ISRO Mechanical & Aerospace 2023 Question Paper & Solutions

Question 1

Topic: Strength of Materials

A beam is subjected to the uniform load Wc per unit length and MA, VA & VB are the reactions as shown in the figure. The expression for shear force V(x) and Bending moment M(x) is given by

• A. V(x) = VA + Wc x, M(x) = MA + VA x - Wc x² / 2
• B. V(x) = VA - Wc x, M(x) = MA - VA x - Wc x² / 2
• C. V(x) = VA + Wc x, M(x) = MA - VA x - Wc x² / 2
• D. V(x) = VA - Wc x, M(x) = MA + VA x - Wc x² / 2

Answer: D. V(x) = VA - Wc x, M(x) = MA + VA x - Wc x² / 2

Explanation: Taking the left end A as the origin and rightward as the positive x-direction, the upward shear force is VA and downward distributed load is Wc. The shear force at distance x is V(x) = VA - Wc*x. The bending moment at distance x, considering counter-clockwise MA as positive, is M(x) = MA + VA*x - Wc*x² / 2.

Question 2

Topic: Manufacturing

A weld with a triangular cross section as shown below is to be produced on a steel work piece by shielded metal arc welding operation using steel electrode. If a 24 V power supply is used and the welding speed is 12 mm/s, what is the current needed? Assume the efficiency as 80% and the specific energy to melt steel as 8 J/mm³.

• A. 50 A
• B. 75 A
• C. 125 A
• D. 175 A

Answer: C. 125 A

Explanation: The figure shows a V-groove or double fillet weld cross-section. Assuming the total area is 25 mm² (base 10 mm and height 5 mm or similar), the volume melted per second = Area * velocity = 25 * 12 = 300 mm³/s. Heat required = 300 * 8 = 2400 J/s (W). Heat supplied = V * I * efficiency = 24 * I * 0.8 = 19.2 * I. Equating them: 19.2 * I = 2400 => I = 125 A.

Question 3

Topic: Theory of Machines

A slider moving outward on a rod OR with a velocity of 3 m/s and acceleration of 9 m/s². The rod has an angular velocity of 7 rad/s counter clockwise about O and an angular acceleration of 12 rad/s² clockwise. The magnitude of absolute acceleration of the slider at location A in m/s² is (where OA = 1m)

• A. 20
• B. 50
• C. 67
• D. 79

Answer: B. 50

Explanation: Radial acceleration ar = d^2r/dt² - r*omega² = 9 - 1*(7^2) = 9 - 49 = -40 m/s². Tangential acceleration at = r*alpha + 2*(dr/dt)*omega = 1*(-12) + 2*3*7 = -12 + 42 = 30 m/s². Absolute acceleration = $\sqrt{ar² + at²}$ = $\sqrt{(-40}$^2 + 30^2) = 50 m/s².

Question 4

Topic: Material Science

The specific stiffness of a material is defined as the ratio of

• A. Mass to Young's modulus
• B. Density to Young's modulus
• C. Young's modulus to Density
• D. Young's modulus to Mass

Answer: C. Young's modulus to Density

Explanation: Specific stiffness, also known as specific modulus, is a materials property consisting of the elastic modulus (Young's modulus) divided by the mass density of the material.

Question 5

Topic: Machine Design

The type of gear used to transmit rotary motion between non-parallel and non-intersecting shaft is

• A. Spur gear
• B. Helical gear
• C. Bevel gear
• D. Worm gear

Answer: D. Worm gear

Explanation: Worm gears, crossed helical gears, and hypoid gears are used to transmit power between non-parallel, non-intersecting shafts. Among the options given, the worm gear is the correct answer.

Question 6

Topic: Machine Design

Which of the bearings given below is not suitable for axial thrust load?

• A. Deep groove ball bearing
• B. Angular contact ball bearing
• C. Single row tapered bearing
• D. Straight Cylindrical roller bearing

Answer: D. Straight Cylindrical roller bearing

Explanation: Straight cylindrical roller bearings are designed to carry heavy radial loads and allow for axial displacement between the inner and outer rings, meaning they cannot support any axial thrust load.

Question 7

Topic: Theory of Machines

The configuration of a planar four bar mechanism with frictionless joints is shown in the figure below. The length of the links are L1 = 40 mm, L2 = 15 mm, L3 = 35 mm and L4 = 30 mm. The suffixes 1, 2, 3 and 4 represents the fixed, input, coupler and output link respectively. If Ti and To are the torque to the input and output link, what is the mechanical advantage (To/Ti) of the mechanism?

• A. 2
• B. 0.5
• C. Infinity
• D. 0

Answer: C. Infinity

Explanation: In the given configuration, links 2 and 3 are collinear (180 degrees). This is a toggle position where a finite input torque creates an infinite output torque (theoretically) because the mechanical advantage becomes infinite. The ratio of output torque to input torque (To/Ti) approaches infinity.

Question 8

Topic: Strength of Materials

A solid bar 50 mm in diameter and 2000 mm long consists of a steel and aluminum part joined together. When axial force P is applied to the system, a strain gauge attached to the aluminum indicates an axial strain of 800 mu m/m. What is the corresponding axial strain on the Steel? Assume Est = 200 GPa and EAl = 70 Gpa.

• A. 28 mu m/m
• B. 140 mu m/m
• C. 280 mu m/m
• D. 14 mu m/m

Answer: C. 280 mu m/m

Explanation: The parts are in series, so the stress is identical in both. Stress = Strain_Al * E_Al = 800 * 10^-6 * 70 GPa = 56 MPa. Strain in steel = Stress / E_st = 56 MPa / 200 GPa = 280 * 10^-6 = 280 microstrain (mu m/m).

Question 9

Topic: Machine Design

Two flanges are connected by a single bolt of nominal diameter 10 mm and is preloaded to a torque of 30 Nm. An external tensile load of 10,000 N is applied to the joint. The torque coefficient is 0.15 and joint stiffness constant is 0.2. The maximum tension in the bolt is

• A. 10,000 N
• B. 20,000 N
• C. 22,000 N
• D. 30,000 N

Answer: C. 22,000 N

Explanation: Preload force Fi = Torque / (K * d) = 30 / (0.15 * 0.010) = 20,000 N. The maximum tension in the bolt Fb = Fi + (C * P) = 20,000 + (0.2 * 10,000) = 20,000 + 2000 = 22,000 N.

Question 10

Topic: Theory of Machines

A mass m is attached to two identical springs having stiffness k as shown in figure below. The natural frequency of the vibrating system is

• A. (1/2pi) * $\sqrt{k/m}$
• B. (1/2pi) * $\sqrt{2k/m}$
• C. (1/2pi) * $\sqrt{k/2m}$
• D. (1/2pi) * $\sqrt{4k/m}$

Answer: B. (1/2pi) * $\sqrt{2k/m}$

Explanation: The two springs are attached to fixed supports on either side of the mass. When the mass is displaced by x, both springs exert a restoring force of kx in the same direction. Thus, they act in parallel. Equivalent stiffness K_eq = k + k = 2k. Natural frequency = (1/2pi) * $\sqrt{2k/m}$.

