GATE 2025 Mechanical Engineering (ME) Question Paper with Solutions

 

GATE 2025 Mechanical Engineering (ME)

Total Marks: 100

The GATE 2025 Mechanical Engineering (ME) examination, conducted , represents one of the most prestigious competitive benchmark assessments for engineering graduates across India. Designed to evaluate a candidate’s conceptual clarity, analytical skills, and practical problem-solving ability, this paper serves as the gateway for admission into postgraduate programs (M.Tech, Ph.D.) at premier institutions like IITs, NITs, and IISc, as well as recruitment in top Public Sector Undertakings (PSUs).

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📌 Paper Structure & Mark Distribution

The examination follows a 3-hour duration format with a total of 65 questions accounting for 100 marks. The questions are structured into two distinct sections:

• General Aptitude (GA): Comprises 10 questions carrying 15 marks (5 one-mark questions and 5 two-mark questions). It evaluates verbal ability, numerical reasoning, analytical reading, and spatial aptitude.
• Subject-Specific Core Engineering (ME): Comprises 55 questions carrying 85 marks. This includes Engineering Mathematics alongside core Mechanical discipline subjects.

🎯 Question Types & Marking Scheme

To rigorously test conceptual rigor, the test paper incorporates three distinct question formats:

• Multiple Choice Questions (MCQs): Objective questions with four options and a single correct choice. Contains negative marking (¹⁄₃ mark deduction for 1-mark questions; ²⁄₃ mark deduction for 2-mark questions).
• Multiple Select Questions (MSQs): Questions with one or more correct options. No negative marking, but full credit is awarded only if all correct choices are selected.
• Numerical Answer Type (NAT): Calculation-based questions requiring exact numerical answers rounded to specified decimals or integers. No negative marking.

⚙️ Key Syllabus Clusters Covered

The core technical section covers fundamental domains of Mechanical Engineering, including:

Domain	Core Subjects Included
Design & Mechanics	Engineering Mechanics, Strength of Materials, Theory of Machines, Machine Design
Thermal & Fluid Sciences	Thermodynamics, Fluid Mechanics, Heat Transfer, Applications (IC Engines, Power Plants, RAC)
Manufacturing & Industrial	Casting, Joining, Machining, Metrology, Operations Research, Production & Inventory Control

A strategic approach to this paper requires a balanced combination of strong fundamental theory, speed, and numerical accuracy. Utilizing the interactive solutions provided below will help analyze step-by-step methodology and refine exam performance.

Question 1

Fish : Shoal :: Lion : ________

Select the correct option to complete the analogy.

Solution

Shoal refers to a group of fish swimming together.

Similarly, a group of lions is called a Pride.

✅ Correct Answer: (A) Pride

Question 2

Identify the grammatically correct sentence:

Solution

Relative pronoun "who" ke saath aane wala verb uske antecedent (pehle aane wale subject) ke anusar change hota hai.

Yahan antecedent "I" hai, isliye iske saath verb "am" use hoga (I am responsible).

✅ Correct Answer: (A) It is I who am responsible for this fiasco.

Question 3

Two cars, P and Q, start from a point X in India at 10 AM. Car P travels North with a speed of 25 km/h and car Q travels East with a speed of 30 km/h. Car P travels continuously but car Q stops for some time after travelling for one hour. If both the cars are at the same distance from X at 11:30 AM, for how long (in minutes) did car Q stop?

Solution

1. Car P dwara tay ki gayi doori:
Time taken (10:00 AM se 11:30 AM) = 1.5 hours.
Distance of P from X = 25 km/h × 1.5 h = 37.5 km.

2. Car Q dwara liya gaya time:
Dono 11:30 AM par same distance par hain, isliye Q ki doori bhi 37.5 km hai.
Q dwara travel karne ka time = Distance / Speed = 37.5 / 30 = 1.25 hours (yani 1 hour 15 minutes = 75 minutes).

3. Stopping time calculation:
Total time elapsed (10:00 AM se 11:30 AM) = 90 minutes.
Stopping time of Q = 90 minutes - 75 minutes = 15 minutes.

✅ Correct Answer: (C) 15

Question 4

The ceiling function of a real number x, denoted by ce(x), is defined as the smallest integer that is greater than or equal to x. Similarly, the floor function, denoted by fl(x), is defined as the largest integer that is smaller than or equal to x. Which one of the following statements is NOT correct for all possible values of x?

Solution

1. Definition ke anusar Check karte hain:
Definition se, hamesha fl(x) ≤ x ≤ ce(x) hota hai.

2. Integers ke liye counter-example:
Agar x ek integer ho (jaise x = 3), toh:
• ce(3) = 3
• fl(3) = 3

Yahan fl(3) = ce(3) ho jata hai, isliye fl(x) < ce(x) sabhi values ke liye sahi nahi hai (ye strictly less than nahi, balki equal bhi ho sakta hai).

✅ Correct Answer: (D) fl(x) < ce(x)

Question 5

P and Q play chess frequently against each other. Of these matches, P has won 80% of the matches, drawn 15% of the matches and lost 5% of the matches.

If they play 3 more matches, what is the probability of P winning exactly 2 of these 3 matches?

Solution

1. Probabilities define karein:
• P ke jeetne ki probability, p = 80% = 80/100 = 4/5
• P ke na jeetne ki probability (draw ya loss), q = 1 - p = 1 - 4/5 = 1/5

2. Binomial Distribution Formula:
Total matches, n = 3
P ko exactly k = 2 matches jeetne hain.

P(X = 2) = ³C₂ × (p)² × (q)^3-2

3. Calculation:
• ³C₂ = 3
• P(X = 2) = 3 × (4/5)² × (1/5)¹
• P(X = 2) = 3 × (16/25) × (1/5) = 48 / 125

✅ Correct Answer: (A) 48 / 125

Question 6

Identify the option that has the most appropriate sequence such that a coherent paragraph is formed:

P. At once, without thinking much, people rushed towards the city in hordes with the sole aim of grabbing as much gold as they could.

Q. However, little did they realize about the impending hardships they would have to face on their way to the city: miles of mud, unfriendly forests, hungry beasts and inimical local lords — all of which would reduce their chances of getting gold to almost zero.

R. All of them thought that easily they could lay their hands on gold and become wealthy overnight.

S. About a hundred years ago, the news that gold had been discovered in Kolar spread like wildfire and the whole State was in raptures.

Solution

1. Opening Sentence (S):
Paragraph S se shuru hoga kyunki ye background set karta hai ("About a hundred years ago...").

2. Immediate Action (P):
Gold ki khabar milte hi logon ne bina soche samjhe kooch kar diya (P: "At once, without thinking much...").

3. Expectation (R):
Sabhi ne socha ki ve aasani se raato-raat ameer ban jayenge (R: "All of them thought that easily they could...").

4. Contrast/Reality (Q):
Q transition-word "However" ke saath aane walii hardships ko batata hai jo unki expectations ke opposite hai.

Is prakar sahi sequence S → P → R → Q hai.

✅ Correct Answer: (D) S → P → R → Q

Question 7

If HIDE and CAGE are coded as 19-23-7-11 and 5-2-17-11 respectively, then what is the code for HIGH?

Solution

1. Coding Pattern samjhein:
Pehle Alphabets ki Reverse Alphabetical Positions (A=26, B=25, ..., Z=1) nikalte hain:

• H = 19, I = 18, D = 23, E = 22
• C = 24, A = 26, G = 20, E = 22

2. Given codes se compare karein:
• HIDE → (19), (18+5=23), (23), (22-11=11) → 19-23-7-11
• direct direct letter-to-code mapping lekar bhi **HIGH** nikal sakte hain:

• H = 19 (HIDE se)
• I = 23 (HIDE se)
• G = 17 (CAGE se)
• H = 19 (HIDE se)

Is prakar **HIGH** ka code 19-23-17-19 banega.

✅ Correct Answer: (D) 19-23-17-19

Question 8

The given figure is reflected about the horizontal dashed line and then rotated clockwise by 90° about an axis perpendicular to the plane of the figure.

