Previous Year Paper

GATE ME 2009 Official Paper (Previous Year Paper)

Held in 2009

60 questions180 minutes

With answers

Paper last updated: 15 October 2026

Questions & answers

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Mechanical (60 questions)

Question 1

For a matrix [M] \(= \left[ {\begin{array}{*{20}{c}} {\frac{3}{5}}&{\frac{4}{5}}\ x&{\frac{3}{5}} \end{array}} \right]\), the transpose of the matrix is equal to the inverse of the matrix [M]^T = [M]^-1. The value of x is given by

AOption (a)
\(- \frac{4}{5}\)
BOption (b)
\(- \frac{3}{5}\)
COption (c)
\(\frac{3}{5}\)
DOption (d)
\(\frac{4}{5}\)

View answer & explanation

Correct answer

\(- \frac{4}{5}\)

Why this answer is correct

Concept:

Orthogonal matrix:

For an orthogonal matrix, transpose is equal to its inverse.

[M]^T = [M^-1]

For an orthogonal matrix:

MM^T = I

Calculation:

Given:

[M] \(= \left[ {\begin{array}{*{20}{c}} {\frac{3}{5}}&{\frac{4}{5}}\ x&{\frac{3}{5}} \end{array}} \right];\),

Given M is the orthogonal matrix

∴ [M][M]^T = I

\(\left[ {\begin{array}{*{20}{c}} {\frac{3}{5}}&{\frac{4}{5}}\ x&{\frac{3}{5}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\frac{3}{5}}&x\ {\frac{4}{5}}&{\frac{3}{5}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0\ 0&1 \end{array}} \right];\)

\(= \left[ {\begin{array}{*{20}{c}} {1}&{\frac{3x}{5}} +\frac{12}{25}\ \frac{3x}{5}+\frac{12}{25}&x^2+{\frac{9}{25}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0\ 0&1 \end{array}} \right];\)

so, \(\frac{{12}}{{25}} + \frac{3}{5}x = 0 \Rightarrow x = -\frac{{4}}{5};\)

The value of x is \(\frac{{ - 4}}{5}\)

Question 2

The divergence of the vector field \(3xz\hat i + 2xy\hat j - y{z^2}\hat k\) at a point (1,1,1) is equal to

AOption (a)
7
BOption (b)
4
COption (c)
3
DOption (d)
0

View answer & explanation

Correct answer

Why this answer is correct

Concept:

The divergence of any vector field \(\vec A\) is defined as:

\(Div= \vec \nabla .\vec A\)

The nabla operator is defined as:

\(\vec \nabla ={\hat i\frac{\partial }{{\partial x}} + \hat j\frac{\partial }{{\partial y}}}+\hat k\frac{\partial }{\partial z}\)

Calculation:

Given:

\(V = 3xz\hat i + 2xy\hat j - y{z^2}\hat k\)

Divergence vector field of V is = ∇.V

\(= \left( {\frac{\partial }{{\partial x}}\hat i + \frac{\partial }{{\partial y}}\hat j + \frac{\partial }{{\partial z}}\hat k} \right).\left( {3xz\hat i + 2xy\hat j - y{z^2}\hat k} \right);\)

= 3z + 2x – 2yz

(∇.V)_(1,1,1) = 3(1) + 2(1) – 2(1)(1) = 3

Question 3

The inverse Laplace transform of \(\frac{1}{{{s^2};+;s}}\) is

AOption (a)
1 + e^t
BOption (b)
1 – e^t
COption (c)
1 – e^-t
DOption (d)
1 + e^-t

View answer & explanation

Correct answer

1 – e^-t

Why this answer is correct

Explanation:

\(F\left( t \right) = {L^{ - 1}}\left[ {\frac{1}{{s\left( {s + 1} \right)}}} \right];\)

\(F\left( t \right) = {L^{ - 1}}\left[ {\frac{(s+1)-s}{{s\left( {s + 1} \right)}}} \right];\)

\(F\left( t \right) = {L^{ - 1}}\left[ \frac{1}{s} - \frac{1}{{s + 1}}\right];\)

Taking partial fraction, we get so,

\(F\left( t \right) = {L^{ - 1}}\left[ {\frac{1}{s}} \right] - {L^{ - 1}}\left[ {\frac{1}{{s + 1}}} \right] \F(t)= 1 - {e^{ - t}}\)

Question 4

If three coins are tossed simultaneously, the probability of getting at least one head

AOption (a)
1/8
BOption (b)
3/8
COption (c)
1/2
DOption (d)
7/8

View answer & explanation

Correct answer

7/8

Why this answer is correct

Explanation:

Number of cases = 23 = 8

Possible outcomes = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Number of favorable cases = 7

{HHH, HHT, HTH, THH, HTT, THT, TTH}

Then probability P = 7/8

Question 5

If a closed system is undergoing an irreversible process, the entropy of the system

AOption (a)
Must increase
BOption (b)
Always remains constant
COption (c)
Must decrease
DOption (d)
Can increase, decrease or remain constant

View answer & explanation

Correct answer

Can increase, decrease or remain constant

Why this answer is correct

Explanation:

If a closed system is undergoing an irreversible process, the change in entropy of the system is given by

dS > 0, or dS = 0, or dS < 0

In an irreversible process in which heat is removed from the system then its entropy can decrease.

The entropy of a closed system is given by

\(ds = \frac{{\delta Q}}{T} + {\left( {\delta s} \right)_{gen}}\)

When the process is irreversible then entropy generation in the system (δs)gen is always positive, the heat transfer will decide whether the entropy will increase or decrease.

When heat is added to the system

\(\frac{{\delta Q}}{T} = + ve,;{\left( {\delta s} \right)_{gen}} = + ve\)

∴ \(\delta s = \frac{{\delta Q}}{T} + {\left( {\delta s} \right)_{gen}} = + ve\) i.e. entropy increases

When Heat is removed from the system

\(\frac{{\delta Q}}{T} = ;-ve,;{\left( {\delta s} \right)_{gen}} = ; + ve\)

∴ \(\delta s = \frac{{\delta Q}}{T} + ;{\left( {\delta s} \right)_{gen}} = ; + ve,; - ve;or;0\) i.e. entropy increases decreases or remains constant

When the process is adiabatic, dQ = 0,

\(\frac{{\delta Q}}{T} = 0,;{\left( {\delta s} \right)_{gen}} = + ve\)

\(\delta s = \frac{{\delta Q}}{T} + {\left( {\delta s} \right)_{gen}} = ; + ve\) i.e. entropy increases

∴ when a closed system is undergoing an irreversible process the entropy may increase, decrease or remain constant.

Question 6

A coolant fluid at 30°C flows over a heated flat plate maintained at a constant temperature of 100°C. The boundary layer temperature distribution at a given location on the plate may be approximated as T = 30 + 70exp(−y) where y (in m) is the distance normal to the plate and T is in °C. If the thermal conductivity of the fluid is 1.0 W/mK, the local convective heat transfer coefficient (in W/m²K) at that location will be

AOption (a)
0.2
BOption (b)
1
COption (c)
5
DOption (d)
10

View answer & explanation

Correct answer

Why this answer is correct

Concept:

At point of contact, heat conducted ie equal to heat convected.

Calculation:

Given:

Flat plate = 100°C, Coolant fluid = 30°C

k = 1.0 W/mK,

According to the question, the formula for temperature distribution is given by:

T = 30 + 70e^-y

Under steady-state condition

Q_cond = Q_conv

\(- kA\frac{{dT}}{{dy}} = hA{\rm{\Delta }}T\)

\(\Rightarrow - kA\left( { - 70{e^{ - y}}} \right) = hA{\rm{\Delta }}T\)

Now, applying boundary conditions, at the surface of the plate:

y = 0

⇒ 70kA = hAΔT

\(\Rightarrow h = \frac{{70k}}{{{\rm{\Delta }}T}} = \frac{{70 \times 1}}{{100 - 30}} = 1;{W/}{{{m^2}k}}\)

Question 7

A frictionless piston-cylinder device contains a gas initially at 0.8MPa and 0.015 m³. It expands quasi-statically at constant temperature to a final volume of 0.030 m³. The work output (in kJ) during this process will be

AOption (a)
8.32
BOption (b)
12.00
COption (c)
554.67
DOption (d)
8320.00

View answer & explanation

Correct answer

8.32

Why this answer is correct

Concept:

Work Done in case of isothermal system is given by:

\(W = {P_1}{V_1}\ln \frac{{{V_2}}}{{{V_1}}}\)

Where, P₁ = Initial Pressure, V1 = Initial Volume, V2 = Final Volume.

Calculation:

Given:

P1 = 0.8 MPa = 800 kPa, V1 = 0.015 m³, V2 = 0.030 m3.

Work done in case of isothermal system is given:

\(W = {P_1}{V_1}\ln \frac{{{V_2}}}{{{V_1}}}\)

\(W=800 \times 0.015 \times \ln \frac{{0.030}}{{0.015}}\)

W = 8.317 kJ ≈ 8.32 kJ.

Question 8

In an ideal vapor-compression refrigeration cycle, the specific enthalpy of refrigerant (in kJ/kg) at the following states is given as:

Inlet of condenser: 283

Exit of condenser: 116

Exit of evaporator: 232

The COP of this cycle is

AOption (a)
2.27
BOption (b)
2.75
COption (c)
3.27
DOption (d)
3.75

View answer & explanation

Correct answer

2.27

Why this answer is correct

Explanation:

The COP of the ideal Vapour Compression Refrigeration System (VCRS):

It is the ratio of refrigerating effect (R.E) to the work done by the compressor (W_c).

\(COP=\frac{Refrigerating;effect}{Work;done;by;compressor}\Rightarrow \frac{R.E}{W_c}\)

Process 1-2: Isentropic work done by Compressor

Process 2-3: Heat removal at constant pressure by Condensor

Process 3-4: Isenthalpic expansion by Throttling valve

Process 4-1: Heat addition at constant pressure by Evaporator.

Calculation:

Given:

h₁ = 232 kJ/kg, h₂ = 283 kJ/kg

h3 = h4 = 116 kJ/kg

\(COP = \frac{{Refrigeration\ effect}}{{Work\ done}} = \frac{{{h_1};- ;{h_4}}}{{{h_2};-;{h_1}}}\)

\(\therefore\frac{{232;- ;116}}{{283;-;232}} \Rightarrow 2.27\)

Question 9

A compressor undergoes a reversible, steady flow process. The gas at inlet and outlet of the compressor is designated as state 1 and state 2 respectively. Potential and kinetic energy changes are to be ignored. The following notations are used:

v = specific volume and P = pressure of the gas.

The specific work required to be supplied to the compressor for this gas compression process is

AOption (a)
\(\mathop \smallint \limits_1^2 Pdv\)
BOption (b)
\(\mathop \smallint \limits_1^2 vdP\)
COption (c)
v₁(P₁ – P₂)
DOption (d)
–P₂(v₁ – v₂)

View answer & explanation

Correct answer

\(\mathop \smallint \limits_1^2 vdP\)

Why this answer is correct

Explanation:

Steady flow energy equation, neglecting changes in K.E and P.E.

\(h_1;+;\frac{{∂Q}}{{∂m}}=h_2;+;\frac{∂W}{∂m}\) ......eq(i)

and First Law for open system

\(\frac{{∂Q}}{{\partial m}}=dh-vdP;\)

\(\frac{{∂Q}}{{\partial m}}=(h_2;-;h_1)-vdP;\)

for a reversible and adaibatic Process (Tds = ∂Q = 0).

⇒ h2 - h1 = vdP

form (i),

⇒ h₁- h₂ = \(\frac{{∂W}}{{dm}}\)

⇒ \(\frac{{∂W}}{{dm}}=-vdP\)

By integrating,

⇒ specific work w = \(-\mathop \smallint \limits_1^2 vdP\) (for turbine)

⇒ Wc = \(\mathop \smallint \limits_1^2 vdP\) (for compressor).

