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401. What is cold working?
ⓐ. Deformation of metals at high temperatures
ⓑ. Deformation of metals at or below recrystallization temperature
ⓒ. Heating metals until they recrystallize
ⓓ. Deformation only in the elastic region
Correct Answer: Deformation of metals at or below recrystallization temperature
Explanation: Cold working refers to plastic deformation of metals at room temperature or below their recrystallization temperature, leading to strain hardening.
402. What happens to dislocation density during cold working?
ⓐ. It decreases
ⓑ. It remains constant
ⓒ. It increases significantly
ⓓ. It disappears completely
Correct Answer: It increases significantly
Explanation: Cold working causes dislocations to multiply and entangle, which strengthens the metal but reduces ductility.
403. Which of the following properties increases after cold working?
ⓐ. Ductility
ⓑ. Tensile strength and hardness
ⓒ. Electrical conductivity
ⓓ. Grain size
Correct Answer: Tensile strength and hardness
Explanation: Cold working strengthens the material by strain hardening, increasing yield strength, tensile strength, and hardness, but reducing ductility.
404. Which of the following properties decreases due to cold working?
ⓐ. Hardness
ⓑ. Yield strength
ⓒ. Ductility
ⓓ. Density
Correct Answer: Ductility
Explanation: Cold working reduces ductility because dislocations block further plastic flow, making the metal less deformable.
405. A copper wire is cold drawn to reduce its diameter. Which changes occur?
ⓐ. Strength increases, ductility decreases
ⓑ. Strength decreases, ductility increases
ⓒ. Both strength and ductility increase
ⓓ. Both strength and ductility decrease
Correct Answer: Strength increases, ductility decreases
Explanation: Cold drawing introduces strain hardening, increasing strength but reducing ductility of copper.
406. Why are metals annealed after cold working?
ⓐ. To increase dislocation density
ⓑ. To increase brittleness
ⓒ. To relieve internal stresses and restore ductility
ⓓ. To lower recrystallization temperature
Correct Answer: To relieve internal stresses and restore ductility
Explanation: Annealing reduces dislocation density, relieving stresses induced during cold working and making the metal more ductile again.
407. Which of the following is NOT an effect of cold working?
ⓐ. Increase in hardness
ⓑ. Increase in yield strength
ⓒ. Reduction in ductility
ⓓ. Increase in grain size
Correct Answer: Increase in grain size
Explanation: Cold working refines the grain structure and increases dislocation density, but grain size does not increase; it usually decreases until recrystallization.
408. Why is cold rolling of steel sheets done in industries?
ⓐ. To reduce strength of steel
ⓑ. To increase thickness of sheets
ⓒ. To improve surface finish and increase strength
ⓓ. To reduce hardness of steel
Correct Answer: To improve surface finish and increase strength
Explanation: Cold rolling refines surface finish and improves mechanical strength by introducing strain hardening.
409. Which mechanical property is most improved by cold working?
ⓐ. Plasticity
ⓑ. Stiffness
ⓒ. Yield strength
ⓓ. Density
Correct Answer: Yield strength
Explanation: Cold working significantly increases the yield strength of metals, making them more resistant to deformation under load.
410. Why is excessive cold working avoided?
ⓐ. It reduces hardness
ⓑ. It reduces yield strength
ⓒ. It leads to brittleness and cracks due to low ductility
ⓓ. It lowers tensile strength
Correct Answer: It leads to brittleness and cracks due to low ductility
Explanation: Excessive cold working reduces ductility drastically. The material becomes brittle and prone to cracking under stress.
411. Why is strain hardening used in the manufacturing of copper wires?
ⓐ. To make them more ductile
ⓑ. To increase their tensile strength without increasing diameter
ⓒ. To reduce electrical conductivity
ⓓ. To reduce yield strength
Correct Answer: To increase their tensile strength without increasing diameter
Explanation: Cold drawing of copper wires introduces strain hardening, which increases tensile strength, allowing wires to carry greater mechanical loads without breaking.
412. Why are strain-hardened aluminum sheets used in aircraft manufacturing?
ⓐ. To increase weight of the aircraft
ⓑ. To improve strength-to-weight ratio
ⓒ. To reduce modulus of elasticity
ⓓ. To reduce hardness
Correct Answer: To improve strength-to-weight ratio
Explanation: Strain hardening strengthens aluminum sheets, making them lightweight yet strong, an essential requirement in aerospace applications.
413. How does strain hardening benefit the production of bolts and fasteners?
ⓐ. It reduces their brittleness
ⓑ. It increases their shear and tensile strength
ⓒ. It decreases their yield strength
ⓓ. It makes them elastic forever
Correct Answer: It increases their shear and tensile strength
Explanation: Cold working of bolts and fasteners increases dislocation density, thereby enhancing strength and resistance against shear and tensile forces.
414. In which of the following processes is strain hardening most beneficial?
ⓐ. Forging
ⓑ. Wire drawing and rolling
ⓒ. Melting and casting
ⓓ. Heat treatment
Correct Answer: Wire drawing and rolling
Explanation: Strain hardening improves the strength of wires and sheets during drawing and rolling, making these processes highly dependent on plastic deformation.