Which one of the following options correctly shows the resultant figure?
Note: The figures shown are representative.

Solution

1. Horizontal Reflection (Flip top-to-bottom):
Jab ham figure ko horizontal dashed line ke along reflect (flip) karenge:

• Top shapes niche chale jayenge aur Bottom shapes upar aa jayenge.
• Sabhi individual shapes vertically invert (ulte) ho jayenge.

2. 90° Clockwise Rotation:
Flip ke baad jab figure ko 90° Clockwise ghumate hain:

• Crescent arc (jo bottom-right me tha) flip hokar top-right me aayega aur 90° rotation ke baad top-left me horizontal curvature ke saath dikhega.
• Arrow element rotate hokar option (A) ki tarah left-pointing position align karega.

Is transformation ke baad resultant figure Option (A) se match karti hai.

? Correct Answer: (A)

Question 9

Which one of the following options has the correct sequence of objects arranged in the increasing number of mirror lines (lines of symmetry)?

Solution

1. Sabhi shapes ke Mirror Lines (Lines of Symmetry) calculate karte hain:

• Isosceles triangle: 1 line of symmetry
• Equilateral triangle: 3 lines of symmetry
• Square: 4 lines of symmetry
• Circle: Infinite lines of symmetry

2. Increasing Order (Badhte kram me):
1 < 3 < 4 < Infinite
Isosceles triangle → Equilateral triangle → Square → Circle

? Correct Answer: (B) Isosceles triangle; Equilateral triangle; Square; Circle

Question 10

A final year student appears for placement interview in two companies, S and T. Based on her interview performance, she estimates the probability of receiving job offers from companies S and T to be 0.8 and 0.6, respectively. Let p be the probability that she receives job offers from both the companies. Select the most appropriate option.

Solution

1. Given Values:
• P(S) = 0.8
• P(T) = 0.6
• P(S ∩ T) = p

2. Maximum Value of p:
Intersection probability kabhi bhi individual sets ki probability se badi nahi ho sakti:
p = min(P(S), P(T)) = min(0.8, 0.6) = 0.6

3. Minimum Value of p:
Probability formula: P(S ∪ T) = P(S) + P(T) - P(S ∩ T) ≤ 1
0.8 + 0.6 - p ≤ 1
1.4 - p ≤ 1 → p ≥ 0.4

Isliye, p ki range 0.4 ≤ p ≤ 0.6 hai.

? Correct Answer: (B) 0.4 ≤ p ≤ 0.6

Question 11

Let A and B be real symmetric matrices of same size. Which one of the following options is correct?

Solution

1. Matrix Transpose Law Check karein:
Kisi bhi do matrices A aur B ke product ka transpose Reversal Law follow karta hai:
(AB)^T = B^TA^T

2. Baaki Options kyu galat hain:
• (A) A^T = A^-1 keval Orthogonal Matrices ke liye hota hai.
• (B) AB = BA keval tabhi hota hai jab matrices Commute karti hon (hamesha zaroori nahi).
• (D) A = A^-1 keval Involutory Matrices ke liye hota hai.

? Correct Answer: (C) (AB)^T = B^TA^T

Question 12

For the differential equation given below, which one of the following options is correct?

$$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \quad (0 \leq x \leq 1, 0 \leq y \leq 1)$$

Solution

Laplace Equation Check karein:
Given equation: $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$

For Option (B): $u = e^x \sin y$
• $\frac{\partial u}{\partial x} = e^x \sin y \implies \frac{\partial^2 u}{\partial x^2} = e^x \sin y$
• $\frac{\partial u}{\partial y} = e^x \cos y \implies \frac{\partial^2 u}{\partial y^2} = -e^x \sin y$

Adding both terms: $e^x \sin y - e^x \sin y = 0$. Hence, Option (B) satisfies the Laplace equation.

✅ Correct Answer: (B) u = e^x sin y is a solution for all x and y

Question 13

The divergence of the curl of a vector field is

Solution

Vector Identity:
Kisi bhi vector field $\mathbf{V}$ ke liye, divergence of curl ki identity hoti hai:
$$\nabla \cdot (\nabla \times \mathbf{V}) = 0$$ Pura divergence of curl humesha zero hota hai.

✅ Correct Answer: (D) zero

Question 14

If two unbiased coins are tossed, then what is the probability of having at least one head?

Solution

Probability Calculation:
• Sample Space = {HH, HT, TH, TT} $\implies$ Total outcomes = 4
• Favorable outcomes (at least one head) = {HH, HT, TH} $\implies$ 3 outcomes

$$\text{Probability} = \frac{3}{4} = 0.75$$

✅ Correct Answer: (D) 0.75

Question 15

Let a spherical block of ice at -7 °C be exposed to atmospheric air at 30 °C with the gravitational direction downwards. What will be the overall direction of air flow in this situation?

Solution

Free Convection Concept:
• Jab 30 °C ki warm air cold ice block (-7 °C) ke contact mein aati hai, toh hava thandi hokar denser (bhari) ho jaati hai.
• Gravity ki wajah se denser thandi hava neeche ki taraf move karti hai.
• Is natural convection process ki wajah se air flow ka overall direction Downward hota hai.

✅ Correct Answer: (B) Downward

Question 16

Consider two identical tanks with a bottom hole of diameter d. One tank is filled with water and the other tank is filled with engine oil. The height of the fluid column h is same in both the cases. The fluid exit velocity in the two tanks are V₁ and V₂. Neglecting all losses, which one of the following options is correct?

Solution

Torricelli's Law:
• Jab saare losses neglect kiye jaate hain, tab exit velocity formula hota hai:
$$V = \sqrt{2gh}$$ • Ye velocity fluid ki density ya viscosity par depend nahi karti.
• Kyunki dono tanks mein fluid height h same hai, isliye exit velocity dono tanks ke liye equal hogi ($V_2 = V_1$).

✅ Correct Answer: (B) V₂ = V₁

Question 17

In a laboratory experiment using a scaled down model to measure scour at a bridge pier, the Froude number is important. The ratio of the prototype length to the model length is 100. If the velocity of the model is 1 m s^-1, the velocity (in m s^-1) of the prototype is

Solution

Froude Model Law:
• Froude number equality: $Fr_p = Fr_m \implies \frac{V_p}{\sqrt{g L_p}} = \frac{V_m}{\sqrt{g L_m}}$

• Velocity Ratio Formula:
$$\frac{V_p}{V_m} = \sqrt{\frac{L_p}{L_m}}$$ • Given $\frac{L_p}{L_m} = 100$ and $V_m = 1 \text{ m/s}$:
$$V_p = V_m \times \sqrt{100} = 1 \times 10 = 10 \text{ m s}^{-1}$$

✅ Correct Answer: (C) 10

Question 18

Air inside a rigid, thermally-insulated tank undergoes stirring using a mechanical stirrer. Which one of the following options is correct?

Solution

Thermodynamic Analysis:
• Enthalpy ($H$): Stirrer ke dwara work input ($W_{in}$) system mein add ho raha hai, aur tank insulated ($Q = 0$) hai. First Law se internal energy aur temperature dono badhenge ($dU = dW_{in}$). Ideal gas ke liye $H = C_p T$, isliye enthalpy **increase** hogi.
• Entropy ($S$): Mechanical stirring ek irreversible process hai ($\Delta S_{gen} > 0$). Thermal insulation ki wajah se $dS = \frac{dQ}{T} + dS_{gen} = 0 + dS_{gen} > 0$, isliye entropy bhi **increase** hogi.

✅ Correct Answer: (C) Both the enthalpy and the entropy of the air increase

Question 19

In a psychrometric chart, one axis represents dry-bulb temperature. The axis, that is perpendicular to the dry-bulb temperature axis, represents

Solution

Psychrometric Chart Axes:
• Horizontal axis (X-axis) Dry-Bulb Temperature ko show karta hai.
• Vertical axis (Y-axis), jo horizontal axis ke perpendicular (perpendicular) hota hai, **Specific Humidity** (Humidity Ratio, $\omega$) ko represent karta hai.

