GATE ME 2018 Official Paper: Shift 2 (Previous Year Paper)

The correct answer is complemented, complimented.

Key Points

• Let's understand the meaning of the marked words:
• Complement: add to (something) in a way that enhances or improves it; make perfect.
• ​Example: B**ecause Jim is a serious man who always thinks before he acts, he is the perfect complement for Lisa who lives by her impulses.
• Compliment: politely congratulate or praise (someone) for something.
• ​Example: I just wanted to compliment you for the wonderful speech you gave tonight.
• From the above-given explanation, we can say that 'complemented' will be used in the first blank and 'complimented' will be used in the second blank.

​Therefore, the correct sentence is: The dress complemented her so well that they all immediately complimented her on her appearance.

The correct answer is undiminished.

Key Points

• Let's understand the meaning of the marked word.
• Undiminished: not diminished, reduced, or lessened.
• Example:  Even in his eighties, the man’s eyesight was undiminished and he could see as well as his grandson.
• Thus, from the above-given explanation, we can say that 'undiminished' will be used in the blank as the sentence is trying to explain that the judge's standing in the legal community is still intact even after false allegations.

Therefore, the correct sentence is: The judge’s standing in the legal community, though shaken by false allegations of wrongdoing, remained undiminished.

Additional Information

• Let's explore other options:

According to the English alphabet series and its positional value:

The logic follows here is:

In Each term, one letter is increased, and in each term consecutive letters.

In First Term: 2 letters

In Second Term: 3 letters

In Third Term: 4 letters

So;

In Fourth Term: 5 letters

Hence, "TUVWX" is the correct answer.

Explanation:

The perimeter of Circle of radius r = 2πr

The perimeter of a square of side s = 4s

The perimeter of an equilateral triangle of side a = 3a

The perimeters of a circle, a square, and an equilateral triangle are equal.

2πr = 4s = 3a

• s = 3/4 a = 0.75 a
• r = 3/2π a = 0.477 a

Area of an equilateral triangle:

\(A_{trg}=\frac{\sqrt 3}{4}a^2=0.433 a^2 \)

Area of square = s² = (0.75 a)² = 0.5625 a²

Area of Circle = 2πr2 = π(0.477 a)2 = 0.714 a2

A_circle > A_Square > A_trg

Concept:

Logarithmic Properties:

\({\log _a}xy = {\log _a}x + {\log _a}y\)

\({\log _a}\frac{x}{y} = {\log _a}x - {\log _a}y\)

\({\log _a}{x^n} = n{\log _a}x\)

\({\log _a}b = \frac{{{{\log }_c}b}}{{{{\log }_c}a}}\)

\({\log _a}b = \frac{1}{{{{\log }_b}a}}\)

\({\log _a}a = 1\)

Calculation:

\(y=\frac{1}{{1 + {{\log }_u}vw}} + \frac{1}{{1 + {{\log }_v}wu}} + \frac{1}{{1 + {{\log }_w}uv}}\)

\(y=\frac{1}{{\log_uu+ {{\log }_u}vw}} + \frac{1}{{\log_vv + {{\log }_v}wu}} + \frac{1}{{\log_ww + {{\log }_w}uv}}=\frac{1}{\log_uuvw}+\frac{1}{\log_vuvw}+\frac{1}{\log_wuvw}\)

Using \({\log _a}b = \frac{1}{{{{\log }_b}a}}\)

\(y={\log _{uvw}}u + {\log _{uvw}}v + {\log _{uvw}}w={\log _{uvw}}uvw=1\)

Each student watched either only one film or all three.

Let a = students who watched only A,

b = students who watched only B,

c = students who watched only C,

x = students who watched A, B and C i.e. all three

Given: P(A) = 13, P(B) = 16, P(C) = 19

a + x = 13

b + x = 16

c + x = 19

a + b + c + 3x = 13 + 16 + 19

a + b + c + 3x = 48

Total strength of class is 40,

a + b + c + x = 40

⇒ 2x = 8

x = 4

So 4 students watched all three films.

Concept:

Area of Square of side 'a' = a²

The perimeter of a square of side 'a' = 4s

Area of a circle of radius 'r' = πr²

Circumference of a circle of radius 'r' = 2πr

Calculation:

A wire would enclose an area of 1936 m2 if it is bent into a square. Side of square = ?

a² = 1936 m² ⇒ a = 44 m

As wire encloses a square so length of wire = Perimeter of a square

L = 4a = 176 m

The wire is cut into two pieces. The longer piece (k) is thrice as long as the shorter piece (s).

k = 3s and k + s = L = 176 m

3s + s = 176 m ⇒ s = 44 m ⇒ k = 3s = 132 m

So the longer piece is of 132 m length and the shorter piece is of 44 m length.

The long and the short pieces are bent into a square and a circle, respectively.

So the perimeter of a square = Length of Longer wire = 132 m

4a' = 132 m ⇒ a' = 33 m

Area of the square = a² = 33² = 1089 m²

The perimeter of a circle =  Length of shorter wire = 44 m

2πr = 44 ⇒ r = 7 m

Area of circle = πr² = 154 m²

Sum of the areas enclosed by the two pieces: A_square + A_circle = 1089 + 154 = 1243 m²

Concept:

Time and Work - Important Formula:

• If A can do a piece of work in n days, work done by A in 1 day = 1/n
• If A does 1/n work in a day, A can finish the work in n days
• If M₁ men can do W₁ work in D₁ days working H₁ hours per day and M₂ men can do W₂ work in D₂ days working H₂ hours per day (where all men work at the same rate), then M₁D₁H₁/W₁ = M₂D₂H₂/W₂
• If A can do a piece of work in p days and B can do the same in q days, A and B together can finish it in pq/(p+q) days
• If A is thrice as good as B in work, then
• The ratio of work done by A and B = 3: 1 and
• The ratio of time taken to finish a work by A and B = 1 : 3

Calculation:

125 robots operating for 7 hours a day are working for 39 days and completing five-seventh of the work

X robots operating for8 hours a day are working for 52 - 39 = 13  days and completing 1 - 5/7 = 2/7^th of the work

\(\frac{{{M_1}{D_1}{H_1}}}{{{W_1}}} = \frac{{{M_2}{D_2}{H_2}}}{{{W_2}}}\)

\(\frac{{125 \times 39 \times 7}}{{\frac{5}{7}}} = \frac{{X \times 13 \times 8}}{{\frac{2}{7}}} \Rightarrow X = 131.25\)

So, the number of additional robots to be required

= 131.25 – 125 = 6.25 or 7 robots

• Condition 1: If the house number is a multiple of 3, then it is a number from 50 to 59.
• Possibilities: 51, 54, 57
• But if one of these is the right answer, then it should be a multiple of 4 and 6 which is not the case here.
• Condition 2: If the house number is NOT a multiple of 4, then it is a number from 60 to 69.
• Possibilities: 61, 63, 65, 66, 67, 69
• But if one of these is the right answer, it should be multiple of six (i.e. 66) and not a multiple of 3 (but 66 is multiple of 3). So it is not a required sample space.
• Condition 3: If the house number is NOT a multiple of 6, then it is a number from 70 to 79.
• Possibilities: 70, 71, 73, 74, 75, 76, 77, 79
• But if one of these is the right answer, it should be multiple of 4 (i.e. 76) and not a multiple of 3 (i.e. 70, 71, 73, 74, 76, 77, 79).

So 76 is the correct answer.

Concept:

Out of six tosses, the result of four tosses is given as HHHT__ __.

For the last two tosses, the event of occurrence is: HH, HT, TH, TT

• The probability that two heads will occur, P(HH) = 1/4
• The probability that two tails will occur, P(TT) = 1/4
• The probability that one head and one tail will occur, P(HT/TH) = 2/4 = 1/2

P(HT/TH) > P(TT) or P(HH)

"One H and one T will occur" has the highest probability of being correct

Concept:

The Fourier series for the function f(x) in the interval α < x < α + 2π is given by

\(f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\cos nx + \mathop \sum \limits_{n = 1}^\infty {b_n}\sin nx\)

where

\({a_o} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)dx;;{a_n} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)\cos nxdx;;{b_n} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)\sin nxdx\)

When f is an even periodic function of period 2L, then its Fourier series contains only cosine (include possibly, the constant term) terms.