Question 11

Topic: Theory of Machines

A damped spring-mass system is forced to oscillate at t=0 by the application of an impulse force. Which of the following are true?

• A. If the damping ratio is between 0 and 1, the system will show no oscillations
• B. If the damping ratio is between 0 and 1, the system will eventually come to rest
• C. If the damping ratio is 0, the system will eventually come to rest
• D. The system will always oscillate

Answer: B. If the damping ratio is between 0 and 1, the system will eventually come to rest

Explanation: For an underdamped system (damping ratio between 0 and 1), it will oscillate with decaying amplitude and eventually come to rest due to energy dissipation. A damping ratio of 0 would mean it oscillates forever.

Question 12

Topic: Engineering Mechanics

In an accident site, tyre marks shows that a car was travelling along straight level street skidded for a total distance of 23 m, after the brakes applied. The coefficient of friction between the tyre and road estimated to be 0.5, what was the probable speed of car when break applied.

• A. 54 kmph
• B. 60.7 kmph
• C. 45.3 kmph
• D. 97 kmph

Answer: A. 54 kmph

Explanation: Deceleration a = -mu * g = -0.5 * 9.81 = -4.905 m/s². Using v² - u² = 2as, 0 - u² = 2 * (-4.905) * 23. u² = 225.63 => u = 15.02 m/s. In kmph, u = 15.02 * 3.6 = 54.07 kmph, which is approximately 54 kmph.

Question 13

Topic: Engineering Mechanics

If a solid cylinder, thin hollow cylinder and a sphere, all with same material and are having same outside diameter are allowed to roll down in same inclination without sliding. All surfaces are having same coefficient of friction. What is the order of the bodies reaching bottom

• A. All reach together
• B. Solid cylinder, sphere, thin hollow cylinder
• C. Sphere, solid cylinder, thin hollow cylinder
• D. Thin hollow cylinder, solid cylinder, sphere

Answer: C. Sphere, solid cylinder, thin hollow cylinder

Explanation: Acceleration of a rolling body down an incline is a = (g*sin(theta)) / (1 + (I/mr²)). The value of I/mr² is 0.4 for a sphere, 0.5 for a solid cylinder, and 1 for a thin hollow cylinder. A lower I/mr² means higher acceleration. Thus, the sphere reaches first, followed by the solid cylinder, and then the thin hollow cylinder.

Question 14

Topic: Strength of Materials

A 0.2 mm thick strip is bent and laid over a frictionless pulley of 25 mm diameter. If E = 100 GPa, the maximum stress in the strip is

• A. 100 MPa
• B. 200 MPa
• C. 400 MPa
• D. 800 MPa

Answer: D. 800 MPa

Explanation: The bending stress is sigma = E * y / R. Here, y is half the thickness of the strip (0.2/2 = 0.1 mm). The radius of the neutral axis R = D/2 + y = 25/2 + 0.1 = 12.6 mm (approximated as 12.5 mm for calculation matching options). sigma = 100,000 MPa * (0.1 / 12.5) = 800 MPa.

Question 15

Topic: Fluid Mechanics

The velocity field of a certain two dimensional flow is given by V(x,y) = k(x i - y j) where k = 2/s and x and y are in meters. If the fluid density is 100 kg/m³ and pressure at origin is 10 kPa, what is the pressure at point (3,3)?

• A. 3.2 kPa
• B. 6.4 kPa
• C. 1.6 kPa
• D. None

Answer: B. 6.4 kPa

Explanation: The flow is steady and irrotational. By Bernoulli's equation, P1/rho + V1^2/2 = P2/rho + V2^2/2. At the origin (0,0), V1 = 0 and P1 = 10,000 Pa. At (3,3), velocity magnitude V2 = $\sqrt{(2*3}$^2 + (-2*3)^2) = $\sqrt{36 + 36}$ = $\sqrt{72}$ m/s. P2 / 100 + 72/2 = 10000 / 100. P2/100 + 36 = 100. P2 = 6400 Pa = 6.4 kPa.

Question 16

Topic: Fluid Mechanics

A spherical water drop of diameter 2 mm splits in to 8 small drops of equal size in air at temperature of 25 °C. The surface tension coefficient of water in air is 0.073 N/m and dynamic viscosity is 0.01 poise. The approximate work done in splitting up the drop is

• A. 0.29 x 10^-6 J
• B. 0.92 x 10^-6 J
• C. 0.86 x 10^-6 J
• D. 0.78 x 10^-6 J

Answer: B. 0.92 x 10^-6 J

Explanation: Initial volume = Final volume, so (4/3)*pi*R³ = 8 * (4/3)*pi*r³, which means r = R/2. The initial radius R = 1 mm. Change in surface area = Final Area - Initial Area = 8*(4*pi*r²) - 4*pi*R² = 8*4*pi*(R/2)^2 - 4*pi*R² = 4*pi*R². Work done = Surface tension * Change in area = 0.073 * 4 * pi * (10^-3)^2 = 0.917 x 10^-6 J, which is approximately 0.92 x 10^-6 J.

Question 17

Topic: Fluid Mechanics

Which is identified as the coefficient of momentum transport?

• A. Density
• B. Kinematic viscosity
• C. Dynamic viscosity
• D. Reynolds number

Answer: B. Kinematic viscosity

Explanation: Kinematic viscosity is also known as the momentum diffusivity or the coefficient of momentum transport, as it represents the ratio of dynamic viscosity to density.

Question 18

Topic: Fluid Mechanics

Which is the most economical depth of flow in an open rectangular channel?

• A. Depth of flow equals to the width of channel
• B. Depth of flow is double the width of channel
• C. Depth of flow is half of the width of channel
• D. Depth of flow is 1/3rd the width of channel

Answer: C. Depth of flow is half of the width of channel

Explanation: For a rectangular channel to be most economical (i.e., yielding maximum discharge for a given cross-sectional area), the hydraulic radius must be half the depth, which translates to the depth of flow being exactly half the width of the channel.

Question 19

Topic: Machine Design

A 50 mm diameter solid shaft is welded to a flat plate all around by fillet weld of 4mm leg size as shown in Fig. If the allowable shear strength of the weld material is 100 MPa, the maximum approximate torque in Nm that the welded joint can sustain under pure torsion is

• A. 500*pi / $\sqrt{2}$
• B. 1000*pi / $\sqrt{2}$
• C. 500*pi*$\sqrt{2}$
• D. 1000*pi*$\sqrt{2}$

Answer: A. 500*pi / $\sqrt{2}$

Explanation: Throat thickness t = 0.707 * s = 4 / $\sqrt{2}$ mm. Polar moment of inertia of a circular weld J = pi * t * D³ / 4. Torsional shear stress tau = (T * r) / J. Therefore, T = tau * J / r = tau * (pi * t * D²) / 2. T = 100 * pi * (4 / $\sqrt{2}$) * (50^2) / 2 = 500,000 * pi / $\sqrt{2}$ N-mm = 500 * pi / $\sqrt{2}$ N-m.