Question 10

A block weighing 981 N is resting on a horizontal surface. The coefficient of friction between the block and the horizontal surface is μ = 0.2. A vertical cable attached to the block provides partial support as shown. A man can pull horizontally with a force of 100 N. What will be the tension, T (in N) in the cable if the man is just able to move the block to the right?

AOption (a)
176.2
BOption (b)
196.0
COption (c)
481.0
DOption (d)
981.0

View answer & explanation

Correct answer

481.0

Why this answer is correct

Concept:

Number of vertical forces:

∑F_y = T + R_N - W = 0

Static friction force is given as,

F_s = μR_N

Calculation:

Given:

W = 981 N, μ = 0.2, F_s = 100 N

Normal reaction:

\(\Rightarrow {R_N} = \frac{{100}}{{0.2}} = 500~N\)

Tension T

⇒ T = 981 - 500 = 481 N

Question 11

If the principal stresses in a plane stress problem, are σ₁ = 100 MPa, σ₂ = 40 MPa, the magnitude of the maximum shear stress (in MPa) will be

AOption (a)
60
BOption (b)
50
COption (c)
30
DOption (d)
20

View answer & explanation

Correct answer

Why this answer is correct

Concept:

In the case of 3-D analysis, maximum shear stress can be calculated as:

maximum of \([\frac{{{σ _1} - {σ _2}}}{2},\frac{{{σ _2} - {σ _3}}}{2},\frac{{{σ _1} - {σ _3}}}{2}]\)

And In the case of plane stress analysis, maximum shear stress can be calculated by,

\({{\bf{\tau }}_{{\bf{max}}}} = \frac{{{{\rm{\sigma }}_1} - {{\rm{\sigma }}_2}}}{2}\), where σ₁ and σ₂ are principal stresses in a plane.

Calculation:

Given:

σ1 = 100 MPa , σ₂ = 40 MPa

Now, In plane stress analysis, we have

\({{\bf{\tau }}_{{\bf{max}}}} = \frac{{{{\rm{\sigma }}_1} - {{\rm{\sigma }}_2}}}{2}\)

\(\therefore {\tau _{max}} = \frac{{{\sigma _1} - {\sigma _2}}}{2} = \frac{{100 - 40;}}{{2;}} = 30;MPa\)

Maximum In-Plane shear stress/Surface shear stress:

\(\tau_{max,inplane}=\frac{{\sigma _1};-;{\sigma _2}}{{2}}\)

Maximum wall shear stress/Out plane shear stress/Absolute maximum shear stress:

\(\tau_{max,abs}=\frac{{\sigma _{max}};-;{\sigma _{min}}}{{2}}=\frac{\sigma_1}{2}\)

Question 12

A simple quick return mechanism is shown in the figure. The forward to return ratio of the quick return mechanism is 2:1. If the radius of the crank O₁P is 125 mm, then the distance’s’ (in mm) between the crank centre to lever pivot centre point should be.

AOption (a)
144.3
BOption (b)
216.5
COption (c)
240.0
DOption (d)
250.0

View answer & explanation

Correct answer

250.0

Why this answer is correct

Concept:

Quick return ratio:

\(\frac{{Time;of;cutting;stroke}}{{Time;of;return;stroke}} = \frac{\beta }{\alpha } = \frac{{360 - \alpha }}{\alpha }\)

O₁P = r = 125 mm

\(QRR = \frac{2}{1} = \frac{\beta }{\alpha } = \frac{{360 - \alpha }}{\alpha }\)

⇒ α = 120°

then \(\angle R{O_1}{O_2} = \frac{\alpha }{2} = 60^\circ\)

from ΔRO₂O₁

\(\begin{array}{l} \sin \left( {90 - \frac{\alpha }{2}} \right) = \frac{{{O_1}R}}{{{O_1}{O_2}}} = \frac{r}{{{O_1}{O_2}}}\ \Rightarrow \sin \left( {90 - 60} \right) = \frac{{125}}{{{O_1}{O_2}}} \Rightarrow {O_1}{O_2} = 250;mm \end{array}\)

Question 13

The rotor shaft of a large electric motor supported between short bearings at both the ends shows a deflection of 1.8 mm in the middle of the rotor. Assuming the rotor to be perfectly balanced and supported at knife edges at both ends, the likely critical speed (in rpm) of the shaft is

AOption (a)
350
BOption (b)
705
COption (c)
2810
DOption (d)
4430

View answer & explanation

Correct answer

705

Why this answer is correct

Concept:

Critical speed:

\(\omega_c=\omega_n=\sqrt{\frac{s}{m}}=\sqrt{\frac{g}{∆}}\)

Calculation:

Given:

Δ = 1.8 mm

\({\omega_c} = \sqrt {\frac{g}{\delta }} \Rightarrow \frac{{2\pi {N_c}}}{{60}} = \sqrt {\frac{g}{\delta }}\)

\(\Rightarrow {N_c} = \frac{{60}}{{2\pi }}\sqrt {\frac{{9.81}}{{1.8 \times {{10}^{ - 3}}}}} = 704.98 \approx 705\ rpm\)

Question 14

A solid circular shaft of diameter d is subjected to a combined bending moment M and torque, T. The material property to be used for designing the shaft using the relation \(\frac{{16}}{{\pi {d^3}}}\sqrt {{M^2} + {T^2}}\) is

AOption (a)
Ultimate tensile strength (S_u)
BOption (b)
Tensile yield strength (S_y)
COption (c)
Torsional yield strength (S_sy)
DOption (d)
Endurance strength (S_e)

View answer & explanation

Correct answer

Torsional yield strength (S_sy)

Why this answer is correct

Explanation:

This is the case of combined bending and torsion.

So equivalent Torque:

\({T_e} = \sqrt {{M^2} + {T^2}}\)

then induced shear stress:

\(\tau = \frac{{16{T_e}}}{{\pi {d^3}}} = \frac{{16}}{{\pi {d^3}}}\sqrt {{M^2} + {T^2}}\)

So, for safe design \(\tau < \frac{{{S_{sy}}}}{{FOS}}\)

Question 15

The effective number of lattice points in the unit cell of simple cubic, body centered cubic, and face centered cubic space lattices, respectively, are

AOption (a)
1,2,2
BOption (b)
1,2,4
COption (c)
2,3,4
DOption (d)
2,4,4

View answer & explanation

Correct answer

1,2,4

Why this answer is correct

Concept:

Diagram
Effective no. of lattice points	\(\Rightarrow \frac{1}{8} × 8 = 1\)	\(\frac{1}{8} × 8 + 1 = 2\)	\(\frac{1}{8} × 8 + \frac{1}{2} × 6 = 4\)

In the unit cell effective number of atoms = corner atoms × (1/8) + face cantered atoms × (1/2) + 1 (Inner atom)

Unit Cell	Coordination No.	No. of Atoms Per Unit Cell	Atomic packing factor
Simple Unit Cell	6	1	52%
Body-centred Cubic	8	2	68%
Face-centred Cubic	12	4	74%
Hexagonal Closest Packed	12	6	74%

Question 16

Friction at the tool-chip interface can be reduced by

AOption (a)
decreasing the rake angle
BOption (b)
increasing the depth of cut
COption (c)
decreasing the cutting speed
DOption (d)
increasing the cutting speed

View answer & explanation

Correct answer

increasing the cutting speed

Why this answer is correct

Explanation:

Parameter	High	Low
Cutting Speed	High cutting speed results in lesser chip tool contact time and better heat dissipation. These two effects make it infavourable condition to form a bond between tool and chip and the coefficient of friction is reduced.	Usually at relatively low cutting speeds, due to higher tool-chip contact time and lower heat dissipation, the friction between chip and tool is very high such that the chip material welds itself to the tool face. Presence of these welded materials eventually results in to Built up Edge (BUE) formation which further increases friction.
Rake Angle	By increasing the rake angle, chip flow will be easier and the friction will reduce.	For machining stronger and brittle materials smaller or lower rake angle is used but this results to higher friction force.
Depth of cut	Relatively higher depth of cut creates favourable condition for built up edge formation due to more tool-chip contact area and hence increases friction.	Relatively lower depth of cut reduces chance of built up edge formation due to less tool-chip contact area and hence reduces friction.
Feed	Relatively higher feed creates favourable condition for built up edge formation and hence increases friction.	Relatively lower feed reduces chance of built up edge formationand hence reduces friction.

Question 17

Two streams of liquid metal, which are not enough to fuse properly, result into a casting defect known as

AOption (a)
cold shut
BOption (b)
swell
COption (c)
sand wash
DOption (d)
scab

View answer & explanation

Correct answer

cold shut

Why this answer is correct

Explanation:

The casting defect and their reasons are mentioned below:

Cold shut: It occurs when two-stream of molten metals approach each other in the mould from opposite directions.

Swell: It is a localized enlargement of casting at its top side and is caused due to pressure of molten metal on soft rammed sand.

Sand wash: It is an irregularity in the surface of the casting resulting from the erosion of the sand mould during pouring and the contour of the erosion is formed in the surface of the final casting.

Scab: It is a hollow cavity on a flat casting surface caused by the liquid metal pressure, which succeeds in pushing back the mould sand at certain places.

Additional Information

Casting defects are those characteristics that generate imperfection in a casting exceeding the quality limits imposed by the design or service condition of the casting. As per standard casting defects, they are of seven types:

Metallic Projection:

Mismatch
Lifts and shifts
Fins and flash
Swell

Cavities

Blow-holes
Pin-holes or gas porosity

Discontinuities

Hot tears

Defective surface

Fusion and metal penetration
Hard spots
Cold shots
Scabs or washes
Scar
Blister

Incomplete castings

Misrun and cold shut
Pour short
Runout

Incorrect dimension and shape

Warping
The drop

Inclusions

Sponginess or honeycombing

Question 18

The expected time (t_e) of a PERT activity in terms of optimistic time (t_o), pessimistic time (t_p) and most likely time (t_m) is given by

AOption (a)
\({t_e} = \frac{{\left( {{t_o} + 4{t_m} + {t_p}} \right)}}{6}\)
BOption (b)
\({t_e} = \frac{{{t_o} + 4{t_p} + {t_m}}}{6}\)
COption (c)
\({t_e} = \frac{{{t_o} + 4{t_m} + {t_p}}}{3}\)
DOption (d)
\({t_e} = \frac{{{t_o} + 4{t_m} + {t_p}}}{4}\)

View answer & explanation

Correct answer

\({t_e} = \frac{{\left( {{t_o} + 4{t_m} + {t_p}} \right)}}{6}\)

Why this answer is correct

Concept:

The PERT (Project Evaluation and Review Technique) technique is used, when activity time estimates are stochastic in nature. For each activity, three values of time (optimistic, most likely, pessimistic) are estimated.

Optimistic time (to) estimate is the shortest possible time required for the completion of the activity
Most likely time (tm) estimate is the time required for the completion of activity under normal circumstances
Pessimistic time (tp) estimate is the longest possible time required for the completion of the activity

In PERT expected time of an activity is determined by using the below-given formula:

\({t_e} = \frac{{\left( {{t_o} + 4{t_m} + {t_p}} \right)}}{6}\)

Additional Information

Variance, σ2 = \({\left[ {\frac{{{t_P} - {t_0}}}{6}} \right]^2}\)

Question 19

Which of the following is the correct data structure for solid models?