415. Why are strain-hardened stainless steels used in making surgical instruments?
ⓐ. To make them less elastic
ⓑ. To increase hardness and resistance to wear
ⓒ. To reduce their corrosion resistance
ⓓ. To make them brittle
Correct Answer: To increase hardness and resistance to wear
Explanation: Strain hardening makes stainless steel harder, improving its durability and sharpness for surgical instruments while retaining corrosion resistance.
416. Why is strain hardening applied in the manufacture of springs?
ⓐ. To reduce elasticity
ⓑ. To improve fatigue resistance and hardness
ⓒ. To increase density
ⓓ. To decrease ductility completely
Correct Answer: To improve fatigue resistance and hardness
Explanation: Strain hardening enhances spring stiffness and resistance to repeated stress cycles, improving service life of springs.
417. What is the drawback of strain-hardened materials in engineering applications?
ⓐ. Increased ductility
ⓑ. Reduced yield strength
ⓒ. Decreased ductility and risk of brittleness
ⓓ. Reduced hardness
Correct Answer: Decreased ductility and risk of brittleness
Explanation: Strain-hardened materials gain strength but lose ductility, making them susceptible to cracking under sudden loads.
418. Why is strain hardening used in making armor plates?
ⓐ. To make them lighter
ⓑ. To improve strength and hardness against penetration
ⓒ. To increase thermal conductivity
ⓓ. To reduce stiffness
Correct Answer: To improve strength and hardness against penetration
Explanation: Armor plates are strengthened by strain hardening so they can resist impact and penetration by absorbing large amounts of energy.
419. How is strain hardening useful in the production of beverage cans?
ⓐ. It reduces brittleness
ⓑ. It allows thin sheets of aluminum to have sufficient strength
ⓒ. It increases thickness of the can walls
ⓓ. It reduces tensile strength
Correct Answer: It allows thin sheets of aluminum to have sufficient strength
Explanation: Strain hardening strengthens thin aluminum sheets, enabling lightweight yet durable cans that withstand internal pressure.
420. Which of the following is an example of strain hardening applied in sports equipment?
ⓐ. Elastic bands in shoes
ⓑ. Cold-worked aluminum bats and bicycle frames
ⓒ. Glass rods for pole vault
ⓓ. Rubber balls
Correct Answer: Cold-worked aluminum bats and bicycle frames
Explanation: Aluminum alloys are strain hardened to improve strength while maintaining low weight, making them ideal for sports equipment like bats and bicycle frames.
421. What is the main difference between strain hardening and annealing?
ⓐ. Strain hardening increases ductility; annealing decreases it
ⓑ. Strain hardening increases strength; annealing restores ductility
ⓒ. Both increase brittleness
ⓓ. Both decrease tensile strength
Correct Answer: Strain hardening increases strength; annealing restores ductility
Explanation: Strain hardening strengthens a material by dislocation accumulation but reduces ductility. Annealing reduces dislocation density, relieving stresses and improving ductility.
422. Which property decreases after annealing compared to strain hardening?
ⓐ. Ductility
ⓑ. Tensile strength and hardness
ⓒ. Corrosion resistance
ⓓ. Thermal conductivity
Correct Answer: Tensile strength and hardness
Explanation: Strain hardening increases tensile strength and hardness, but annealing reverses this effect by allowing dislocations to rearrange, reducing strength and increasing ductility.
423. Why is annealing often done after cold working?
ⓐ. To reduce strength permanently
ⓑ. To relieve internal stresses and increase ductility
ⓒ. To make the material brittle
ⓓ. To reduce thermal conductivity
Correct Answer: To relieve internal stresses and increase ductility
Explanation: Cold working induces high dislocation density. Annealing rearranges dislocations, restoring ductility and toughness while relieving residual stresses.
424. Which heat treatment method increases both strength and hardness similar to strain hardening?
ⓐ. Normalizing
ⓑ. Quenching
ⓒ. Tempering
ⓓ. Case hardening
Correct Answer: Quenching
Explanation: Quenching increases hardness and strength by rapid cooling, like strain hardening, but through microstructural changes rather than dislocation density.
425. What is the effect of tempering on quenched steel compared to strain hardening?
ⓐ. Increases brittleness further
ⓑ. Reduces brittleness while maintaining strength
ⓒ. Reduces ductility drastically
ⓓ. Increases dislocation density
Correct Answer: Reduces brittleness while maintaining strength
Explanation: Quenched steel is brittle. Tempering reduces brittleness while retaining most of the strength. Strain hardening, on the other hand, sacrifices ductility for strength.
426. Which of the following correctly compares strain hardening and annealing in terms of dislocation density?
ⓐ. Both increase dislocation density
ⓑ. Strain hardening increases dislocation density; annealing reduces it
ⓒ. Strain hardening reduces dislocation density; annealing increases it
ⓓ. Neither affects dislocation density
Correct Answer: Strain hardening increases dislocation density; annealing reduces it
Explanation: Dislocation multiplication occurs during plastic deformation (strain hardening). Annealing allows recovery and recrystallization, reducing dislocation density.