✅ Correct Answer: (B) specific humidity

Question 20

Among the following surface hardening processes, steel is heated to the lowest temperature in

Solution

Process Temperatures Comparison:
• Carburizing: 870 °C – 950 °C
• Cyaniding: 800 °C – 870 °C
• Carbonitriding: 750 °C – 850 °C
• Nitriding: 500 °C – 550 °C

Nitriding process Ferritic region (sub-critical temperature) mein perform hota hai, isliye iska temperature sabse kam hota hai.

✅ Correct Answer: (C) nitriding

Question 21

The welding process commonly used for fabricating tailor-welded blanks of dissimilar thickness for automotive applications is

Solution

Automotive Tailor-Welded Blanks (TWBs):
• Automotive industry mein alag-alag thickness aur materials ki sheets ko join karne ke liye **Laser Beam Welding (LBW)** ka use hota hai.
• Laser welding high precision, low heat input, aur narrow weld seam provide karti hai jo automated sheet metal forming ke liye ideal hai.

✅ Correct Answer: (B) laser welding

Question 22

The yield stress of a metal in uniaxial tension is 200 MPa. According to von Mises yield criterion, the yield stress (in MPa) of the metal in pure shear is closest to

Solution

von Mises Yield Criterion:
According to von Mises theory, pure shear yield strength ($\tau_y$) aur uniaxial tensile yield strength ($\sigma_y$) ke beech ka relation hai:
$$\tau_y = \frac{\sigma_y}{\sqrt{3}}$$ Given: $\sigma_y = 200 \text{ MPa}$

$$\tau_y = \frac{200}{\sqrt{3}} \approx \frac{200}{1.732} \approx 115.47 \text{ MPa} \approx 115.5 \text{ MPa}$$

✅ Correct Answer: (A) 115.5

Question 23

In computer aided design (CAD), solid models can be constructed using

Solution

Solid Modeling Techniques:
• CAD mein solid modeling ke do primary standard approaches hote hain: **Boundary Representation (B-rep)** aur **Constructive Solid Geometry (CSG)**.
• B-rep solid ko uski outer boundaries (vertices, edges, faces) se define karta hai.
• Baaki options (Bezier curves, B-splines, NURBS) curves/surfaces representation ke liye use hote hain, primary solid modeling schemes ke liye nahi.

✅ Correct Answer: (A) boundary representation (B-rep)

Question 24

Ceramics and glass are machined by

Question 25

When assembled, the hole 30^+0.030_+0.000 mm and shaft 30^+0.020_-0.020 mm will result in

Solution

Hole Limits:
• Maximum Hole Size = 30.030 mm
• Minimum Hole Size = 30.000 mm

Shaft Limits:
• Maximum Shaft Size = 30.020 mm
• Minimum Shaft Size = 29.980 mm

Analysis:
• Maximum Clearance = Max Hole - Min Shaft = 30.030 - 29.980 = +0.050 mm (Clearance)
• Minimum Clearance = Min Hole - Max Shaft = 30.000 - 30.020 = -0.020 mm (Interference)

Cunki yahan par Clearance aur Interference dono ki possibility hai, isliye yeh Transition Fit hai.

✅ Correct Answer: (C) transition fit

Question 26

A rigid circular disc of radius r (in m) is rolling without slipping on a flat surface. The angular velocity of the disc is ω (in rad s^-1). The velocities (in m s^-1) at points O and A, respectively, are

Solution

Pure Rolling Motion Analysis:
• Center of Mass (Point O) ki linear velocity rightward direction mein: V_O = rω î hoti hai.
• Pure rolling (without slipping) ki condition mein, contact point (Point A) par instantaneous velocity zero hoti hai: V_A = 0 î.

Isliye, point O aur point A ki velocities respectively rω î and 0 î honge.

✅ Correct Answer: (A) rω î and 0 î

Question 27

A truss structure is loaded as shown in the figure below. Among the options given, which member in the truss is a zero-force member?

Solution

Joint B ka Equilibrium:
• Joint B par teen members milte hain: AB, BC, aur BD.
• Members AB aur BC collinear hain (dono same straight line AC par hain).
• Member BD is straight line ke perpendicular hai (angle = 90 degree) aur joint B par koi external force apply nahi ho raha hai.
• Zero-force member rule ke mutabiq: Agar teen members ek joint par milte hain aur unme se do collinear hote hain jabki teesre member par koi external load nahi lag raha hota, toh teesra member **zero-force member** hota hai.

Isliye, member BD me force F_BD = 0 hoga.

✅ Correct Answer: (A) BD

Question 28

A metallic square plate is subjected to a uniform hydrostatic pressure (P). Choose the correct Mohr's circle representing the state of stress at any point in the plate from the options given below.

For the Mohr's circle, normal stress is positive towards right and shear stress is positive in the upward direction.

Solution

Hydrostatic Pressure Analysis:
• Hydrostatic pressure me sabhi directions me compressive stress (normal stress = -P) lagta hai aur koi shear stress nahi hota.
• Principal stresses: sigma_x = -P, sigma_y = -P, aur tau_xy = 0.

Mohr's Circle Parameters:
• Center = ((sigma_x + sigma_y) / 2, 0) = (-P, 0)
• Radius = sqrt(((sigma_x - sigma_y) / 2)^2 + tau_xy^2) = 0

Cunki radius zero hai, isliye Mohr's circle ek **point** ban jata hai jo negative normal stress axis par **-P** position par lie karta hai.

✅ Correct Answer: (B) A point on the negative normal stress axis at -P

Question 29

In the context of balancing of rotating and reciprocating masses, which one of the following options is true?

Solution

Balancing Concept Analysis:
• **Dynamic Balancing:** Ek rigid rotor par lagne wale sabhi unbalanced forces aur couples ko complete balance karne ke liye kam se kam do distinct planes mein do balancing masses ki zaroorat hoti hai.
• **Reciprocating Engine:** Single-cylinder engine mein reciprocating mass ko revolving masses se poori tarah (completely) balance nahi kiya ja sakta, kyunki partially balance karne par perpendicular direction mein primary force unbalanced reh jata hai.

✅ Correct Answer: (B) An unbalanced rigid rotor can be completely balanced using two balancing masses attached in two distinct planes

Question 30

A shaft carries a helical spur gear. Which one of the following bearings can NOT be used to support it?

Solution

Helical Gear Forces Analysis:
• Helical gear power transmission ke dauran Radial load ke sath-sath **Axial (Thrust) load** bhi generate karta hai.
• Angular contact bearing, Double-row ball bearing, aur Tapered roller bearing dono axial aur radial loads bear kar sakte hain.
• **Straight roller bearing (Cylindrical roller bearing)** sirf heavy radial load bear kar sakta hai aur axial load withstand nahi kar sakta.

Isliye, Straight roller bearing ka use helical gear shaft ke liye nahi kiya ja sakta.

✅ Correct Answer: (C) Straight roller bearing

Question 31

The values of a function f obtained for different values of x are shown in the table below.

x	0	0.25	0.5	0.75	1.0
f(x)	0.9	2.0	1.5	1.8	0.4

Using Simpson's one-third rule,
Integral from 0 to 1 of f(x) dx ≈ ______ (rounded off to 2 decimal places).

Enter your answer:

Solution

Simpson's 1/3 Rule Formula:
I = (h / 3) * [ (y0 + y4) + 4*(y1 + y3) + 2*(y2) ]

• Step height, h = 0.25 - 0 = 0.25
• Values: y0 = 0.9, y1 = 2.0, y2 = 1.5, y3 = 1.8, y4 = 0.4

• Sum of extreme terms (y0 + y4) = 0.9 + 0.4 = 1.3
• Sum of odd terms 4*(y1 + y3) = 4 * (2.0 + 1.8) = 4 * 3.8 = 15.2
• Sum of even terms 2*(y2) = 2 * 1.5 = 3.0

I = (0.25 / 3) * [ 1.3 + 15.2 + 3.0 ]
I = (0.25 / 3) * 19.5 = 4.875 / 3 = 1.4583 ≈ 1.46

✅ Correct Answer: 1.44 to 1.47 (Exact: 1.46)

Question 32

The thermal efficiency of an ideal air-standard Otto cycle is 0.5. The value of specific heat ratio of air is 1.4. The compression ratio of the cycle is ______ (rounded off to 1 decimal place).