AOption (a)
solid part → faces → edges → vertices
BOption (b)
solid part → edges → faces → vertices
COption (c)
vertices → edges → faces → solid parts
DOption (d)
vertices → faces → edges → solid parts

View answer & explanation

Correct answer

vertices → edges → faces → solid parts

Why this answer is correct

Explanation:

The correct data structure for modelling solids is vertices → edges → faces → solid parts.
A solid part can be defined by the number and combination of certain features. These features are as follow:
Vertices
Edges
Faces
Edges are the line segments that make up the backbone of 3D parts.
Faces are flat surfaces that are surrounded by edges. A plane or two-dimensional figure represents a face of a solid shape.
The vertices are the points where the edges meet.

Question 20

Which of the following forecasting methods takes a fraction of forecast error into account for the next period forecast?

AOption (a)
simple average method
BOption (b)
moving average method
COption (c)
weighted moving average method
DOption (d)
exponential smoothening method

View answer & explanation

Correct answer

exponential smoothening method

Why this answer is correct

Explanation:

Forecast error (ei) for a period is defined as the difference between actual and actual and forecasted demand.

e_i = Actual demand - Forecast demand ⇒ D_i - F_i

Exponential forecasting:

\({F_T} = {F_{T - 1}} + α ({D_{T - 1}} - {F_{T - 1}})\)

where

F_T is the forecast for the next period

\(({D_{T - 1}} - {F_{T - 1}})\) is the forecast error and

α is the smoothing constant.

Thus exponential smoothing takes into account the forecast error of the previous period, for the forecast of the next period.

Additional Information

Simple Average Method:

In the simple moving average, we take the average of the past data points for future demand.

For 'n' period moving average forecast will be given by:

\({F_{n+1}} = \frac{{{D_1};+;{D_2};+;{D_3};+ ;{D_4};+;\ldots \ldots \ldots \ldots \ldots \ldots ;+;{D_n}}}{n}\)

Weighted Moving Average Method:

In weighted moving average, the highest weightage is given to recent data & it decreases for older data points.

For n period weighted moving average, weightage is as follows:

\(\frac{n}{{{\rm{\Sigma }}n}},;\frac{{n - 1}}{{{\rm{\Sigma }}n}},;\frac{{n - 2}}{{{\rm{\Sigma }}n}}, ;- - - - - - - ,\frac{1}{{{\rm{\Sigma }}n}}\)

\({F_{n+1}} = \left[ ({{w_{1}} \times {D_{1}});+;({w_{2}\times {D_{2}}});+;..........+;({w_{n}} \times {D_{n}}})\right]\)

Question 21

An analytic function of a complex variable z = x + iy is expressed as f (z) = u(x,y) + iv(x, y) where i = √-1 . If u = xy, the expression for v should be

AOption (a)
\(\frac{{{{\left( {x + y} \right)}^2}}}{2} + k\)
BOption (b)
\(\frac{{{x^2} - {y^2}}}{2} + k\)
COption (c)
\(\frac{{{y^2} - {x^2}}}{2} + k\)
DOption (d)
\(\frac{{{{\left( {x - y} \right)}^2}}}{2} + k\)

View answer & explanation

Correct answer

\(\frac{{{y^2} - {x^2}}}{2} + k\)

Why this answer is correct

Concept:

Z – x +iy is a analytic function

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

and u = xy

from the Cauchy – Riemam equation

\(\frac{{\partial u}}{{\partial x}} = \frac{{\partial v}}{{\partial y}}\ and\ \frac{{\partial u}}{{\partial y}} = - \frac{{\partial v}}{{\partial x}}\)

then \(\frac{{\partial u}}{{\partial x}} = y = \frac{{\partial v}}{{\partial y}}\ and\ \frac{{\partial u}}{{\partial y}} = x=\frac{{ - \partial v}}{{\partial x}}\)

Let v(x,y) be the conjugate function of u(x,y)

\(dv = \frac{{\partial v}}{{\partial x}}.dx + \frac{{\partial v}}{{\partial y}}.dy = \left( { - x} \right).dx + ydy\)

integrating both the side \(\smallint dv = - \smallint xdx + \smallint ydy\)

\(\Rightarrow v = - \frac{{{x^2}}}{2} + \frac{{{y^2}}}{2} + k = \frac{{{y^2} - {x^2}}}{2} + k\)

Question 22

The solution of \(x\frac{{dy}}{{dx}} + y = {x^4}\) with the condition \(y\left( 1 \right) = \frac{6}{5}\) is

AOption (a)
\(y = \frac{{{x^4}}}{5} + \frac{1}{x}\)
BOption (b)
\(y = \frac{{4{x^4}}}{5} + \frac{4}{{5x}}\)
COption (c)
\(y = \frac{{{x^4}}}{5} + 1\)
DOption (d)
\(y = \frac{{{x^5}}}{5}\)

View answer & explanation

Correct answer

\(y = \frac{{{x^4}}}{5} + \frac{1}{x}\)

Why this answer is correct

Explanation:

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

\(\frac{{dy}}{{dx}} + \frac{y}{x} = {x^3}\)

This is single order differential equation of:

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

then P = 1/x and Q = x³

now \(IF = {e^{\mathop \smallint pdx}} = {e^{ \smallint \frac{1}{x}dx}} = {e^{\ln x}} = x\)

then solution of equation

⇒ y(IF)= ∫Q(IF)dx + C

\(\begin{array}{l} \Rightarrow y\left( x \right) = \smallint {x^3}.xdx + c\ \Rightarrow yx = \frac{{{x^5}}}{5} + c \end{array}\)

Applying boundary condition y(1) = 6/5

\(\Rightarrow \frac{6}{5}\left( 1 \right) = \frac{{{{\left( 1 \right)}^5}}}{5} + c \Rightarrow C = 1\)

then \(yx = \frac{{{x^5}}}{5} + 1 \Rightarrow y = \frac{{{x^4}}}{5} + \frac{1}{x}\)

Question 23

A path AB in the form of one quarter of a circle of unit radius is shown in the figure. Integration of (x+y)² on path AB traversed in a counter-clockwise sense is

AOption (a)
\(\frac{x}{2} - 1\)
BOption (b)
\(\frac{\pi }{2} + 1\)
COption (c)
\(\frac{\pi }{2}\)
DOption (d)
1

View answer & explanation

Correct answer

\(\frac{\pi }{2} + 1\)

Why this answer is correct

equation of circle ⇒ x² + y² = 1 (radius = 1)

Let x = cosθ , y =sinθ , r = 1

then

\(\begin{array}{l} I = \mathop \smallint \limits_0^{\frac{\pi }{2}} {\left( {cos\theta + sin\theta } \right)^2}d\theta \ = \mathop \smallint \limits_0^{\frac{\pi }{2}} \left( {{{\cos }^2}\theta + {{\sin }^2}\theta + 2cos\theta sin\theta } \right)dx \= \mathop \smallint \limits_0^{\frac{\pi }{2}} \left( {1 + sin2\theta } \right)d\theta \ = \left( {\theta - \frac{{cos2\theta }}{2}} \right)_0^{\frac{\pi }{2}}\ \Rightarrow I = \left( {\frac{\pi }{2} - 0} \right) - \left( {\frac{{cos\pi }}{2} - \frac{{cos0}}{2}} \right) \= \frac{\pi }{2} + 1 \end{array}\)

Question 24

The distance between the origin and the point nearest to it on the surface z² = 1 + xy is

AOption (a)
1
BOption (b)
\(\frac{{\sqrt 3 }}{2}\)
COption (c)
√3
DOption (d)
2

View answer & explanation

Correct answer

Why this answer is correct

Explanation:

Surface z² = 1 + xy ...(i)

Let P(x, y, z) is nearest point to surface (i)

so, distance from origin

\(d = \sqrt {{{\left( {x - 0} \right)}^2} + {{\left( {y - 0} \right)}^2} + {{\left( {z - 0} \right)}^2}}\)

\(\Rightarrow {d^2} = {x^2} + {y^2} + {z^2} \Rightarrow {z^2} = {d^2} - \left( {{x^2} + {y^2}} \right)\) ...(ii)

from (i) and (ii)

1 + xy = d² – x² – y²

⇒ d² = x² + y² + xy + 1 = f (x,y) ...(iii)

Applying maximum and minimum condition to z²:

\(\frac{{\partial f}}{{\partial x}} = 2x + y,\frac{{\partial f}}{{\partial y}} = 2y + x\)

then \(\frac{{\partial f}}{{\partial x}} = 0 = 2x + y,\frac{{\partial f}}{{\partial y}} = 02y + x\)

put x = 0, y = 0 (stationary point)

Now, \(P = \frac{{{\partial ^2}f}}{{\partial {x^2}}} = 2,q = \frac{{{\partial ^2}f}}{{\partial x\partial y}} = 1,r = \frac{{{\partial ^2}f}}{{\partial {y^2}}} = 2\)

or pr – q² = 4 – 1 = 3 > 0 and r = +ve

so f(x,y) is minimum at (0,0)

Hence, minimum value of z² at (0,0)

z² = x² + y² + xy + 1 = (0)² + (0)² + (0)(0) + 1 = 1

Then the nearest point is

z² = 1 + xy = 1+ (0)(0) = 1

or z = ± 1

The distance between the two points = 1

Question 25

The area enclosed between the curves y² = 4x and x² = 4y is

AOption (a)
\(\frac{{16}}{3}\)
BOption (b)
8
COption (c)
\(\frac{{32}}{3}\)
DOption (d)
16

View answer & explanation

Correct answer

\(\frac{{16}}{3}\)

Why this answer is correct

Explanation:

y² = 4x, x² = 4y

or \({y^2} = 4x = 4\sqrt {4y} = 8\sqrt y;\)

⇒ y⁴ – 64y = 0

then y = 0, 4

Similarly put y = 0, 4

In curve x² = 4y

we get x = 4, 0

So, point of intersection are (0,0) and (4,4)

\(\begin{array}{l} Area = \mathop \smallint \limits_0^4\ \mathop \smallint \limits_{2\sqrt x }^{\frac{{{x^2}}}{4}} dydx = \mathop \smallint \limits_0^4 \left[ y \right]_{2\sqrt x }^{\frac{{{x^2}}}{4}}dx\ = \mathop \smallint \limits_0^4 \left[ {\frac{{{x^2}}}{4} - 2\sqrt x } \right]dx = \left[ {\frac{{{x^3}}}{{12}} - \frac{{4{x^{\frac{3}{2}}}}}{3}} \right]_0^4\ = \left| {\frac{{{4^3}}}{{12}} - \frac{{4{{\left( 4 \right)}^{\frac{3}{2}}}}}{3}} \right| = \frac{{16}}{3} \end{array};\)

Question 26

The standard deviation of a uniformly distributed random variable between 0 and 1 is

AOption (a)
\(\frac{1}{{\sqrt {12} }}\)
BOption (b)
\(\frac{1}{{\sqrt 3 }}\)
COption (c)
\(\frac{5}{{\sqrt {12} }}\)
DOption (d)
\(\frac{7}{{\sqrt {12} }}\)

View answer & explanation

Correct answer

\(\frac{1}{{\sqrt {12} }}\)

Why this answer is correct

Concept:

Standard deviation of a uniformly distributed random variable is given by:

F(x) = \(\left{ {\begin{array}{*{20}{c}} {\frac{1}{{\beta - \alpha }}}\ {;else;0;;;} \end{array}} \right.\) If α < x < β,

Variance =\( :\frac{{{{\left( {\beta - \alpha } \right)}^2}}}{{12}}\)

∴ Standard Deviation,

\(;\sigma= \sqrt {\frac{{{{\left( {\beta; - ;\alpha } \right)}^2}}}{{12}};} ;\)

Calculations:

Given:

β = 1, α = 0

Standard deviation of a uniformly distributed random variable is given by:

\(;\sigma= \sqrt {\frac{{{{\left( {\beta; - ;\alpha } \right)}^2}}}{{12}};} ;\)

\(\sigma {\rm{}} = {\rm{}}\sqrt {{\rm{;}}\frac{{{{\left( {1 - 0} \right)}^2}}}{{12}};} \)

\(\sigma {\rm{;}} = {\rm{}}\sqrt {{\rm{}}\frac{1}{{12}};} \)

\(\sigma {\rm{}} = {\rm{}}\frac{1}{{\sqrt {12} }}\)

Additional Information

Mean of a uniformly distributed random variable is given by:

Mean = \(\left(\frac{{\alpha ; + ;\beta }}{2}\right)\)

Question 27

Consider steady, incompressible and irrotational flow through a reducer in a horizontal pipe where the diameter is reduced from 20 cm to 10 cm. The pressure in the 20 cm pipe just upstream of the reducer is 150 kPa. The fluid has a vapor pressure of 50 kPa and a specific weight of 5 kN/m³. Neglecting frictional effects, the maximum discharge (in m³/s) that can pass through the reducer without causing cavitation is

AOption (a)
0.05
BOption (b)
0.16
COption (c)
0.27
DOption (d)
0.38

View answer & explanation

Correct answer

0.16

Why this answer is correct

Concept:

The reducer is shown in the figure.