427. Why is annealing important for materials after strain hardening in manufacturing?
ⓐ. To increase hardness further
ⓑ. To remove brittleness and allow further working
ⓒ. To make material non-elastic
ⓓ. To reduce corrosion resistance
Correct Answer: To remove brittleness and allow further working
Explanation: After strain hardening, ductility is reduced. Annealing restores ductility, making further forming processes possible without cracking.
428. Which property is improved more effectively by annealing than strain hardening?
ⓐ. Ductility and toughness
ⓑ. Hardness
ⓒ. Yield strength
ⓓ. Tensile strength
Correct Answer: Ductility and toughness
Explanation: Strain hardening increases hardness and strength but reduces ductility. Annealing improves ductility and toughness by relieving stresses.
429. Which heat treatment method involves heating followed by slow cooling to soften metals?
ⓐ. Quenching
ⓑ. Annealing
ⓒ. Tempering
ⓓ. Nitriding
Correct Answer: Annealing
Explanation: Annealing is the process of heating a metal above recrystallization temperature and then slowly cooling, making it softer and more ductile.
430. Which statement best summarizes the comparison between strain hardening and annealing?
ⓐ. Both increase strength
ⓑ. Strain hardening increases strength but reduces ductility, while annealing restores ductility but lowers strength
ⓒ. Both restore ductility
ⓓ. Both reduce hardness
Correct Answer: Strain hardening increases strength but reduces ductility, while annealing restores ductility but lowers strength
Explanation: The two processes are complementary. Strain hardening is used to strengthen, and annealing is used to restore ductility for further processing.
431. What is fracture toughness of a material?
ⓐ. Resistance of a material to elastic deformation
ⓑ. Ability of a material to resist crack propagation under stress
ⓒ. The maximum stress a material can withstand before breaking
ⓓ. The stress required to initiate plastic deformation
Correct Answer: Ability of a material to resist crack propagation under stress
Explanation: Fracture toughness is the capacity of a material to resist crack growth. A material with high fracture toughness can prevent sudden failure even with existing flaws.
432. Which material has very low fracture toughness?
ⓐ. Mild steel
ⓑ. Copper
ⓒ. Glass
ⓓ. Aluminum
Correct Answer: Glass
Explanation: Glass is brittle with low fracture toughness. Cracks propagate rapidly once formed, leading to sudden failure without plastic deformation.
433. Which parameter is commonly used to describe fatigue strength of a material?
ⓐ. Stress concentration factor
ⓑ. Endurance limit
ⓒ. Modulus of resilience
ⓓ. Poisson’s ratio
Correct Answer: Endurance limit
Explanation: Endurance limit is the maximum stress amplitude a material can withstand for an infinite number of cycles without failure due to fatigue.
434. Why does fatigue failure occur in metals?
ⓐ. Due to gradual crack initiation and propagation under cyclic loading
ⓑ. Due to sudden overload in one cycle
ⓒ. Because dislocations move only elastically
ⓓ. Due to thermal expansion only
Correct Answer: Due to gradual crack initiation and propagation under cyclic loading
Explanation: Fatigue failure begins with micro-cracks that grow under repeated stress cycles until sudden fracture occurs.
435. In an S–N curve (stress vs. number of cycles), what does the horizontal region represent?
ⓐ. Breaking stress
ⓑ. Endurance limit of the material
ⓒ. Yield strength
ⓓ. Fracture toughness
Correct Answer: Endurance limit of the material
Explanation: The horizontal asymptote on the S–N curve indicates the endurance limit, where the material can withstand infinite cycles without fatigue failure.
436. Which material property is most important for aircraft components subjected to cyclic loading?
ⓐ. High fracture toughness and fatigue resistance
ⓑ. Low Young’s modulus
ⓒ. High brittleness
ⓓ. High density
Correct Answer: High fracture toughness and fatigue resistance
Explanation: Aircraft parts must resist crack propagation (fracture toughness) and endure cyclic stresses without fatigue failure.
437. Which of the following equations estimates the stress intensity factor $K$ for fracture mechanics?
ⓐ. $K = \sigma \sqrt{\pi a}$
ⓑ. $K = \frac{\sigma}{E}$
ⓒ. $K = \tau \cdot \phi$
ⓓ. $K = \frac{F}{A}$
Correct Answer: $K = \sigma \sqrt{\pi a}$
Explanation: In fracture mechanics, stress intensity factor $K$ depends on applied stress $\sigma$ and crack length $a$. Failure occurs when $K$ exceeds fracture toughness $K_c$.
438. Why is fracture toughness higher in ductile materials than brittle materials?
ⓐ. Ductile materials have high density
ⓑ. Ductile materials can undergo plastic deformation, blunting cracks
ⓒ. Brittle materials resist elastic strain more
ⓓ. Brittle materials always have higher yield strength
Correct Answer: Ductile materials can undergo plastic deformation, blunting cracks
Explanation: Plastic deformation absorbs energy and prevents crack growth, giving ductile materials higher fracture toughness compared to brittle ones.
439. Which of the following best describes fatigue failure?
ⓐ. Occurs after single overload
ⓑ. Gradual crack initiation and sudden fracture after cyclic stress
ⓒ. Sudden shattering without crack growth
ⓓ. Purely elastic behaviour under cyclic loads
Correct Answer: Gradual crack initiation and sudden fracture after cyclic stress
Explanation: Fatigue failure develops slowly under repeated stresses much below the ultimate strength. Once cracks grow large, sudden fracture occurs.
