}}{E} = \frac{{100}}{{200}} = 0.5\)

\( \Rightarrow {\epsilon_2} = \frac{{{\sigma _2}}}{E} - \mu \left( {\frac{{{\sigma _1}}}{E} + \frac{{{\sigma _3}}}{E}} \right) = - \mu \frac{{{\sigma _1}}}{E}\)

\( \Rightarrow {\epsilon_2} = - 0.3\left( {\frac{{100}}{{200}}} \right) = - 0.15\therefore \left( {{\epsilon_1},\epsilon{_2}} \right) = \left( {0.5, - 0.15} \right)\)

Explanation:

Circular pitch: It is the distance measured on the circumference of the pitch circle from a point of one tooth to the corresponding point on the next tooth. It is usually denoted by pc. Mathematically,

If the circular diameter is D, and No. of teeth is T

Circular pitch, \(p_c = \pi m=\frac{{\pi D}}{T}\)

where m is the module of gear. (m = D/T)

Explanation:

Corrected endurance strength is defined as

σ_e = K_a K_b K_c K_d.σ^’_e

where, σ^’_e = Endurance strength and K_a = size factor, K_b = surface factor, K_c = load factor, K_d = Temperature factor

{K_a, K_b, K_c, K_d} < 1

So, with the increases in surface roughness and size of the beam, endurance strength will decrease.

Concept:

Euler Equation of Motion:

\({\vec F_p} + {\vec F_y} = {\vec F_i}\)

\({\vec F_p} + {\vec F_y} = m\vec a\)

\(\frac{{\partial u}}{{\partial t}} + u\frac{{\partial u}}{{\partial x}} + v\frac{{\partial u}}{{\partial y}} + w\frac{{\partial u}}{{\partial z}} = X - \frac{1}{\rho }\frac{{\partial p}}{{\partial x}}\)

\(\frac{{\partial v}}{{\partial t}} + u\frac{{\partial v}}{{\partial x}} + v\frac{{\partial v}}{{\partial y}} + w\frac{{\partial v}}{{\partial z}} = Y - \frac{1}{\rho }\frac{{\partial p}}{{\partial y}}\)

\(\frac{{\partial w}}{{\partial t}} + u\frac{{\partial w}}{{\partial x}} + v\frac{{\partial w}}{{\partial y}} + w\frac{{\partial w}}{{\partial z}} = Z - \frac{1}{\rho }\frac{{\partial p}}{{\partial z}}\)

Where X, Y, and Z are body forces.

\( - ∇ P + \rho ∇ ϕ = \rho \left[ {\frac{{D\vec u}}{{Dt}}} \right]\)

\(\frac{D}{{Dt}} = \frac{\partial }{{\partial t}} + u\frac{\partial }{{\partial x}} + v\frac{\partial }{{\partial y}} + w\frac{\partial }{{\partial z}}\)

Assume: flow is incompressible and friction less

Euler’s equation for vector form:

\( - ∇ P + \rho ∇ ϕ = \rho \left[ {\frac{{D\vec u}}{{Dt}}} \right]\)

\( - ∇ P + \rho ∇ ϕ = \rho \left[ {u\frac{{d\vec u}}{{dx}}\hat i + v\frac{{d\vec u}}{{dy}}\hat j + w\frac{{d\vec u}}{{dz}} + \frac{{d\vec u}}{{dt}}} \right]\)

\(\vec u = x\hat i - y\hat j\)

-∇P + ρ∇ϕ = ρ [xî + (-y) (-1)ĵ + 0 + 0]

Zero body force in x and y direction.

-∇P + ρ [0î + 0ĵ] = ρ [xî + yî] ⇒ ∇P = -ρ [xî + yĵ]

Concept:

View Factor F _1-2  means the fraction of radiation leaving the surface 1 and striking the surface 2.

Following points are important about View Factor

• Summation rule if a body is exchanging radiation by n surfaces by

          F11 +F12 + ------ F1n= 1

• Reciprocity theorem A1 F12 = A2 F21​

​Calculation:

By summation rule

F_1-1 + F_1-2 = 1

∵ F_1-1 = 0 because no radiation leaving the surface 1 will strike the surface 1.

∴ F_1-2 = 1

Explanation:

Cross-sectional area varies linearly in the direction of heat transfer:

A = cx + B

1 – D steady state with no heat generation

\(\frac{d}{{dt}}\left( { - kA\frac{{dT}}{{dx}}} \right) = 0\)

\( \Rightarrow - kA\frac{{dT}}{{dx}} = {c_1} \Rightarrow \smallint dT = - \smallint \frac{{{c_1}}}{{kA}}dx = - \smallint \frac{{{c_1}}}{{K\left( {cx + B} \right)}}dx = - \frac{{{c_1}}}{{cK}}\ln \left( {cx + B} \right)\left. \right|_{{x_1}}^{{x_2}}\)

\( \Rightarrow {T_2} - {T_1} = - \frac{C}{K}\ln \left( {\frac{{c{x_2} + B}}{{c{x_1} + B}}} \right);\)

i.e. temperature variation is logarithmic.

Concept:

For a pure substance undergoing an infinitesimal reversible process.

1. dU = Tds – pdV
2. dH = dU + pdV + Vdp = Tds + Vdp
3. df = dU – Tds – SdT = –pdv – SdT
4. dG = dH – Tds – SdT = Vdb – Sdt

for dz = Mdx + Ndy

\({\left( {\frac{{\partial M}}{{\partial y}}} \right)_x} = {\left( {\frac{{\partial N}}{{\partial x}}} \right)_y}\)

Applying this to the four equations:

\(\begin{array}{l} dU = Tds - pdv \Rightarrow {\left( {\frac{{\partial T}}{{\partial V}}} \right)_s} = - {\left( {\frac{{\partial P}}{{\partial S}}} \right)_V}\ dH = Tds +vdP \Rightarrow {\left( {\frac{{\partial T}}{{\partial P}}} \right)_s} = {\left( {\frac{{\partial V}}{{\partial S}}} \right)_P} \end{array}\)

\(\begin{array}{l} df = - pdv - SdT \Rightarrow {\left( {\frac{{\partial P}}{{\partial T}}} \right)_V} = {\left( {\frac{{\partial S}}{{\partial V}}} \right)_T}\ dG = Vdp - SdT \Rightarrow {\left( {\frac{{\partial V}}{{\partial T}}} \right)_p} = - {\left( {\frac{{\partial S}}{{\partial P}}} \right)_T} \end{array}\)

Calculation:

According to Maxwell’s relation:

\({\left( {\frac{{\partial v}}{{\partial s}}} \right)_p} = {\left( {\frac{{dT}}{{dP}}} \right)_s}\)

Explanation:

As seen from the diagram when temperature increase from 1 – 1’ then quality of steam increases from 2 – 2’.