At both ends of reducer, applying bernaulli’s equation,

\(\frac{{{P_1}}}{{\rho g}} + \frac{{{V_1}^2}}{{2g}} + {z_1} = \frac{{{P_2}}}{{\rho g}} + \frac{{{V_2}^2}}{{2g}} + {z_2}\)

And applying continuity equation,

A₁V₁ = A₂V₂

Calculation:

Given:

d₁ = 20 cm, d₂ = 10 cm, P₁ = 150 kPa, P₂ = 50 kPa, w = 5 kN/m² = ρg

Applying continuity equation

A₁V₁ = A₂V₂

_{\(\frac{\pi }{4}{d_1}^2 \times {V_1} = \frac{\pi }{4}{d_2}^2 \times {V_2}\)}

\(\frac{\pi }{4}{20^2} \times {V_1} = \frac{\pi }{4}{10^2} \times {V_2}\)

⇒ V₂ = 4V₁ ________________(1)

For maximum discharge P_V= P₂= 50 kPa

Applying bernaulli’s equation

\(\frac{{{P_1}}}{{\rho g}} + \frac{{{V_1}^2}}{{2g}} + {z_1} = \frac{{{P_2}}}{{\rho g}} + \frac{{{V_2}^2}}{{2g}} + {z_2}\)

and z₁ = z₂ , w = ρg = 5 kN/m²

then \(\frac{{150}}{5} + \frac{{V_1^2}}{{2g}} = \frac{{50}}{5} + \frac{{{{\left( {4{V_1}} \right)}^2}}}{{2g}} \Rightarrow {V_1} = 5.114\ m/sec\)

then V₂ = 4V₁ = 4 × 5.114 = 20.46 m/sec

Now, maximum discharge \({Q_{max}} = {A_2}{V_2} = \frac{\pi }{4}d_2^2{V_2}\)

\(= \frac{\pi }{4}{\left( {10 \times {{10}^{ - 2}}} \right)^2} \times 20.46 = 0.16~{m^3}/sec\)

Question 28

In a parallel flow heat exchanger operating under steady state, the heat capacity rates (product of specific heat at constant pressure and mass flow rate) of the hot and cold fluid are equal. The hot fluid, flowing at 1 kg/s with C_p = 4 kJ/kgK, enters the heat exchanger at 102°C while the cold fluid has an inlet temperature of 15°C. The overall heat transfer coefficient for the heat exchanger is estimated to be 1 kW/m²K and the corresponding heat transfer surface area is 5 m². Neglect heat transfer between the heat exchanger and the ambient. The heat exchanger is characterized by the following relation:

2ϵ = 1 – exp (−2NTU) .

The exit temperature (in °C) for the cold fluid is

AOption (a)
45
BOption (b)
55
COption (c)
65
DOption (d)
75

View answer & explanation

Correct answer

Why this answer is correct

Heat capacity

\({\dot C_h} = {\dot M_h}{C_{ph}} = 1 \times 4 = 4\frac{{kJ}}{{sec - k}}\)

And \({\dot C_h} = {\dot C_c}\)

Also effectiveness \(\epsilon = \frac{{1 - {e^{ - NTU\left( {1 + R} \right)}}}}{{1 + R}}\) _____________(i)

Where \(R = \frac{{{C_c}}}{{{C_h}}} = \frac{{{C_{min}}}}{{{C_{max}}}} = 1\)

Also from energy balance \({C_h}\left( {{T_{{h_1}}} - {T_{{h_2}}}} \right) = {C_c}\left( {{T_c}_2 - {T_{{c_1}}}} \right)\)

\(\Rightarrow \frac{{{C_c}}}{{{C_h}}} = \frac{{{T_{{h_1}}} - {T_{{h_2}}}}}{{{T_{{c_2}}} - {T_{{c_1}}}}} = 1 \Rightarrow {T_{{h_1}}} - {T_{{h_2}}} = {T_{{c_2}}} - {T_{{c_1}}}\)

And \(NTU = \frac{{UA}}{{{C_{min}}}} = \frac{{1 \times 5}}{4} = 1.25\)

Then from equation (1), we get

\(\epsilon = \frac{{1 - {\rm{exp}}\left( { - 1.25 \times 2} \right)}}{2} = 0.46\)

The maximum possible heat transfer

\({Q_{max}} = {C_{min}}\left( {{T_h}_1 - {T_{{c_1}}}} \right) = 4\left( {102 - 15} \right) = 348\ kW\)

Actual transfer Q_act = ϵ.Q_max = 0.46 × 348

\(\begin{array}{l} \Rightarrow {C_c}\left( {{T_{{c_2}}} - {T_{{c_1}}}} \right) = 0.46 \times 348\ \Rightarrow {T_{{c_2}}} = \frac{{0.46 \times 348}}{4} + 15 = 55^\circ C \end{array}\)

Question 29

In an air-standard Otto cycle, the compression ratio is 10. The condition at the beginning of the compression process is 100 kPa and 27 °C. Heat added at constant volume is 1500 kJ/kg, while 700 kJ/kg of heat is rejected during the other constant volume process in the cycle. The specific gas constant for air is 0.287 kJ/kg-K. The mean effective pressure (in kPa) of the cycle is:

AOption (a)
103
BOption (b)
310
COption (c)
515
DOption (d)
1032

View answer & explanation

Correct answer

1032

Why this answer is correct

Concept:

Otto Cycle

Compression ratio \(r=\frac{V_1}{V_2}\)

Swept Volume \(V_s=V_1-V_2⇒ V_2(r-1)\)

W_net = Q_s - Q_r

where Q_s and Q_r is heat supplied and rejected and constant volume process.

Mean effective pressure (P_m):

The Mean Effective Pressure of the otto cycle is given by,

\(P_m = \frac{{Net~work~output}}{{Swept~volume}}=\frac{W_{net}}{V_s}⇒ \frac{W_{net}}{V_2(r;-;1)}\)

Calculation:

Given:

r = 10, P₁ = 100 kPa, T₁ = 27 °C ⇒ 300 K, Qs = 1500 kJ/kg, Qr = 700 kJ/kg and R = 0.287 kJ/kg-K

\({P_1}{V_1} = R{T_1} ⇒ {V_1} = \frac{{R{T_1}}}{{{P_1}}} \Rightarrow \frac{{0.287;\times;300}}{{100}} = 0.861;m^3/kg\)

\(r = \frac{{{V_1}}}{{{V_2}}} = 10 ⇒ V_1 = 10{V_2}\)

\({V_2}= \frac{V_1}{{10}} = 0.0861;m^3/kg\)

\({W_{net}} = {Q_s} - {Q_r} = 1500 - 700 = 800;kJ/kg\)

Now Mean Effective Pressure (Pm):

\({P_m} = \frac{{{W_{net}}}}{{{V_2(r;-;1)}}} = \frac{{800}}{{0.0861\left( {10 ;- ;1} \right)}} = 1032.4\ kPa\)

Question 30

An irreversible heat engine extracts heat from a high temperature source at a rate of 100 kW and rejects heat to a sink at a rate of 50 kW. The entire work output of the heat engine is used to drive a reversible heat pump operating between a set of independent isothermal heat reservoirs at 17°C and 75°C. The rate (in kW) at which the heat pump delivers heat to its high temperature sink is

AOption (a)
50
BOption (b)
250
COption (c)
300
DOption (d)
360

View answer & explanation

Correct answer

300

Why this answer is correct

Concept:

\({\left( {COP} \right)_{H.P}} = \frac{{{Q_3}}}{{{Q_3}~- ~{Q_4}}} = \frac{{{Q_3}}}{W_{in}}\)

for a reversible process

\(\frac{{{Q_3}}}{{{T_3}}} = \frac{{{Q_4}}}{{{T_4}}} ⇒ \frac{{{Q_3}}}{{{Q_4}}} = \frac{{{T_3}}}{{{T_4}}}\)

then \({\left( {COP} \right)_{HP}} = \frac{{{T_3}}}{{{T_3}; -; {T_4}}} = \frac{{{Q_3}}}{W}\)

Calculation:

Given:

Q₁ = 100 kW, Q₂ = 50 kW, T₃ = 75 °C = 348 K and T₄ = 17 °C = 290 K

W = Q₁ - Q₂ = 100 - 50 ⇒ 50 kW.

\(⇒ {Q_3} = W\left( {\frac{{{T_3}}}{{{T_3}; -; {T_4}}}} \right) = 50\left( {\frac{{348}}{{348 ;- ;290}}} \right) = 300\ kW\)

Question 31

You are asked to evaluate assorted fluid flows for their suitability in a given laboratory application. The following three flow choices, expressed in terms of the two-dimensional velocity fields in the xy – plane, are made available.

P. u = 2y, v = – 3x

Q. u = 3xy, v = 0

R. u = – 2x, v = 2y

Which flow(s) should be recommended when the application requires the flow to be incompressible and irrotational?

AOption (a)
P and R
BOption (b)
Q
COption (c)
Q and R
DOption (d)
R

View answer & explanation

Correct answer

Why this answer is correct

Concept:

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

foe irrotational fluids \({{\rm{\Sigma }}_z} = 0 = \frac{1}{2}\left( {\frac{{\partial v}}{{\partial x}} - \frac{{\partial u}}{{\partial y}}} \right) \Rightarrow \frac{{\partial v}}{{\partial x}} - \frac{{\partial u}}{{\partial y}} = 0\)

Calculation:

For P:

\(u = 2y \Rightarrow \frac{{\partial u}}{{\partial x}} = 0,\frac{{\partial u}}{{\partial y}} = 2\)

\(v = - 3x \Rightarrow \frac{{\partial v}}{{\partial x}} = -3,\frac{{\partial v}}{{\partial y}} = 0\)

\(\frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} = 0 \Rightarrow 0 + 0 = 0\) (Flow is incompressible)

Likewise and \(\frac{{\partial v}}{{\partial x}} - \frac{{\partial u}}{{\partial y}} = - 5 \ne 0\) (Rotational)

For Q:

\(u = 3xy \Rightarrow \frac{{\partial u}}{{\partial x}} = 3y,\frac{{\partial u}}{{\partial y}} = 3x\)

and \(v = 0 \Rightarrow \frac{{\partial v}}{{\partial y}} = \frac{{\partial u}}{{\partial y}} = 0\)

hence \(\frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} = 3y \ne 0\) (compressible flow)

and \(\frac{{\partial v}}{{\partial x}} - \frac{{\partial u}}{{\partial y}} = 0 \Rightarrow 0 - 3x = - 3x \ne 0\) (Rotational)

For R:

\(u = - 2x \Rightarrow \frac{{\partial u}}{{\partial x}} = - 2,\frac{{\partial u}}{{\partial y}} = 0\)

\(v = 2y \Rightarrow \frac{{\partial v}}{{\partial y}} = 2,\frac{{\partial u}}{{\partial x}} = 0\)

then \(\frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} = 0 \Rightarrow - 2 + 2 = 0\) (incompressible)

and \(\frac{{\partial v}}{{\partial x}} - \frac{{\partial u}}{{\partial y}} = 0 \Rightarrow 0 - 0 = 0\) (Irrotational flow)

Question 32

Water at 25°C is flowing through a 1 Km long pipe of 200 mm diameter at the rate of 0.07 m³/s. If the value of Darcy friction factor for this pipe is 0.02 and density of water is 1000 kg/m³, the pumping power (in KW) required to maintain the flow is

AOption (a)
1.8
BOption (b)
17.4
COption (c)
20.5
DOption (d)
41.0

View answer & explanation

Correct answer

17.4

Why this answer is correct

Concept:

Head Loss \({h_f} = \frac{{fl{V^2}}}{{2gD}} = \frac{{fL}}{{2gD}}{\left( {\frac{Q}{A}} \right)^2}\)

Calculation:

Given:

f = 0.02, L = 1000 kg/m³, Q = 0.07 m³/s, D = 200 mm.