\(f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\frac{{\cos n\pi x}}{L}\)

When f is an odd periodic function of period 2L, then its Fourier series contains only sine terms.

\(f\left( x \right) = \mathop \sum \limits_{n = 1}^\infty {b_n}\sin \frac{{n\pi x}}{L}\)

Calculation:

 f(x) = cos2(x)

\(f(x) = \frac{{1 + \cos 2x}}{2} = \frac{1}{2} + \frac{{\cos 2x}}{2}\)

On comparing with: \(f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\frac{{\cos n\pi x}}{L}\)

a₀ = 1

a₁ = 0

a₂ = 1/2

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:

vector \(\vec u = {e^x}\left( {\cos y\hat i + \sin y\hat j} \right)\)

Divergence of u will be

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

\(= \left( {i\frac{\partial }{{\partial x}} + j\frac{\partial }{{\partial y}}} \right).\left( {{e^x}.\cos y\hat i + {e^x}.{\rm{siny}}\hat j} \right)\)

\( = \frac{\partial }{{\partial x}}\left( {{e^x}.\cos y} \right) + \frac{\partial }{{\partial y}}\left( {{e^x}.\sin y} \right)\)

\( = {e^x}.\cos y + {e^x}.\cos y\)

\(\vec \nabla .\vec u = 2{e^x}\cos y\)

Explanation:

3-D heat equation is given as below

\(\left( {\frac{{{\partial ^2}T}}{{d{x^2}}} + \frac{{{\partial ^2}T}}{{\partial {y^2}}} + \frac{{{\partial ^2}T}}{{\partial {z^2}}}} \right) + \frac{{Q\left( {x,t} \right)}}{K} = \frac{1}{\alpha }\frac{{\partial T}}{{\partial t}}\)

For 1 – D & without heat generation:

\(\frac{{{\partial ^2}T}}{{\partial {x^2}}} = \frac{1}{\alpha }\frac{{\partial T}}{{\partial t}}\)

Where α ÷ thermal diffusivity.

Wave equation is given by:

\(\frac{{{\partial ^2}u}}{{\partial {t^2}}} = {C^2}.\frac{{{\partial ^2}u}}{{\partial {x^2}}}\)      (1-D)

Laplace equation:

\(\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}} + \frac{{{\partial ^2}u}}{{\partial {z^2}}} = 0\)     (3-D)

\({\nabla ^2}u = 0\)

Concept:

\({y^3}\frac{{dy}}{{dx}} + {x^3} = 0\)

Given y (0) = 1,      y(-1) = ?

\({y^3}\frac{{dy}}{{dx}} + {x^3} = 0\)

y³ dy = -x³dx

\(\mathop \smallint \nolimits{y^3}dy = - \mathop \smallint \nolimits {x^3}dx + c\)

\(\frac{{{y^4}}}{4} = - \frac{{{x^4}}}{4} + C\)

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

Now given y (0) = 1

\(C = \left( {\frac{1}{4}} \right) + \frac{{{{\left( 0 \right)}^2}}}{4}\)

\(C = \frac{1}{4}\)

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

Y (-1) =?

\(\frac{{{y^4}}}{4} + \frac{{{{\left( { - 1} \right)}^4}}}{4} = \frac{1}{4}\)

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

\(\frac{{{y^4}}}{4} = 0 \Rightarrow y = 0\)

y (-1) = 0

Concept:

Ends of the columns:

             

In this question:

\({P_{min}} = \frac{{{\pi ^2}EI}}{{L{e^2}}}\)

\({P_{min}} = \frac{{{\pi ^2}EI}}{{{L^2}}}\)

Concept:

Power transmitted through the gear remains constant 

\(PowerP = \frac{{2\pi NT}}{{60}}\)

For two mating gears A and B 

NATA = NBTB

where N is the angular speed in rpm and T is the torque

also, 

NAZA = NBZB 

Where Z is the number of teeth on the gear. 

Calculation:

Given, Z_A = 12 teeth, Z_B = 48 teeth

T_A = 100 N-m, Z_C = 12, Z_D = 60 teeth

T_D = ?

∵ Now N_AZ_A = N_BZ_B ⇒ N_A × 12 = N_B × 48

N_A = 4 N_B            ----(1)

\(PowerP = \frac{{2\pi NT}}{{60}}\)

P_A = P_B ⇒ N_AT_A = N_BT_B

4N_B T_A = N_B T_B                                                    [N_A = 4 N_B    By equation (1)]

T_B = 4 × 100 = 400 N-m

Now gear B & C are on the same shaft, therefore torque on gear B & C will be the same.

Thus T_C = 400 N-m

Now, N_CZ_C = N_DZ_D

N_C × 12 = N₀ × 60 ⇒ N_C = 5 N­_D        ----(2)

N_CT_C = N_DT_D

5 N_D× 400 = N_D T_D                                                [N_C = 5 N­_D By equation (2)]

T_D = 2000 N-m

Concept:

Transmissibility is given by:

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

\(C = 2\zeta \sqrt {KM} \Rightarrow \zeta = \frac{C}{{2\sqrt {KM} }}\)

Calculation:

After additional damper in parallel \(\zeta ' = \frac{{C + C'}}{{2\sqrt {KM} }}\)

Thus  \(\zeta ' > \zeta \)

Now,  \({\omega _d} = {\omega _n}\sqrt {1 - {\zeta ^2}} \)  

 \(\zeta ' > \zeta \)indicating that \(\omega _d' < {\omega _d}\)

\(time\ period\left( {{T_d}} \right) = \frac{{2\pi }}{{{\omega _d}}}\)

Thus, the time period of free oscillation will increase.

External tensile load applied to a no preloaded joint transmits entirely to the bolt. The stiffness of a bolted joint without a preload equals the bolt stiffness K_e = K_b. For a preloaded bolted joint, joint stiffness is the sum of bolt & clamped member stiffness k_e = K_b + K_c. Therefore, preloading increases the stiffness of the bolted joints, which also increases the resonant frequency of an assembly.

Note:

However, once a joint is preloaded, increasing the magnitude of the preload can not make a joint stiffer.

Concept:

From Wein's displacement law 

\({\lambda _{max}}T = 2898;\mu m - k\left( {constant} \right)\)

Thus, \({\lambda _{peak}}T = \lambda _{peak}'T'\)

Calculation:

Given, 

Black body λpeak = 1.45 μm at 2000 K.

Now, \(\lambda _{peak}' = 2.90;\mu m\)

1.45 × 2000 = 2.90 × T'

T' = 1000 K

Explanation:

Entropy change for an ideal gas with constant properties undergoing a quasi-static process:

\({s_2} - {s_1} = {C_p}\ln \left( {\frac{{{V_2}}}{{{V_1}}}} \right) + {C_v}\ln \left( {\frac{{{P_2}}}{{{P_1}}}} \right)\)

\({s_2} - {s_1} = {C_v}\ln \left( {\frac{{{T_2}}}{{{T_1}}}} \right) + R\ln \left( {\frac{{{V_2}}}{{{V_1}}}} \right)\)

\({s_2} - {s_1} = {C_p}\ln \left( {\frac{{{T_2}}}{{{T_1}}}} \right) - R\ln \left( {\frac{{{P_2}}}{{{P_1}}}} \right)\)

• 50% reaction stage steam turbine also known as “Parsons turbine”.
• The velocity triangles for inlet & outlet conditions are symmetrical having.

 

\(\left{ {\begin{array}{*{20}{c}} {{\alpha _1} = {\beta _2}}\ {{\alpha _2} = {\beta _1}} \end{array}} \right};;;;;;;& ;;;;\left{ {\begin{array}{*{20}{c}} {{V_{{r_1}}} = {V_2}}\ {{V_{{r_2}}} = {V_1}} \end{array}} \right}\)

Explanation:

On cooling reaction will tend in forward direction and on heating cycle reaction will reverse in direction.