Efficiency and Rankine cycle increases with increasing effective mean temperature of heat addition.

Effective mean temperature of heat addition increases by

(a)  High dryness fraction

(b)  High temperature of heat addition

(c)  High pressure of heat addition

Note: This is the advantage of reheating cycle.

Explanation:

Hardness is defined as resistance to plastic deformation or penetration.

Hardenability is defined as the ease with which hardness may be attained by quenching.

It is also defined as the ability to develop maximum hardness by quenching.

It is the process to have a hardened layer of martensite after quenching and also to have a high hardness at the same given depth

Note: Hardness and hardenability is two different properly of material, not consider the same.

Explanation:

• Fluidity is defined as ability of a metal to flow and fill the mold.
• The higher the pouring temperature, higher the fluidity.
• Pouring temperature increases with increase in degree of superheat and hence fluidity increases.

Explanation:

Knurling: Knurling is a process of impressing a diamond shape pattern on the surface of a workpiece to provide a better gripping surface.

Knurling is the operation of producing straight-lined, diamond-shaped patterns or cross lined patterns on a cylindrical external surface by pressing a tool called knurling tool. Knurling is not a cutting operation but it is a forming operation.

 

Chamfering: Chamfering is a finishing process to make a bevel, groove or furrow in the machined part.

Roll forming: Roll forming is a continuous bending process in which opposite rolls are used to produce long sections of formed shapes from coil or strip stock.

Explanation:

Plug gauge is a cylindrical bar with highly finished ends of different diameters is used to check hole diameter.

Important points:

• Snap gauge – For gauging external dimensions
• Plug gauge – For gauging internal dimensions
• Taper plug gauge – For gauging taper holes
• Ring gauge – For gauging external dimensions
• Gap gauge – For gauging gaps and grooves
• Radius gauge – For gauging radii
• Thread pitch gauge – For gauging external dimensions

 

Calculation:

The transformation matrix for mirroring a point about the line Y = X in X – Y plane is \(\left[ {\begin{array}{*{20}{c}} 0&1\ 1&0 \end{array}} \right]\)

Concept:

• Mean is the average of the number.

Mean = Sum of all observations/Number of observations

• Mode of a set of observation is the observation that occurs the most often (most frequently).
• Median refers to the value which lies in the middle of the data (when arranged in an increasing or decreasing order).

Calculation:

Given data are 3, x, 2, and 4

\(\therefore Mean = \frac{{{\rm{\Sigma }}{x_i}}}{N} = \frac{{3 + x + 2 + 4}}{4}\) 

\( \Rightarrow x = \frac{{9 + x}}{4} \Rightarrow x = 3\) 

Now the data are: 3, 3, 2, 4

Mode = 3 (∵ 3 occurs most frequently)

Concept:

If in truss problem there are 3 members in which 2 are in the same line and there is no external force, then the force in the 3^rd member will be zero

Calculation:

Consider point B;

F_BC = 0

Consider point C;

∵ F_BC is already zero

F_AC and F_CE is in the same line so, F_DC = 0

Concept:

Calculation:

σ₁ = 300 MPa, σ₂ = 100 MPa

Radius of Mohr’s circle \( = \frac{{{\sigma _1} - {\sigma _2}}}{2} = \frac{{300 - 100}}{2} = \frac{{200}}{2} = 100\)

ON represent the normal stress = 260 MPa

MN is the shear stress at the same plane where the normal stress is 260 MPa

CM represents the maximum shear stress on the plane where the normal stress is 260 MPa

CM = 100 MPa

CN = ON - OC = 260 - 200 = 60 MPa

For ΔCMN,

(CM)² = (CN)² + (MN)²

(100)² = (60)² + MN²

MN = 80

∴ The magnitude of shear stress τs = 80 MPa

Concept:

\(\frac{M}{I} = \frac{\sigma }{y} = \frac{E}{R}\)

where, M= bending moment, I = Moment of inertia of the area of cross section, σ = Bending stress, y = distance of extreme fibre from the neutral axis, E = Young’s modulus, R = radius of curvature

Calculation:

Given: d = 1 mm, R = 1 m, E = 100 GPa

\({\left( {{\sigma _b}} \right)_{max}} = \frac{{E{Y_{max}}}}{R}\)

\({\left( {{\sigma _b}} \right)_{max}} = \frac{{E\left( {\frac{d}{2}} \right)}}{R} = \frac{{100 \times {{10}^9} \times \left( {0.5} \right) \times {{10}^{ - 3}}}}{1} = 50;MPa\)

Concept:

Reynolds number for pipe flow is given as

\({R_e} = \frac{{\rho VD}}{\mu }\)

Where V = mean velocity

Assume V₁ and V₂ is entrance and exit velocity of the fluid respectively.

By conservation of mass

ṁ_in = ṁ_out

Calculation:

ṁ_in = ṁ_out

ρA₁V₁ = ρA₂V₂ [water is incompressible fluid so ρ will be same]

A₁V₁ = A₂V₂

⇒ V₁ = V₂ [A₁ = A₂ (Given)]

∴ Reynold number is constant because ρ, V, D, μ remain same

Concept:

\({N_u} = \frac{{hD}}{K}\)

Where D is characteristic dimension of body.

For vertical plate case D = L

Rayleigh Number (R_a) = Gr.Pr

\( = \frac{{g\beta {\rm{\Delta }}T{L^3}}}{{{v^2}}} \times Pr\) 

Where Gr = Grashoff Number, Pr = Prandtl Number

Calculation:

Nu = K(Ra)^1/4

\(\frac{{hL}}{K} = K{\left[ {\frac{{g\beta {\rm{\Delta }}T{L^3}}}{{{v^2}}} \times Pr} \right]^{1/4}}\) 

\(so,;h \propto \frac{1}{{{L^{1/4}}}}\) 

\(\frac{{{h_1}}}{{{h_2}}} = {\left( {\frac{{{L_2}}}{{{L_1}}}} \right)^{1/4}}\) 

As L₂ = 16L₁

_{\(\left( {\frac{{{h_1}}}{{{h_2}}}} \right) = {\left( {16} \right)^{1/4}} = 2\)}

Concept:

Energy released per spark across the IEG is

\(E = \frac{1}{2}CV_d^2\) 

Where C is capacitance of conductor and V_b is breakdown voltage.