\(\Rightarrow {h_f} = \frac{{0.02 \times 1000 \times {{\left( {0.07} \right)}^2}}}{{2 \times 9.81 \times 0.2 \times {{\left[ {\frac{\pi }{4}{{\left( {0.2} \right)}^2}} \right]}^2}}} = 25.30\ m\)

Then power P = ρgQ × h_f = 1000 × 9.81 × 0.07 × 25.30

P = 17.37 kW

Question 33

Consider steady-state heat conduction across the thickness in a plane composite wall (as shown in the figure) exposed to convection conditions on both sides.

Given

h_i = 20 W / m²K; h₀ = 50 W / m²K; T_∞,i = 20°C; T_∞,0 = - 2°C ; k₁ = 20 W / mk; k₂ = 50 W / mk; L₁ = 0.30 m and L₂ = 0.15 m

Assuming negligible contact resistance between the wall surfaces, the interface temperature, T (in °C), of the two walls will be

AOption (a)
-0.50
BOption (b)
2.75
COption (c)
3.76
DOption (d)
4.50

View answer & explanation

Correct answer

3.76

Why this answer is correct

Concept:

The equivalent diagram is given by

Under the steady-state condition, Conduction-Convection heat transfer rate across 1^st slab will be equal to Conduction-Convection heat transfer rate across 2^nd slab, i.e.,

\(\Rightarrow\frac{{{T_∞ }_i - T}}{{\frac{1}{{{h_1}}} + \frac{{{L_1}}}{{{k_1}}}}} = \frac{{T - {T_∞ }_o}}{{\frac{1}{{{h_2}}} + \frac{{{L_2}}}{{{k_2}}}}}\)

Calculation:

Given:

hi = 20 W / m2K, h0 = 50 W / m2K, T∞,i = 20°C, T∞,0 = - 2°C, k1 = 20 W/mK, k2 = 50 W/mK, L1 = 0.30 m and L2 = 0.15 m

Therefore, \(\frac{{20 - T}}{{\frac{1}{{20}} + \frac{{0.3}}{{20}}}} = \frac{{T + 2}}{{\frac{1}{{50}} + \frac{{0.15}}{{50}}}}\)

⇒ \(\frac{(20-T)20}{1.3}=\frac{(T+2)50}{1.15}\)

T = 3.75°C

Question 34

The velocity profile of a fully developed laminar flow in a straight circular pipe, as shown in the figure, is given by the expression \(u\left( r \right) = \frac{{ - {R^2}}}{{4\mu }}\left( {\frac{{dp}}{{dx}}} \right)\left( {1 - \frac{{{r^2}}}{{{R^2}}}} \right)\) , where \(\frac{dp}{dx}\) is a constant. The average velocity if fluid in the pipe is

AOption (a)
\(-\frac{{{R^2}}}{{8\mu}}\left( {\frac{{dp}}{{dx}}} \right)\)
BOption (b)
\(- \frac{{{R^2}}}{{4\mu}}\left( {\frac{{dp}}{{dx}}} \right)\)
COption (c)
\(- \frac{{{R^2}}}{{2\mu}}\left( {\frac{{dp}}{{dx}}} \right)\)
DOption (d)
\(-\frac{{{R^2}}}{\mu}\left( {\frac{{dp}}{{dx}}} \right)\)

View answer & explanation

Correct answer

\(-\frac{{{R^2}}}{{8\mu}}\left( {\frac{{dp}}{{dx}}} \right)\)

Why this answer is correct

\(u\left( r \right) = \frac{{ - {R^2}}}{{4\mu }}\left( {\frac{{dp}}{{dx}}} \right)\left( {1 - \frac{{{r^2}}}{{{R^2}}}} \right)\)

Velocity profile in fully developed laminar how in pipe is parabolic with a maximum at center line & minimum at pipe wall then average velocity

\(\begin{array}{l} {V_{avg}} = {\frac1{\pi R^2}}\mathop \smallint \limits_0^R u\left( r \right)dr = \frac{{ - 1}}{{{\pi R^2}}}\mathop \smallint \limits_0^R \frac{{{R^2}}}{{4\mu }}\left( {\frac{{dp}}{{dx}}} \right){\left( {1 - \frac{{{r^2}}}{{{R^2}}}} \right)2\pi r{dr}}\ = - \frac{1}{{2\mu }}\left( {\frac{{dp}}{{dx}}} \right)\mathop \smallint \limits_0^R \left( {r - \frac{{{r^3}}}{{{R^2}}}} \right)dr\ = - \frac{1}{{2\mu }}\left( {\frac{{dp}}{{dx}}} \right)\left[ {\frac{{{r^2}}}{2} - \frac{{{r^4}}}{{4{R^2}}}} \right]_0^R = - \frac{1}{{2\mu }}\left( {\frac{{dp}}{{dx}}} \right)\left[ {\frac{{{R^2}}}{2} - \frac{{{R^4}}}{{4{R^2}}}} \right]\ = - \frac{{{R^2}}}{{8\mu }}\left( {\frac{{dp}}{{dx}}} \right) \end{array}\)

Question 35

A solid shaft of diameter d and length L is fixed at both the ends. A torque, T₀ is applied at a distance, L/4 from the left end as shown in the figure given below

The maximum shear stress in the shaft is

AOption (a)
\(\frac{{16{T_0}}}{{\pi {d^3}}}\)
BOption (b)
\(\frac{{12{T_0}}}{{\pi {d^3}}}\)
COption (c)
\(\frac{{8{T_0}}}{{\pi {d^3}}}\)
DOption (d)
\(\frac{{4{T_0}}}{{\pi {d^3}}}\)

View answer & explanation

Correct answer

\(\frac{{12{T_0}}}{{\pi {d^3}}}\)

Why this answer is correct

T₀ = T₁ + T₂ _________________(i)

but from torsional equation

\(\frac{T}{J} = \frac{\tau }{r} = \frac{{G\theta }}{l} \Rightarrow T = \frac{{GJ\theta }}{l}\)

or \(T \propto \frac{1}{l} \Rightarrow {T_1}{l_1} = {T_2}{l_2}\)

\(\Rightarrow {T_1} \times \frac{L}{4} = {T_2} \times \frac{{3L}}{4} \Rightarrow {T_1} = 3{T_2}\)

from equation (i) \({T_0} = 3{T_2} + {T_2} = 4{T_2} \Rightarrow {T_2} = \frac{{{T_0}}}{4}\)

and then T₁ = 3T₀/4

hence T₁ > T₂

So, maximum shear stress is developed due to T1

\(\frac{{{T_1}}}{J} = \frac{\tau _{max}}{r} \Rightarrow {\tau _{max}} = \frac{{{T_1}}}{J}r\)

\(\Rightarrow {\tau _{max}} = \frac{{\frac{{3{T_0}}}{4}}}{{\frac{\pi }{{32}}{d^4}}} \times \frac{d}{2} = \frac{{32 \times 3{T_0}}}{{8\pi \times {d^3}}} = \frac{{12{T_0}}}{{\pi {d^3}}}\)

Question 36

An epicyclic gear train is shown schematically in the adjacent figure. The sun gear 2 on the input shaft is a 20 teeth external gear. The planet gear 3 is a 40 teeth external gear. The ring gear 5 is a 100 teeth internal gear. The ring gear 5 is fixed and the gear 2 is rotating at 60 rpm ccw (ccw = counter – clockwise and cw = clockwise) The arm 4 attached to the output shaft will rotate at

AOption (a)
10 rpm ccw
BOption (b)
10 rpm cw
COption (c)
12 rpm cw
DOption (d)
12 rpm ccw

View answer & explanation

Correct answer

10 rpm ccw

Why this answer is correct

T₂ = 20, T₃ = 40

T₅ = 100, N₅ = 0, N₂ = 60 (CW)

Taking CCW as + ve

No.	Condition & Motion	Revolution of Elements
Sun gear 2	Planet gear 3	Arm 4	Ring gear 5
1.	Arm-fixed and gear 2 rates 1 rpm (CCW)	1	\(- \frac{{{T_2}}}{{{T_3}}}\)	0	\(- \frac{{{T_2}}}{{{T_3}}} \times \frac{{{T_3}}}{{{T_5}}}\)
2.	+x rpm to gear 2	+x	\(- x\frac{{{T_2}}}{{{T_3}}}\)	0	\(- x\frac{{{T_2}}}{{{T_5}}}\)
3.	add + y rev to all elements	+y	+y	+y	+y
4.	Total Motion	x+y	\(y + x\frac{{{T_2}}}{{{T_3}}}\)	y	\(y - x\frac{{{T_2}}}{{{T_5}}}\)

\({N_5} = 0 = y - x\frac{{{T_2}}}{{{T_5}}} \Rightarrow y - x\left( {\frac{{20}}{{100}}} \right) = 0 = \Rightarrow y = \frac{x}{5}\)

and x + y = 60

or \(x + \frac{x}{5} = 60 \Rightarrow x = 50\ rpm\)

then y =10 rpm = N₄ (CCW)

Question 37

A forged steel link with uniform diameter of 30 mm at the centre is subjected to an axial force that varies from 40 kN in compression to 160 kN in tension. The tensile (S_u), yield (S_y) and corrected endurance (S_e) strengths of the steel material are 600 MPa, 420 MPa and 240 MPa respectively. The factor of safety against fatigue endurance as per Soderberg’s criterion is

AOption (a)
1.26
BOption (b)
1.37
COption (c)
1.45
DOption (d)
2.00

View answer & explanation

Correct answer

1.26

Why this answer is correct

σ_u = 600 MPa, σ_y = 420 MPa, σ_e = 240 MPa

d = 30 mm, Fmax = 160 kN (T), fmin = -40kN (c)

Maximum stress \(= {\sigma _{max}} = \frac{{{F_{max}}}}{A} = \frac{{160 \times {{10}^3}}}{{\frac{\pi }{4}{{\left( {30} \right)}^2}}} = 226.47\ MPa\)

Minimum stress \(= {\sigma _{min}} = \frac{{{F_{min}}}}{A} = \frac{{ - 40 \times {{10}^3}}}{{\frac{\pi }{4}{{\left( {30} \right)}^2}}} = - 56.62\ MPa\)

Mean stress, \({\sigma _m} = \frac{{{\sigma _{max}} + {\sigma _{min}}}}{2} = \frac{{226.47 - 56.62}}{2} = 84.925\ MPa\)

Variable stress \({\sigma _v} = \frac{{{\sigma _{max}} - {\sigma _{min}}}}{2} = \frac{{226.47 - \left( {-56.62} \right)}}{2} = 141.545\ MPa\)