Explanation:

During solidification of pure molten metal, the grains in the casting near the mould wall are fine and randomly oriented because the rate of solidification is high at the surface of mould during solidification.

Explanation:

Centrifugal casting:

It is used to manufacture hollow products (such as pipes) having a rotationally symmetric hollow space & on-axis coinciding with the axis of rotation of the centrifugal casting machine.

Casting materials: Especially cast iron, cast steel, heavy & light metals.

Blow moulding:

• Can form intricate shapes, hollow shapes with thin walls.
• Ideal for mass production.
• Plastics used: HDPE, LDPE, PET, PP, PS, PVC.
• End-uses: Plastic bottles & containers of all sizes and shapes; e.g. soft drinks bottles & shampoo bottles.

Rolling: Is the process of reducing the thickness or changing the cross-section of a long workpiece by compressive forces applied through a set of rolls.

• Products: Structures, rails, threaded bolts by thread rolling.

Impact extrusion:

• It is similar to indirect extrusion, it is often included in the cold-extrusion category. The descends rapidly on the blank (slug), which is extruded backward.
• Performed at higher speeds & shorter strokes as compared to conventional extrusion.

Products:

• Made by this process included toothpaste tubes & battery cases.
• Can produce thin walled tubular sections, ones having thickness to diameter ratios as small as 0.005.

Explanation:

Feed rate in slab milling operation is given by, 

f_m = f_t × N × Z 

Where f_t is the feed per tooth.

N = Spindle rotational speed (in rpm)

Z = Number of teeth in cutter (teeth per rev)

Explanation:

The unconventional machining process and their characteristics and the application areas are discussed in the table below:

Explanation:

lathe chuck:

• A lathe chuck is a holding device which is used for holding the job firmly against the cutting tool.

There are two types of chuck:

Four Jaw Chuck:

• The four-jaw chuck is also called an independent chuck since each jaw can be adjusted independently.
• Four jaw chucks are used for a wide range of regular and irregular shapes.
• A work can be trued to within 0.02 mm accuracy, using this chuck.
• This type of chuck is much more heavily constructed than the self-centering chuck, and has much greater holding power.
• Each jaw is moved independently by a square thread screw.
• The jaws are reversible for holding large diameter jobs.
• The independent four-jaw chuck has four jaws each working independently of the others in its own slot in the chuck body and actuated by its own separate square threaded screw.
• By suitable adjustment of the jaws, a workpiece can be set to run either true or eccentric with the machine centre.
• Finished jobs when held in a four-jaw chuck can be trued with the help of a dial test indicator.

Three Jaw Chuck:

• It is also known as three jaws universal chuck, self-centring chuck and concentric chuck having three jaws which work at the same time.
• Three jaw chucks are used to hold only perfect round and regular jobs

Concept:

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

Where Ch is the holding cost

D is the annual demand

Q is the quantity ordered

Co is the setup or ordering cost

Calculation:

Given, D = 9600 tyres/year, C_h = Rs. 16 tyre/year, C_o = Rs. 75

\(Economic;order\left( {EOQ} \right) = \sqrt {\frac{{2D{C_o}}}{{{C_h}}}} = \sqrt {2 \times 600 \times 75} \) = 300 units

Concept:

The determinant of the inverse of an invertible matrix is the inverse of the determinant.

det (A^-1) = 1 / det (A)

Calculation:

\(A = \left[ {\begin{array}{*{20}{c}} 1&2&3\ 0&4&5\ 0&0&1 \end{array}} \right]\)

Expanding along a₁₁

|A| = 1 (4 - 0) = 4

|A^-1| = 1/|A| =1/4 = 0.25

Concept:

Stiffness of torsional shaft = (Used as a spring)

∴ Torsional stiffness \(\left( {{k_t}} \right) = \frac{T}{\theta }\)

From torsion equation, 

\(\because\frac{T}{{{I_p}}} = \frac{{G\theta }}{L}\)

Where IP is the polar moment of inertia, T is the applied torque, G is the modulus of rigidity, L is the length of the shaft

\(∴\frac{T}{\theta } = \frac{{G{I_p}}}{L}\)

\(I_P=\frac{\pi}{32}(d_o^4-d_i^4)\)

Calculation:

Given, R_i = 10 mm, R_o = 20 mm

L = 1 m, G = 150 GPa

\(\frac{T}{\theta } = \frac{{150 \times {{10}^9} \times \pi \left( {d_o^4 - d_i^4} \right)}}{{32 \times 1}}\)

\(k_t= \frac{{150 \times {{10}^9} \times \pi }}{{32}} \times \left{ {{{40}^4} - {{20}^4}} \right} \times {10^{ - 12}}\)

\(k_t= \frac{{150 \times {{10}^9} \times \pi \times \left( {256 - 16} \right) \times {{10}^4} \times {{10}^{ - 12}}}}{{32}}\)

\(k_t= \frac{{15 \times \pi \times 24 \times {{10}^3}}}{{32}}\)

⇒ kt  = 35.34 × 10³ N-m/rad

_kt = 35.34 kN-m/rad

Given,

σ_a = 1100 N^-0.15

σ_a = stress amplitude (in MPa)

N = Fatigue lift in cycles

N = 1 × 10⁵ cycles, then

σ_a = 1100 (1 × 10⁵)^-0.15

σ_a = 195.61 MPa

Explanation

\({\sigma _x} = \frac{{5x}}{{\sqrt {R{e_x}} }}\)

Local Reynolds number \(\left( {R{e_x}} \right) = \frac{{\rho vx}}{\mu }\) 

\({\delta _{x}} \propto \sqrt x\)

When, x = L

\({\delta _L} \propto \sqrt L\)

\({\delta _L} = k\sqrt L\)       ----(1)

If the length is increased to 2L:

\({\delta _{2L}} = k\sqrt {2L} \)       ----(2)

By equation (1) and (2):

\(% ;change = \frac{{{\delta _{2L}} - {\delta _L}}}{{{\delta _L}}} \times 100%\)

\(% ;change = \frac{{\left( {\sqrt {2 - 1} } \right)k\sqrt L }}{{k\sqrt L }} \times 100 = 41.42%\)

Concept:

The Work output = Area enclosed by P-V diagram

Calculation:

The area enclosed by the P-V diagram is nothing but the area of triangle ABC

 \(\Rightarrow A = W = \frac{1}{2} \times \left( {2.5 - 2} \right) \times \left( {650 - 400} \right)\)

\(W_{output}= \frac{1}{2} \times 250 \times 0.5 = 62.5 ; kJ/Cycle\)

Average number of customers are given as

λ = 30/hr  = 0.5 /min

Probability of customer arrival in 1 minute or less, is given as

P(≤t) = 1 - e^-λt

P(≤1) = 1 - e^{-0.5 × 1}

p(≤t) = 0.393

Probability of customers arrival in 3 minutes or less, is given as

P(≤t) = 1 - e^-λt

P(≤3) = 1 - e-0.5 × 3

p(≤t) = 0.7768

Probability of customer arrival between 1 and 3 minute, is given as

P(≤3) - P(≤1) = 0.7768 - 0.393

P(≤3) - P(≤1) = 0.38

Concept involved: As the tube is frictionless so the initial potential energy and final potential energy will be the same. Thus, when the ball is dropped from 1 m height it will again move 1 m vertically in the slant portion of the tube. The total distance moved will cover both downward and upward movement of the ball.

Total distance moved = AB + BC

Total distance = (1 + √2) m = 2.414 m

Mistake Point:

A common mistake that can happen here is that getting confused between the vertical and slant height. As the gravitational potential depends on the vertical distance, not the slant height.

Mistake is taking

BC = 1 m

Total distance = AB + BC = (1 + 1) m = 2 m

This is wrong. This will indicate that some energy is lost which is incorrect.