Calculation:

Given: Vb = 200 V, C = 50 μF = 50 × 10^-6 F

\(E = \frac{1}{2} \times 50 \times {10^{ - 6}} \times {\left( {200} \right)^2} = 1;J\)

Given vector  \(\vec v = \frac{1}{3}\left( { - {y^3}\hat i + {x^3}\hat j + {z^3}\hat k} \right)\)

Bounded by open surface of hemisphere x² + y² + z² = 1

Such that z ≥ 0, and closed curve (x² + y² = 1)

We have to find

\(I = \smallint \left( {\nabla \times u} \right).\hat nds\)

⇒ from stokes theorem, we have

\(\mathop \int!!!\int \nolimits_s \left( {\nabla \times u} \right).\hat nds = \mathop \oint \nolimits_c u.dr\)

\( \Rightarrow I = \frac{1}{3}\left{ { - \oint {y^3}dx + \oint {x^3}dy} \right}\)

Now, changing the integral with substitution

x = cos θ ⇒ dx = -sin θ dθ

y = sin θ ⇒ dy = cos θ dθ

\(I = \frac{1}{3}\mathop \smallint \nolimits_0^{2\pi } {\sin ^4}\theta d\theta + \mathop \smallint \nolimits_0^{2\pi } \frac{1}{3}{\cos ^4}\theta d\theta \)

\(I = \frac{4}{3}\left( {\frac{3}{4} \times \frac{1}{2} \times \frac{\pi }{2} + \frac{3}{4} \times \frac{1}{2} \times \frac{\pi }{2}} \right) = \frac{\pi }{2}\)

Given differential equation:

\({x^2}\frac{{{d^2}y}}{{d{x^2}}} - 2x\frac{{dy}}{{dx}} + 2y = 4\)

is the standard form of Euler-Cauchy DE.

So, let x = e^z

⇒ dx = e^zdz = x.dz

\( \Rightarrow \frac{1}{{dz}} = \frac{x}{{dx}}i.e.\frac{d}{{dz}} = x\frac{d}{{dx}}\)

Similarly, we can obtain \({x^2}\frac{{{d^2}}}{{d{x^2}}} = \frac{{{d^2}}}{{d{z^2}}} - \frac{d}{{dz}}\)

⇒ xD = θ

⇒ x²D² = θ (θ - 1)

Making these substitutions in given DE, we get

\({x^2}\frac{{{d^2}y}}{{d{x^2}}} - 2x\frac{{dy}}{{dx}} + 2y = 4\)

(θ (θ - 1) – 2θ + 2) y = 4

θ (θ - 1) – 2(θ – 1) = 0

⇒ Solution for this will constitute of CF & PI.

⇒ CF = (θ – 1) (θ – 2) = 0 ⇒ θ = 1 & θ = 2

⇒ y = c₁e^z + c₂e^2z = c₁x + c₂x² 

\(also,;PI = \frac{1}{{f\left( \theta \right)}}{e^{az}} = \frac{1}{{f\left( a \right)}}{e^{az}} \Rightarrow \frac{{4 \times {e^{0z}}}}{{\left( {\theta - 1} \right)\left( {\theta - 2} \right)}} = \frac{{4 \times {e^{0z}}}}{{\left( { - 1} \right)\left( { - 2} \right)}} = \frac{4}{2} = 2\)

⇒ y = CF + PI = c₂x² + c₁x + 2

{We can also solve this problem by differentiating option also, if we don’t remember the process.}

Concept:

\(Error% = \frac{{Exact - Approximate}}{{Exact}} \times 100\)

Calculation:

\(f\left( x \right) = \cos x \Rightarrow f\left( x \right) = - \sin x \Rightarrow f'\left( {\frac{\pi }{6}} \right) = - \sin \left( {\frac{\pi }{6}} \right) = - 0.5;\left( {Exact} \right)\)

\(f'\left( x \right) = \frac{{\cos \left( {x + h} \right) - \cos \left( {x - h} \right)}}{{2h}} = \frac{{\cos \left( {\frac{\pi }{6} + 0.1\times 180/\pi} \right) - \cos \left( {\frac{\pi }{6} - 0.1\times 180/\pi} \right)}}{{2\left( {0.1} \right)}}\)

Using: cos a – cos b = - 2 sin [(a + b)/2] × sin [(a - b)/2]

\(f'\left( x \right) = \frac{{ - 2\sin \left( {\frac{\pi }{6}} \right)\sin \left( {0.1\times 180/\pi} \right)}}{{0.2}} = - 0.4992\)

\(Error;% = \frac{{ - 0.5 - \left( { - 0.4992} \right)}}{{ - 0.5}} \times 100 = 0.16% \Rightarrow option;b)\)

Concept:

For perfectly elastic collision momentum and kinetic energy is conserved and the coefficient of restitution is equal to 1.

Calculation:

Given:

mass of ball = 3 kg

Initial velocity = 4 m/s

Ball undergoes elastic collision, perfectly, with a second ball of mass m.

KE of 3 kg ball after the collision is 6 J.

To find possible value of m.

For perfect elastic collision.

Coefficient of restitution = 1

\(\frac{{Velocity;of;separation}}{{Velocity;of;aproach}} = 1\)

\( \Rightarrow \frac{{\left( {{V_2} - {V_1}} \right)}}{{{u_1} - {u_2}}} = \frac{{\left( {{V_2} - {V_1}} \right)}}{{4 - 0}} = 1 \Rightarrow {V_2} - {V_1} = 4 \Rightarrow {V_1} = {V_2} - 4\)

Also KE before collision = KE after collision

\( \Rightarrow \frac{1}{2}{m_1}u_1^2 + \frac{1}{2}{m_2}u_2^2 = \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_2}v_2^2\)

\( \Rightarrow \frac{{3 \times \left( {16} \right)}}{2} + 0 = 6 + \frac{1}{2}mv_2^2\)

⇒ 48 = 12 + mv²₂ ⇒ mv²₂ = 36

Also, from conservation of linear momentum we have

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(3 × 4) + 0 = 3v₁ + mv_2­

⇒ 12 = 3v₁ + mv₂ 

\( \Rightarrow 12 = 3\left( {{v_2} - 4} \right) + m{v_2} \Rightarrow 12 + 12 = \left( {3 + m} \right){v_2} \Rightarrow {v_2} = \frac{{24}}{{3 + m}}\)

mv₂² = 36

\( \Rightarrow m{\left( {\frac{{24}}{{3 + m}}} \right)^2} = 36\)

⇒ m (48) = 3 (3 + m)² ⇒ 9 + 6m + m² = 16 m

⇒ m² – 10 m + 9 = 0 ⇒ (m - 9) (m - 1) = 0

⇒ m = 9 or m = 1

⇒ correct option is d).