Question 20

Topic: Robotics

What is the revolving joint of the robot called?

• A. V-Joint
• B. L-Joint
• C. O-Joint
• D. T-Joint

Answer: A. V-Joint

Explanation: In robotics, a revolving joint where the input link is parallel to the axis of rotation and the output link is perpendicular to it is standardly referred to as a V-joint (reVolving joint).

Question 21

Topic: Machine Design

The diagram below is showing the various criteria of fatigue failure. Which option is the correct representation of these criteria? (Subscripts e, y and u represent endurance, yield and ultimate respectively.)

• A. 1 - Soderberg line, 2 - Goodman line, 3 - Gerber line
• B. 1 - Goodman line, 2 - Soderberg line, 3 - Gerber line
• C. 1 - Soderberg line, 2 - Gerber line, 3 - Goodman line
• D. 1 - Gerber line, 2 - Soderberg line, 3 - Goodman line

Answer: A. 1 - Soderberg line, 2 - Goodman line, 3 - Gerber line

Explanation: The Soderberg line (1) is the most conservative, connecting the endurance limit (Se) to the yield strength (Sy). The Goodman line (2) connects the endurance limit (Se) to the ultimate tensile strength (Su). The Gerber line (3) is a parabola connecting Se to Su.

Question 22

Topic: Manufacturing

A 20 mm diameter Aluminium rod is turned to 19 mm diameter in a single pass. If the axial speed of the tool is 200 mm/min and the spindle rotate at 600 rpm, the material removal rate is nearly

• A. 6100 mm³/min
• B. 12250 mm³/min
• C. 3050 mm³/min
• D. 9150 mm³/min

Answer: A. 6100 mm³/min

Explanation: Material Removal Rate (MRR) for turning = (pi/4) * (D_initial^2 - D_final^2) * feed_velocity. MRR = (pi/4) * (20^2 - 19^2) * 200 = 50 * pi * (400 - 361) = 1950 * pi = 6126.1 mm³/min, which is nearly 6100 mm³/min.

Question 23

Topic: Theory of Machines

A planar four bar mechanism with all revolute joints has link lengths AB=30 mm, BC=120 mm, CD=100 mm and AD=80 mm. The link to be fixed to obtain a double crank mechanism is

• A. AB
• B. BC
• C. CD
• D. AD

Answer: A. AB

Explanation: By Grashof's law, Shortest + Longest (S+L) = 30 + 120 = 150 mm. The sum of the other two links (P+Q) = 100 + 80 = 180 mm. Since S+L < P+Q, it is a Grashof linkage. To obtain a double crank mechanism, the shortest link (AB) must be fixed.

Question 24

Topic: Engineering Mechanics

A small ring Z carries vertical load A of mass 27.3 kg by two strings ZX and ZY. String ZY carries a load B at its free end through a frictionless pulley. Find the value of B (rounded to nearest integer), if the string is stable at the configuration shown in fig. (Angle ZX makes 45 deg with horizontal/vertical, ZY makes 60 deg).

• A. 14 kg
• B. 20 kg
• C. 10 kg
• D. 5 kg

Answer: B. 20 kg

Explanation: Applying Lami's Theorem at node Z: Tension B / sin(180 - 45) = Load A / sin(45 + 60). Therefore, B / sin(135) = 27.3 / sin(105). B = 27.3 * sin(45) / sin(105) = 27.3 * 0.707 / 0.9659 = 19.98 kg, which rounds to 20 kg.

Question 25

Topic: Machine Design

A flat belt run over a pulley and transmitting the power of 2.5 KW. The linear velocity of the belt is 2.5 m/s. The angle of lap is 165 deg and the coefficient of friction between the belt and pulley is 0.3. If the initial tension of the belt is increased by 10%, what is the effect on power transmission? (neglecting the slip and centrifugal effect of the belt)

• A. The power transmission is increased by 10%
• B. The power transmission is reduced by 10%
• C. The power transmission is doubled
• D. The power transmission is reduced to half

Answer: A. The power transmission is increased by 10%

Explanation: Power transmitted P = (T1 - T2) * V. The ratio T1/T2 = e^(mu*theta) remains constant. The initial tension T0 = (T1 + T2)/2. If T0 is increased by 10%, both T1 and T2 increase proportionally by 10%. Consequently, the difference (T1 - T2) increases by 10%, leading to a 10% increase in power transmission.

Question 26

Topic: Strength of Materials

An aluminium ring of mean diameter of 5m at ambient temperature is dipped into a liquid nitrogen bath. What will be the mean diameter of ring in the LN2 bath. The coefficient of linear expansion, of aluminium alloy in the temperature range may be taken as 20x10^-6 /K (Ambient temperature may be taken as 24 °C. Boiling temperature of liquid nitrogen: 77 K)

• A. 4.978 m
• B. 4.958 m
• C. 5.138 m
• D. 4.865 m

Answer: A. 4.978 m

Explanation: Initial temperature T1 = 24 °C = 297 K. Final temperature T2 = 77 K. Delta T = 77 - 297 = -220 K. Change in diameter dD = D * alpha * Delta T = 5 * 20x10^-6 * (-220) = -0.022 m. New diameter = 5 - 0.022 = 4.978 m.

Question 27

Topic: Engineering Mechanics

A travelling crane as shown is used to carry a load of two tonnes at point D. A counter mass CW is to be added as shown to avoid toppling of the crane. Find the limiting value of counter mass and distance, X from F, such that it will not topple the crane even in the absence of load at D. The mass of the crane may be taken as 2 tonnes and is acting 2 meters away from E towards D.

• A. CW = 2 t at X = 8 m
• B. CW = 16 t at X = 10 m
• C. CW = 1.5 t at X = 6 m
• D. CW = 16.67 t at X = 12 m

Answer: A. CW = 2 t at X = 8 m

Explanation: To avoid toppling left (no load at D), overturning moment by CW about the left support must not exceed restoring moment by crane weight: CW * X <= 2 tonnes * 8 m = 16 t-m. To avoid toppling right (load at D), restoring moment by CW must exceed overturning moment: CW * (X + 6) >= 2 * 12 + 2 * 2 = 28 t-m. Testing Option (a): CW=2, X=8 yields CW * X = 16 and CW * (X+6) = 28, perfectly satisfying both limits.

Question 28

Topic: Machine Design

A journal bearing of radius 'r1' and radial clearance 'c1' has a sommerfeld number 's1'. Another journal bearing of radius 'r2' and radial clearance 'c2' has a sommerfeld number 's2'. All the other parameters remain the same for both bearings. If r2 = 2 r1 and c2 = c1 / 2, then s2 / s1 is

• A. 4
• B. 8
• C. 16
• D. 32

Answer: C. 16

Explanation: The Sommerfeld number S = (r/c)^2 * (mu * N / P). Since only radius and clearance are changing, S2 / S1 = (r2/c2)^2 / (r1/c1)^2. Substituting the values gives ((2*r1) / (c1/2))^2 / (r1/c1)^2 = (4)^2 = 16.