From soderberg’s criterion \(\frac{1}{{Fs}} = \frac{{{\sigma _m}}}{{{\sigma _y}}} + \frac{{{\sigma _v}}}{{{\sigma _e}}}\)

\(\Rightarrow \frac{1}{{Fs}} = \frac{{84.925}}{{420}} + \frac{{141.545}}{{240}} = 0.791\)

⇒ FS = 1.26

Question 38

An automotive engine weighing 240 kg is supported on four springs with linear characteristics. Each of the front two springs have a stiffness of 16 MN/m while the stiffness of each rear spring is 32 MN/m. The engine speed (in rpm), at which resonance is likely to occur, is

AOption (a)
6040
BOption (b)
3020
COption (c)
1424
DOption (d)
955

View answer & explanation

Correct answer

6040

Why this answer is correct

k₁ = k₂ = 16 MN/m

k₃ = k₄ = 32 MN/m, m = 240 kg

springs are parallel, so \({k_{eq}} = {k_1} + {k_2} + {k_3} + {k_4} = 16 + 16 + 32 + 32 \Rightarrow {K_{eq}} = 96\frac{{MN}}{m}\)

we know that at resonance \(\omega = {\omega _n} = \sqrt {\frac{k}{m}} \Rightarrow \frac{{2\pi N}}{{60}} = \sqrt {\frac{{{k_{eq}}}}{m}}\)

\(\Rightarrow N = \frac{{60}}{{2\pi }}\times \sqrt {\frac{{96 \times {{10}^6}}}{{240}}} = 6039.5 \simeq 6040\ rpm\)

Question 39

A vehicle suspension system consists of a spring and a damper. The stiffness of the spring is 3.6 kN/m and the damping constant of the damper is 400 Ns/m. If the mass is 50 kg, then the damping factor ζ and damped natural frequency (f_d), respectively, are

AOption (a)
0.471 and 1.19 Hz
BOption (b)
0.471 and 7.48 Hz
COption (c)
0.666 and 1.35 Hz
DOption (d)
0.666 and 8.50 Hz

View answer & explanation

Correct answer

0.471 and 1.19 Hz

Why this answer is correct

Concept:

Then damped natural frequency is given by \({\omega _d} = \left( {\sqrt {1 - {\zeta ^2}} } \right){\omega _n}\)

Where ζ is the damping factor and ω_n is the natural frequency of the system

Natural Frequency \({\omega _n} = \sqrt {\frac{k}{m}} \)

damping factor \(\zeta = \frac{C}{{{C_c}}} = \frac{C}{{2\sqrt {km} }}\)

Where C is the damping constant, k is the stiffness of the spring and m is mass of the system

Calculation:

Given, k = 3.6 kN/m = 3.6 ×10³ N/m, C = 400 Ns/m, m = 50 kg

Natural frequency \({\omega _n} = \sqrt {\frac{k}{m}} = \sqrt {\frac{{3.6 \times {{10}^3}}}{{50}}} = 8.485;rad/s\)

and damping factor \(ζ = \frac{C}{{{C_c}}} = \frac{C}{{2\sqrt {km} }} = \frac{{400}}{{2\sqrt {3.6 \times {{10}^3} \times 50} }} = 0.471\)

Then damped natural frequency \({\omega _d} = \left( {\sqrt {1 - {ζ ^2}} } \right){\omega _n} = \left( {\sqrt {1 - {{0.471}^2}} } \right)8.485 = 7.48\)

\({f_d} = \frac{{{\omega _d}}}{{2\pi }} = \frac{{7.48}}{{2\pi }} = 1.19;Hz\)

Question 40

A frame of two arms of equal length L is shown in the adjacent figure. The flexural rigidity of each arm of the frame is EI. The vertical deflection at the point of application of load P is

AOption (a)
\(\frac{{P{L^3}}}{{3EI}}\)
BOption (b)
\(\frac{{2P{L^3}}}{{3EI}}\)
COption (c)
\(\frac{{P{L^3}}}{{EI}}\)
DOption (d)
\(\frac{{4P{L^3}}}{{3EI}}\)

View answer & explanation

Correct answer

\(\frac{{4P{L^3}}}{{3EI}}\)

Why this answer is correct

Concept:

\(\Rightarrow \)

The strain energy in BC due to point load P

Strain energy in bar BC is \({U_{BC}} = \smallint \frac{{{M^2}dx}}{{2EI}}\)

M = Px

\({U_{BC}} = \smallint_0^L \frac{{{M^2}dx}}{{2EI}}\)

\({U_{BC}} = \smallint_0^L \frac{{{P^2x^2}dx}}{{2EI}}\)

\({U_{BC}} =\frac{P^2}{2EI} \smallint_0^L{{{x^2}dx}}{{}}\)

\({U_{BC}} =\frac{P^2}{2EI} \left[{\frac{x^3}{3}}\right]_0^L\)

\({U_{BC}} =\frac{P^2}{2EI} \left[{\frac{L^3}{3}}\right]=\frac{P^2L^3}{6EI}\)

The vertical deflection of point 'C' is due to point load P on C and due to the rotation of joint 'B'

Strain energy in bar AB is \({U_{AB}} = \smallint \frac{{{M^2}dx}}{{2EI}}\)

M = PL

\({U_{AB}} = \smallint_0^L \frac{{{M^2}dx}}{{2EI}}\)

\({U_{AB}} = \smallint_0^L \frac{{{(PL)^2}dx}}{{2EI}}\)

\({U_{AB}} =\frac{P^2L^3}{2EI}\)

U = U_AB + U_BC

\({U} =\frac{P^2L^3}{6EI}+ \frac{P^2L^3}{2EI} = \frac{2PL^3}{3EI}\)

\(\frac{\partial U}{\partial P} =\frac{2\times 2PL^3}{3EI}=\frac{4PL^3}{3EI}\)

Question 41

A uniform rigid rod of mass M and length L is hinged at one end as shown in the adjacent figure. A force P is applied at a distance of 2L/3 from the hinge so that the rod swings to the right. The horizontal reaction at the hinge is

AOption (a)
–P
BOption (b)
0
COption (c)
P/3
DOption (d)
2P/3

View answer & explanation

Correct answer

Why this answer is correct

Explanation:

When rod swings to right, Linear acceleration & angular acceleration (∝) comes into picture

Let R = reaction at hinge

Linear acceleration

\(a = \propto r = \frac{L}{2} \times \propto \Rightarrow \propto = \frac{{2a}}{L}\)

and \({\rm{\Sigma }}{M_G} = {I_G} \times \propto\)

\(R\left( {\frac{L}{2}} \right) + P\left( {\frac{L}{6}} \right) = \frac{{M{L^2}}}{{12}}\left( {\frac{{2a}}{L}} \right)\ \Rightarrow a = \frac{{3R}}{M} + \frac{P}{M}\) ______________(1)

\(P-R = Ma = M\left( {\frac{{3R}}{M} + \frac{P}{M}} \right)\)

⇒ P – R = 3R + P ⇒ R = 0

Question 42

Match the approaches given below to perform stated kinematics / dynamics analysis of machine.

Analysis	Approach
P. Continuous relative rotation	1. D’ Alembert’s principle
Q. Velocity and acceleration	2. Grubler’s criterion
R. Mobility	3. Grashoff’s law
S. Dynamic-static analysis	4. Kennedy’s theorem

AOption (a)
P – 1, Q – 2, R – 3, S – 4
BOption (b)
P – 3, Q – 4, R – 2, S –1
COption (c)
P – 2, Q – 3, R – 4, S –1
DOption (d)
P – 4, Q – 2, R – 1, S –3

View answer & explanation

Correct answer

P – 3, Q – 4, R – 2, S –1

Why this answer is correct

The Grubler’s criterion applies to mechanisms with only single degree of freedom joints where the overall movability of the mechanism is unity. 3L-2j-4=0

According to Grashof ’s law for a four bar mechanism,the sum of the shortest and longest link lengths should not be greater than the sum of the remaining two link lengths if there is to be continuous relative motion between the two links. S+L≤P+Q

The Aronhold Kennedy’s theorem states that if three bodies move relatively to each other,they have three instantaneous centres and lie on a straight line. This is used for calculating velocity and acceleration of the mechanism.

D-Alembert’s principle states that the resultant force acting on a body together with the reversed effective force (or inertia force), are in equilibrium.This principle is used to reduce a dynamic problem into an equivalent static problem.

Continuous relative motion ⇒ Grashoff law

Velocity and acceleration → Kennedy’s theorem

Mobility → Grubler’s criterion

Dynamic static Analysis → D’Alembert’s Principle

Question 43

A company uses 2555 units of an item annually. Delivery lead time is 8 days. The reorder point (in number of units) to achieve optimum inventory is

AOption (a)
7
BOption (b)
8
COption (c)
56
DOption (d)
60

View answer & explanation

Correct answer

Why this answer is correct

Concepts:

Lead Time:

The time gap between the placing of an order and its actual arrival in the inventory is known as lead time.
Lead time can be greater, less, or equal to the order cycle.

Order Cycle:

The time period between two successive orders is called Order Cycle.

Re-order Level (ROL):

The quantity in hand while placing the order.
ROL = Lead Time × Demand.

[NOTE: If lead time is in days then demand should be units/day]

Calculation:

Given:

D = 2555 units/year, T_L = 8 days ⇒ 8/365 year.

ROL = Lead Time × Demand

\(ROL=2555×\frac{8}{365}\Rightarrow56;units\)

Additional Information

When lead time T_L is greater than cycle time T, then:

ROL = (T_L - T) × D

Question 44

Consider the following Linear Programming Problem (LPP):

Maximize Z = 3x₁ + 2x₂ Subject to

x₁ ≤ 4

x₂ ≤ 6

3x₁ + 2x₂ ≤ 18

x₁ ≥ 0, x₂ ≥ 0

AOption (a)
The LPP has a unique optimal solution
BOption (b)
The LPP is infeasible
COption (c)
The LPP is unbounded
DOption (d)
The LPP has multiple optimal solutions

View answer & explanation

Correct answer

The LPP has multiple optimal solutions

Why this answer is correct

Explanation:

Z_max = 3x₁ + 2x₂

Subjected to

x₁ ≤ 4 ___________(i)

x₂ ≤ 6 ___________(ii)

3x₁ + 2x₂ ≤ 18 ___________(iii)

and x₁, x₂ ≥ 0

Now the intersections of the lines are denoted by E, F

for E, 3x₁+ 2x₂ = 18

and x₂ = 6

So, 3x₁ + (2 × 6) = 18 ⇒ x₁ = 2 ⇒ E (2,6)

For F, 3x₁ + 2x₂ = 18

and x₁ = 4

⇒ (3 × 4) + 2x₂ = 18 ⇒ x₂ = 3 ⇒ F (4,3)

So the feasible points are (0, 0), (4, 0), (4, 3), (2, 6), (0, 6)

Z(0, 0) = 0

Z(4, 0) = 12

Z(4, 3) = (3 × 4) + (2 × 3) = 18

Z(2, 6) = (3 × 2) + (2 × 6) = 18

Z(0, 6) = 12

So the objective function has multiple solution.