Concept:

Given \(\mathop \oint \limits_C^; \frac{1}{{5z - 4}}dz = A\pi i\) 

\(\mathop \oint \limits_C^; \frac{1}{{5z - 4}}dz = \frac{1}{5}\mathop \oint \limits_C^; \frac{1}{{z - \frac{4}{5}}}\)

There is a pole at \(z = \frac{4}{5 }\)

Thus, \(\mathop \oint \limits_C^; \frac{1}{{5z - 4}}dz = \frac{{2\pi i}}{5}\lim z \to \frac{4}{5}\left{ {\frac{{z - \frac{4}{5}}}{{z - \frac{4}{5}}}} \right}\)

Given,

\(A\pi i = \frac{{2\pi i}}{5} \times 1\)

\(A = \frac{2}{5}\)

Explanation:

x₁ & x₂: two independent exponentially distributed random variables with means 0.5 and 0.25.

A continuous random variable x is said to have an exponential (λ) distribution if it has probability density function.

\({f_x}\left( {x/\lambda } \right) = \left{ {\begin{array}{*{20}{c}} {\lambda {e^{ - \lambda x}}}&{for;x > 0}\ 0&{for;x \le 0} \end{array}} \right.\)

Where, λ > 0, is called the rate of distribution. The mean of the exponential (λ) distribution is calculated using integration by parts as:

\(E\left[ x \right] = \mathop \smallint \limits_0^\infty x\lambda .{e^{ - \lambda x}}dx\)

\(= \lambda \left{ {\left[ {\frac{{ - x.{e^{ - \lambda x}}}}{\lambda }} \right]_0^\infty + \frac{1}{\lambda }\mathop \smallint \limits_0^\infty {e^{ - \lambda x}}.dx} \right}\)

\(= \lambda \left{ {\lambda \left[ {0 + 0} \right] + \frac{1}{\lambda }\left[ {\frac{{ - {e^{ - \lambda x}}}}{\lambda }} \right]_0^\infty } \right}\)

\(= \lambda \left{ {0 + \frac{1}{\lambda }\left[ {0 + \frac{1}{\lambda }} \right]} \right}\)

\(= \lambda .\frac{1}{{{\lambda ^2}}}\)

\(E\left[ x \right] = \frac{1}{\lambda }\)

Let x₁, x₂ …..., x_n be independent random variables, with x_i having exponential (λ_i) distribution. Then the distribution of min (x₁, x₂ ……, x_n) is exponential (λ₁ + λ₂ + …. + λ_n)

Thus, mean of min \(\left( {{x_1},{x_2}} \right) = \frac{1}{{{\lambda _1} + {\lambda _2}}}\)

\({\lambda _1} = \frac{1}{{E\left[ {{x_1}} \right]}} = \frac{1}{{0.5}} = 2\)

\({\lambda _2} = \frac{1}{{E\left[ {{x_2}} \right]}} = \frac{1}{{0.25}} = 4\)

Mean of min \(\left( {{x_1},{x_2}} \right) = \frac{1}{{2 + 4}}\)

\({\lambda _2} = \frac{1}{{E\left[ {{x_2}} \right]}} = \frac{1}{{0.25}} = 4\)

Thus, y = min (x₁, x₂) is exponentially distributed with mean \(\left( {\frac{1}{6}} \right)\).

Explanation:

Position vector \(\vec r = x\hat i + y\hat j + zk\)

\(\left| {\vec r} \right| = \sqrt {{x^2} + {y^2} + {z^2}}\)

\(\phi = \ln \left| {\vec r} \right|\)

\(Gradient;\nabla \phi = \left( {i\frac{\partial }{{\partial x}} + j\frac{\partial }{{\partial y}} + k\frac{\partial }{{\partial z}}} \right).\left{ {\ln \sqrt {{x^2} + {y^2} + {z^2}} } \right}\)

\(= \frac{1}{2}\left{ {i\frac{\partial }{{\partial x}}\ln \left( {{x^2} + {y^2} + {z^2}} \right) + j.\frac{\partial }{{\partial y}}\ln \left( {{x^2} + {y^2} + {z^2}} \right) + k\frac{\partial }{{\partial z}}\ln \left( {{x^2} + {y^2} + {z^2}} \right)} \right}\)

\(= \frac{1}{2}\left{ {i.\frac{{2x}}{{{x^2} + {y^2} + {z^2}}} + j\frac{{2y}}{{{x^2} + {y^2} + {z^2}}} + k\frac{{2z}}{{{x^2} + {y^2} + {z^2}}}} \right}\)

\(\nabla \phi = \left{ {\frac{{xi + yj + zk}}{{{x^2} + {y^2} + {z^2}}}} \right}\)

\(Gradient;\nabla \phi = \frac{{\vec r}}{{\vec r.\vec r}}\)

Explanation:

As the acceleration of point Q is zero, so assuming PQR a rigid body hinged at Q.

\({\vec a_{RP}} = {\vec a_R} - {\vec a_p}\)

Given,

Given: The point R is accelerating with respect to point P at 10∠180° m/s².

i.e. 10 m/s² acceleration at 180° of (P) so only radial acceleration exist i.e. ω = constant; α = 0

∴ a_RP = (RP) ω² = 10

⇒ ω² × 20 = 10

\(\omega = \left( {\frac{1}{{\sqrt 2 }}} \right)*\)

Now as α is zero so R w.r.t Q will also have a radial acceleration in the x-axis

\({a_R} = \left( {RQ} \right){\omega ^2} = 16 \times {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} = 8;m/{s^2}\)

Now our reference is (PR) so the angle will be (180 + θ) *

\(\sin \theta = \left( {\frac{{12}}{{20}}} \right) \Rightarrow \theta = {\sin ^{ - 1}}\left( {\frac{3}{5}} \right) = 37^\circ \)

180 + 37 = 217°

a_R = 8 ∠217°

Concept:

A translating body or a body in general plane motion is in pure rotation about its instantaneous centre of rotation.

and the velocity is given by 

V = ω × Distance of instantaneous centre of rotation

Calculation:

O is the instantaneous centre of the rigid rod, so rod will have pure rotation along the instantaneous centre.

Let rotational speed be = ω (about I.C.)

\(OP = 1\sin 45^\circ = \frac{1}{{\sqrt 2 }}\)

\(OQ = 1\cos 45^\circ = \frac{1}{{\sqrt 2 }}\)

V_P = rω = OP × ω

\(5 = \frac{1}{{\sqrt 2 }}ω\)     ----(1)

V_Q = OQ × ω

\({V_Q} = \frac{1}{{\sqrt 2 }}ω\)     ----(2)

By equation (1) & (2):

5 = V_Q

V_Q = 5 m/s

Concept:

The angle of twist on a shaft is given by 

\(\theta=\frac{{{T} \times L}}{{G\times{I_P}}} \)

Where T is the torque, L is the length of the shaft

G is the modulus of the rigidity and

I_P is the polar moment of inertia

Calculation:

Given T = 150 N.m

The torsional reactions at point P are TP  and at point Q it is TQ

TP + TQ = 150      ---- (1)

Since torque is being applied on the same shaft, thus twist of rod will be same.

Thus, \(\frac{{{T_P} \times 100}}{{G{I_P}}} = \frac{{{T_Q} \times 200}}{{G{I_P}}} \)

T_P = 2T_Q      ----  (2)

By equation (1) & (2):

T_Q = 50 N.m

T_P = 100 N.m

(T_P, T_Q) = (100, 50)

Concept:

Uniform Acceleration and Retardation

Under the given situation one can apply the laws of kinematics.