Question 29

Topic: Machine Design

Match the following: Theories of failure with Shape of region of safety. (A) Rankine Theory, (B) Tresca Theory, (C) St. Venants theory, (D) Von mises Theory. (i) Hexagon, (ii) Ellipse, (iii) Square, (iv) Rhombus.

• A. (A)-(ii), (B)-(iv), (C)-(i), (D)-(iii)
• B. (A)-(iii), (B)-(iv), (C)-(ii), (D)-(i)
• C. (A)-(iii), (B)-(i), (C)-(iv), (D)-(ii)
• D. (A)-(iv), (B)-(i), (C)-(iii), (D)-(ii)

Answer: C. (A)-(iii), (B)-(i), (C)-(iv), (D)-(ii)

Explanation: Rankine theory (Maximum Normal Stress) boundary is a Square (iii). Tresca theory (Maximum Shear Stress) boundary is a Hexagon (i). St. Venants theory (Maximum Principal Strain) boundary is a Rhombus (iv). Von Mises (Maximum Distortion Energy) boundary is an Ellipse (ii).

Question 30

Topic: Engineering Mechanics

The moment of resultant of concurrent forces with respect to a center in their plane is equal to the algebraic sum of the moments of components with respect to same center. This theorem is known as

• A. Theorem of Varignon
• B. Method of moments
• C. Theorem of equilibrium of active forces
• D. Theorem of three forces

Answer: A. Theorem of Varignon

Explanation: Varignon's Theorem states that the moment of a resultant of two or more concurrent forces about any point is equal to the algebraic sum of the moments of the individual forces about that same point.

Question 31

Topic: Material Science

Stress strain curve of three types of material is given in the figure below. Please identify the correct combination.

• A. E- brittle, F - ductile, G - plastic
• B. E- plastic, F - brittle, G - ductile
• C. E- brittle, F - plastic, G - ductile
• D. E- ductile, F - plastic, G - brittle

Answer: A. E- brittle, F - ductile, G - plastic

Explanation: Curve E is steep and fractures without much plastic deformation, which is characteristic of a brittle material. Curve F shows significant strain before fracture, characteristic of a ductile material. Curve G is completely flat (horizontal), indicating continuous deformation at a constant stress, which is perfectly plastic.

Question 32

Topic: Machine Design

A machine member is subjected to fluctuating stress, $sigma = sigma_0 cos(5pi t)$. The endurance limit of the material is 350 MPa and the yield strength is 500 MPa. If the factor of safety used in the design is 3.5, then the maximum allowable stress $sigma_0$ in MPa according to soderberg fatigue criteria is

• A. 10
• B. 200
• C. 100
• D. 50

Answer: C. 100

Explanation: The stress varies sinusoidally from $+sigma_0$ to $-sigma_0$. Therefore, the mean stress $sigma_m = 0$, and the variable (amplitude) stress $sigma_v = sigma_0$. Using the Soderberg criteria: $(sigma_m / S_y) + (sigma_v / S_e) = 1 / FOS$. Since $sigma_m = 0$, the equation becomes $sigma_0 / 350 = 1 / 3.5$. Solving for $sigma_0$ gives 100 MPa.

Question 33

Topic: Strength of Materials

A plane stress element in a structural member has under loading $sigma_x = 4P$, $sigma_y = 2P$ and $tau_{xy} = sqrt{3}P$, where $P>0$. The yield strength of the material is 300 MPa. If the member is designed using the maximum shear stress theory, then the value of P at which yielding starts is

• A. 60 MPa
• B. 75 MPa
• C. 100 MPa
• D. 120 MPa

Answer: B. 75 MPa

Explanation: Principal stresses are calculated as $sigma_{1,2} = (sigma_x + sigma_y)/2 \pm sqrt{((sigma_x - sigma_y)/2)^2 + tau_{xy}^2}$. Substituting the values: $(4P + 2P)/2 \pm sqrt{((4P - 2P)/2)^2 + (sqrt{3}P)^2} = 3P \pm sqrt{P² + 3P^2} = 3P \pm 2P$. Thus, $sigma_1 = 5P$ and $sigma_2 = P$. Maximum shear stress $tau_{max} = (sigma_1 - sigma_2)/2 = (5P - P)/2 = 2P$. According to the Maximum Shear Stress Theory (Tresca), yielding starts when $tau_{max} = S_y / 2$. Therefore, $2P = 300 / 2 = 150$, giving $P = 75$ MPa.

Question 34

Topic: Strength of Materials

Radius of the Mohr circle of stress is numerically equal to

• A. Maximum normal stress
• B. Maximum shear stress
• C. Average normal stress
• D. Average shear stress

Answer: B. Maximum shear stress

Explanation: In a Mohr's circle for a 2D stress state, the radius is calculated as $sqrt{((sigma_x - sigma_y)/2)^2 + tau_{xy}^2}$, which is identically the formula for the maximum in-plane shear stress.

Question 35

Topic: Manufacturing

In a casting process, a horizontal channel through which molten metal flows from pouring basin to reach mold cavity is called

• A. Runner
• B. Thinner
• C. Corridor
• D. Raiser

Answer: A. Runner

Explanation: In casting, the runner is the horizontal channel in the mold that distributes the molten metal from the sprue/pouring basin into the mold cavity.

Question 36

Topic: Manufacturing

3D Printing Techniques are

• A. Selective laser sintering
• B. Selective laser melting
• C. Electron beam melting
• D. All the Above

Answer: D. All the Above

Explanation: Selective Laser Sintering (SLS), Selective Laser Melting (SLM), and Electron Beam Melting (EBM) are all popular additive manufacturing (3D printing) techniques used particularly for metals and polymers.

Question 37

Topic: Machine Design

A round wire helical spring is loaded by the axial force. The maximum shear stress in the wire cross section will be at

• A. Outer fibre
• B. Inner fibre
• C. Center of the cross section
• D. Both outer and inner fibres

Answer: B. Inner fibre

Explanation: In a helical spring, the direct shear stress and the torsional shear stress are additive on the inner edge of the coil. Additionally, the curvature effect (accounted for by the Wahl factor) further concentrates the stress at the innermost fiber.

Question 38

Topic: Machine Design

The power is transmitted through a spur gear with 20 deg pressure angle mounted at the mid span of shaft supported on bearings at the ends. The nature of stress induced in the shaft is

• A. Normal stress due to bending only.
• B. Normal stress due to bending and shear stress due to torsion.
• C. Normal stress due to bending, axial loading and shear stress due to torsion
• D. Normal stress due to axial loading and shear stress due to torsion

Answer: B. Normal stress due to bending and shear stress due to torsion.

Explanation: A spur gear transmits both a tangential force component (which causes torsional shear stress in the shaft) and a radial force component (which causes bending moment, hence normal bending stress). Unlike helical gears, spur gears do not produce axial/thrust forces.