Question 45

Six jobs arrived in a sequence as given below:

Jobs	Processing Time (days)
I	4
II	9
III	5
IV	10
V	6
VI	8

Average flow time (in days) for the above jobs using Shortest Processing Time rule is

AOption (a)
20.83
BOption (b)
23.16
COption (c)
125.00
DOption (d)
139.00

View answer & explanation

Correct answer

20.83

Why this answer is correct

by SPT rule job sequence is

I	III	V	VI	II	IV

Jobs	Process Time	Flow Time
I	4	4
III	5	9
V	6	15
VI	8	23
II	9	32
IV	10	42

Total job flow tine = 4 + 9 + 15 +23 + 32 + 42 = 125

Average flow time \(= \frac{{125}}{6} = 20.83\ days\)

Question 46

Minimum shear strain in orthogonal turning with a cutting tool of zero rake angle is

AOption (a)
0.0
BOption (b)
0.5
COption (c)
1.0
DOption (d)
2.0

View answer & explanation

Correct answer

2.0

Why this answer is correct

Concept:

Shear strain during turning is given by

\(s=cot\ \phi +tan\ (\phi-α)\)

Calculation:

Given:

α = 0°

Shear strain

\(∴ s=cot\ \phi +tan\ \phi\)

for minimum value, \({\frac{ds}{d\phi}} = 0\)

\(\Rightarrow - cose{c^2}\ \phi + {\sec ^2}\ \phi = 0 \Rightarrow - \frac{1}{{{{\sin }^2}\phi}} + \frac{1}{{{{\cos }^2}\phi}} = 0 \Rightarrow {\cos ^2}\phi - {\sin ^2}\phi = 0 \Rightarrow {\tan ^2}\phi = 1 \Rightarrow \phi= \frac{{\rm{\pi }}}{4}\)

again \(\frac{{{d^2}s}}{{d\phi^2}} = - 2cosec\ \phi \times \left( { - {\rm{cosec\ \phi \times cot\ \phi}}} \right) + 2\sec\ \phi\times \left( {\sec\ \phi \tan\ \phi } \right)\)

\(At\ \phi= \frac{{\rm{\pi }}}{4},\)

\(∴\frac{{{d^2}s}}{{{d\phi^2}}} = 2cose{c^2}\left( {\frac{\pi }{4}} \right)\cot \left( {\frac{\pi }{4}} \right) + 2\ {\sec ^2}\frac{\pi }{4}\ \tan \frac{\pi }{4}\)

∴ (2 × 2 × 1) + (2 × 2 × 1) = 8 > 0

So, minimum value at \(\phi = \frac{\pi }{4}\)

\(\therefore{s_{min}} = \cot \frac{\pi }{4} + \tan \frac{\pi }{4} = 1 + 1 = 2\)

Question 47

Electrochemical machining is performed to remove material from an iron surface of 20mm × 20mm under the following conditions:

Inter electrode gap = 0.2mm

Supply voltage DC = 12V

Specific resistance of electrolyte = 2Ω cm

Atomic weight of Iron = 55.85

Valency of Iron = 2

Faraday 's constant = 96540 Coulombs

The material removal rate (in g/s) is

AOption (a)
0.3471
BOption (b)
3.471
COption (c)
34.71
DOption (d)
347.1

View answer & explanation

Correct answer

0.3471

Why this answer is correct

We know that \(R = \frac{{PL}}{A} = \frac{{2 \times 10 \times 0.2}}{{20 \times 20}} = 0.01{\rm{\Omega }}\)

then \(I = \frac{V}{R} = \frac{{12}}{{0.01}} = 1200\ Amp\)

So, Rate or removal in \(= \frac{I}{F} \times \frac{z}{v} = \frac{{1200}}{{96540}} \times \frac{{55.85}}{2}\)

\(\Rightarrow \dot m = 0.3471\ g/s\)

Question 48

Match the following

NC Code	Definition
P. M05	1. Absolute coordinate system
Q. G01	2. Dwell
R. G04	3. Spindle stop
S. G90	4. Linear interpolation

AOption (a)
P – 2, Q – 3, R – 4, S – 1
BOption (b)
P – 3, Q – 4, R – 1, S – 2
COption (c)
P – 3, Q – 4, R – 2, S –1
DOption (d)
P – 4, Q – 3, R – 2, S – 1

View answer & explanation

Correct answer

P – 3, Q – 4, R – 2, S –1

Why this answer is correct

Explanation:

Important G codes:

G 00 – Rapid Transverse
G 01 – Linear Interpolation
G 02 – CW Circular Interpolation
G 03 – CCW Circular Interpolation
G 04 – Dwell
G 90 – Absolute coordinate system

Important M codes:

M 00 – Program Stop
M 03 – Spindle (CW)
M 04 – Spindle (CCW)
M 05 – Spindle Stop
M 08 – Coolant on
M 09 – Coolant off
M 10 – Clamp-on
M 11 – Clamp off
M 02 or M 30 – Program stop, reset to start.

Question 49

What are the upper and lower limits of the shaft represented by 60 f₈?

Use the following data:

Diameter 60 lies in the diameter step of 50-80 mm

Fundamental tolerance unit, i, in μm=0.45D^1/3+0.001D, where D is the representative size in mm;

Tolerance value for IT8 = 25i.

Fundamental deviation for ‘f’ shaft = -5.5D^0.41

AOption (a)
Lower limit = 59.924mm, Upper Limit = 59.970mm
BOption (b)
Lower limit = 59.954mm, Upper Limit = 60.000mm
COption (c)
Lower limit = 59.970mm, Upper Limit = 60.016mm
DOption (d)
Lower limit = 60.000mm, Upper Limit = 60.046mm

View answer & explanation

Correct answer

Lower limit = 59.924mm, Upper Limit = 59.970mm

Why this answer is correct

Since D = 60 mm less between 50 – 80 mm

therefore geometric mean dia

\(D = \sqrt {50 \times 80} = 63.246\ mm\)

tolerance unit i = 0.45 D ^1/3 – 0.001D

⇒ i = 0.45 (63.246)2/3 + 0.001 (63.246) = 0.00186 mm

for the hole of grade &, standard tolerance

= 25i = 25 × 0.00186 = 0.0465 mm

Fundamental deviation for ‘f’ shaft = ρf = - 5.5 D^0.41

= -5.5 (63.246)0.41 = - 0.030115 mm

Upper Limit of shaft = Basic size + fundamental deviation

= 60 – 0.030115 = 59.970 mm

Lower limit of shaft = Upper limit – Tolerance

= 59.970 – 0.0465 = 59.924 mm

Question 50

Match the items in Column I and Column II.

Column I	Column II
P. Metallic Chills	1. Support for the core
Q. Metallic Chaplets	2. Reservoir of the molten metal
R. Riser	3. Control cooling of critical sections
S. Exothermic Padding	4. Progressive solidification

AOption (a)
P – 1, Q – 3, R – 2, S – 4
BOption (b)
P – 1, Q – 4, R – 2, S – 3
COption (c)
P – 3, Q – 4, R – 2, S –1
DOption (d)
P – 4, Q – 1, R – 2, S – 3

View answer & explanation

Correct answer

P – 4, Q – 1, R – 2, S – 3

Why this answer is correct

Metallic chills → progressive solidification

Metallic chaplets ⇒ support for the core

Riser → reservoir of the molten metal

Exothermic padding → control cooling of critical sections

The inlet and the outlet conditions of stream for an adiabatic steam turbine are as indicated in the notations are as usually followed

Question 51

Based on Passage: View directions & passage

If mass flow rate of steam through the turbine is 20kg/s, the power output of the turbine (in MW) is

AOption (a)
12.157
BOption (b)
12.941
COption (c)
168.001
DOption (d)
168.785

View answer & explanation

Correct answer

12.157

Why this answer is correct

S.F.E.E at inlet and outlet of the turbine \(\Rightarrow {h_1} +{\frac{V_1^2}{2}}+{z_1}g+{\frac{dQ}{dm}}= {h_2} +{\frac{V_2^2}{2}}+{z_2}g+ \frac{{d{W}}}{{dm}}\)

and for adiabatic dQ = 0

then, \(\Rightarrow \frac{{d{W}}}{{dm}}=({h_1} -{h_2} )+({\frac{V_1^2}{2}}-{\frac{V_2^2}{2}})+({z_1}- {z_2)}g\)

\(\Rightarrow \frac{{d{W}}}{{dm}}=(3200 -2600)+({\frac{160^2}{2}}-{\frac{100^2}{2}})\times {10^{-3}}+({10}- {6)}\times 9.81\times {10^{-3}} =607.84\ kJ/kg\)

Power output of turbine

\(P = mass\ flow\ rate\times {\frac{dW}{dm}} = 20\times 607.84\times 10^{-3}=12.157 \ MW\)

Question 52

Based on Passage: View directions & passage

Assume the above turbine to be part of a simple Rankine cycle. The density of water at the inlet to the pump is 1000 kg/m³. Ignoring kinetic and potential energy effects, the specific work (in kJ/kg) supplied to the pump is

AOption (a)
0.293
BOption (b)
0.351
COption (c)
2.930
DOption (d)
3.510

View answer & explanation

Correct answer

2.930

Why this answer is correct

S.F.E.E for pump \(\Rightarrow {h_4} = {h_3} + \frac{{d{W_P}}}{{dm}}\)

\(\Rightarrow \frac{{d{W_P}}}{{dm}} = {h_4} - {h_3} = dh\) __________(i)

and reversible adiabatic compression as = dh – vdp = 0

⇒ dh = vdp

now from equation (1)

\(\begin{array}{l} \frac{{d{W_P}}}{{dm}} = dh = vdp = \frac{1}{\rho }\left( {{P_1} - {P_2}} \right)\ \frac{{d{W_P}}}{{dm}} = \frac{{3000 - 70}}{{1000}} = 2.93;kJ/kg \end{array}\)

Radiative heat transfer is intended between the inner surfaces of two very large isothermal parallel metal plates. While the upper plate (designated as plate 1) is a black surface and is the warmer one being maintained at 727ºC, the lower plate (plate 2) is a diffuse and gray surface with an emissivity of 0.7 and is kept at 227ºC. Assume that the surfaces are sufficiently large to form a two-surface enclosure and steady state conditions to exist. Stefan Boltzmann constant is given as 5.67×10^-8W/m²K⁴

Question 53

Based on Passage: View directions & passage

The irradiation (in kW/m²) for the upper plate (plate 1) is

AOption (a)
2.5
BOption (b)
3.6
COption (c)
17.0
DOption (d)
19.5

View answer & explanation

Correct answer

19.5

Why this answer is correct

Concept

Irradiation – It is denoted by G.

Irradiation is defined as the total thermal radiation incident upon a surface per unit time per unit area.

Radiosity – it is denoted by J.

Radiosity is defined as the total thermal radiation leaving a surface per unit time per unit area.

J = Emitted thermal energy + Reflected part of incident thermal energy.

J = E + ρ G

J = ϵ E_b + ρ G

where E = Energy emitted by Nonblack body.

E_b = Energy emitted by the Black body.

ρ = Reflectivity of the surface.

For any surface,

α +ρ +τ = 1

where α = Absorptivity of the surface.

ρ = Reflectivity of the surface.

τ = Transmitivity of the surface.

For opaque surface transmitivity τ = 0

∴ For opaque surface, α +ρ = 1 or ρ = 1-α

Kirchhoff’s law states that when a body is in thermal equilibrium with its surroundings, its emissivity is equal to its absorptivity.

α =ϵ therefore ρ = 1-ϵ

Thus, Radiosity is given by

J = ϵ E_b + (1 - ϵ ) × G

Calculation:

Given:

The temperature of the upper plate,plate₁ T₁ = 727 ° C = 1000 K

The temperature of the lower plate,plate₂, T₂ = 227 ° C = 500 K

The emissivity of the plate₁, ϵ ₁ = 0.7

Irradiation of plate₁ = Total thermal radiation incident on the plate₁.

= Fraction of thermal energy emitted by the plate₁ per unit time per unit area reaching the plate₁ after reflection from plate₂ + Thermal radiation emitted by plate₂ per unit time per unit area and reaching the plate₁.