440. Why are surface defects dangerous for fatigue life?
ⓐ. They act as stress concentrators where cracks can initiate
ⓑ. They increase Young’s modulus of the material
ⓒ. They reduce hardness of the surface
ⓓ. They decrease thermal conductivity
Correct Answer: They act as stress concentrators where cracks can initiate
Explanation: Fatigue cracks usually start at the surface because surface irregularities act as stress concentration points, reducing fatigue life.
441. Which of the following is the primary cause of brittle fracture in materials?
ⓐ. Excessive plastic deformation
ⓑ. Rapid crack propagation without significant plastic flow
ⓒ. High ductility
ⓓ. Increase in temperature
Correct Answer: Rapid crack propagation without significant plastic flow
Explanation: Brittle fracture occurs suddenly with very little plastic deformation. Cracks propagate rapidly, often perpendicular to applied stress.
442. Why does ductile fracture occur more slowly than brittle fracture?
ⓐ. Because ductile materials have lower Young’s modulus
ⓑ. Because ductile materials undergo large plastic deformation before breaking
ⓒ. Because ductile materials have higher density
ⓓ. Because ductile materials cannot sustain stress
Correct Answer: Because ductile materials undergo large plastic deformation before breaking
Explanation: Plastic deformation in ductile materials absorbs energy and slows crack propagation, unlike brittle fracture which occurs suddenly.
443. Which environmental factor commonly contributes to fracture in metals?
ⓐ. High humidity and corrosion
ⓑ. High reflectivity
ⓒ. Low conductivity
ⓓ. Low density
Correct Answer: High humidity and corrosion
Explanation: Environmental effects like corrosion can initiate cracks, reducing toughness and making metals more prone to fracture.
444. What is stress concentration, and how does it cause fracture?
ⓐ. Even distribution of stress across a material
ⓑ. Localized increase in stress near flaws or notches
ⓒ. Sudden reduction in stress at surface
ⓓ. Elastic recovery after loading
Correct Answer: Localized increase in stress near flaws or notches
Explanation: Stress concentration occurs near defects or sharp corners, greatly increasing local stress, which accelerates crack initiation and fracture.
445. Which of the following methods improves fracture resistance in metals?
ⓐ. Introducing more surface flaws
ⓑ. Refining grain size
ⓒ. Reducing ductility
ⓓ. Increasing brittleness
Correct Answer: Refining grain size
Explanation: Smaller grains hinder crack propagation and increase toughness. Grain refinement is a common method to improve fracture resistance.
446. Why are alloys generally tougher than pure metals?
ⓐ. Alloys always have higher density
ⓑ. Alloying introduces obstacles to dislocation and crack motion
ⓒ. Alloys have lower Young’s modulus
ⓓ. Alloys are more brittle
Correct Answer: Alloying introduces obstacles to dislocation and crack motion
Explanation: Alloying elements create lattice distortions, impeding dislocation and crack movement, which improves fracture resistance.
447. Which process reduces the risk of fracture by relieving internal stresses?
ⓐ. Quenching
ⓑ. Annealing
ⓒ. Cold working
ⓓ. Strain hardening
Correct Answer: Annealing
Explanation: Annealing reduces dislocation density and internal stresses, improving ductility and resistance to fracture.
448. Which of the following is a common cause of fracture in ceramics?
ⓐ. High ductility
ⓑ. Presence of micro-cracks and porosity
ⓒ. Plastic deformation of grains
ⓓ. High thermal conductivity
Correct Answer: Presence of micro-cracks and porosity
Explanation: Ceramics are brittle and contain flaws such as micro-cracks and pores, which serve as initiation sites for fracture.
449. Which engineering method is used to prevent crack growth in aircraft structures?
ⓐ. Using brittle materials
ⓑ. Stress concentration at rivet holes
ⓒ. Designing with crack arresters and rounded corners
ⓓ. Allowing sharp notches
Correct Answer: Designing with crack arresters and rounded corners
Explanation: Sharp corners act as stress concentrators. Rounded corners and crack arrest features reduce stress concentration and delay crack propagation.
450. Why does fracture toughness improve with higher temperature in many metals?
ⓐ. Because ductility decreases
ⓑ. Because dislocation motion becomes easier, allowing plastic deformation to absorb energy
ⓒ. Because brittleness increases
ⓓ. Because density increases
Correct Answer: Because dislocation motion becomes easier, allowing plastic deformation to absorb energy
Explanation: At higher temperatures, dislocations move more easily, plastic deformation increases, and the material can absorb more energy before fracture.
451. Why is fracture toughness important in aerospace engineering?
ⓐ. Aircraft materials are exposed only to static loads
ⓑ. Aircraft structures must resist crack growth under cyclic stresses and high loads
ⓒ. Aircraft metals should always fail suddenly
ⓓ. Aircraft require materials with low ductility
Correct Answer: Aircraft structures must resist crack growth under cyclic stresses and high loads
Explanation: Aerospace components face fluctuating stresses during flight. High fracture toughness ensures cracks do not propagate rapidly, preventing catastrophic failure.