To analyse the boundary condition of this problem we will consider (assume) that the interface remains in contact after deformation also.

i) Since the pairs are in contact at interface, so the expansion/compression in the cylinders at interface will be such that the interface in contact must remain in contact so u^M_r = u^N_r otherwise joint will separate.

ii) Now it is given that the contact is friction less, so there will be no any resistance to relative slip at interface. So, we cannot arrive at any relation between u^N_θ & u^M_θ

iii) Now the radial stress caused in cylinder at interface are shown below separately.

From newtons third law, F_N = F_M as no external forces are at junction. Also the junction area for M & N are same

\( \Rightarrow \frac{{{F_N}}}{A} = \frac{{{F_M}}}{A} \Rightarrow \sigma _r^N = \sigma _r^M\)

iv) σ_θ is stress as shown, it will depend on material properly which is different for M & N, so they are not equal.

So correct option is 1

Explanation:

To find maximum deflection expression.

Students are advised to memorise the standard results. This \(\delta = \frac{{q{L^4}}}{{30EI}}\) is a standard result for cantilever beam with UVL.

For derivation of formula refer this

Here \(w = \frac{{px}}{L}\) per unit length at a distance x from O

\(\therefore EI\frac{{{d^4}y}}{{d{x^4}}} = - \frac{{px}}{L}\)

\(EI\frac{{{d^3}y}}{{d{x^3}}} = - \frac{{p{x^2}}}{{2L}} + {C_1}\)

S.F. is zero at x = 0, giving C₁ = 0

\(\therefore EI\frac{{{d^3}y}}{{d{x^3}}} = - \frac{{p{x^2}}}{{2L}}\)

\(EI\frac{{{d^2}y}}{{d{x^2}}} = - \frac{{p{x^3}}}{{6L}} + {C_2}\)

B.M. is zero at x = 0, giving C₂ = 0

\(\therefore EI\frac{{{d^2}y}}{{d{x^2}}} = - \frac{{p{x^3}}}{{6L}}\)

Integrate again

\(EI\frac{{dy}}{{dx}} = - \frac{{p{x^4}}}{{24L}} + {C_3}\)

\(But;\frac{{dy}}{{dx}} = 0;at;x = L,;giving\)

\(0 = - \frac{{p{L^4}}}{{24L}} + {C_3}\)

\(i.e.,;{C_3} = \frac{{p{L^3}}}{{24}}\)

\(\therefore \frac{{dy}}{{dx}} = \frac{1}{{EI}}\left( { - \frac{{p{x^4}}}{{24L}} + \frac{{p{L^3}}}{{24}}} \right) = \frac{p}{{24LEI}}\left( {{L^4} - {x^4}} \right)\)

\(y = \frac{p}{{24LEI}}\left( {{L^4}x - \frac{{{x^5}}}{5}} \right) + {C_4}\)

Now y = 0 at x = L, giving

\(0 = \frac{p}{{24EIL}}\left( {{L^5} - \frac{{{L^5}}}{5}} \right) + {C_4}\)

\({C_4} = - \frac{{4p{L^5}}}{{5 \times 24EIL}} = - \frac{{p{L^4}}}{{30EI}}\)

\(\therefore y = \frac{p}{{24LEI}}\left( {{L^4}x - \frac{{{x^5}}}{5}} \right) - \frac{{p{L^4}}}{{30EI}}\)

\(y = - \frac{p}{{120EIL}}\left( {4{L^5} - 5{L^4}x + {x^5}} \right)\)

Since the deflection will be maximum at the end O i.e. at x = 0, put this in the above obtained expression to get \(\delta = \frac{{p{L^4}}}{{30EI}}\), where negative indicates the deflection is towards negative y axis.

For small angular rotation θ of the rod, compression in the spring (3k) is

\({\delta _1} = \left( {\frac{L}{4}} \right)\theta \Rightarrow {F_1} = {F_{3k}} = {k_1}{\delta _1} = 3k\frac{L}{4}\theta = \frac{3}{4}kL\theta \)

Expansion of damper:

\({\delta _2} = \left( {\frac{L}{4}} \right)\theta \)

\( \Rightarrow {\dot \delta _2} = \frac{L}{4}\dot \theta \)

\( \Rightarrow {F_2} = {F_c} = C;{\dot \delta _2} = \frac{{CL}}{4}\dot \theta \)

Expansion of spring with stiffness k is

\({\delta _3} = \left( {\frac{L}{2} + \frac{L}{4}} \right)\theta = \frac{{3L}}{4}\theta \)

\( \Rightarrow {F_3} = {F_k} = \frac{{3L}}{4}\theta k\)

Now taking moment of all forces about O and inertia forces to be zero, we get

\(I\overset{\ddot{\ }}{\mathop{\theta }},+\frac{3}{4}kL\theta \left( \frac{L}{4} \right)+\frac{CL}{4}\dot{\theta }\frac{L}{4}+\frac{3L}{4}\theta k\left( \frac{3L}{4} \right)=0\)

\(I = {I_{about;centre}} + m{\left( {\frac{L}{4}} \right)^2} = \frac{{m{L^2}}}{{12}} + \frac{{m{L^2}}}{{16}} = \frac{{7m{L^2}}}{{48}}\)

\(\frac{7}{48}m{{L}^{2}}\overset{\ddot{\ }}{\mathop{\theta }},+\frac{3}{16}k{{L}^{2}}\theta +\frac{9}{16}k{{L}^{2}}\theta +\frac{C{{L}^{2}}}{16}\dot{\theta }=0\)

\(\Rightarrow \frac{7}{3}m{{L}^{2}}\overset{\ddot{\ }}{\mathop{\theta }},+12k{{L}^{2}}\theta +C{{L}^{2}}\dot{\theta }=0\)

Comparing with: m_eqθ̈ + c_eqθ̇ +k_eqθ = 0

\(\Rightarrow \xi =\frac{{{C}_{eq}}}{2\sqrt{{{k}_{eq}}{{m}_{eq}}}}=\frac{C{{L}^{2}}}{2\sqrt{12k{{L}^{2}}\times \frac{7}{3}m{{L}^{2}}}}=\frac{C{{L}^{2}}}{2\sqrt{12k{{L}^{2}}\times \frac{7}{3}m{{L}^{2}}}}=\frac{1}{2\left( 2\sqrt{7} \right)}\frac{C}{\sqrt{mk}}\)

Also for critical damping ξ = 1

\(\Rightarrow \frac{C}{\sqrt{mk}}\times \frac{1}{4\sqrt{7}}=1\)

\(\Rightarrow \frac{C}{\sqrt{mk}}=4\sqrt{7}\)

Concept:

Heat engine:

The efficiency (η) of a heat engine is defined as:

\(η = \frac{\text{Work done by heat engine}}{\text{Heat absorbed from the source}}\)

\(\therefore η = \frac{W}{Q_1}\)

Heat pump:

The COP of Heat Pump is given by:

\(CO{P_{HP}} = \frac{{Desired;Output}}{{Required;Input}} = \frac{{{Q_1}}}{W} \)

Calculation:

Given:

To find a relationship between (η) of a heat engine and COP of a heat pump.