Uniform acceleration and uniform declaration

‘ω’ angular velocity, ‘δ’ angle turned by the cam

Total time of flight \(\left( {{t_0}} \right) = \left( {\frac{\delta }{\omega }} \right)\)

\(S = ut + \frac{1}{2}a{t^2}\)

Calculation:

Applying for the acceleration half

\(\left( {S = \frac{h}{2}} \right)\) and (u = 0); \(t = \left( {\frac{{{t_0}}}{2}} \right)\)

\(\Rightarrow \frac{h}{2} = \frac{1}{2}a \times {\left( {\frac{{{t_0}}}{2}} \right)^2};\frac{h}{2} = \frac{1}{2} \times a \times {\left( {\frac{\delta }{{2\omega }}} \right)^2}\)

\(\Rightarrow {a = \frac{{4h{\omega ^2}}}{{{\delta ^2}}}}\)

Now maximum velocity will occur at the end of the acceleration half and after that declaration starts hence velocity will decrease.

[V² – u² = 2aS]

Using this equation again for acceleration half.

\({v^2} = 2as = 2 \times \frac{{4h{\omega ^2}}}{{{\delta ^2}}} \times \frac{h}{2} \Rightarrow v = \frac{{2h\omega }}{\delta }\)

\({v_{maximum}} = \frac{{2h\omega }}{\delta }\)

Concept:

The axial strain is given by 

\(ϵ=\frac{{{P}}}{{{A}{E}}}\)

Where P is the load acting in the axial direction, A is the cross-sectional area and E is the Modulus of elasticity

Calculation:

ES = 210 GPa

EAI = 70 GPa

Since the magnitude of axial strains in both bars same. Thus,

\(\frac{{{P_s}}}{{{A_s}{E_s}}} = \frac{{{P_{Al}}}}{{{A_{Al}}{E_{Al}}}}\)

Since, \({A_S} = {A_{Al}} = 1;{m^2}\)

\(\frac{{{P_S}}}{{210}} = \frac{{{P_{Al}}}}{{70}}\)

P_S = 3 P_Al

P = P_S + P_AI      ----(1)

Now, ϵ = 10^-6

\(\frac{{{P_S}}}{{{A_S}{E_S}}} = {10^{-6}}\)

P_S = 10^-6 × 1 × 210 × 10⁹

P_S = 210 KN

P_AI = 70 KN

P = P_AI + P_S = 210 + 70 = 280 KN

Concept:

Bernoulli equation between section (1) & (2)

\(\frac{{{P_1}}}{{\rho g}} + \frac{{V_1^2}}{{2g}} + {Z_1} = \frac{{{P_2}}}{{\rho g}} + \frac{{V_2^2}}{{2g}} + {Z_2} + {h_L}\)

h_L = 0 (Gradually reducing cross-section)

ρ = Constant (incompressible fluid)

Calculation:

ρair = 1.2 kg/m3

\(\frac{{{P_2} - {P_1}}}{{\rho g}} = \frac{{V_1^2}}{{2g}} - \frac{{V_2^2}}{{2g}}\)

\({P_2} - {P_1} = \frac{\rho }{2}\left( {V_1^2 - V_2^2} \right)\)

∴ Q = 1.5 m³/s

A₁V₁ = A₂V₂ = 1.5 m³/s

\({A_1} = \frac{\pi }{4}d_1^2\)

\({A_1} = \frac{\pi }{4}{\left( {0.4} \right)^2} = \frac{{4\pi }}{{100}}{m^2}\)

\({A_2} = \frac{\pi }{4}{\left( {0.2} \right)^2} = \frac{\pi }{{100}}\)

Thus, \({V_1} \times \frac{{4\pi }}{{100}} = 1.5\)

\({V_1} = \frac{{150}}{{4\pi }}\)

\({V_2} = \frac{{150}}{{\pi }}\)

\({P_2} - {P_1} = \frac{{1.2}}{2}\left{ {{{\left( {\frac{{150}}{{4\pi }}} \right)}^2} - {{\left( {\frac{{150}}{\pi }} \right)}^2}} \right}\)

\(= 0.6 \times {150^2}\left( {\frac{1}{{16{\pi ^2}}} - \frac{1}{{{\pi ^2}}}} \right)\)

= -1282.346 N/m² (or pascal)

P₂ - P₁ = -1.28 KPa

Explanation:

Maximizing z = x₁ - x₂

Constraints x₁ + x₂ ≤ 10

x₁ ≥ 0

x₂ ≥ 0

x₂ ≤ 5

Corner points are: (0, 0) (5, 0), (5, 5), (0, 10).

Z (0, 0) = 0 - 0 = 0

Z (0, 5) = 0 - 5 = -5

Z (5, 5) = 5 - 5 = 0

Z (10, 0) = 10 - 0 = 10

Maximum Z = Z (10, 0) = 10

Thus, problem has one optimum solution.

\(\frac{{{d^2}y}}{{d{x^2}}} + \frac{{dy}}{{dx}} - 6y = 0\)

With, \(y\left( 0 \right) = 0,;and\frac{{dy}}{{dx}}\left( 0 \right) = 1\)

Y (1) = -?

∵ D²y + Dy - 6y = 0

Characteristic equation will be -

m² + m - 6 = 0

m² + 3m - 2m - 6 = 0

m = -3, 2

Since roots are real & different thus:

\(y\left( x \right) = {C_1}{e^{ - 3x}} + {C_2}{e^{2x}}\)

Given y (0) = 0

0 = C₁ + C₂      ----(1)

\(\frac{{dy}}{{dx}} = - 3{C_1}{e^{ - 3x}} + 2{C_2}{e^{2x}}\)

\(\frac{{dy}}{{dx}}\left( 0 \right) = - 3{C_1} + 2{C_2}\)

1 = -3C₁ + 2C₂      ----(2)

By equation (1) & (2)

5C₁ = -1

\({C_1} = - \frac{1}{5}\)

\({C_2} = \frac{1}{5}\)

Thus, \(y\left( x \right) = - \frac{1}{5}{e^{ - 3x}} + \frac{1}{5}{e^{2x}}\)

\(y\left( 1 \right) = - \frac{1}{5}{e^{ - 3 \times 1}} + \frac{1}{5}{e^{2 \times 1}}\)

= 1.467

Y (1) = 1.47

Concept:

Let ‘P’ be the pressure, circumferential stress developed \({{\rm{\sigma }}_{\rm{h}}} = \frac{{PD}}{{2t}} = \frac{{P.R}}{t}\)

Axial stress developed due to pressure \({{\rm{\sigma }}_{\rm{L}}} = \frac{{PD}}{{4t}} = \frac{{P.R}}{{2.t}}\) (tensile)

t = wall thickness, R = mean radius of the cylindrical can, P = pressure inside the can 

Total axial tensile stress = Axial stress developed due to pressure + additional tensile stress acting in the axial direction

Calculation:

Given R = 100 mm, t = 5 mm

External axial tensile stress = 50 MPa

Since, circumferential = Axial stress

\(\Rightarrow \frac{{PR}}{t} = \frac{{PR}}{{2t}} + 50\)

\(\Rightarrow \frac{{PR}}{{2t}} = 50\)

\(\Rightarrow \frac{{P \times 100}}{{2 \times 5}} = 50 \Rightarrow P = 5;MPa\)

Mistake point:

Here axial stress developed due to pressure is (tensile) not (compressive)

So, if by mistake

\(\frac{{PR}}{t} = 50 - \frac{{PR}}{{2t}}\)

\(\Rightarrow \frac{3}{2}\frac{{PR}}{t} = 50\)

\(\Rightarrow \frac{{3 \times P \times 100}}{{2 \times 5}} = 50\)

⇒ P = 1.67 MPa which is incorrect.