452. Which design principle is applied in aircraft fuselages to prevent crack propagation?
ⓐ. Introducing sharp corners
ⓑ. Using crack arrest features and rounded windows
ⓒ. Maximizing stress concentration
ⓓ. Reducing ductility of materials
Correct Answer: Using crack arrest features and rounded windows
Explanation: Sharp corners concentrate stress, promoting cracks. Aircraft windows are rounded to reduce stress concentration, enhancing fracture resistance.
453. In automotive crash design, why are ductile materials preferred?
ⓐ. They absorb energy through plastic deformation before failure
ⓑ. They always resist all fractures
ⓒ. They break suddenly without deformation
ⓓ. They are cheaper to produce
Correct Answer: They absorb energy through plastic deformation before failure
Explanation: Ductile materials undergo large plastic deformation during crashes, absorbing energy and improving passenger safety.
454. Which property is most critical for materials used in car chassis?
ⓐ. High electrical resistivity
ⓑ. High fracture toughness and ductility
ⓒ. High brittleness
ⓓ. Low Young’s modulus
Correct Answer: High fracture toughness and ductility
Explanation: Car chassis must resist crack propagation and withstand impact loads. High fracture toughness and ductility prevent sudden fracture.
455. Why are composites like carbon-fiber reinforced polymers used in aerospace structures?
ⓐ. They are brittle and heavy
ⓑ. They provide high strength-to-weight ratio and good fracture resistance
ⓒ. They reduce toughness
ⓓ. They have very high density
Correct Answer: They provide high strength-to-weight ratio and good fracture resistance
Explanation: Composites combine low density with high strength and fracture toughness, making them ideal for aerospace applications.
456. Which fracture mechanics principle is applied in designing safe automobile tires?
ⓐ. Stress concentration must be maximized
ⓑ. Crack propagation must be controlled under cyclic loads
ⓒ. Yield strength must be ignored
ⓓ. Elasticity is not required
Correct Answer: Crack propagation must be controlled under cyclic loads
Explanation: Tires experience millions of stress cycles. Crack-resistant rubber and design considerations reduce fatigue and fracture risks.
457. Why is fracture mechanics applied in the design of bridges?
ⓐ. Bridges never experience dynamic loads
ⓑ. To ensure cracks do not propagate under repeated traffic and environmental stresses
ⓒ. To reduce ductility of steel
ⓓ. To maximize stress concentration at joints
Correct Answer: To ensure cracks do not propagate under repeated traffic and environmental stresses
Explanation: Bridges experience cyclic loading and environmental effects. High fracture toughness prevents crack propagation, ensuring structural safety.
458. In jet engines, why is fatigue resistance as important as fracture toughness?
ⓐ. Because engines operate only once
ⓑ. Because jet engine parts face high-speed cyclic stresses and thermal loads
ⓒ. Because metals must be brittle at high temperatures
ⓓ. Because ductility is unimportant
Correct Answer: Because jet engine parts face high-speed cyclic stresses and thermal loads
Explanation: Jet engines undergo continuous cyclic loading at extreme temperatures. High fatigue resistance and fracture toughness are critical to avoid failure.
459. Which method is commonly used in automotive and aerospace industries to improve fracture resistance of steel?
ⓐ. Grain refinement by controlled heat treatment
ⓑ. Increasing porosity
ⓒ. Adding sharp notches in design
ⓓ. Cold rolling without annealing
Correct Answer: Grain refinement by controlled heat treatment
Explanation: Fine-grained steels have higher fracture toughness. Controlled cooling and heat treatment refine grains, enhancing resistance to crack propagation.
460. Why is damage tolerance analysis important in structural engineering?
ⓐ. To reduce ductility of materials
ⓑ. To predict how cracks initiate and grow under service conditions
ⓒ. To eliminate elasticity in design
ⓓ. To maximize stress concentration at flaws
Correct Answer: To predict how cracks initiate and grow under service conditions
Explanation: Damage tolerance ensures structures remain safe even with small cracks. Engineers use fracture mechanics to predict crack growth and set inspection intervals.
461. What is creep?
ⓐ. Sudden fracture of a material under stress
ⓑ. Time-dependent plastic deformation of a material under constant stress at high temperature
ⓒ. Purely elastic deformation under load
ⓓ. Vibrations in a solid under dynamic load
Correct Answer: Time-dependent plastic deformation of a material under constant stress at high temperature
Explanation: Creep is the gradual, permanent deformation of materials subjected to constant stress, especially when operating at temperatures above \~0.4 × melting temperature (in K).
462. Which equation approximately describes steady-state creep rate?
ⓐ. $\dot{\epsilon} = \frac{\sigma}{E}$
ⓑ. $\dot{\epsilon} = A \sigma^n e^{-Q/RT}$
ⓒ. $\dot{\epsilon} = \sigma \cdot t$
ⓓ. $\dot{\epsilon} = kT$
Correct Answer: $\dot{\epsilon} = A \sigma^n e^{-Q/RT}$
Explanation: The Norton’s law (power law creep) relates creep strain rate $\dot{\epsilon}$ to applied stress $\sigma$, temperature $T$, and activation energy $Q$.