For heat engine \(\eta =\frac{W}{{{Q}_{1}}}\) and for heat pump \(COP=\frac{{{Q}_{3}}}{W}\)

From energy balance for heat pump ⇒ Q₃ = W + Q₂

And From energy balance for heat engine ⇒ Q₂ = Q₁ – W

Q₃ = W + Q₁ – W = Q₁ 

\(\therefore COP_{HP}=\frac{{{Q}_{3}}}{W}=\frac{{{Q}_{1}}}{W}=\frac{1}{\eta }\)

Concept:

For finding the wt % of liquid present in alloy, we use the lever rule.

The weight fraction of solid-phase:

\({X_s} = \frac{{{w_o} - {w_l}}}{{{w_s} - {w_l}}}\)

The weight fraction of liquid phase:

\({X_l} = \frac{{{w_s} - {w_o}}}{{{w_s} - {w_l}}}\)

where ws, wo, wl denote the solid, overall and liquid composition.

Calculation:

Given an alloy x containing 60% P and 40% Q. To find % of solid and liquid fractions at 1250°C from lever rule, we have 

\({{m}_{s}}=\left( \frac{{{m}_{y}}-{{m}_{L}}}{{{m}_{s}}-{{m}_{L}}} \right)\)

\({{m}_{s}}=\frac{40-32}{68-32}=22.22%\)

∵ m_y, m_L & m_s are in % ⇒ direct substitution will give result in %.

\(also,~{{m}_{liq}}=-\frac{\left( {{m}_{y}}-{{m}_{s}} \right)}{\left( {{m}_{s}}-{{m}_{L}} \right)}=\frac{-40+68}{68-32}=77.8%\)

We are asked to calculate the time along critical path. For complete analysis of critical path we need to do forward pass and backward pass computation. In this question only time is asked, and we know that the time along the critical path is maximum of all the paths in our network so we will directly evaluate the maximum time along any path and mark the answer to save time. Also forward and backward pass computation is necessary if we are asked to find critical path or float related parameters.

In this question “<” & “>” symbols are used to just describe the precedence relationship.

i.e. A < B means A occurs before B.

⇒ On constructing the network diagram with the given precedence relationship.

Concept:

Velocity diagram and Acceleration diagram, \({V_t} = r\omega \) and \({a_r} = r{\omega ^2}\)

Trigonometric relations: for an isosceles triangle two sides are equal.

To find acceleration of the slider crank using exact analysis.

V_A → velocity of A w.r.t. O

V_BA → velocity of B w.r.t. A.

Velocity component of B along x direction = V_BAcos 45

Also due to the constraint on slider V_BA along x = 0 (i.e. V_BA is along y direction only)

⇒ V_BA cos 45 = 0

⇒ V_BA = 0

Also, V_BA = V_B - V_A = 0 ⇒ V_B = V_A

Now, constructing the velocity diagram for given case V_A = rω

Now Constructing the acceleration diagram:

∵ Point a will have two accelerations:

(i) Radial = rω² towards 0.

(ii) Tangential a_BA = unknown

Also, the tangential, component velocity of A = B, and acceleration will also be equal to y component of B. Also, the tangential acceleration of B will be in the direction shown by V_BA in configuration diagram.

⇒ from trigonometry o’b’ = a’o’

⇒ a_y = rω² (upwards, so positive).

Concept:

Deflection of a cantilever beam under a twisting moment is given by \(\delta = \frac{{M{L^2}}}{{2EI}}\)

Calculation:

Given A horizontal cantilever beam of circular cross – section with length (L) = 1.0 m & EI (Flexural rigidity) = 200 N.m².

Moment applied M_A = 1.0 N.m

To find δ in (mm) at end

\({\rm{\delta }} = \frac{{{\rm{M}}{{\rm{L}}^2}}}{{2{\rm{EI}}}} = \frac{{\left( 1 \right){{\left( 1 \right)}^2}}}{{2 \times 200}};{\rm{m}} = \frac{1}{{400}}{\rm{;m}} = 0.25 \times {10^{ - 2}}{\rm{m}}\)

⇒ δ = 2.5 × 10^-3 m = 2.5 mm.

Concept:

The centrifugal force due to a rotating mass is mrω²

For equilibrium ∑F = 0 along any direction

Trigonometric identity cos²θ + sin²θ = 1

Given two masses

m_a = 10 kg, m_b = 20 kg, r_a = 200 mm, r_b = 400 mm

∵ when these masses rotate, they will experience centrifugal forces in the direction as shown by arrow

In order to counter this resulting force our balance mass must be in a position where it produces a horizontal and vertical force component, which is possible when it is in (III) quadrant as shown by arrow.

Now applying force balance along horizontal direction

∑F_X = 0 ⇒ (m_a) (r_a) (ω)² = mrω² cos θ

∵ ω is same for all the masses as all mass are on some shaft.

⇒ m_ar_a = mr cos θ

(10) (200) = m (200) cos θ ⇒ m cos θ = 10    _____ (1)

Again, applying force balance along vertical direction

∑F_Y = 0 ⇒ m_br_b= mr sin θ

⇒ 20 × (400) = m (200) sin θ

⇒ m sin θ = 40       _____ (2)

∵ we are asked to find masses only

⇒ Squaring and adding equation (1) and (2) we get

\({\rm{;}}{\left( {40} \right)^2} + {\left( {10} \right)^2} = {m^2}{\sin ^2}\theta + {m^2}{\cos ^2}\theta \)

\(\Rightarrow 1700 = {m^2}\)

⇒ m = 41.231 kg = 41.23 (rounding off to two digits)

Concept:

Instantaneous Centre, Kennedy’s Theorem: 

For an Isosceles triangle, two sides and angles are equal.