Concept:

Modified Goodman line: 

The shaded region represents Modified Goodman's Criteria 

The equation for Modified Goodman line:

It depends upon the condition below, 

\(\frac{{σ _a^*}}{{σ _m^*}} = \frac{{{σ _e}({σ _{ut}} - {σ _{yt}})}}{{{σ _{ut}}({σ _{yt}} - {σ _e})}}\)

If, \(\frac{{{σ _a}}}{{{σ _m}}} > \frac{{σ _a^*}}{{σ _m^*}} ⇒ Line;AP;will;be;chosen;\)

If, \(\frac{{{σ _a}}}{{{σ _m}}} < \frac{{σ _a^*}}{{σ _m^*}} ⇒ Line;PB;will;be;chosen\)

Where,

σa = limiting safe stress amplitude

Se = endurance limit of the component

σm = limiting safe mean stress

Sut = ultimate tensile strength

Syt = Yield strength

FOS = Factor of safety

Calculation:

Given:

Pmax = 90 N/mm2, Pmin = -10 N/mm2, σe = 150 N/mm2, σut = 600 N/mm2 and σyt = 480 N/mm2

\({σ _m} =\sigma _s+ \frac{{{σ _{max}};+;{σ _{min}}}}{2} ⇒\frac{60}{A}+ \frac{90;+;(-10)}{2} = \frac{100}{A};kN/mm^2\)

Now, check the condition,

\(\frac{{σ _a^*}}{{σ _m^*}} = \frac{{{σ _e}({σ _{ut}} - {σ _{yt}})}}{{{σ _{ut}}({σ _{yt}} - {σ _e})}} = \frac{{150\left( {600 - 480} \right)}}{{600\left( {480 - 150} \right)}} = 0.091\)

\(\frac{{{σ _a}}}{{{σ _m}}} = \frac{{50}}{{100}} = 0.5\)

\( ⇒ \frac{{{σ _a}}}{{{σ _m}}} = \frac{{σ _a^*}}{{σ _m^*}} = 0.5\)

⇒ So, AP will be be taken as the boundary for modified Goodman's Criteria

The equation for the modified Goodman line for line PB,

\(\frac{{{σ _a}}}{{{σ _{e}}}} + \frac{{{σ _m}}}{{{σ _{ut}}}} = \frac{1}{N}\)

\([\frac{{50}}{{150A}} + \frac{{100}}{{600A}} ]\times 1000= \frac{1}{2}\)

A = 1000 mm²

⇒ a2 = 1000

⇒ a = 31.62 mm

Concept:

Let us consider ‘m’ as mass of the disc, ‘I’ as moment of inertial of the disc, ‘R’ as radius of the disc, ‘a_cm’ as acceleration of the disc center, ‘α’ as angular acceleration of the disc

‘f’ as the friction force acting on the disc

FBD of the disc

As no-slip condition Rα = a_cm

And f × R = Iα

\(I = \frac{{m{R^2}}}{2}\) for disc

Calculation:

Now 100 – f = m a_cm 

And f × R = Iα

\(I = \frac{{m{R^2}}}{2}\) for disc

\(\Rightarrow f \times R = \frac{{m{R^2}}}{2}{\rm{\alpha }}\)

\(\Rightarrow f = \frac{{mR \alpha }}{2} = \left( {\frac{{m{a_{cm}}}}{2}} \right)\)

\(\Rightarrow 100 - \frac{{m{a_{cm}}}}{2} = m{a_{cm}}\)

\(\Rightarrow 100 = \frac{3}{2}m \times {a_{cm}}\)

\(\Rightarrow 100 = \frac{3}{2} \times 10 \times {a_{cm}}\)

⇒ a_cm = 6.667 m/s²

Mistake Point:

The possible mistakes that can happen here is wrong formula for the moment of inertia of the circular disc. So, by mistake I = mR²

And 100 – f = ma_cm

f × R = Iα ⇒ f = mRα ⇒ f = ma_cm

100 – ma_cm = ma_cm

⇒ 100 = 2ma_cm

⇒ a_cm = 5 m/s²

This is incorrect.

Concept:

Bernoulli’s equation is given by 

\(\frac{P}{{\rho g}} + \frac{1}{2}\frac{{{v^2}}}{g} + z = constant\)

and any fluid flows satisfy continuity equation

continuity equation

A1v1 = A2v2

Calculation:

The given problem can be solved by applying Bernoulli’s equation between the piston surface and at the bottom of the container.

let, ‘ρ’ be the density of water, ‘A’ be the area of the piston, ‘a’ be the area of the hole, ‘v₁’ be the velocity of the piston, ‘v₂’ be the velocity of water coming out of the hole, ‘W’ be the weight of the piston.

W = 100 g

Bernoulli’s equation:

\(\frac{P}{{\rho g}} + \frac{1}{2}\frac{{{v^2}}}{g} + z = constant\)

\(\frac{{{P_A}}}{{\rho g}} + \frac{{v_1^2}}{{2g}} + 0.5 = \frac{{{P_B}}}{{\rho g}} + \frac{{v_2^2}}{{2g}}\)

\(\Rightarrow \frac{{{P_{atm}}}}{{\rho g}} + \frac{W}{{A.\rho g}} + \frac{{v_1^2}}{{2g}} + 0.5 = \frac{{{P_{atm}}}}{{\rho g}} + \frac{{v_2^2}}{{2g}}\)

\(\Rightarrow \left[ {\frac{W}{{A.\rho g}} + 0.5 = \frac{{v_2^2}}{{2g}} - \frac{{v_1^2}}{{2g}}} \right]\)      ----(1)

Now applying continuity equation

Av₁ = av₂

⇒ 10^-2v₁ = 10^-3v₂

⇒ 10v₁ = v₂

From eqn(1)

\(\Rightarrow \frac{W}{{A.\rho g}} + 0.5 = \frac{{{{\left( {10{v_1}} \right)}^2} - v_1^2}}{{2g}}\)

\(\Rightarrow \frac{{100 \times 10}}{{{{10}^{ - 2}} \times {{10}^3} \times 10}} + 0.5 = \frac{{99v_1^2}}{{20}}\)

\(\Rightarrow 10.5 = \frac{{99v_1^2}}{{20}}\)

⇒ v₁ = 1.456 m/s

Mistake Point:

The possible mistake that can happen here is that considering the system as a free container without any weight and by mistake applying.

\({v_2} = \sqrt {2gH} \)

H = 0.5 m

v₂ = √10 m/s = 3.162 m/s

and Av₁ = av₂

⇒ 10^-2 × v₁ = 10^-3 × 3.162

V₁ = 0.316 m/s

This is incorrect as the piston has mass thus pressure at the top will be more than the (atm) pressure

Concept:

Under the steady state; the total heat flux coming inside is equal to the total heat flux going outside.

Here we can find out the heat flux coming out.

Equivalently

Since there is vacuum between the plates the heat transfer will take place between the black plates due to radiation only

Radiation heat transfer between black plates is given by 

\(Q=σ;(T_1^4-T_2^4)\) (Heat flux due to radiation), where σ = Stefan-Boltzmann constant

Calculation:

Given, σ = 5.67 × 10-8 W/m2 K4.

T₁ = 350 K, T₂ = 300 K

Q = 5.67 × 10^-8 [(350)⁴ – (300)⁴] W/m² = 5.67 × 69.0625 W/m² = 391.584 W/m²

Mistake Point:

The common mistake that can happen here is that we take the whole system to be in the vacuum.

Q = σ. (T⁴₁ – T⁴₂) = 5.67 × 10^-8 [(400)⁴ - (300)⁴] = 992.25 W/m²

This is incorrect as the whole heat transfer doesn’t take place sole by radiation but also due to conduction.

But from (350 K) plate to (300 K) plate it takes only due to radiation.

Concept:

Under the given situation the piston will move downwards only when the total downward pressure is more than or equal to the total upward pressure.

FBD of the piston

Total downward pressure = [external pressure] + [pressure due to weight of piston]

When the piston is just about to move so the air inside undergoes a constant volume process.

\( {\frac{{{P_1}}}{{{T_1}}} = \frac{{{P_2}}}{{{T_2}}}} \)

Calculation:

Total downward pressure \(= \left( {100kPa + \frac{W}{A}} \right)\)

 

Total vertical pressure = Pressure of inside air = P₀

Limiting condition

\({P_0} = 100 + \frac{W}{A} = 100 + \frac{{Mg}}{A}\)

\(\Rightarrow {P_0} = \left( {100 + \frac{{25 \times g \times {{10}^{ - 3}}}}{{100 \times {{10}^{ - 4}}}}} \right)kPa\) [Take care of the units]

⇒ P₀ = 125 KPa

Given, P_i = 200 KPa, T_i = 400°C

Let the final temperature be (T₀)

As the piston is just about to move so the air inside undergoes a constant volume process.