Enter your answer:

Solution

Otto Cycle Thermal Efficiency Formula:
eta = 1 - (1 / r^(gamma - 1))

• Given: eta = 0.5, gamma = 1.4
• 0.5 = 1 - (1 / r^(1.4 - 1))
• 1 / r^0.4 = 0.5
• r^0.4 = 1 / 0.5 = 2
• r = 2^(1 / 0.4) = 2^2.5
• r = 5.6568 ≈ 5.7

✅ Correct Answer: 5.5 to 5.8 (Exact: 5.7)

Question 33

During a welding operation, thermal power of 2500 W is incident normally on a metallic surface. As shown in the figure, the heated area is circular. Out of the incident power, 85% of the power is absorbed within a circle of radius 5 mm while 65% is absorbed within an inner concentric circle of radius 3 mm. The power density in the shaded area is ______ W mm^-2 (rounded off to 2 decimal places).

Enter your answer:

Solution

Power Density Calculation:
• Total Power, P = 2500 W
• Power absorbed in outer circle (r = 5 mm): P_outer = 85% of 2500 = 0.85 * 2500 = 2125 W
• Power absorbed in inner circle (r = 3 mm): P_inner = 65% of 2500 = 0.65 * 2500 = 1625 W
• Power in the shaded annular ring: P_shaded = P_outer - P_inner = 2125 - 1625 = 500 W

• Area of shaded ring: A_shaded = pi * (R^2 - r^2) = pi * (5^2 - 3^2) = pi * (25 - 9) = 16 * pi mm^2 ≈ 50.2655 mm^2

• Power density in shaded area: Power Density = P_shaded / A_shaded = 500 / 50.2655 = 9.947 ≈ 9.95 W mm^-2

✅ Correct Answer: 9.80 to 10.10 (Exact: 9.95)

Question 34

A liquid metal is poured in a mold cavity of size 200 mm × 200 mm × 200 mm. The cooling is uniform in all directions with NO additional compensation for shrinkage. Considering the volumetric shrinkage during solidification and solid contraction as 7% and 8%, respectively, the length of the cube edge after cooling down to the room temperature is ______ mm (rounded off to 1 decimal place).

Enter your answer:

Solution

Step-by-step Calculation:
• Initial cavity volume, V_0 = 200 × 200 × 200 = 8,000,000 mm^3

• Solidification Shrinkage (7%):
Solidification ke baad volume = V_0 × (1 - 0.07) = 8,000,000 × 0.93 = 7,440,000 mm^3

• Solid Contraction (8%):
Room temperature tak cool hone ke baad final volume = 7,440,000 × (1 - 0.08) = 7,440,000 × 0.92 = 6,844,800 mm^3

• Final Edge Length (a):
a^3 = 6,844,800 mm^3
a = (6,844,800)^(1/3) ≈ 189.9 mm

✅ Correct Answer: 189.0 to 190.5 (Exact: 189.9)

Question 35

A block of mass 1 kg connected to a spring of stiffness 10 N m^-1 is operating in a viscous medium such that the damping ratio (or damping factor) is equal to the ratio of the damped frequency to the natural frequency. The magnitude of the damping ratio for this system is ______ (rounded off to 2 decimal places).

Enter your answer:

Solution

Damping Ratio Derivation:
• Damped frequency formula: omega_d = omega_n * sqrt(1 - zeta^2)
• Ratio of damped to natural frequency: omega_d / omega_n = sqrt(1 - zeta^2)

• Question ke mutabiq, Damping ratio (zeta) = omega_d / omega_n
• Therefore: zeta = sqrt(1 - zeta^2)

• Both sides square karne par:
zeta^2 = 1 - zeta^2
2 * zeta^2 = 1
zeta^2 = 1 / 2
zeta = 1 / sqrt(2) ≈ 0.707 ≈ 0.71

✅ Correct Answer: 0.70 to 0.72 (Exact: 0.71)

Question 36

In the closed interval [0, 3], the minimum value of the function f given below is:

f(x) = 2x^3 - 9x^2 + 12x

Solution

Step 1: Critical Points find karein:
f'(x) = d/dx (2x^3 - 9x^2 + 12x) = 6x^2 - 18x + 12
f'(x) = 0 put karne par:
6(x^2 - 3x + 2) = 0
(x - 1)(x - 2) = 0 ⇒ x = 1, x = 2

Step 2: Sabhi points par f(x) evaluate karein (x = 0, 1, 2, 3):
• f(0) = 2(0)^3 - 9(0)^2 + 12(0) = 0
• f(1) = 2(1)^3 - 9(1)^2 + 12(1) = 2 - 9 + 12 = 5
• f(2) = 2(2)^3 - 9(2)^2 + 12(2) = 16 - 36 + 24 = 4
• f(3) = 2(3)^3 - 9(3)^2 + 12(3) = 54 - 81 + 36 = 9

Closed interval [0, 3] mein sabse minimum value 0 hai (jo ki x = 0 par milti hai).

✅ Correct Answer: (A) 0

Question 37

Considering the actual demand and the forecast for a product given in the table below, the mean forecast error and the mean absolute deviation, respectively, are

Period	1	2	3	4	5	6	7	8	9	10
Actual demand	425	415	420	430	427	418	422	416	426	421
Forecast	427	422	416	422	423	420	419	418	430	415

Solution

Forecast Error e_i = Actual (A_i) - Forecast (F_i):
• Period 1: 425 - 427 = -2 (Absolute = 2)
• Period 2: 415 - 422 = -7 (Absolute = 7)
• Period 3: 420 - 416 = +4 (Absolute = 4)
• Period 4: 430 - 422 = +8 (Absolute = 8)
• Period 5: 427 - 423 = +4 (Absolute = 4)
• Period 6: 418 - 420 = -2 (Absolute = 2)
• Period 7: 422 - 419 = +3 (Absolute = 3)
• Period 8: 416 - 418 = -2 (Absolute = 2)
• Period 9: 426 - 430 = -4 (Absolute = 4)
• Period 10: 421 - 415 = +6 (Absolute = 6)

1. Mean Forecast Error (MFE):
Sum of errors = (-2) + (-7) + 4 + 8 + 4 + (-2) + 3 + (-2) + (-4) + 6 = 8
MFE = 8 / 10 = 0.8

2. Mean Absolute Deviation (MAD):
Sum of absolute errors = 2 + 7 + 4 + 8 + 4 + 2 + 3 + 2 + 4 + 6 = 42
MAD = 42 / 10 = 4.2

✅ Correct Answer: (B) 0.8 and 4.2

Question 38

Match the mold elements in the casting process with the most suitable function:

Mold element	Function
P. Blind riser	I. Casting with internal cavity
Q. Chill	II. Molten metal reservoir
R. Skim bob	III. Nucleating agent
S. Core	IV. Assisting in faster heat removal from melt
T. Insulating sleeve	V. Removal of impurities
U. Inoculant	VI. Increasing the solidification time

Solution

Correct Matching Details:
• P. Blind riser ➔ II. Molten metal reservoir: Riser liquid metal ka reservoir hota hai jo shrinkage compensation ke liye use hota hai.
• Q. Chill ➔ IV. Assisting in faster heat removal from melt: Chills rapid cooling / directional solidification ke liye lagaye jaate hain.
• R. Skim bob ➔ V. Removal of impurities: Skim bob slag aur impurities ko trapping karke remove karta hai.
• S. Core ➔ I. Casting with internal cavity: Core hollow internal cavity banana ke kaam aata hai.
• T. Insulating sleeve ➔ VI. Increasing the solidification time: Insulation heat loss kam karta hai jisse solidification time badhta hai.
• U. Inoculant ➔ III. Nucleating agent: Inoculants nucleation rate badhate hain fine grain structure ke liye.