For an equilateral triangle, all sides are equal, and all angles are also equal (60°)

For a right-angled triangle \(\tan \theta = \frac{{Perpendicular}}{{Hypotenuse}}\)

Calculation:

Given four-bar linkage

O₂O₄ = O₂A = L

AB = O₄B = √2 L

From Kennedy’s Theorem

w₂ (I₂₄ I₂₁) = w₄ (I₂₄ I₄₁)

where I₂₄ I-centre of link 2 – 4, I₂₁ is I-centre of link 2 – 1, I₄₁ is the I-centre of link 4 – 1

O₂O₄ = O₂A = L

Two sides are equal, and one angle is 90°.

⇒ This triangle is right-angled isosceles.

\(\Rightarrow A{O_4} = \sqrt {{L^2} + {L^2}} = \sqrt 2 L;and;\angle {O_2};A{O_4} = 45^\circ = \angle ;AB{O_4}\)

AB = BO₄ = AO₄ = √2 L → Equilateral triangle

⇒ ∠O₄AB = 60°

Also, I₂₄ A B is a straight line

⇒ ∠I₂₄ A O₂ + 45° + 60° = 180.

⇒ ∠ I₂₄ AO₂ = 75°

Also, In Δ I₂₄A O₂

\(\tan 75^\circ = \frac{{{I_{24}}{I_{12}}}}{{A{O_2}}} \Rightarrow 3.732 = \frac{{{I_{24}}{I_{12}}}}{L}\)

⇒ I₂₄ I₁₂ = 3.732 L

⇒ I₂₄ I₁₄ = I₂₄ I₁₂ + I₁₂I₁₄

⇒ I₂₄ I₁₄ = 3.732 L + L = 4.732 L.

\(\Rightarrow \frac{{{\omega _4}}}{{{\omega _2}}} = \frac{{{I_{24}}{I_{21}}}}{{{I_{24}}{I_{41}}}} = \frac{{3.732;L}}{{4.732;L}} = 0.7887\)

\(\Rightarrow \frac{{{\omega _4}}}{{{\omega _2}}} = 0.79\) (Round to two decimal places).

Concept:

Binomial probability distribution: \(P\left( x \right) = {}_{}^n{C_x}{p^x}{q^{n - x}} = \frac{{n!}}{{\left( {n - x} \right)! \times x!}}{p^x}{q^{n - x}}\)

Probability of defective (P_d) = p = 0.05

q = 1 – p = 0.95

15 parts are selected at random i.e. n = 15

To find probability that at least two defected

So, this is equivalent to find more than 0 or 1 defective

⇒ Probability of 0 defective = P (0) = ¹⁵C₀p⁰q¹⁵ = 1 × 0.95¹⁵ = 0.4633.

Probability of 1 defective = P (1) = ¹⁵C₁p¹q¹⁴ = 15 × 0.05¹ × 0.95¹⁴ = 0.3657

⇒ Probability of 0 or 1 defective = 0.4633 + 0.3657 = 0.829

⇒ P (defective ≥ 2) = 1 – P (defective ≤ 1)

⇒ P = 1 – 0.829 = 0.171

P = 0.17 (Rounded to two decimals).

Concept:

Gyroscopic couple is given by Iωω_p

Given a disc of mass m and radius r mounted on a shaft of length 1.5r as shown

\({\omega _p} = \frac{\omega }{{10}}\)

First let us understand the direction of Gyroscopic torque and its resulting reaction.

The ω and ω_p directions are as shown:

Apply right hand thumb rule pointing figure towards ω and closing towards w_p, we get torque in direction of thumb, which is as shown.

Now the torque applied on shaft due to bearings to restrict its motion must be opposite to that of obtained.

⇒ To produce this torque R_Q must be downward and R_P upward.

Also. Torque (Gyroscopic) = \(I;w{w_p} = \frac{{m{r^2}}}{2}\omega \times \frac{\omega }{{10}} = \frac{{m{r^2}{\omega ^2}}}{{20}}\)

∵ No other forces are acting on shaft.

⇒ R_P + R_Q = 0 ⇒ R_P = - R_Q.

Also couple due to R_P, R_Q = (R_P) (1.5r)

\(\Rightarrow \left( {{R_P}} \right)\left( {1.5r} \right) = \frac{{m{r^2}{\omega ^2}}}{{20}} \Rightarrow \left( {{R_P}} \right)\left( {1.5} \right) = \frac{{m{\omega ^2}r}}{{20}} = \frac{{m\left( {300} \right)}}{{20}}\)

\({R_P} = \frac{{m\left( {300} \right)}}{3} = 10;m.; \Rightarrow {R_Q} = - 10m.;\)

Also, the mass of disc will be supported by both the bearing equally (as load is in mid, leading to symmetric loading)

\({R_{{P_2}}} = \frac{{mg}}{2} = 5m\)

& \({R_{{Q_2}}} = 5m\)

⇒ Net reaction at Q and P are:

⇒ R_Q,net = - 10 m + 5 m = - 5m

R_P,net = 10m + 5m = 15 m

\(\Rightarrow \frac{{Larger;Reaction}}{{Smaller;Reaction}} = \frac{{15m}}{{ - 5m}} = - 3\)

Concept:

Torque T = r × F

Force due to friction F = μN

Refer the figure shown. 

Consider the pinned lever and Brake-shoe as one rigid member.

Since drum is rotating counter - clockwise so braking torque must act clockwise.

For braking torque clockwise friction force on drum must be towards (+ x)

⇒ From newtons third law, force on brake shoe will be in -x direction.

Now taking moment about 0 equal to zero.

⇒ N(b) – Fa – (μN) C = 0

⇒ N (0.5) – 100 (1) – (0.35) N (0.2) = 0

⇒ N (0.43) – 100 = 0

⇒ N = 232.558 N

⇒ Braking torque, T = (μN) r = (0.35) (232.558) (0.25) = 20.3488 Nm.

⇒ Braking torque = 20.35 Nm (Rounded to two decimal)

Concept:

The frictional head loss for a fully developed laminar pipe flow is \({h_f} = \frac{{32\mu Vl}}{{\rho g{d^2}}} = \frac{{32\mu Ql}}{{\rho gA{d^2}}}\)

Note: In this problem, no information regarding the friction factor in two pipes A and B are given so we cannot use the equation \({h_f} = \frac{{fl{v^2}}}{{2gd}}\)

Area of A = Area of B and L_A = 1 and L_B = 2

Flow is laminar and fully developed.

Frictional losses across the length of pipes A & B are the same.