Question 39

Topic: Engineering Mechanics

As shown in the figure, the plane, having a mass of 8000 kg, is hoisted back until $theta = 60^circ$ and then the pull-back cable AC is released when the plane is at rest. Determine the speed of the plane just before it crashes into the ground at $theta = 30^circ$. (Neglect the size of the plane and effect of lift caused by the wings during the motion, assume $g=10 m/s²$)

• A. 23.98 m/s
• B. 12.1 m/s
• C. 9.56 m/s
• D. 6.76 m/s

Answer: B. 12.1 m/s

Explanation: By conservation of energy, the loss in potential energy equals the gain in kinetic energy. The vertical drop $Delta h = L cos(30^circ) - L cos(60^circ) = 20(sqrt{3}/2) - 20(1/2) = 17.32 - 10 = 7.32$ m. Equating energies: $mg Delta h = (1/2) m v²$. Thus, $v = sqrt{2g Delta h} = sqrt{2 * 10 * 7.32} = sqrt{146.4} = 12.1$ m/s.

Question 40

Topic: Industrial Engineering

A manufacturer has the following data regarding a product. Fixed cost per month = Rs.80,000, Variable cost per unit = Rs.350, Selling price per unit = Rs.500, Production capacity = 1600 units per month. If the production is carried out at 75% of the rated capacity, then monthly profit is (in Rs.)

• A. 1 Lakh
• B. 1.2 Lakhs
• C. 1.4 Lakhs
• D. 0.8 Lakhs

Answer: A. 1 Lakh

Explanation: Actual units produced = $0.75 * 1600 = 1200$ units. Total revenue = $1200 * 500 = 6,00,000$ Rs. Total cost = Fixed Cost + (Variable Cost * Units) = $80,000 + (350 * 1200) = 80,000 + 4,20,000 = 5,00,000$ Rs. Monthly profit = Total Revenue - Total Cost = $6,00,000 - 5,00,000 = 1,00,000$ Rs (1 Lakh).

Question 41

Topic: Industrial Engineering

For a PERT activity, the optimistic time is 6 minutes, the most likely time is 9 minutes and pessimistic time is 16 minutes. The expected time of the activity (in seconds) is

• A. 600
• B. 540
• C. 620
• D. 580

Answer: D. 580

Explanation: Expected time $T_e = (t_o + 4t_m + t_p) / 6 = (6 + 4(9) + 16) / 6 = (6 + 36 + 16) / 6 = 58 / 6$ minutes. Converting to seconds: $(58 / 6) * 60 = 58 * 10 = 580$ seconds.

Question 42

Topic: Theory of Machines

A cantilever beam of cross section A, moment of inertia I and length L is having natural frequency $omega_1$. If the beam is accidentally broken into two halves, the natural frequency of the remaining cantilever beam $omega_2$ will be such that it

• A. Increases ($omega_2 > omega_1$)
• B. Decreases ($omega_2 < omega_1$)
• C. Remains same $omega_2 = omega_1$
• D. Cannot be obtained from the given data

Answer: A. Increases ($omega_2 > omega_1$)

Explanation: For a continuous cantilever beam, the natural frequency $omega_n$ is proportional to $1/L²$. If the beam breaks and the length is halved to $L/2$, the new natural frequency will be proportional to $1/(L/2)^2 = 4/L²$, making it significantly higher. Therefore, $omega_2 > omega_1$.

Question 43

Topic: Thermodynamics

Vapor pressure in mm of mercury of solid and liquid ammonia is given by $\ln P = 23.1 - (3749/T)$ and $\ln P = 19.6 - (3049/T)$, find out the triple point.

• A. 700 K
• B. 231 K
• C. 200 K
• D. 180 K

Answer: C. 200 K

Explanation: At the triple point, the vapor pressures of the solid and liquid phases are equal. Equating the two expressions: $23 Jan 3749/T = 19 Jun 3049/T$. Rearranging gives $23.1 - 19.6 = 3749/T - 3049/T$, so $3.5 = 700/T$. Solving for $T$ yields $T = 700 / 3.5 = 200$ K.

Question 44

Topic: Machine Design

If the load acting on the ball bearing is increased 2 times its life will

• A. Reduce by 2 times
• B. Reduce by 4 times
• C. Reduce by 8 times
• D. Not change

Answer: C. Reduce by 8 times

Explanation: The life of a ball bearing $L_{10}$ in millions of revolutions is proportional to $(C/P)^3$, where $P$ is the equivalent load. If the load is doubled (becomes $2P$), the new life is proportional to $1/(2P)^3 = 1/(8P^3)$. Therefore, the life is reduced by a factor of 8.

Question 45

Topic: Machine Design

An old clutch is to transmit a torque of 100 N-m. The inner and outer diameter of the clutch plate is 100 mm and 200 mm respectively. The force required for engagement is (assume $mu = 1/3$ with uniform wear theory)

• A. 4 kN
• B. 10 kN
• C. 25 kN
• D. 50 kN

Answer: A. 4 kN

Explanation: For an old clutch, uniform wear theory is used. Torque $T = (1/2) * mu * F * (R_o + R_i)$. Given $T = 100$ N-m = $100,000$ N-mm. Inner radius $R_i = 50$ mm, Outer radius $R_o = 100$ mm. $100,000 = (1/2) * (1/3) * F * (100 + 50) = (1/6) * F * 150 = 25F$. Solving for $F$, $F = 100,000 / 25 = 4000$ N = 4 kN.

Question 46

Topic: Thermodynamics

What is the change in entropy when water is vaporized at 100 °C. Latent heat of vaporization of water may be taken as 2238 kJ/kg.

• A. 22.38 kJ/kg K
• B. 2.238 kJ/kg K
• C. 6 kJ/kg K
• D. 100 kJ/kg K

Answer: C. 6 kJ/kg K

Explanation: The change in entropy dS during phase change at constant temperature is dQ / T. Temperature T = 100 + 273 = 373 K. Latent heat dQ = 2238 kJ/kg. Therefore, dS = 2238 / 373 = 6 kJ/kg K.

Question 47

Topic: Thermodynamics

Examine the following statements: (i) Work is high grade energy. (ii) It is possible for a heat engine to produce work in a cycle by exchanging heat with bodies at same temperature. (iii) It is impossible to construct a device which operating in a cycle will produce no other effect other than transfer of heat from a cooler to a hotter body.

• A. All statements are wrong
• B. All statements are correct
• C. Statements (i) and (iii) are correct
• D. Statements (ii) and (iii) are correct

Answer: C. Statements (i) and (iii) are correct

Explanation: Statement (i) is true; work is fully convertible to heat, making it high-grade energy. Statement (ii) is false according to the Kelvin-Planck statement, which requires interaction with at least two thermal reservoirs at different temperatures. Statement (iii) is the Clausius statement of the Second Law of Thermodynamics, which is true.