\(\left[ {\frac{{{P_1}}}{{{T_1}}} = \frac{{{P_2}}}{{{T_2}}}} \right]\)

\(\Rightarrow \frac{{{P_i}}}{{{T_i}}} = \frac{{{P_0}}}{{{T_0}}}\)

\(\Rightarrow \frac{{200}}{{673}} = \frac{{125}}{{{T_0}}};{T_0} = 420.625;K\)

T₀ = 147.625 °C

Mistake Point: The possibility of mistake here is not changing the temperature from (°C) → (K).

\(\left( {\frac{{{P_1}}}{{{T_1}}} = \frac{{{P_2}}}{{{T_2}}}} \right) \Rightarrow \frac{{{P_0}}}{{{T_0}}} = \frac{{{P_i}}}{{{T_i}}}\)

\(\Rightarrow \frac{{125}}{{{T_0}}} = \frac{{200}}{{400}}\)

⇒ T₀ = 250°C

This is incorrect as temperature are in absolute scale.

Concept:

The given p-h diagram is for VCRS cycle, 

1 - 2 → Compression

4 - 1 → Evaporation

Let (ṁf) be the mass flow rate of the refrigerant in (kg/s)

Cooling takes place in the evaporator.

Cooling power of evaporation = ṁ_f × (h1 – h4)

Let ‘η_c’ be the isentropic efficiency of the compressor.

Compressor power (magnitude) \(= \frac{{{{\dot m}_f}}}{η_c } \times \left[ {{h_2} - {h_1}} \right]\)

Calculation:

Given, η_c = 0.75

Cooling power of evaporation = ṁf × (400 - 250) = 150 ṁ_f

Given cooling power = 15 kW

⇒ 15 = 150 ṁf (take care of units)

⇒ ṁf = 0.1 kg/s

Compressor power (magnitude) \(= \frac{{{{\dot m}_f}}}{η_c } \times \left[ {{h_2} - {h_1}} \right]\)

\(P = \frac{{0.1}}{{0.75}} \times \left( {475 - 400} \right)KW\)

P = 10 KW

Mistake: Here we commit a mistake by multiplying the efficiency rather than dividing it

Compressor power = η × ṁf × (h2 – h1)

= 0.75 × 0.1 × (475 - 400)

= 5.625 KW

This is incorrect as the compressor is a power consumer.

Calculation:

Let ‘P’ be the atmospheric pressure in (kPa)

‘ϕ’ be the relative humidity, ‘P_v’ be the partial pressure of the air in (KPa), ‘P_vs’ be the saturation pressure of water in (KPa), ‘ω’ be the specific humidity of air in (kg/kg of dry air)

\(ϕ = \frac{{{P_v}}}{{{P_{vs}}}} \)

\(\omega = \frac{{0.622{P_v}}}{{\left( {P - {P_v}} \right)}} \)

Calculation:

Given, ϕ = 60%, P = 100 kPa, P_vs = 4.24 kPa

\(\Rightarrow 0.6 = \frac{{{P_v}}}{{4.24}};{P_v} = 2.54;kPa\)

\(\omega = \frac{{0.622{P_v}}}{{\left( {P - {P_v}} \right)}} = \frac{{0.622 \times 2.54}}{{\left( {100 - 2.54} \right)}}\)

ω = 0.01623 kg/kg of dry air = 16.23 g/kg of dry air

Mistake: By applying the wrong formula.

\(ϕ = \frac{{{P_{vs}}}}{{{P_v}}} \Rightarrow 0.6 = \frac{{4.24}}{{{P_v}}}\)

P_v = 7.06 KPa

\(\omega = \frac{{0.622 \times {P_v}}}{{\left( {P - {P_v}} \right)}} = \frac{{0.622 \times 7.06}}{{\left( {100 - 7.06} \right)}} = ;0.04724;kg/kg;of;dry;air\)

ω = 47.24 kg/kg of dry air

This is wrong as the partial pressure is always less than the saturation pressure. This is due to the wrong formula.

The given problem can be solved by model laws.

We (know) power (P) = ρQgH

‘ρ’ is the density of fluid, ‘Q’ is the discharge volumetric, ‘H’ is the head, ‘P₀’ be the power of full-sized turbine.

Power of the model (P_m) = ρQgH = 10³ × 1 × 10 × 2 = 0.02 MW

\({\left. {\frac{{{P_0}}}{{D_0^5N_0^3}}} \right|_{prototype}} = {\left. {\frac{{{P_m}}}{{D_m^5N_m^3}}} \right|_{model}}\)

Also given: H_m = 2 m, D_m = D_o/5, N_m = 450 rpm, H₀ = 30 m

\(\frac{{{H_0}}}{{D_0^2N_0^2}} = \frac{{{H_m}}}{{D_m^2N_m^2}}\)

\(\Rightarrow \frac{{30}}{{D_0^2N_0^2}} = \frac{2}{{{{\left( {\frac{1}{5}} \right)}^2}D_0^2}} \times {\left( {450} \right)^2}\)

⇒ N₀ = 348.56 πpm

Using this in equation:

\({\left. {\frac{{{P_0}}}{{D_0^5N_0^3}}} \right|_{prototype}} = {\left. {\frac{{{P_m}}}{{D_m^5N_m^3}}} \right|_{model}}\)

\(\frac{{{P_0}}}{{D_0^5 \times {{\left( {348.56} \right)}^3}}} = \frac{{0.02}}{{{{\left( {\frac{1}{5}} \right)}^5}D_0^5 \times {{\left( {450} \right)}^3}}}\)

⇒ P₀ = 29.045 MW

Points to remember: To remember and apply all the formulas correctly. Any mistakes in the formula will result into incorrect answer.

Concept:

The relationship between true stress and strain is governed by the equation 

σ = σ0 ϵη

At UTS, ϵ = n

ϵ = True strain \(= \ln \left( {\frac{{{L_f}}}{{{L_i}}}} \right)\)

Where (Lf) is the final length; (Li) is the initial length.

Calculation:

Here σ = 1020 ε^0.4

So ϵ = 0.4 at (UTS)

\(\Rightarrow 0.4 = \ln \left( {\frac{{{L_f}}}{{{L_i}}}} \right) \Rightarrow \frac{{{L_f}}}{{{L_i}}} = {e^{0.4}}\)

\(\Rightarrow \frac{{{L_f}}}{{{L_i}}} = 1.491\)

As volume remains constant throughout

(A_iL_i = A_f L_f) Given A_i = 100 mm²

⇒ 100 L_i = A_f L_f

\(\Rightarrow {A_f} = 100 \times \frac{{{L_i}}}{{{L_f}}} \Rightarrow {A_f} = \frac{{100}}{{1.49}}\)

A_f = 67.03 mm²

Concept:

Formula for drawing stress is given as

\({\sigma _f} = {\left( {{\sigma _\gamma }} \right)_{avg}}\left{ {1 + \frac{1}{{\mu .\cot \propto }}} \right}\left[ {1 - {{\left( {\frac{{{d_f}}}{{{d_i}}}} \right)}^{2\mu \cot \propto }}} \right]\)

Calculation:

Here the formula for drawing stress is already given we just need to put the values to get the answer.

\({\sigma _f} = {\left( {{\sigma _\gamma }} \right)_{avg}}\left{ {1 + \frac{1}{{\mu .\cot \propto }}} \right}\left[ {1 - {{\left( {\frac{{{d_f}}}{{{d_i}}}} \right)}^{2\mu \cot \propto }}} \right]\)

Given (σ_γ)_avg = 350 MPa

μ = 0.1, α = 5°, d_f = 7.5 mm, d_i = 10 mm

\({\sigma _f} = 350\left{ {1 + \frac{1}{{0.1 \times \cot 5}}} \right}\left[ {1 - {{\left( {\frac{{7.5}}{{10}}} \right)}^{2 \times 0.1 \times \cot 5}}} \right] = 316.24;MPa\)

Points to remember: Put the values correctly and while using virtual calculator try to solve bracket wise to avoid unnecessary calculation.