To find \(\frac{{{Q_B}}}{{{Q_A}}}\)

Calculation:

Since \({h_f} = \frac{{32\mu VL}}{{\rho g{d^2}}} = \frac{{32\mu QL}}{{\rho g{d^2}A}}\)

\(;{h_{{f_A}}} = \frac{{32\mu {Q_A}{L_A}}}{{\rho g;d_A^2A}};& ;{h_{{f_B}}} = \frac{{32\mu {Q_B}{L_B}}}{{\rho gd_B^2A}}\)

The frictional head loss across the length of the pipes is the same:

\(;{h_{{f_A}}} = \frac{{32\mu {Q_A}{L_A}}}{{\rho g;d_A^2A}}; = \frac{{32\mu {Q_B}{L_B}}}{{\rho gd_B^2A}}\)

\(\Rightarrow \frac{{{Q_B}}}{{{Q_A}}} = \frac{{{L_A}}}{{{L_B}}} = \frac{1}{2} = 0.5;\)

Concept:

Drag force \(F = \frac{{{C_d}\rho {v^2}}}{2}\)

Power P = Fv

Calculation:

Drag coefficient of car = 0.1 (C_d) {window roof closed}

C_d = 0.8 (windows and roof open)

V₁ = 44 km/h {windows and roof closed}

To find V₂ {windows and roof open} for same amount of power.

\({F_d} = \frac{1}{2};\left( {\rho A{V^2}} \right){C_d}\)

\({F_{{d_1}}} = \frac{1}{2}\rho A;{\left( {44} \right)^2}\left( {0.1} \right)\)

\({F_{{d_2}}} = \frac{1}{2};\left( {\rho A} \right){\left( {{V_2}} \right)^2}\left( {0.8} \right)\)

Since Power is same:

\(\Rightarrow {V_1};\left( {{F_{{d_1}}}} \right) = \left( {{F_{{d_2}}}} \right){V_2}\)

\(\Rightarrow \frac{1}{2}\rho A;{\left( {44} \right)^2};\left( {0.1} \right)\left( {44} \right) = \frac{1}{2};\left( {\rho A} \right){\left( {{V_2}} \right)^3}\left( {0.8} \right)\)

⇒ V₂ = 22 km/h

Concept:

F_LM = 0.8, F_NR = 0.4

To find \({F_{PQ}}\)

F_PQ = F₇₈ + F₇₉

Also due to identical shape F₁₂ = F₁₄ = F₇₉.

And F₁₃ = F₅₆ = F₇₈

Also, F_NR = 0.4 = F₅₆ = F₇₈ = F₁₃

F_LM = 0.8 ⇒ F₁₂ + F₁₄ + F₁₃ = 2F₁₂ + 0.4 = 0.8 ⇒ F₁₂ = 0.2 = F₇₉

F_PQ = F₇₈ + F₇₉ = 0.4 + 0.2 = 0.6

Concept:

\(LMTD = \frac{{{\theta _2} - {\theta _1}}}{{\ln \left( {\frac{{{\theta _2}}}{{{\theta _1}}}} \right)}} = \frac{{{\theta _1} - {\theta _2}}}{{\ln \left( {\frac{{{\theta _1}}}{{{\theta _2}}}} \right)}} = {\theta _m}\)

Energy conservation between hot fluid and cold fluid,

C_h (ΔT_h) = C_c (ΔT_c)

Calculation:

Given: Th_i = Inlet temperature of hot fluid = 100°C, Tc_i = Inlet temperature of cold fluid = 15°C, Tc_e = Exit temperature of cold fluid = 45°C

C_h = 2000 W/K, C_c = 1200 W/K

C_h (Δ T_h) = C_c (Δ T_c)

⇒ 2000 (100 – T­_he) = 1200 (45° - 15°)

⇒ T_he = 82°C

θ₁ = Th_i – Tc_i = 85°C; θ₂ = T_he – T_ce = 37°C

\(LMTD\left( {{\theta _m}} \right) = \frac{{{\theta _1} - {\theta _2}}}{{\ln \left( {\frac{{{\theta _1}}}{{{\theta _2}}}} \right)}} = \frac{{85 - 37}}{{\ln \left( {\frac{{85}}{{37}}} \right)}} = 57.71^\circ C\)

∴ LMTD (θ_m) = 57.71 K

Concept:

According to “Gouy-stodola” Theorem.

Irreversibility I = T₀ (Δs)_universe = T₀ S_gen

Where T₀ = Ambient temperature

Calculation:

 

Increase in specific enthalpy = Δh = h₂ – h₁ = 2.50 kJ/kg

Power input = Increase in enthalpy

2.50 = ṁ × Δh = ṁ × 2.50

ṁ = 1 kg/s

ΔS = S₂ – S₁ = 0.007 kJ/kg

I = T₀ S_gen = T₀ (Ṡ_exit – Ṡ_inlet)

I = T₀ ṁ(S₂ – S₁) = 300 × 1 × 0.007 = 2.10 kW

Concept:

The velocity triangles at inlet and outlet are shown:

Where, β1, β2 are vane angles at inlet and outlet

V1, V2 are absolute velocities at inlet and outlet

Vr1, Vr2 are relative velocities at inlet and outlet

Discharge is given by Q = πD2B2 × Vf2

Where, D2, B2 are diameter and width of the impeller at outlet and Vf2 is outlet flow velocity.

Power input to the 'Pump' is given as ‘P’ = ṁ (Vw2u2 – Vw1u1)

Thus, \({V_{f2}} = \frac{Q}{{\pi {D_2}{B_2}}}\)

\({\eta _{man}} = \frac{{g{H_m}}}{{{V_{w2}}{U_2}}};and;{U_2} = \frac{{\pi {D_2}N}}{{60}}\)

And from the outlet Δ,

\(\tan {\beta _2} = \frac{{{V_{f2}}}}{{{U_2} - {V_{w2}}}}\)

Calculation:

Given:

Inlet flow is axial, so whirl component at inlet is zero

V_w1 = 0

Because discharge is radial, so whirl component at exit is

V_w2 = u₂

Given: P = 2 kW, N = 3000 rpm and r₂ = 50 mm

\(\therefore V{w_2} = {u_2} = \frac{{\pi DN}}{{60}} = \frac{{\pi \times 0.1 \times 3000}}{{60}} = 15.71;m/sec\)

Power input to the pump ‘P’ = ṁ (V_w2u₂ – V_w1u₁)

Where, ṁ = mass flow rate of liquid in kg/sec.

2000 = ṁ (15.71 × 15.71 - 0)

⇒ ṁ = 8.1057 kg/sec.