✅ Correct Answer: (A) P-II, Q-IV, R-V, S-I, T-VI, U-III

Question 39

Wire drawing operation is performed on a perfectly plastic metal without any strain hardening. Assuming no friction and no redundant work, the maximum possible percentage reduction in area in a single pass is closest to

Solution

Step-by-step Derivation:
• Ideal wire drawing formula: sigma_d = sigma_y * ln(A_0 / A_f)
• Maximum drawing stress without yielding (wire breakage) occurs when sigma_d = sigma_y.

• Therefore:
sigma_y = sigma_y * ln(A_0 / A_f)
1 = ln(A_0 / A_f)
A_0 / A_f = e^1 ≈ 2.718

• Fractional Reduction in Area (R):
R = (A_0 - A_f) / A_0 = 1 - (A_f / A_0) = 1 - (1 / e)
R = 1 - (1 / 2.718) = 1 - 0.3678 = 0.6322

• Percentage Reduction:
Percentage Reduction = 0.6322 × 100% ≈ 63.2%

✅ Correct Answer: (B) 63.2

Question 40

In relation to additive manufacturing, match the following:

Process	Layer creation technique	Material
P. Stereolithography	1. Injection of powder stream	I. Paper
Q. Fused deposition modeling	2. Extrusion of melted polymer	II. Epoxy
R. Laminated object manufacturing	3. Liquid layer curing	III. Titanium
S. Laser-engineered net shaping	4. Sheet material deposition	IV. Acrylonitrile butadiene styrene (ABS)

Solution

Correct Matching Explanation:
• P. Stereolithography ➔ 3. Liquid layer curing ➔ II. Epoxy: UV laser liquid resin (photopolymer/epoxy) ko cure karta hai.
• Q. Fused Deposition Modeling (FDM) ➔ 2. Extrusion of melted polymer ➔ IV. ABS: Thermoplastic filament (jaise ABS) melt karke extrude ki jati hai.
• R. Laminated Object Manufacturing (LOM) ➔ 4. Sheet material deposition ➔ I. Paper: Laminated sheets/paper layers ko cut aur bond kiya jata hai.
• S. Laser-Engineered Net Shaping (LENS) ➔ 1. Injection of powder stream ➔ III. Titanium: Metal powder stream ko laser energy source dwara melt/deposit kiya jata hai.

✅ Correct Answer: (B) P-3-II, Q-2-IV, R-4-I, S-1-III

Question 41

A plate of 30 mm thickness is fed through a rolling mill with two powered rolls. Each roll has a diameter of 500 mm. The plate thickness is to be reduced to 27 mm in a single pass. Assume no change in width. The process feasibility and the maximum draft (in mm) can be represented, respectively, as

Use the coefficient of friction as 0.12

Solution

Step-by-step Calculation:
• Given data:
Initial thickness (h_1) = 30 mm
Final thickness (h_2) = 27 mm
Required draft (Δh_req) = h_1 - h_2 = 30 - 27 = 3 mm
Roll diameter (D) = 500 mm ⇒ Radius (R) = 250 mm
Coefficient of friction (mu) = 0.12

• Maximum Possible Draft (Δh_max):
Δh_max = mu^2 * R
Δh_max = (0.12)^2 * 250 = 0.0144 * 250 = 3.6 mm

• Feasibility Check:
Chunki required draft (3 mm) < maximum draft (3.6 mm), isliye rolling process Bina kisi external force ke feasible hai.

✅ Correct Answer: (A) feasible and 3.6

Question 42

The system shown in the figure below consists of a cantilever beam (with flexural rigidity EI and negligible mass), a spring (with spring constant K and negligible mass) and a block of mass m. Assuming a lumped parameter model for the system, the fundamental natural frequency (ω_n) of the system is

Solution

Step-by-step Solution:
• Stiffness of the cantilever beam at the free end: k_beam = 3EI / L³
• Beam aur spring mass 'm' ke sath parallel connection mein hain, kyunki dono ka displacement same hai.
• Equivalent stiffness: k_eq = k_beam + K = (3EI / L³) + K
• Natural frequency (ω_n):
ω_n = √(k_eq / m) = √[ (3EI / L³ + K) / m ]

✅ Correct Answer: (A) √[ (3EI / L³ + K) / m ]

Question 43

The endurance limit of a specific grade of steel is same as its yield strength. The ultimate strength of this grade of steel is twice of its yield strength. A component made of this steel is loaded in tension and unloaded periodically. It is required that the component does NOT fail for at least 10⁶ loading cycles, as per the Soderberg law. Considering a factor of safety of 2, the maximum applied tensile principal stress is

Solution

Step-by-step Calculation:
• Given conditions:
Endurance limit S_e = Yield strength S_yt
Factor of safety (N) = 2
Loading: Tensile loading and unloading periodically (sigma_min = 0, sigma_max = sigma)

• Mean stress (sigma_m) and Stress amplitude (sigma_a):
sigma_m = (sigma_max + sigma_min) / 2 = sigma / 2
sigma_a = (sigma_max - sigma_min) / 2 = sigma / 2

• Soderberg Law:
(sigma_a / S_e) + (sigma_m / S_yt) = 1 / N

• Since S_e = S_yt:
(sigma / 2S_e) + (sigma / 2S_e) = 1 / 2
sigma / S_e = 1 / 2
sigma = S_e / 2

Isliye maximum applied stress endurance limit ki **half** (aadhi) hogi.

✅ Correct Answer: (B) half of the endurance limit

Question 44

For a fully-developed pipe flow, which of the following options is/are correct?

(A) For the same maximum velocity, the average velocity is higher in the turbulent regime than that of the laminar regime (B) Compressibility effects are important if Mach number is less than 0.3 (C) For laminar flow, the friction factor is independent of surface roughness (D) For laminar flow, friction factor decreases with decrease in Reynolds number

Solution

Statement Analysis:
• (A) Correct: Laminar flow mein V_avg / V_max = 0.5 hota hai, jabki Turbulent flow mein V_avg / V_max ≈ 0.81 (1/7th power law) hota hai. Isliye same maximum velocity par turbulent flow ki average velocity laminar se adhik hoti hai.
• (B) Incorrect: Flow ko incompressible mana jata hai agar Mach number M < 0.3 ho.
• (C) Correct: Laminar flow mein Darcy friction factor f = 64 / Re hota hai, jo strictly Reynolds number par depend karta hai aur surface roughness se independent hota hai.
• (D) Incorrect: Chunki f = 64 / Re, Reynolds number kam hone par friction factor badhta hai, kam nahi hota.

✅ Correct Options: (A) and (C)

Question 45

If C is the unit circle in the complex plane with its center at the origin, then the value of n in the equation given below is ________ (rounded off to 1 decimal place).

∮_C [ z³ / ( (z² + 4)(z² - 4) ) ] dz = 2 π i n

Enter your answer:

Solution

Step-by-step Calculation:
• Given contour C: |z| = 1 (unit circle).
• Singularities of the integrand f(z) = z³ / ( (z² + 4)(z² - 4) ):
z² + 4 = 0 ➔ z = ± 2i
z² - 4 = 0 ➔ z = ± 2

• Magnitudes of all poles: |z| = 2.
• Chunki contour |z| = 1 hai, isliye **saare poles contour C ke bahar hai** (none of the poles lie inside C).

• Cauchy-Goursat Theorem ke according, agar integrand contour ke andar analytic hai, toh integral = 0.
∮_C f(z) dz = 0

• Equation comparison:
2 π i n = 0 ➔ n = 0

✅ Correct Answer: 0 (or 0.0)

Question 46

The directional derivative of the function f given below at the point (1, 0) in the direction of (1/2)(i + √3 j) is ________ (rounded off to 1 decimal place).

f(x, y) = x² + xy²

Enter your answer:

Solution

Step-by-step Solution:
• Gradient vector ∇f = (∂f/∂x) i + (∂f/∂y) j
∂f/∂x = 2x + y²
∂f/∂y = 2xy

• Point (1, 0) par Gradient evaluation:
∇f(1, 0) = (2(1) + 0²) i + (2(1)(0)) j = 2 i + 0 j

• Direction vector: u = (1/2) i + (√3 / 2) j
Magnitude |u| = √[ (1/2)² + (√3/2)² ] = √[ 1/4 + 3/4 ] = 1 (ye unit vector hai).