463. A material has creep rate described by $\dot{\epsilon} = A\sigma^3$. If stress is doubled, how does creep rate change?
ⓐ. Increases 2 times
ⓑ. Increases 4 times
ⓒ. Increases 6 times
ⓓ. Increases 8 times
Correct Answer: Increases 8 times
Explanation: With exponent $n=3$, $\dot{\epsilon} \propto \sigma^3$. Doubling stress increases creep rate by $2^3 = 8$.
464. A metal sample experiences a steady-state creep strain rate of $1 \times 10^{-6} \, s^{-1}$ at $200 \, MPa$. If stress is increased to $300 \, MPa$ with stress exponent $n = 4$, what will be the new creep rate?
ⓐ. $2.0 \times 10^{-6} \, s^{-1}$
ⓑ. $3.2 \times 10^{-6} \, s^{-1}$
ⓒ. $5.1 \times 10^{-6} \, s^{-1}$
ⓓ. $5.1 \times 10^{-5} \, s^{-1}$
Correct Answer: $5.1 \times 10^{-5} \, s^{-1}$
Explanation: $\frac{\dot{\epsilon}_2}{\dot{\epsilon}_1} = \left(\frac{\sigma_2}{\sigma_1}\right)^n = (300/200)^4 = (1.5)^4 = 5.06$.
So, $\dot{\epsilon}_2 = 1 \times 10^{-6} \times 5.06 \times 10^{1} = 5.1 \times 10^{-5}$.
465. In the creep curve, which stage shows nearly constant strain rate?
ⓐ. Primary creep
ⓑ. Secondary creep
ⓒ. Tertiary creep
ⓓ. Instantaneous elastic strain
Correct Answer: Secondary creep
Explanation: Secondary or steady-state creep is characterized by a nearly constant strain rate, governed by balance between strain hardening and recovery.
466. A turbine blade operates at $1000 \, K$. If the activation energy for creep is $200 \, kJ/mol$, and gas constant $R = 8.314 \, J/mol·K$, what factor reduces creep rate when temperature drops to $900 \, K$?
ⓐ. $e^{-200000/(8.314 \times 100)}$
ⓑ. $e^{200000/RT}$
ⓒ. Ratio of $e^{-Q/RT}$ at 1000 K and 900 K
ⓓ. Independent of temperature
Correct Answer: Ratio of $e^{-Q/RT}$ at 1000 K and 900 K
Explanation: According to Norton’s law, creep rate depends exponentially on temperature via $e^{-Q/RT}$. Lowering $T$ reduces creep rate drastically.
467. If creep strain is given by $\epsilon = \epsilon_0 + \dot{\epsilon}t$, where $\epsilon_0 = 0.002$ and $\dot{\epsilon} = 2 \times 10^{-6} \, s^{-1}$, find strain after 10 hours.
ⓐ. 0.0052
ⓑ. 0.0092
ⓒ. 0.074
ⓓ. 0.20
Correct Answer: 0.074
Explanation: $t = 10 \, h = 36000 \, s$.
$\epsilon = 0.002 + (2 \times 10^{-6})(36000) = 0.002 + 0.072 = 0.074$.
468. Which factor primarily controls creep mechanism at very high temperatures (close to melting point)?
ⓐ. Dislocation slip
ⓑ. Diffusion of atoms (Nabarro-Herring creep)
ⓒ. Elastic recovery
ⓓ. Fatigue
Correct Answer: Diffusion of atoms (Nabarro-Herring creep)
Explanation: At very high temperatures, creep is dominated by atomic diffusion through the lattice or grain boundaries.
469. A specimen shows primary, secondary, and tertiary creep stages. Which stage is most dangerous for failure?
ⓐ. Primary
ⓑ. Secondary
ⓒ. Tertiary
ⓓ. All equally dangerous
Correct Answer: Tertiary
Explanation: Tertiary creep occurs due to necking and micro-crack formation, leading to rapid increase in strain rate and eventual fracture.
470. In designing high-temperature components like boiler tubes, creep resistance is improved by:
ⓐ. Reducing grain size
ⓑ. Using alloys with stable carbides and larger grains
ⓒ. Increasing brittleness
ⓓ. Reducing melting point of materials
Correct Answer: Using alloys with stable carbides and larger grains
Explanation: Large grains reduce grain boundary sliding and stable carbides strengthen the material, improving creep resistance at high temperatures.
471. Which of the following is the most important factor affecting creep rate in metals?
ⓐ. Density of the material
ⓑ. Temperature relative to melting point
ⓒ. Electrical conductivity
ⓓ. Poisson’s ratio
Correct Answer: Temperature relative to melting point
Explanation: Creep is significant when operating temperature exceeds about 0.4 $T_m$ (in Kelvin). The higher the temperature relative to melting point, the higher the creep rate.
472. According to Norton’s law, steady-state creep rate is proportional to:
ⓐ. $\sigma$
ⓑ. $\sigma^n$
ⓒ. $E$
ⓓ. $t$
Correct Answer: $\sigma^n$
Explanation: Steady-state creep rate $\dot{\epsilon} = A \sigma^n e^{-Q/RT}$, where $n$ is the stress exponent. Thus, creep rate depends strongly on stress raised to a power.