Concept:

If ‘t’ = uncut chip thickness and ‘tc’ = chip thickness 

\(r = \left( {\frac{t}{{{t_c}}}} \right)\) (thickness ratio)

\(\tan \phi = \frac{{r \cos α }}{{1 - r \sin α }}\)

‘ϕ’ is the shear angle

Calculation:

This question formula based.

Given ‘α’ (rake angle) = +15°

‘t’ (unc ut chip thickness) = 0.5 mm

‘t_c’ (chip thickness) = 1.25 mm

\(r = \left( {\frac{t}{{{t_c}}}} \right)=\frac{0.5}{1.25}=0.4\)

‘ϕ’ is the shear angle

\(\tan \phi = \frac{{r \cos α }}{{1 - r \sin α }} \Rightarrow \frac{{0.4 \times \cos 15}}{{1.25 - \left( {0.4 \times \sin 15} \right)}} = 0.431\)

ϕ = tan^-1(0.431) = 23.31°

Concept:

Taylor’s tool life equation is given by 

VTn = constant

As drills are identical hence, they will have the same size. i.e. (diameter)

As feed is also the same hence time to drill a hole will also be the same.

Let ‘t’ be the time to drill a hole.

Also

[velocity ∝ rpm];

[Time ∝ number of holes]

Calculation:

Taylor’s tool life: VTⁿ = constant

Test 1

V₁ ∝ 150 rpm, T₁ ∝ 300 (number of holes)

Test 2

V₂ ∝ 300 rpm, T₂ ∝ 200 (number of holes)

So, V₁T₁ⁿ = V₂T₂ⁿ

150 (300)ⁿ = 300 (200)ⁿ

\(\Rightarrow {\left( {\frac{3}{2}} \right)^n} = 2;n = \frac{{\ln \left( 2 \right)}}{{\ln \left( {1.5} \right)}} = 1.709\)

Test 3,

V₃ ∝ 200 rpm, 

Again

V₁T₁ⁿ = V₃T₃ⁿ

⇒ 150 (300)ⁿ = 200(T₃)ⁿ [n = 1.709]

\(\Rightarrow {\left( {\frac{{{T_3}}}{{300}}} \right)^{1.709}} = \left( {\frac{{15}}{{20}}} \right)\)

⇒ T₃ = 253.52 holes

Mistake Point:

Understand the concept of velocity and time while using Taylor’s equation. Don’t use the velocity ∝ number of holes and Time ∝ rpm

Concept:

Chvorinov’s law for solidification time.

\(time = k{\left( {\frac{{Volume}}{{Area}}} \right)^2}\)

\(\left( {\frac{{Volume}}{{Area}} = Module} \right)\)

Considering riser to be a side-riser.

Module of riser \(= \frac{{\frac{{\pi {D^2}}}{4} \times D}}{{\left( {2 \times \frac{{\pi {D^2}}}{4}} \right) + \pi D \times 1}} = \frac{D}{6}\)

Calculation:

Given, steel rectangular plate with dimensions 80 mm × 120 mm × 20 mm,

Module of steel plate \(= \frac{{80 \times 120 \times 20}}{{2\left[ {80 \times 120 + 20 \times 120 + 80 \times 20} \right]}} = \left( {\frac{{120}}{{17}}} \right)\) 

Time ∝ (Module)²

\(\Rightarrow \frac{{time;of;riser}}{{time;of;plate}} = {\left( {\frac{{Module;of;riser}}{{Module;of;plate}}} \right)^2}\)

\(\Rightarrow \frac{3}{2} = {\left( {\frac{{\frac{D}{6}}}{{\frac{{120}}{{17}}}}} \right)^2}\)

⇒ D = 51.87 mm

Take care of the units and correctly apply the formula to get the answer.

Explanation:

Here the shortest path length from node 1 to node 6 is asked.

So, the resultant is

\(\left( 1 \right)\mathop \to \limits^2 \left( 2 \right)\mathop \to \limits^2 \left( 5 \right)\mathop \to \limits^1 \left( 4 \right)\mathop \to \limits^2 \left( 6 \right)\)

Length = 2 + 2 + 1 + 2 = 7

Mistake Point: Here don’t confuse with the critical path

\(\left( 1 \right)\mathop \to \limits^4 \left( 3 \right)\mathop \to \limits^3 \left( 5 \right)\mathop \to \limits^2 \left( 6 \right)\)

4 + 3 + 4 = 11

Explanation:

The formula for material removal is already given

Given I = 300 A; T = 1600°C

Material removal rate = 4 × 10⁴ × I × T^-1.23

= 4 × 10⁴ × 300 × (1600)^-1.23 = 1374.401 mm³/min

Volume to be removed:

\(V = \frac{\pi }{4}{D^2} \times h = \frac{\pi }{4} \times {\left( {25} \right)^2} \times 20;m{m^3} = ;9817.47;m{m^3}\)

Time taken \(= \frac{{Volume;removed}}{{material;removal;rate}}\)

\(T = \frac{{9817.47}}{{1374.401}}min = 7.14;min\)

Mistake: Don’t change temperature here from C° → K

T = 1600°C already given

T ≠ (1600 + 273) = 1873 K

By doing this material removal rate will come different and hence the time will change which will load to incorrect answer.

Concept:

Power (P) = Voltage (V) × Current (I) = V × I

Calculation:

This problem can simply be done by balancing the dimension in (LHS) and (RHS).

Power (P) = Voltage (V) × Current (I) = 30 × 100 Watt

Consumed power = η × P

Consumed power = 0.70 × 30 × 100 W = 2100 W

Now

Consumed power = specific energy × Area × welding speed

⇒ 2100 = 14 × 20 × V [Taking care of units]

⇒ V = 7.5 mm/s [all are in mm]

Points to remember: here is that units must be same along both (LHS) and (RHS). As all the specific energy and area are in terms of (mm) so velocity will also come in the terms of (mm).

Concept:

The heat transfer is given by (Q)

Q = U Acs (LMTD)

U = overall heat transfer coefficient, Acs = surface area

LMTD of the system:

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

Calculation:

Here, 

θ1 = 30 – 14 = 16°

θ2 = 30 – 22 = 8°

\({\theta _m} = \frac{{{\theta _1} - {\theta _2}}}{{\ln \left( {\frac{{{\theta _1}}}{{{\theta _2}}}} \right)}} = \frac{{16 - 8}}{{\ln \left( {\frac{{16}}{8}} \right)}} = \frac{8}{{\ln 2}} = 11.54\)

The heat transfer is given by (Q)

Q = 2000 × 50 × 11.54 = 1.154 MW

Concept:

A spark-ignition engine works upon Otto cycle

Air standard, efficiency for an Otto cycle is given by \({\eta = 1 - \frac{1}{{{r^{\gamma - 1}}}}} \)

Where ‘r’ is compression ratio, γ = 1.4 for air

Efficiency is also given by 

\(\eta = \frac{{Power;output}}{{Heat;input}}\)

Heat input = Q = (mass flow rate of fuel) × (calorific value of fuel) = ṁ × cv

Calculation:

Given, ṁ = 10.3 kg/hr, cv =  44,000 kJ/kg, Power output = 70 kW

\(Q = \left( {\frac{{10.3}}{{3600}} × {44,000}} \right)kW = ;125.89;KW\)

\(\eta = \frac{{70}}{{125.89}} = 0.556\)

Now, 

\(0.556 = 1 - \frac{1}{{{r^{\left( {1.4 - 1} \right)}}}} \Rightarrow \frac{1}{{{{\text{r}}^{0.4}}}} = 0.444\)

⇒ r = 7.61