• Directional Derivative = ∇f · u
= (2 i + 0 j) · (0.5 i + 0.866 j)
= 2 × 0.5 + 0 = 1

✅ Correct Answer: 1 (or 1.0)

Question 47

Let y be the solution of the differential equation with the initial conditions given below. If y(x = 2) = A ln 2, then the value of A is ________ (rounded off to 2 decimal places).

x² (d²y / dx²) + 3x (dy / dx) + y = 0

Initial conditions: y(x = 1) = 0  and  (dy / dx)(x = 1) = 1

Enter your answer:

Solution

Step-by-step Solution:
• Cauchy-Euler equation: Substitute x = e^z (z = ln x).
D(D - 1)y + 3Dy + y = 0
D² + 2D + 1 = 0
(D + 1)² = 0 ➔ Roots: D = -1, -1 (Repeated roots)

• General Solution:
y = (C₁ + C₂ z) e^-z = (C₁ + C₂ ln x) / x

• Applying initial condition 1: y(1) = 0
0 = (C₁ + 0) / 1 ➔ C₁ = 0
So, y = (C₂ ln x) / x

• Derivative dy/dx:
dy/dx = C₂ (1 - ln x) / x²

• Applying initial condition 2: dy/dx(1) = 1
1 = C₂ (1 - 0) / 1 ➔ C₂ = 1

• Particular Solution: y = (ln x) / x

• At x = 2:
y(2) = (ln 2) / 2 = 0.5 ln 2

• Given y(2) = A ln 2 ➔ A = 0.5

✅ Correct Answer: 0.5 (or 0.50)

Question 48

Consider a cylindrical furnace of 5 m diameter and 5 m length with bottom, top and curved surfaces maintained at uniform temperatures of 800 K, 1500 K and 500 K, respectively. The view factor between the bottom and top surfaces, F₁₂ is 0.2. The magnitude of net radiation heat transfer rate between the bottom surface and the curved surface is ________ kW (rounded off to 1 decimal place).

All surfaces of the furnace can be assumed as black.
The Stefan-Boltzmann constant, σ = 5.67 × 10^-8 W m^-2 K^-4.

Enter your answer:

Solution

Step-by-step Calculation:
• Given values:
Diameter D = 5 m, Radius r = 2.5 m
Area of bottom surface (A₁) = π × (2.5)² = 19.635 m²
Temperatures: T₁ = 800 K, T₂ = 1500 K, T₃ = 500 K
View factor F₁₂ = 0.2

• Summation Rule for Bottom Surface (1):
F₁₁ + F₁₂ + F₁₃ = 1
Flat surface ke liye F₁₁ = 0:
0 + 0.2 + F₁₃ = 1 ➔ F₁₃ = 0.8

• Net Radiation Exchange between Surface 1 and Surface 3 (Q₁₃):
Q₁₃ = σ × A₁ × F₁₃ × (T₁⁴ - T₃⁴)
Q₁₃ = (5.67 × 10^-8) × 19.635 × 0.8 × (800⁴ - 500⁴)
Q₁₃ = (5.67 × 10^-8) × 15.708 × (4.096 × 10¹¹ - 0.625 × 10¹¹)
Q₁₃ = 8.9064 × 10^-7 × 3.471 × 10¹¹
Q₁₃ = 309142.1 W = 309.1 kW

✅ Correct Answer: 309.1 (Acceptable range: 308.5 to 309.8)

Question 49

A pitot tube connected to a U-tube mercury manometer measures the speed of air flowing in the wind tunnel as shown in the figure below. The density of air is 1.23 kg m^-3 while the density of water is 1000 kg m^-3. For the manometer reading of h = 30 mm of mercury, the speed of air in the wind tunnel is ________ m s^-1 (rounded off to 1 decimal place).

Assume: Specific gravity of mercury = 13.6
Acceleration due to gravity = 10 m s^-2

Enter your answer:

Solution

Step-by-step Calculation:
• Given Data:
Density of air (ρ_air) = 1.23 kg/m³
Density of mercury (ρ_m) = 13.6 × 1000 = 13600 kg/m³
Manometer height (h) = 30 mm = 0.03 m
g = 10 m/s²

• Pressure Difference (ΔP):
ΔP = (ρ_m - ρ_air) × g × h
Chunki ρ_air << ρ_m, ΔP ≈ ρ_m × g × h
ΔP = 13600 × 10 × 0.03 = 4080 Pa

• Air Speed (V):
V = √[ (2 × ΔP) / ρ_air ]
V = √[ (2 × 4080) / 1.23 ]
V = √[ 8160 / 1.23 ] = √[ 6634.146 ] ≈ 81.4 m/s

✅ Correct Answer: 81.4 (Acceptable range: 81.0 to 82.0)

Question 50

Consider a velocity field V = 3z i + 0 j + Cx k, where C is a constant. If the flow is irrotational, the value of C is ________ (rounded off to 1 decimal place).

Enter your answer:

Solution

Step-by-step Calculation:
• Velocity components:
u = 3z
v = 0
w = Cx

• Irrotational flow ke liye curl of velocity zero hona chahiye (∇ × V = 0):
ω_y = (1/2) * (∂u/∂z - ∂w/∂x) = 0

• Partial derivatives calculate karte hain:
∂u/∂z = ∂(3z)/∂z = 3
∂w/∂x = ∂(Cx)/∂x = C

• Irrotational condition apply karne par:
∂u/∂z - ∂w/∂x = 0
3 - C = 0 ➔ C = 3

✅ Correct Answer: 3 (or 3.0)

Question 51

Water enters a tube of diameter, D = 60 mm with mass flow rate of 0.01 kg s^-1 as shown in the figure below. The inlet mean temperature is T_m,i = 293 K and the uniform heat flux at the surface of the tube is 2000 W m^-2. For the exit mean temperature of T_m,o = 353 K, the length of the tube, L is ________ m (rounded off to 1 decimal place).

Use the specific heat of water as 4181 J kg^-1 K^-1.

Enter your answer:

Solution

Step-by-step Calculation:
• Given values:
Diameter D = 60 mm = 0.06 m
Mass flow rate (ṁ) = 0.01 kg/s
Heat flux (q") = 2000 W/m²
T_m,i = 293 K, T_m,o = 353 K
Specific heat C_p = 4181 J/(kg·K)

• Energy Balance Equation:
Total heat transfer Q = ṁ × C_p × (T_m,o - T_m,i)
Q = 0.01 × 4181 × (353 - 293)
Q = 0.01 × 4181 × 60 = 2508.6 W

• Heat Transfer via Surface Area:
Q = q" × Surface Area = q" × (π × D × L)
2508.6 = 2000 × π × 0.06 × L
2508.6 = 376.991 × L
L = 2508.6 / 376.991 ≈ 6.7 m

✅ Correct Answer: 6.7 (Acceptable range: 6.5 to 6.8)

Question 52

A thermal power plant is running with no reheat or regeneration. The specific enthalpy and specific entropy of steam at the turbine inlet are 3344 kJ kg^-1 and 6.5 kJ kg^-1 K^-1, respectively. The turbine isentropic efficiency is 0.9, and the mass flow rate of steam at the turbine inlet is 102 kg s^-1. The turbine power output is ________ MW (rounded off to 1 decimal place).

Saturated liquid water		Saturated water vapor
Specific enthalpy (kJ kg-1)	Specific entropy (kJ kg-1 K-1)	Specific enthalpy (kJ kg-1)	Specific entropy (kJ kg-1 K-1)
341	1.1	2645	7.6

Enter your answer:

Question 53

Consider a Pelton wheel of 1 m diameter. The magnitude of relative velocity of water at the bucket inlet is same as the magnitude of relative velocity of water at the bucket exit. The absolute speed of water at the bucket inlet is 125.66 m s^-1. For maximum power output from the Pelton wheel, the rpm of the Pelton wheel is ________ (rounded off to 1 decimal place).