473. A material shows creep exponent $n = 5$. If stress doubles, by what factor does creep rate increase?
ⓐ. 2
ⓑ. 4
ⓒ. 16
ⓓ. 32
Correct Answer: 32
Explanation: $\dot{\epsilon} \propto \sigma^n$. For $n=5$, doubling stress multiplies creep rate by $2^5 = 32$.
474. How does grain size affect creep resistance at high temperatures?
ⓐ. Smaller grains always increase creep resistance
ⓑ. Larger grains reduce grain boundary sliding and improve creep resistance
ⓒ. Grain size has no effect
ⓓ. Smaller grains always reduce creep
Correct Answer: Larger grains reduce grain boundary sliding and improve creep resistance
Explanation: Grain boundaries are weak zones for creep. Larger grains mean fewer boundaries, reducing grain boundary creep.
475. If creep rate is given by $\dot{\epsilon} = A \sigma^4 e^{-Q/RT}$, and activation energy $Q$ increases, what happens to creep rate?
ⓐ. Increases
ⓑ. Decreases
ⓒ. Remains constant
ⓓ. First increases then decreases
Correct Answer: Decreases
Explanation: Higher activation energy reduces atomic diffusion, lowering creep rate.
476. Why are alloys used instead of pure metals in high-temperature applications?
ⓐ. Pure metals always resist creep better
ⓑ. Alloying increases dislocation pinning and slows creep
ⓒ. Alloys reduce hardness but improve ductility
ⓓ. Pure metals are more ductile
Correct Answer: Alloying increases dislocation pinning and slows creep
Explanation: Solute atoms and precipitates in alloys hinder dislocation motion and diffusion, thus reducing creep rate.
477. Which type of creep dominates at very high temperatures near melting point?
ⓐ. Dislocation creep
ⓑ. Diffusional creep
ⓒ. Elastic creep
ⓓ. Fatigue creep
Correct Answer: Diffusional creep
Explanation: At extreme temperatures, atoms move through lattice or along grain boundaries, causing diffusional creep (Nabarro-Herring or Coble creep).
478. A steel component operates at $600 \, ^\circ C$ with creep rate $5 \times 10^{-7} \, s^{-1}$. If temperature rises to $650 \, ^\circ C$, creep rate increases to $2 \times 10^{-6} \, s^{-1}$. Which factor is most responsible?
ⓐ. Increase in density
ⓑ. Reduction in Young’s modulus
ⓒ. Exponential dependence on temperature via $e^{-Q/RT}$
ⓓ. Increase in Poisson’s ratio
Correct Answer: Exponential dependence on temperature via $e^{-Q/RT}$
Explanation: Even small increases in temperature lead to large increases in creep rate due to the exponential Arrhenius term.
479. Why does increasing applied stress accelerate creep rate?
ⓐ. It reduces Young’s modulus
ⓑ. It increases dislocation density and movement
ⓒ. It lowers density
ⓓ. It eliminates grain boundaries
Correct Answer: It increases dislocation density and movement
Explanation: Higher stress provides stronger driving force for dislocation motion and diffusion, thus increasing creep strain rate.
480. Which microstructural feature is designed into superalloys to improve creep resistance?
ⓐ. Large amount of porosity
ⓑ. Coherent precipitates (like $\gamma’$)
ⓒ. Small grain size at high temperature
ⓓ. Reduced hardness
Correct Answer: Coherent precipitates (like $\gamma’$)
Explanation: Superalloys contain coherent precipitates (e.g., Ni$_3$Al, $\gamma’$) that hinder dislocation motion, enhancing high-temperature creep resistance.
481. Why is creep analysis critical in turbine blades?
ⓐ. Blades never operate at high temperatures
ⓑ. Turbine blades operate under constant high stress and temperature for long durations
ⓒ. Turbine blades are brittle at room temperature
ⓓ. Turbine blades are only subjected to elastic strain
Correct Answer: Turbine blades operate under constant high stress and temperature for long durations
Explanation: Turbine blades face high centrifugal stress and gas temperatures. Creep analysis ensures blades do not deform permanently, maintaining efficiency and safety.
482. Which property of alloys is most critical for preventing creep in jet engines?
ⓐ. Low density
ⓑ. High fracture toughness
ⓒ. High creep resistance at elevated temperatures
ⓓ. High thermal conductivity
Correct Answer: High creep resistance at elevated temperatures
Explanation: Jet engine parts must withstand continuous high temperature and cyclic stress. Superalloys are designed with precipitate strengthening to resist creep.
483. A boiler tube operates under stress of $80 \, MPa$ at $600^\circ C$. If the creep rate follows Norton’s law $\dot{\epsilon} = A \sigma^4$, what happens if stress doubles?
ⓐ. Creep rate doubles
ⓑ. Creep rate increases 4 times
ⓒ. Creep rate increases 8 times
ⓓ. Creep rate increases 16 times
Correct Answer: Creep rate increases 16 times
Explanation: With stress exponent $n = 4$, doubling stress increases creep rate by $2^4 = 16$.