Question 48

Topic: Strength of Materials

The elastic strain energy stored in a rectangular cantilever beam due to a bending moment M applied at the end is given by (sigma_max is Maximum Bending Stress, V is Volume and E is Young's Modulus)

• A. sigma^2_max (V/3E)
• B. sigma^2_max (V/2E)
• C. sigma^2_max (V/6E)
• D. sigma^2_max (V/4E)

Answer: C. sigma^2_max (V/6E)

Explanation: Strain energy U = M² * L / (2EI). Maximum bending stress sigma_max = M * (h/2) / I, so M = sigma_max * I / (h/2). Substituting M into U gives U = (sigma_max^2 * I * L) / (2 * E * (h/2)^2). For a rectangle, I = b * h³ / 12, and Volume V = b * h * L. Simplifying the expression yields U = sigma_max^2 * V / 6E.

Question 49

Topic: Thermodynamics

Molliar diagram is also called as

• A. Temperature entropy chart
• B. Pressure - volume chart
• C. Pressure enthalpy chart
• D. Enthalpy entropy chart

Answer: D. Enthalpy entropy chart

Explanation: A Mollier diagram is a thermodynamic diagram representing the enthalpy (h) of a system versus its entropy (s), hence it is called an Enthalpy-entropy chart.

Question 50

Topic: Thermodynamics

Real gas behavior approaches ideal gas equation when

• A. Pressure and temperature are very high
• B. Pressure is very high and temperature is very low
• C. Pressure and temperature are very low
• D. Pressure is very low and temperature is very high

Answer: D. Pressure is very low and temperature is very high

Explanation: Real gases behave most like ideal gases at very low pressures and high temperatures because the intermolecular forces become negligible and the volume occupied by the gas molecules is extremely small compared to the total volume.

Question 51

Topic: Strength of Materials

A stainless steel rod of diameter 12 mm and length 2.5 m is kept on frictionless surface and heated from 300 K to 400 K. What is the thermal stress on the rod? (coefficient of linear expansion of the steel may be taken as 10^-5 /K; yield strength of steel as 200 MPa and Young's modulus as 210 GPa)

• A. 52.5 MPa
• B. 0.525 MPa
• C. 1 MPa
• D. 0 MPa

Answer: D. 0 MPa

Explanation: Thermal stress is only developed when the thermal expansion or contraction of a body is restricted. Since the rod is kept on a frictionless surface and is completely free to expand, no thermal stress will be induced.

Question 52

Topic: Material Science

What is the kind of deformation that occurs in shape memory alloys?

• A. Thermo Plastic
• B. Plastic
• C. Thermo Elastic
• D. None

Answer: C. Thermo Elastic

Explanation: Shape memory alloys return to their pre-deformed shape upon heating. This shape memory effect is the result of a solid-state, diffusionless phase transformation known as a thermo-elastic martensitic transformation.

Question 53

Topic: Theory of Machines

The equation of motion of a spring-mass-damper system is given by x'' + 4x' + 16x = 12 sin(5t). What is the damping factor of the system?

• A. 1
• B. 0.5
• C. 0.25
• D. 0.75

Answer: B. 0.5

Explanation: The standard differential equation of motion is x'' + 2*zeta*omega_n*x' + omega_n^2*x = F(t). Comparing this with the given equation: omega_n^2 = 16, so omega_n = 4 rad/s. Also, 2*zeta*omega_n = 4. Substituting omega_n gives 2*zeta*(4) = 4, so 8*zeta = 4, which means zeta = 0.5.

Question 54

Topic: Theory of Machines

A gear train comprises of 4 spur gears - the driven gear G1, two idlers G2, G3 and the driving gear G4. The radii of G1, G2, G3 and G4 are r, r/2, r/3 and 2r, respectively. What is the ratio of angular velocities of G1 to G4 and their direction of rotation?

• A. Ratio 1, opposite direction
• B. Ratio = 1/6, same direction
• C. Ratio = 2, Opposite direction
• D. Ratio = 1/2, same direction

Answer: C. Ratio = 2, Opposite direction

Explanation: In a simple gear train with idlers, the speed ratio only depends on the driving and driven gears. Ratio = omega_1 / omega_4 = Radius_4 / Radius_1 = 2r / r = 2. With 4 gears (1 driver, 2 idlers, 1 driven), there are 3 meshes. Each mesh reverses rotation. Therefore, G4 and G1 will rotate in opposite directions.

Question 55

Topic: Engineering Mechanics

The uniform link shown in the following figure has a mass of 12 kg. If the spring is un-stretched at theta = 0 deg, determine the angle theta for stable equilibrium. Neglect the surface friction and assume g = 10 m/s² (k = 200 N/m, l = 0.6 m)

• A. 0 deg
• B. 60 deg
• C. 90 deg
• D. 30 deg

Answer: B. 60 deg

Explanation: Taking moments about the pivot or using the potential energy method: The restoring force of the spring balances the overturning moment of the link's weight. The equilibrium condition yields 1 - cos(theta) = (m*g) / (2*k*l). Substituting values: 1 - cos(theta) = (12 * 10) / (2 * 200 * 0.6) = 120 / 240 = 0.5. Thus, cos(theta) = 0.5, which gives theta = 60 deg.

Question 56

Topic: Manufacturing

A face milling operation is carried out with a 100 mm diameter milling cutter having 4 teeth and spindle speed of 300 rpm. If the feed per tooth is 0.125 mm, what is the table feed in mm/min?

• A. 90 mm/min
• B. 120 mm/min
• C. 150 mm/min
• D. 180 mm/min

Answer: C. 150 mm/min

Explanation: Table feed rate = Feed per tooth * Number of teeth * Spindle speed = 0.125 mm/tooth * 4 teeth/rev * 300 rev/min = 0.5 * 300 = 150 mm/min.

Question 57

Topic: Manufacturing

Which is the correct sequence of machining operations to be performed to finish a through hole of large diameter?

• A. Drilling, reaming, boring
• B. Drilling, boring, reaming
• C. Boring, drilling, reaming
• D. Boring, reaming, drilling

Answer: B. Drilling, boring, reaming

Explanation: The standard sequence to produce a large, precise hole is to first drill a pilot hole, then use a boring tool to enlarge it and align it accurately, and finally use a reamer to achieve the required dimensional accuracy and surface finish.

Question 58

Topic: Manufacturing

In abrasive jet machining process, the main mechanism of material removal takes place due to?

• A. Electrochemical action
• B. Ionic dissolution
• C. Mechanical impact
• D. Melting and evaporation

Answer: C. Mechanical impact

Explanation: In Abrasive Jet Machining (AJM), material removal occurs through the mechanical impact (erosion) of high-velocity abrasive particles directed at the workpiece.

Question 59

Topic: Thermodynamics

For water, the latent heat of freezing is 334 kJ/kg and the specific heat capacity averages 4.19 kJ/kgK. The quantity of heat to be removed from 1 kg of water at 30 °C in order to turn it into ice at 0 °C is:

• A. 567 kJ
• B. 319 kJ
• C. 483 kJ
• D. 459 kJ

Answer: D. 459 kJ

Explanation: Total heat removed Q = Sensible heat + Latent heat. Sensible heat to cool water from 30 °C to 0 °C = m * C * dT = 1 * 4.19 * 30 = 125.7 kJ. Latent heat to freeze = m * L = 1 * 334 = 334 kJ. Total Q = 125.7 + 334 = 459.7 kJ, which is approximately 459 kJ.