Enter your answer:

Solution

Step-by-step Calculation:
• Given values:
Wheel diameter (D) = 1 m
Jet velocity (V₁) = 125.66 m/s

• Maximum Power Condition:
Pelton wheel mein maximum power ke liye bucket speed (u) = V₁ / 2
u = 125.66 / 2 = 62.83 m/s

• RPM Calculation:
u = (π × D × N) / 60
62.83 = (π × 1 × N) / 60
N = (62.83 × 60) / π
N = 3769.8 / 3.14159 ≈ 1200 rpm (or 1199.9 rpm)

✅ Correct Answer: 1200.0 (Acceptable range: 1198.0 to 1202.0)

Question 54

A thermodynamically closed system contains 1 kg of hydrogen. The system undergoes a reversible polytropic process with polytropic index 1.3. The work output during the process is 400 kJ. During the process, hydrogen behaves as an ideal gas with constant specific heats. The absolute value of heat transfer during the process is ________ kJ (rounded off to 1 decimal place).

Specific heat of hydrogen at constant pressure, C_p = 14.56 kJ kg^-1 K^-1
Specific heat of hydrogen at constant volume, C_v = 10.4 kJ kg^-1 K^-1

Enter your answer:

Solution

Step-by-step Calculation:
• Specific heat ratio (γ):
γ = C_p / C_v = 14.56 / 10.4 = 1.4

• Polytropic process heat transfer relation:
Q = [ (γ - n) / (γ - 1) ] × W
yaha n = 1.3 aur W = 400 kJ

• Substituting values:
Q = [ (1.4 - 1.3) / (1.4 - 1.0) ] × 400
Q = (0.1 / 0.4) × 400
Q = 0.25 × 400 = 100.0 kJ

✅ Correct Answer: 100.0 (Acceptable range: 99.5 to 100.5)

Question 55

A heat pump, operating in reversed Carnot cycle, maintains a steady air temperature of 300 K inside an auditorium. The heat pump receives heat from the ambient air. The ambient air temperature is 280 K. Heat loss from the auditorium is 15 kW. The power consumption of the heat pump is ________ kW (rounded off to 2 decimal places).

Enter your answer:

Solution

Step-by-step Calculation:
• Given values:
Auditorium Temperature (T_H) = 300 K
Ambient Temperature (T_L) = 280 K
Heat supplied to auditorium (Q_H) = 15 kW

• COP of Carnot Heat Pump:
COP_HP = T_H / (T_H - T_L)
COP_HP = 300 / (300 - 280) = 300 / 20 = 15

• Power Consumption (W):
COP_HP = Q_H / W
15 = 15 / W ➔ W = 1.00 kW

✅ Correct Answer: 1 (or 1.00)

Question 56

The state of stress at a point is shown in the figure given below. Under plane stress assumption, the normal strain along the thickness direction (ε_zz) is ________ (rounded off to 2 decimal places).

Use the Young's Modulus E = 200 GPa and Poisson's ratio ν = 0.27.

Enter your answer:

Solution

Step-by-step Calculation:
• Given Stresses:
σ_xx = 350 GPa (Tension)
σ_yy = 150 GPa (Tension)
σ_zz = 0 (Plane Stress condition)
E = 200 GPa, ν = 0.27

• Generalized Hooke's Law for ε_zz:
ε_zz = (1 / E) * [ σ_zz - ν(σ_xx + σ_yy) ]
ε_zz = - (ν / E) * (σ_xx + σ_yy)

• Substituting values:
ε_zz = - (0.27 / 200) * (350 + 150)
ε_zz = - (0.27 / 200) * 500
ε_zz = - 0.27 × 2.5 = -0.675 (or approximately -0.68)

✅ Correct Answer: -0.68 (or -0.675)

Question 57

A simply supported beam of length 1 m is subjected to a uniformly distributed bending moment of 1 N m per m throughout the length as shown in the figure given below. The bending moment at the mid-point of the beam is ________ N m (rounded off to the nearest integer).

Enter your answer:

Solution

Step-by-step Calculation:
• Total distributed moment m = 1 N·m/m over L = 1 m.
• Total applied couple on the beam = m × L = 1 × 1 = 1 N·m.
• Beam vertical reactions (R_A and R_B) balance this couple:
R_A × 1 + 1 = 0 ➔ R_A = -1 N (downward), R_B = +1 N (upward).

• Bending Moment at Mid-point (x = 0.5 m from left end A):
M(x) = R_A × x + (m × x)
M(0.5) = (-1) × 0.5 + (1 × 0.5) = -0.5 + 0.5 = 0 N m

✅ Correct Answer: 0

Question 58

An isotropic brittle material is tested in the universal testing machine. The stress-strain diagram for the material shows a bi-linear elastic behavior as shown in the figure given below. The strain energy density is ________ MJ m^-3 (rounded off to 2 decimal places).

Enter your answer:

Solution

Step-by-step Calculation:
Strain energy density = Area under the stress-strain (σ-ε) curve.

• Area 1 (Triangle from ε = 0 to 0.01):
Area₁ = 0.5 × base × height
Area₁ = 0.5 × 0.01 × 100 MPa = 0.5 MPa = 0.50 MJ m^-3

• Area 2 (Trapezoid from ε = 0.01 to 0.015):
Width = 0.015 - 0.01 = 0.005
Average height = (100 + 120) / 2 = 110 MPa
Area₂ = 0.005 × 110 = 0.55 MPa = 0.55 MJ m^-3

• Total Strain Energy Density:
U = Area₁ + Area₂ = 0.50 + 0.55 = 1.05 MJ m^-3

✅ Correct Answer: 1.05

Question 59

A pair of spur gears is required to maintain a velocity ratio of 1:2. The module of the gears is 10 mm and the addendum is 10 mm. If the operating pressure angle is 15°, the minimum number of teeth required on the pinion to ensure NO interference/undercutting is ________ (answer in integer).

Enter your answer:

Solution

Step-by-step Calculation:
• Given values:
Gear ratio G = 2 / 1 = 2
Module m = 10 mm
Addendum a_w = 10 mm ➔ Fractional addendum coefficient A_w = a_w / m = 10 / 10 = 1
Pressure angle φ = 15° ➔ sin(15°) ≈ 0.2588, sin²(15°) ≈ 0.066987

• Formula for minimum teeth on pinion (t_min):
t_min = (2 × A_w) / [ G × { √(1 + (1/G)(1/G + 2)sin²φ) - 1 } ]

• Calculation:
(1/G)(1/G + 2) = (1/2)(1/2 + 2) = 0.5 × 2.5 = 1.25
1.25 × sin²(15°) = 1.25 × 0.066987 = 0.083734
√(1 + 0.083734) = √(1.083734) ≈ 1.041025
Denominator term = 2 × (1.041025 - 1) = 2 × 0.041025 = 0.08205
t_min = 2 / 0.08205 ≈ 24.37

• Rounding up to the next integer to prevent interference gives 25 teeth.

✅ Correct Answer: 25

Question 60

An offset slider-crank mechanism is shown in the figure below. The length of the stroke of the slider is ________ mm (rounded off to nearest integer).

Enter your answer:

Solution

Step-by-step Calculation:
• Given parameters:
Crank length (r) = 30 mm
Connecting rod length (l) = 50 mm
Offset distance (e) = 10 mm

• Extreme Position 1 (Outer Dead Center):
Total distance from center line = l + r = 50 + 30 = 80 mm
Horizontal distance x₁ = √[ (l + r)² - e² ] = √[ 80² - 10² ] = √[ 6400 - 100 ] = √6300 ≈ 79.37 mm

• Extreme Position 2 (Inner Dead Center):
Total distance from center line = l - r = 50 - 30 = 20 mm
Horizontal distance x₂ = √[ (l - r)² - e² ] = √[ 20² - 10² ] = √[ 400 - 100 ] = √300 ≈ 17.32 mm

• Stroke Length (S):
S = x₁ - x₂ = 79.37 - 17.32 = 62.05 mm ≈ 62 mm

✅ Correct Answer: 62

Question 61

Two plates of thickness 10 mm each are to be joined by a transverse fillet weld on one side and the resulting structure is loaded as shown in the figure below. If the ultimate tensile strength of the weld material is 150 MPa and the factor of safety to be used is 3, the minimum length of the weld required to ensure that the weld does NOT fail is ________ mm (rounded off to 2 decimal places).