Concept:

An ideal Otto Cycle PV Diagram:

Mean Effective Pressure:

\(\rm M.E.P=\frac{Net-Work;Done}{Swept;Volume}\)

Efficiency of Otto Cycle:

\({η _{otto}} = 1 - \frac{1}{{{{\left( r \right)}^{\gamma - 1}}}}\)

where r = compression ratio = V1/V2.

Calculation:

Given:

η_otto = 0.5

∵ P₁V₁ = RT₁

(100) V₁ = 0.287 × 300 ⇒ V₁ = 0.861 m³/kg

\(\because {\eta _{otto}} = 1 - \frac{1}{{{{\left( r \right)}^{\gamma - 1}}}}\)

\(0.5 = 1 - \frac{1}{{{{\left( r \right)}^{1.4 - 1}}}} \Rightarrow r = 5.65\)

\(\because r = \frac{{{V_1}}}{{{V_2}}} = 5.65\)

⇒ V₂ = 0.1524 m³/kg

∴ Specific work output = P_mep × V_s

= 1000 × (V₁ – V₂) = 1000 × (0.861 – 0.1524) = 708.6 kJ/kg

∵ point 1 and point 3 both are open to atmosphere

∴ P₁ = P₃ = P_atm

Volume of mould cavity = (40 × 30 × 15) = 18000 cm³

Apply Bernoulli’s equation at point 1 and 3.

\(\frac{{{P_{atm}}}}{{\rho g}} + \frac{{V_1^2}}{{2g}} + {z_1} = \frac{{{P_{atm}}}}{{\rho g}} + \frac{{V_3^2}}{{2g}} + 0\) 

V₁ = 0 (Assume)

\({V_3} = \sqrt {2g{z_1}} = \sqrt {2 \times 981 \times 20} = 198.04;cm/s\) 

Time to fill the mould is\(;\left( {{t_f}} \right) = \frac{{{V_m}}}{{{A_g}{V_g}}} = \frac{{Volume}}{{{V_2} \times {A_2}}}\) 

\({t_f} = \frac{{18000}}{{198.09 \times \left( {\frac{\pi }{4} \times {2^2}} \right)}} = 28.92;sec\)

Concept:

\({H_s} = \frac{{VI}}{{Av}}\eta ;J/m{m^3}\)

Calculation:

Given:

Welding current, I = 250 A, Voltage, V = 20 V, Welding speed, v = 5 mm/sec

Arc heat transfer efficiency η = 70%

∴ Heat required per unit length

\({H_s} = \frac{{VI}}{v}\eta = \frac{{250 \times 20 \times 0.7}}{5} = 700;J/mm = 0.7;kJ/mm\)

Note: Heat required per unit volume,

\({H_s} = \frac{{VI}}{{Av}}\eta ;J/m{m^3}\)

Where A is welding Area.

Concept:

Difference of Thickness is given by:

Δh = μ2R

where, Δh = \(h_o - h_f\)

\(h_o\) = Initial Thickness, \(h_f\) = Final Thickness

μ = Coefficient of friction

D = Diameter of Roll

Calculation:

Given:

D = 600 mm, μ = 0.05, h₀ = 2 mm

∵ Δh = μ²R

⇒ (2 - h_f) = 0.05² × 300 ⇒ h_f = 1.25 mm

Concept:

For Drilling

\(\because MRR = \frac{\pi }{4}{d^2} \times fN\)

Power required = Sp. Energy consumption × MRR

Calculation:

Given:

Thickness of plate is 15 mm

Diameter of drill bit D = 10 mm

Feed rate = 0.25 mm/rev

Spindle speed = 1200 rpm

Specific energy required = 0.7 N-m/mm³

\(\because MRR = \frac{\pi }{4}{d^2} \times fN\)

\(= \frac{\pi }{4} \times {10^2} \times 0.25 \times 1200 = 23561.94;m{m^3}/min\)

= 392.69 mm³/sec

Power required = Sp. Energy consumption × MRR

⇒ P = 0.7 × 392.69 = 274.89 J/s = 274.89

Concept:

Shear angle: 

The angle made by the shear plane with respect to the cutting velocity vector is known as the shear angle

The expression of the shear angle is given by:

\(ϕ = ;{\tan ^{ - 1}}\left[ {\frac{{{\bf{r}}\cos {\bf{α }}}}{{1 - {\bf{r}} \sin {\bf{α }}}}} \right]\)

From merchant theory

2ϕ + β - α₀ = 90°  

and μ = tan β 

where, ϕ  = shear angle, α  = Orthogonal rake angle, β = friction angle, r = chip thickness ratio = \(\frac{t_1}{t_2}\), \(t_1\) = uncut chip thickness, \(t_2 \) = deformed chip thickness.

Calculation:

Given:

α = 10°, t = 0.125 mm, t_c = 0.22 mm

Chip thickness ratio \(r = \frac{t}{{{t_c}}} = \frac{{0.125}}{{0.22}} = 0.568\)

\(\because \tan ϕ = \frac{{r\cos \propto }}{{1 - r\sin \propto }} = \frac{{0.568\cos 10^\circ }}{{1 - 0.568\sin 10}}\)

ϕ = 31.83°

From merchant theory,

\(ϕ = \frac{\pi }{4} + \frac{ \propto }{2} - \frac{β }{2}\)

\(\Rightarrow 31.83= 45^\circ + \frac{{10}}{2} - \frac{β }{2} \Rightarrow β = 36.34^\circ \)

∵ μ = tan β = tan (36.34°)

∴ μ = 0.735≈ 0.74

Concept:

\(EOQ = \sqrt {\frac{{2D{C_0}}}{{{C_h}}}} \)

D = Annual demand, C0 = order cost/setup cost, Ch = holding cost per unit per year

Total inventory cost (TIC) = Ordering cost + Holding cost

For EOQ 

\(TIC* = \sqrt {2 \times D \times {C_0} \times {C_h}} \)

Calculation:

Given:

Annual demand, D = 10,000 units, Q = Rs. 400 per order, Holding cost (C_h) = 24/- unit/year, Ordering cost (C₀) = 400/- per order

(TIC)_I = Ordering cost + Holding cost

\(= \frac{D}{Q} \times {C_0} + \frac{Q}{2} \times {C_h}\)

\(= \frac{{10,000}}{{400}} \times 400 + \frac{{400}}{2} \times 24 = Rs.;14800/yea\)

For EOQ order quantity,

\(TIC* = \sqrt {2 \times D \times {C_0} \times {C_h}} \)

\(= \sqrt {2 \times 10,000 \times 400 \times 24} = Rs.;13856.4/year\)

Saving in the total inventory cost per year

= (TIC)_I – TIC*

= 14800 – 13856.4 = Rs. 943.594/year