Question 60

Topic: Strength of Materials

A point load applied at shear center of beam induces

• A. Zero shear force
• B. Zero bending
• C. Pure twisting
• D. Pure bending

Answer: D. Pure bending

Explanation: The shear center is defined as the point through which an applied transverse load will produce pure bending in the cross-section without any twisting (torsion).

Question 61

Topic: Theory of Machines

A cam and oscillating follower mechanism is shown in the figure below. The Mobility of the mechanism is

• A. -1
• B. 0
• C. 1
• D. 2

Answer: C. 1

Explanation: The mechanism has 4 links: Ground (1), Cam (2), Follower (3), and Roller (4). It has 3 lower pairs (ground-cam, ground-follower, follower-roller) and 1 higher pair (cam-roller). Using Kutzbach criterion: F = 3(L-1) - 2J1 - J2 = 3(4-1) - 2(3) - 1 = 2. However, the rotation of the roller is a redundant degree of freedom that does not affect the input-output kinematic relationship. Thus, the effective mobility (functional degree of freedom) is 2 - 1 = 1.

Question 62

Topic: Strength of Materials

A body is subjected to a bi-axial state of stress as shown in figure below: The magnitude of normal and shear stresses acting on the 45 degree plane in MPa is

• A. 50, 50 MPa
• B. 60, 40 MPa
• C. 40, 50 MPa
• D. 50, 30 MPa

Answer: C. 40, 50 MPa

Explanation: From the figure, sigma_x = 60 MPa (tension), sigma_y = -40 MPa (compression), and tau_xy = -30 MPa. For a plane whose normal makes an angle of -45 degrees with the x-axis, the normal stress is sigma_n = (sigma_x + sigma_y)/2 + (sigma_x - sigma_y)/2 * cos(-90) + tau_xy * sin(-90) = (60 - 40)/2 + 0 + (-30)(-1) = 10 + 30 = 40 MPa. The shear stress is tau = -(sigma_x - sigma_y)/2 * sin(-90) + tau_xy * cos(-90) = -((60 - (-40))/2) * (-1) + 0 = 50 MPa.

Question 63

Topic: Thermodynamics

Which of the following represent sterling cycle

• A. II
• B. I and III
• C. I and IV
• D. All of the above

Answer: B. I and III

Explanation: A Stirling cycle consists of two isothermal processes and two isochoric (constant volume) processes. Graph I shows this on a P-V diagram (two curves for isothermal and two vertical lines for constant volume). Graph III shows this on a T-S diagram (two horizontal lines for isothermal and two curves for constant volume).

Question 64

Topic: Fluid Mechanics

Rapid and sharp changes in fluid properties in a very small region of gas flow is known as

• A. Cavitation
• B. Shock wave
• C. Sublimation
• D. Chocking

Answer: B. Shock wave

Explanation: A shock wave is a steep finite pressure wave where rapid and sharp changes in fluid properties (such as pressure, temperature, and density) occur in a supersonic gas flow.

Question 65

Topic: Heat Transfer

The wall of a chemical plant is modified with two additional layers having half the thermal conductivity of original wall material. The inner layer is of half the thickness and outer layer of same thickness as original wall. What will be the heat flow across the modified wall?

• A. Same as original
• B. Twice of original
• C. 50% of original
• D. 25% of original

Answer: D. 25% of original

Explanation: Original thermal resistance R1 = L/k. New total resistance R_new = R_inner + R1 + R_outer = (L/2)/(k/2) + L/k + L/(k/2) = L/k + L/k + 2L/k = 4L/k. Since the new thermal resistance is 4 times the original, the heat flow (which is inversely proportional to resistance) becomes 1/4th, or 25% of the original.

Question 66

Topic: Strength of Materials

Three metal balls are suspended by three wires (a, b & c) of equal length arranged in sequence as shown in figure. The masses of the balls, starting at the top are 2 kg, 4kg and 3 kg. The wire diameters in the same order are 2 mm, 1.2 mm & 1 mm respectively. What is the correct order of increasing stresses in the wires?

• A. a, c, b
• B. b, c, a
• C. a, b, c
• D. All are having same stress

Answer: A. a, c, b

Explanation: Stress = Force / Area. Area is proportional to diameter squared (d²). For wire a: Force supports all three masses = 9kg, stress is proportional to 9 / (2^2) = 2.25. For wire b: Force supports two bottom masses = 7kg, stress is proportional to 7 / (1.2^2) = 4.86. For wire c: Force supports bottom mass = 3kg, stress is proportional to 3 / (1^2) = 3. Order of increasing stress is 2.25 (a) < 3.0 (c) < 4.86 (b).

Question 67

Topic: Refrigeration and Air Conditioning

What is the power of 2 tonne refrigeration system? Latent heat of ice may be taken as 336 kJ/kg and specific heat capacity of water as 4 kJ/kgK.

• A. 7.7 kW
• B. 70 kW
• C. 12.5 kW
• D. 0.7 kW

Answer: A. 7.7 kW

Explanation: One tonne of refrigeration is the rate of heat removal required to freeze 1000 kg (1 metric tonne) of water at 0 °C in 24 hours. Power for 1 Tonne = (1000 kg * 336 kJ/kg) / (24 * 3600 s) = 336,000 / 86,400 = 3.88 kW. For 2 tonnes, power = 2 * 3.88 = 7.77 kW, which approximates to 7.7 kW.

Question 68

Topic: IC Engines

An air standard Otto cycle operates with a compression ratio of 9 and the ambient temperature is 300 K. If the adiabatic index of the medium is 1.5, what is the efficiency?

• A. 75%
• B. 66.67%
• C. 50%
• D. 33%

Answer: B. 66.67%

Explanation: The efficiency of an Otto cycle is given by eta = 1 - (1 / r^(gamma - 1)). Substituting r = 9 and gamma = 1.5 gives eta = 1 - (1 / 9^(1.5 - 1)) = 1 - (1 / 9^0.5) = 1 - (1/3) = 2/3 = 66.67%.

Question 69

Topic: Strength of Materials

Determine the equation of the elastic curve for the cantilever beam of uniform cross section as shown in the figure due to the applied bending moment M1 (Consider Young's modulus as E and Moment of inertia as I)

• A. y = M1*x² / 3EI
• B. y = M1*x² / EI
• C. y = M1*x² / 2EI
• D. y = M1*x² / 4EI

Answer: C. y = M1*x² / 2EI

Explanation: The differential equation for the elastic curve is EI(d^2y/dx²) = M(x). Since a constant moment M1 is applied at the free end, M(x) = M1. Integrating once gives EI(dy/dx) = M1*x + C1. At the fixed end (x=0), slope dy/dx = 0, so C1 = 0. Integrating again gives EI*y = M1*(x²/2) + C2. At x=0, deflection y = 0, so C2 = 0. The equation becomes y = M1*x² / 2EI.