(1 - ϵ ₂) × E_b1 + ϵ ₂× σ × T₂⁴

(1 - ϵ ₂) × σ × T₁⁴ + ϵ ₂× σ × T₂⁴

19.5 kW/m²

Question 54

Based on Passage: View directions & passage

If plate 1 is also a diffuse and gray surface with an emissivity value of 0.8, the net radiation heat exchange (in kW/m2) between plate 1 and plate 2 is

AOption (a)
17.0
BOption (b)
19.0
COption (c)
23.0
DOption (d)
31.7

View answer & explanation

Correct answer

31.7

Why this answer is correct

Concept:

The radiation heat transfer between two infinitely large grey parallel planes are given by,

\({\left( {\frac{Q}{A}} \right)_{1 - 2}} = \frac{{\sigma \times \left( {T_1^4 - T_2^4} \right)}}{{\frac{{1}}{{{\varepsilon _1}}} + \frac{1}{{{\varepsilon _2}}} - 1}}\)

where T₁ = Temperature of plate₁

T₂ = Temperature of plate₂

ϵ ₁ = Emissivity of plate₁

ϵ₂ = Emissivity of plate₂

Calculation:

Given:

Emissivity of plate₁, ϵ ₁ = 0.8

Emissivity of plate₂ , ϵ ₂ = 0.7

Temperature of plate₁ , T₁ = 727 ° C = 1000 K

Temperature of plate₂ , T₂ = 227 ° C = 500 K

The radiation heat exchange between plate₁ and plate₂

_{\({\left( {\frac{Q}{A}} \right)_{1 - 2}} = \frac{{\sigma \times \left( {T_1^4 - T_2^4} \right)}}{{\frac{{1}}{{{_1}}} + \frac{1}{{{_2}}} - 1}}\)}

\(\frac{{5.67 \times {{10}^{ - 8}} \times \left( {{{1000}^4} - {{500}^4}} \right)}}{{\frac{1}{{0.8}} + \frac{1}{{0.7}} - 1}}\)

31.68 kW/m²

Consider the following PERT network: The optimistic time, most likely time and pessimistic time of all the activities are given in the table below:

Activity	Optimistic time (days)	Most likely time (days)	Pessimistic time (days)
1 – 2	1	2	3
1 – 3	5	6	7
1 – 4	3	5	7
2 – 5	5	7	9
3 – 5	2	4	6
5 – 6	4	5	6
4 – 7	4	6	8
6 – 7	2	3	4

Question 55

Based on Passage: View directions & passage

The critical path duration of the network (in days) is

AOption (a)
11
BOption (b)
14
COption (c)
17
DOption (d)
18

View answer & explanation

Correct answer

Why this answer is correct

Activity	Optimistic time (T₀)	Most likely Time(T_m)	Ppessimistic Time (T_P)	\({T_e} - {t_0} + 4{t_m} + {t_p}\) expected completion time (T_e)	Variance \({\sigma ^2} = {\left( {\frac{{{t_p} - {t_o}}}{6}} \right)^2}\)
1 – 2	1	2	3	\(\frac{{1 + 8 + 3}}{6} = 2\)	\({\left( {\frac{{3 - 1}}{6}} \right)^2} = \frac{1}{9}\)
1 – 3	5	6	7	\(\frac{{5 + 24 + 7}}{6} = 6\)	\({\left( {\frac{{7 - 5}}{6}} \right)^2} = \frac{1}{9}\)
1 – 4	3	5	7	\(\frac{{3 + 20 + 7}}{6} = 5\)	\({\left( {\frac{{7 - 3}}{6}} \right)^2} = \frac{4}{9}\)
2 – 5	5	7	9	\(\frac{{5 + 28 + 9}}{6} = 7\)	\({\left( {\frac{{9 - 5}}{6}} \right)^2} = \frac{4}{9}\)
3 – 5	2	4	6	\(\frac{{2 + 16 + 6}}{6} = 4\)	\({\left( {\frac{{6 - 2}}{6}} \right)^2} = \frac{4}{9}\)
5 – 6	4	5	6	\(\frac{{4 + 20 + 6}}{6} = 5\)	\({\left( {\frac{{6 - 4}}{6}} \right)^2} = \frac{1}{9}\)
4 – 7	4	6	8	\(\frac{{4 + 24 + 8}}{6} = 6\)	\({\left( {\frac{{3 - 4}}{6}} \right)^2} = \frac{4}{9}\)
6 – 7	2	3	4	\(\frac{{2 + 12 + 4}}{6} = 3\)	\({\left( {\frac{{4 - 2}}{6}} \right)^2} = \frac{1}{9}\)

hence path 1 – 3 – 5 – 6 – 7 is the critical path, which has 18 days of expected completion time

Question 56

Based on Passage: View directions & passage

The standard deviation of the critical path is

AOption (a)
0.33
BOption (b)
0.55
COption (c)
0.77
DOption (d)
1.66

View answer & explanation

Correct answer

0.77

Why this answer is correct

Activity	Optimistic time (T₀)	Most likely Time(T_m)	pessimistic Time (T_P)	\({T_e} - {t_0} + 4{t_m} + {t_p}\) expected completion time (Te)	Variance \({\sigma ^2} = {\left( {\frac{{{t_p} - {t_o}}}{6}} \right)^2}\)
1 – 2	1	2	3	\(\frac{{1 + 8 + 3}}{6} = 2\)	\({\left( {\frac{{3 - 1}}{6}} \right)^2} = \frac{1}{9}\)
1 – 3	5	6	7	\(\frac{{5 + 24 + 7}}{6} = 6\)	\({\left( {\frac{{7 - 5}}{6}} \right)^2} = \frac{1}{9}\)
1 – 4	3	5	7	\(\frac{{3 + 20 + 7}}{6} = 5\)	\({\left( {\frac{{7 - 3}}{6}} \right)^2} = \frac{4}{9}\)
2 – 5	5	7	9	\(\frac{{5 + 28 + 9}}{6} = 7\)	\({\left( {\frac{{9 - 5}}{6}} \right)^2} = \frac{4}{9}\)
3 – 5	2	4	6	\(\frac{{2 + 16 + 6}}{6} = 4\)	\({\left( {\frac{{6 - 2}}{6}} \right)^2} = \frac{4}{9}\)
5 – 6	4	5	6	\(\frac{{4 + 20 + 6}}{6} = 5\)	\({\left( {\frac{{6 - 4}}{6}} \right)^2} = \frac{1}{9}\)
4 – 7	4	6	8	\(\frac{{4 + 24 + 8}}{6} = 6\)	\({\left( {\frac{{3 - 4}}{6}} \right)^2} = \frac{4}{9}\)
6 – 7	2	3	4	\(\frac{{2 + 12 + 4}}{6} = 3\)	\({\left( {\frac{{4 - 2}}{6}} \right)^2} = \frac{1}{9}\)

1 – 3 – 5 – 6 – 7 is the critical path

variance along critical path

\(\begin{array}{l} {\sigma ^2} = \sigma _{1 - 3}^2 + \sigma _{3 - 5}^2 + \sigma _{5 - 6}^2 + \sigma _{6 - 7}^2\ \frac{1}{9} + \frac{4}{9} + \frac{1}{9} + \frac{1}{9} = \frac{7}{9} \end{array}\)

then standard deviation=(Sum of variances along critical path)^0.5 \(= \sqrt {Variance}\)

\(= \sqrt {\frac{7}{9}} = 0.88\)

the most appropriate answer is 0.77

In a machining experiment, tool life was found to vary with the cutting speed in the following manner:

Cutting speed (m/min)	Tool life (minutes)
60	81
90	36

Question 57

Based on Passage: View directions & passage

The exponent (n) and constant (k) of the Taylor’s tool life equation are

AOption (a)
n = 0.5 and k = 540
BOption (b)
n = 1 and k = 4860
COption (c)
n = −1 and k = 0.74
DOption (d)
n = − 0.5 and k = 1.155

View answer & explanation

Correct answer

n = 0.5 and k = 540

Why this answer is correct

V₁ = 60 m/min, T₁= 81 min

V₂ = 90 m/min, T₂ = 36 min

Taylor’s tool life equation \({V_1}{\rm{T}}_1^n = {{\rm{V}}_2}{\rm{T}}_2^n = {\rm{k}}\)

⇒ 60(81)ⁿ = 90 (36)ⁿ

\(\Rightarrow n\ln \left( {\frac{{81}}{{36}}} \right) - \ln \left( {\frac{{90}}{{60}}} \right) \Rightarrow n = 0.5\)

So, \({V_1}{\rm{T}}_1^n = k \Rightarrow 60{\left( {81} \right)^{0.5}} = {\rm{k}}\)

⇒ k = 540

Question 58

Based on Passage: View directions & passage

What is the percentage increase in tool life when the cutting speed is halved?

AOption (a)
50%
BOption (b)
200%
COption (c)
300%
DOption (d)
400%

View answer & explanation

Correct answer

300%

Why this answer is correct

n = 0.5, V2 = V1/2

\(\begin{array}{l} V{T^n} = C \Rightarrow V{T^{0.5}} = C \Rightarrow V \propto \frac{1}{{\sqrt T }}\ So,\frac{{{V_2}}}{{{V_1}}} = \sqrt {\frac{{{T_1}}}{{{T_2}}}} \Rightarrow \frac{1}{2} = \sqrt {\frac{{{T_1}}}{{{T_2}}}} \Rightarrow \frac{{{T_1}}}{{{T_2}}} = \frac{1}{4} \Rightarrow {T_2} = 4{T_1} \end{array}\)

% increases in tool life \(= \frac{{{T_2} - {T_1}}}{{{T_1}}} \times 100 = 300%\)

A 20º full depth involute spur pinion of 4 mm module and 21 teeth is to transmit 15 kW at 960 rpm. Its face width is 25 mm.

Question 59

Based on Passage: View directions & passage

The tangential force transmitted (in N) is

AOption (a)
3552
BOption (b)
2611
COption (c)
1776
DOption (d)
1305

View answer & explanation

Correct answer

3552

Why this answer is correct

Concept:

Power transmitted by the gear = torque × Angular speed

Tangential force × Radius × Angular speed

F_t × r × ω

Radius, \(r = \frac{D}{2} = \frac{{mT}}{2}\)

where m = Module of the gear.

T = Number of teeth on the gear.

Calculation:

Given:

Module, m = 4 mm

Number of teeth = 21

Speed of rotation, N = 960 rpm

Pressure angle, ϕ = 20 °

Face width, b = 25 mm

Power transmitted, P = 15 kW

Radius, \(r = \frac{D}{2} = \frac{{mT}}{2} = \frac{{4 \times 21}}{2} = 42mm\)

P = F_t × r × ω

\(15 \times {10^3} = {F_t} \times 42 \times {10^{ - 3}} \times \frac{{2\pi \times 960}}{{60}}\)

F_t = 3552.26 N

Question 60

Based on Passage: View directions & passage

Given that the tooth geometry factor is 0.32 and the combined effect of dynamic load and allied factors intensifying the stress is 1.5; the minimum allowable stress (in MPa) for the gear material is

AOption (a)
242.0
BOption (b)
166.5
COption (c)
121.0
DOption (d)
74.0

View answer & explanation

Correct answer

166.5

Why this answer is correct

Concept

The force transmitted by the gear is given by,

\({F_t} = \frac{{{\sigma _b} \times b \times Y \times m}}{{{f_s}}}\)

where σ_b = Minimum allowable stress for gear material

Y = Tooth geometry factor

f_s Combined effect of dynamic load and allied factors intensifying the stress.

b = Face width of the tooth

m = Module of the gear

Calculation:

Given:

Tooth geometry factor, Y = 0.32

The combined effect of dynamic load and allied factors intensifying the stress, f_s = 1.5

Face width of the tooth, b = 25 mm

The module of the gear, m = 4 mm

Force transmitted,

\({F_t} = \frac{{{\sigma _b} \times b \times Y \times m}}{{{f_s}}}\)

\(3552 = {\rm{}}\frac{{{\sigma _b} \times 25 \times 0.32 \times 4}}{{1.5}}\)

σ _b = 166.5 MPa

60 questions in the full paper