Enter your answer:

Solution

Step-by-step Calculation:
• Given values:
Load P = 5 kN = 5000 N
Leg size h = 10 mm
Ultimate strength σ_ut = 150 MPa
Factor of Safety (FOS) = 3

• Permissible Stress (σ_permissible):
σ_permissible = σ_ut / FOS = 150 / 3 = 50 MPa = 50 N/mm²

• Throat Thickness (t):
t = 0.707 × h = 0.707 × 10 = 7.07 mm

• Load Capacity Formula:
P = σ_permissible × Throat Area = σ_permissible × (0.707 × h × L)
5000 = 50 × 0.707 × 10 × L
5000 = 353.5 × L
L = 5000 / 353.5 ≈ 14.14 mm

✅ Correct Answer: 14.14 (Acceptable range: 14.00 to 14.30)

Question 62

Two metal parts (a cylinder and a cube) of same volume are cast under identical conditions. The diameter of the cylinder is equal to its height. The ratio of the solidification time of the cube to that of the cylinder is ________ (rounded off to 2 decimal places).

Assume that solidification time follows Chvorinov's rule with an exponent of 2.

Enter your answer:

Solution

Step-by-step Calculation:
• Chvorinov's Rule:
Solidification time, t ∝ (V / A)²
Therefore, t_cube / t_cyl = [ (V_cube / A_cube) / (V_cyl / A_cyl) ]²
Since V_cube = V_cyl = V:
t_cube / t_cyl = (A_cyl / A_cube)²

• For Cube (side length a):
V = a³ ➔ a = V^1/3
A_cube = 6a² = 6 × V^2/3

• For Cylinder (diameter d, height h = d):
V = (π/4) × d² × d = (π/4) × d³ ➔ d = (4V / π)^1/3
A_cyl = 2 × (π/4)d² + πdh = (π/2)d² + πd² = (3π/2)d²
Substituting d:
A_cyl = (3π/2) × (4V / π)^2/3 = (3π/2) × (4/π)^2/3 × V^2/3

• Area Ratio (A_cyl / A_cube):
A_cyl / A_cube = [ (3π/2) × (4/π)^2/3 ] / 6
A_cyl / A_cube = (π/4) × (4/π)^2/3 = (π/4)^1/3 = (3.14159 / 4)^1/3 ≈ (0.7854)^1/3 ≈ 0.9224

• Solidification Time Ratio:
t_cube / t_cyl = (A_cyl / A_cube)² = (0.9224)² ≈ 0.85

✅ Correct Answer: 0.85 (Acceptable range: 0.83 to 0.87)

Question 63

Cylindrical workpieces of diameter 60 mm and length 400 mm are machined on a lathe at a cutting speed of 25 m min^-1 and a feed of 0.2 mm rev^-1. The Taylor's tool life parameters C and n for this setup are 75 and 0.25, respectively. The tool changing time is 3 minutes. With a labor and overhead cost of ₹ 5 per minute, the tool changing cost per piece is ₹ ________ (rounded off to 2 decimal places).

Enter your answer:

Solution

Step-by-step Calculation:
• Given parameters:
D = 60 mm = 0.06 m, L = 400 mm
Cutting speed V = 25 m/min, Feed f = 0.2 mm/rev
C = 75, n = 0.25, Tool change time T_c = 3 min
Cost rate C_m = ₹ 5/min

• Machining Time per piece (T_m):
N = (1000 × V) / (π × D) = (1000 × 25) / (π × 60) ≈ 132.63 rpm
T_m = L / (f × N) = 400 / (0.2 × 132.63) = 400 / 26.526 ≈ 15.08 min

• Tool Life (T):
V × Tⁿ = C ➔ 25 × T^0.25 = 75
T^0.25 = 3 ➔ T = 3⁴ = 81 min

• Tool Changing Cost per piece:
Cost = (T_m / T) × T_c × C_m
Cost = (15.08 / 81) × 3 × 5
Cost = 0.18617 × 15 ≈ ₹ 2.79

✅ Correct Answer: 2.79 (Acceptable range: 2.70 to 2.85)

Question 64

A company uses 3000 units of a part annually. The units are priced as given in the table below. It costs ₹ 150 to place an order. Carrying costs are 40 percent of the purchase price per unit on an annual basis. The minimum total annual cost is ₹ ________ (rounded off to 1 decimal place).

Order quantity	Unit price (₹)
1 to 499	9.0
500 to 999	8.5
1000 or more	8.0

Enter your answer:

Solution

Step-by-step Calculation:
• Given values:
Annual Demand (D) = 3000 units
Ordering Cost (S) = ₹ 150 per order
Carrying Cost percentage (i) = 0.40 (40%)

• Total Cost Formula:
TC(Q) = D × C + (D/Q) × S + (Q/2) × (i × C)

• Case 1: Price C = ₹ 8.0 (for Q ≥ 1000)
EOQ = √[ (2 × D × S) / (i × C) ] = √[ (2 × 3000 × 150) / (0.40 × 8.0) ] = √[ 900000 / 3.2 ] ≈ 530.3 units.
Since 530.3 < 1000, EOQ is not feasible in this tier. To get this discount, we take minimum feasible quantity Q = 1000:
TC(1000) = (3000 × 8) + (3000/1000)×150 + (1000/2)×(0.4 × 8)
TC(1000) = 24000 + 450 + 1600 = ₹ 26050.0

• Case 2: Price C = ₹ 8.5 (for 500 ≤ Q ≤ 999)
EOQ = √[ (2 × 3000 × 150) / (0.40 × 8.5) ] = √[ 900000 / 3.4 ] ≈ 514.5 units.
Since 514.5 lies in [500, 999], Q = 514.5 is feasible:
TC(514.5) = (3000 × 8.5) + (3000/514.5)×150 + (514.5/2)×(0.4 × 8.5)
TC(514.5) = 25500 + 874.64 + 874.65 = ₹ 27249.3

• Case 3: Price C = ₹ 9.0 (for 1 ≤ Q ≤ 499)
EOQ = √[ (2 × 3000 × 150) / (0.40 × 9.0) ] = √[ 900000 / 3.6 ] = 500 units.
Minimum feasible quantity is Q = 499:
TC(499) = (3000 × 9) + (3000/499)×150 + (499/2)×(0.4 × 9)
TC(499) = 27000 + 901.8 + 898.2 = ₹ 28800.0

• Comparing all feasible total costs, the minimum total annual cost is ₹ 26050.0 (at Q = 1000).

✅ Correct Answer: 26050 (or 26050.0)

Question 65

A project involves eight activities with the precedence relationship and duration as shown in the table below. The slack for the activity D is ________ hours (answer in integer).

Activity	Immediate predecessor	Duration (hours)
A	-	4
B	A	8
C	A	5
D	B	2
E	B	7
F	C	6
G	D	3
H	E, F, G	9

Enter your answer:

Solution

Step-by-step Calculation:
• Paths to activity H (starting after A and B):
- Path 1 through E: A ➔ B ➔ E ➔ H (Length = 4 + 8 + 7 = 19 hrs)
- Path 2 through D & G: A ➔ B ➔ D ➔ G ➔ H (Length = 4 + 8 + 2 + 3 = 17 hrs)
- Path 3 through C & F: A ➔ C ➔ F ➔ H (Length = 4 + 5 + 6 = 15 hrs)

• Critical Path before H:
Path 1 (A-B-E) is the longest with duration 19 hours.
So, Latest Finish Time for D's path segment before H = 19 hrs.

• Slack for Activity D:
Slack = Duration of longest path (A-B-E) - Duration of D's path (A-B-D-G)
Slack = 19 - 17 = 2 hours

✅ Correct Answer: 2