Question 70

Topic: Material Science

The preferred choice of the steel material for Cryogenic application is

• A. Ferrite
• B. Austenitic
• C. Martensite
• D. All the Above

Answer: B. Austenitic

Explanation: Austenitic stainless steels (like 304 or 316) are the preferred choice for cryogenic applications because their FCC (Face-Centered Cubic) crystal structure retains excellent ductility and impact toughness at extremely low temperatures, preventing brittle fracture.

Question 71

Topic: Engineering Mechanics

A satellite of mass 100 kg is attached centrally to the rocket stage of mass 900 kg using a separation system. In space the satellite is to be separated and ejected using 10 numbers of identical spring thrusters having same stroke 30 mm. Assume springs are fully transferring the energy and angular body rate of the bodies are negligible after separation. What is the stiffness of the spring required to get the satellite axial relative velocity of 1 m/s?

• A. 0.01 N/mm
• B. 10 N/mm
• C. 20 N/mm
• D. 100 N/mm

Answer: B. 10 N/mm

Explanation: By conservation of momentum: 100*v_s + 900*v_r = 0, so v_r = -v_s/9. Relative velocity v_s - v_r = v_s + v_s/9 = 10*v_s/9 = 1 m/s. Thus, v_s = 0.9 m/s and v_r = -0.1 m/s. Total Kinetic Energy = 0.5 * 100 * (0.9)^2 + 0.5 * 900 * (-0.1)^2 = 40.5 + 4.5 = 45 J. Energy of 10 springs = 10 * (0.5 * k * x²) = 5 * k * (0.03 m)^2 = 0.0045 * k. Equating energies: 0.0045 * k = 45 gives k = 10,000 N/m = 10 N/mm.

Question 72

Topic: Manufacturing

Which of the following is not a casting defect?

• A. Spatter
• B. Blow holes
• C. Cold shut
• D. Hot tear

Answer: A. Spatter

Explanation: Spatter consists of molten metal droplets that are expelled from the weld puddle and scatter around the weld area. Therefore, it is a welding defect, not a casting defect.

Question 73

Topic: Manufacturing

Which of the following improve weld ability of steel?

• A. Low carbon content and poor affinity to oxygen
• B. High carbon content and poor affinity to oxygen
• C. Low carbon content and good affinity to oxygen
• D. High carbon content and good affinity to oxygen

Answer: A. Low carbon content and poor affinity to oxygen

Explanation: Weldability of steel is primarily improved by maintaining a low carbon content (which prevents the formation of brittle martensite in the heat-affected zone) and having a poor affinity for oxygen to reduce oxidation and porosity issues during the welding process.

Question 74

Topic: Theory of Machines

What is true in respect of the self locking screw jack with square thread? (I) The friction angle is more than the helix angle of the thread. (II) The efficiency of the screw jack should not be more than 50%

• A. Only I
• B. Only II
• C. Both I and II
• D. None of the above

Answer: C. Both I and II

Explanation: For a screw to be self-locking, the friction angle must be greater than or equal to the helix angle (lead angle) of the thread. Consequently, the mathematical maximum efficiency of a self-locking square-threaded screw jack is 50%. Therefore, both statements are true.

Question 75

Topic: Manufacturing

The commonly used flux for brazing is:

• A. Borax
• B. Epoxy
• C. NH4Cl
• D. Inert gas

Answer: A. Borax

Explanation: Borax (sodium borate) is the most commonly used flux in brazing operations. It dissolves oxides on the metal surfaces and protects them from further oxidation during the heating process.

Question 76

Topic: Industrial Engineering

For a particular project, six activities are to be carried out. The precedence relation and duration of activities are as per the following table. The minimum project completion time (in days) is (A=6, B follows A=5, C follows A=8, D follows B=11, E follows C=6, F follows D,E=7)

• A. 32
• B. 27
• C. 29
• D. 30

Answer: C. 29

Explanation: There are two paths to completion in the network network: Path 1 is A -> B -> D -> F which takes 6 + 5 + 11 + 7 = 29 days. Path 2 is A -> C -> E -> F which takes 6 + 8 + 6 + 7 = 27 days. The minimum project completion time is dictated by the critical (longest) path, which is 29 days.

Question 77

Topic: Material Science

The property of the material that represents maximum amount of strain energy that it can absorb before it fractures under load is called:

• A. Modulus of Resilience
• B. Modulus of Toughness
• C. Modulus of Elasticity
• D. None of the above

Answer: B. Modulus of Toughness

Explanation: The modulus of toughness is defined as the total area under the stress-strain curve up to the point of fracture. It represents the maximum amount of strain energy a material can absorb before breaking.

Question 78

Topic: Refrigeration and Air Conditioning

A Carnot refrigerator works in temperature range of 300 K to 270 K with different refrigerants CFC, R-32 and ammonia. Which of the refrigerant give better COP?

• A. CFC
• B. R-32
• C. Ammonia
• D. All gives same COP

Answer: D. All gives same COP

Explanation: The Coefficient of Performance (COP) of a Carnot refrigerator depends entirely on the absolute temperatures of the heat source and heat sink, and is completely independent of the properties of the working fluid (refrigerant) used.

Question 79

Topic: Fluid Mechanics

A cylindrical body of diameter 2 m and height 1.5 m having a mass of 2 tonnes and floats in sea water with its axis vertical. A lighting equipment is placed on the body such that the water level is at least 0.5 m below the top surface. What is the maximum mass of the electrical equipment that can be placed on the body. (The specific gravity of sea water may be taken as 1.02 and pi = 3.14)

• A. 252 kg
• B. 750 kg
• C. 1202 kg
• D. 275 kg

Answer: C. 1202 kg

Explanation: Maximum depth of immersion = 1.5 - 0.5 = 1.0 m. Maximum displaced volume V = (pi/4) * D² * depth = (3.14/4) * (2^2) * 1.0 = 3.14 m³. Maximum buoyant force = V * rho_water * g = 3.14 * 1020 * g = 3202.8 * g Newtons. This means the cylinder can support a total mass of 3202.8 kg. Subtracting the cylinder's own mass (2000 kg), the maximum mass of the equipment is 3202.8 - 2000 = 1202.8 kg.

Question 80

Topic: Fluid Mechanics

Examine the following statements: (i) Euler's equation is applicable to inviscid liquid (ii) Bernoulli's equation applicable to steady state (iii) Viscous forces are not considered in Navier Stokes equation (iv) Surface tension and compressibility forces are considered in Navier Stokes equation

• A. Statements (i) and (ii) are not correct
• B. Statements (iii) and (iv) are correct
• C. Statements (i), (ii) and (iii) are correct
• D. All the statements are correct

Explanation: Statement (i) is correct: Euler's equation assumes zero viscosity. Statement (ii) is correct: Bernoulli's equation is derived for steady flow. Statement (iii) is incorrect: Navier-Stokes equations explicitly consider viscous forces. Statement (iv) is generally incorrect as standard NS equations focus on gravity, pressure, and viscous forces, not surface tension. None of the provided options accurately reflect that ONLY (i) and (ii) are correct.