484. Why are single-crystal superalloys used in turbine blades?
ⓐ. They are cheaper
ⓑ. They have no grain boundaries, preventing grain boundary creep
ⓒ. They reduce strength at high temperatures
ⓓ. They have lower melting points
Correct Answer: They have no grain boundaries, preventing grain boundary creep
Explanation: Grain boundaries are weak spots for creep. Single-crystal superalloys eliminate grain boundary sliding, improving creep resistance.
485. Which design feature helps prevent creep in power plant boiler tubes?
ⓐ. Sharp notches
ⓑ. Use of larger grain size alloys
ⓒ. Use of brittle materials
ⓓ. Lowering melting point of alloys
Correct Answer: Use of larger grain size alloys
Explanation: Larger grains reduce grain boundary sliding, improving creep resistance in high-temperature boiler tubes.
486. Why is creep prevention essential in nuclear reactors?
ⓐ. Reactors never operate at high temperature
ⓑ. Creep can distort fuel rods and reactor components over time
ⓒ. Nuclear fuel does not undergo stress
ⓓ. Creep has no role in nuclear structures
Correct Answer: Creep can distort fuel rods and reactor components over time
Explanation: Nuclear reactors operate at high temperature and constant load. Creep prevention ensures dimensional stability and safety of reactor components.
487. Which of the following techniques increases creep resistance of metals?
ⓐ. Increasing dislocation density only
ⓑ. Precipitation strengthening and alloying with refractory metals
ⓒ. Reducing grain size drastically
ⓓ. Annealing frequently
Correct Answer: Precipitation strengthening and alloying with refractory metals
Explanation: Alloying with refractory elements (Mo, W, Nb) and coherent precipitates hinder dislocation motion, increasing creep resistance.
488. Which component in automobiles is most affected by creep at high temperatures?
ⓐ. Tires
ⓑ. Brake pads
ⓒ. Exhaust valves
ⓓ. Car body frame
Correct Answer: Exhaust valves
Explanation: Exhaust valves operate under high temperature and stress for long periods, making creep resistance essential to prevent failure.
489. A component at $1000 K$ shows creep strain rate of $2 \times 10^{-7} \, s^{-1}$. If operating temperature increases to $1100 K$, creep rate becomes $8 \times 10^{-7} \, s^{-1}$. What is the ratio of creep rates?
ⓐ. 2
ⓑ. 3
ⓒ. 4
ⓓ. 5
Correct Answer: 4
Explanation: The creep rate increased from $2 \times 10^{-7}$ to $8 \times 10^{-7}$, i.e., 4 times, showing exponential sensitivity to temperature.
490. Which preventive measure is widely adopted in designing jet engine blades to combat creep?
ⓐ. Frequent annealing during operation
ⓑ. Using directional solidification or single-crystal growth techniques
ⓒ. Making blades thinner
ⓓ. Reducing alloy strength
Correct Answer: Using directional solidification or single-crystal growth techniques
Explanation: Directionally solidified or single-crystal blades eliminate grain boundary sliding, drastically improving creep resistance in jet engines.
491. What is stress relaxation?
ⓐ. Gradual increase of stress under constant strain
ⓑ. Gradual decrease of stress under constant strain with time
ⓒ. Sudden fracture under impact load
ⓓ. Recovery of elastic strain only
Correct Answer: Gradual decrease of stress under constant strain with time
Explanation: Stress relaxation is the reduction of internal stress in a material that is held at a constant strain, often observed in polymers and metals at high temperature.
492. Which equation represents stress relaxation behavior in viscoelastic materials (Maxwell model)?
ⓐ. $\sigma = E \epsilon$
ⓑ. $\sigma = \sigma_0 e^{-t/\tau}$
ⓒ. $\sigma = \sigma_0 + E t$
ⓓ. $\sigma = \frac{\sigma_0}{1 + t}$
Correct Answer: $\sigma = \sigma_0 e^{-t/\tau}$
Explanation: In Maxwell’s model, stress decays exponentially with time at constant strain, where $\tau = \eta/E$ is the relaxation time.
493. A material initially stressed to $100 \, MPa$ shows a stress of $37 \, MPa$ after 100 s at constant strain. What is the relaxation time $\tau$?
ⓐ. 50 s
ⓑ. 100 s
ⓒ. 200 s
ⓓ. 300 s
Correct Answer: 100 s
Explanation: $\sigma = \sigma_0 e^{-t/\tau}$.
Here, $37 = 100 e^{-100/\tau}$.
So, $e^{-100/\tau} = 0.37 \approx e^{-1}$. Thus, $\tau = 100 \, s$.
494. In which type of materials is stress relaxation most significant at room temperature?
ⓐ. Brittle ceramics
ⓑ. Elastic metals
ⓒ. Polymers and rubbers
ⓓ. Glass
Correct Answer: Polymers and rubbers
Explanation: Polymers and rubbers exhibit stress relaxation even at room temperature due to their viscoelastic nature.
495. Why is stress relaxation important in designing bolts and fasteners?
ⓐ. Bolts should always increase stress with time
ⓑ. Stress relaxation can reduce clamping force, leading to loosening
ⓒ. Stress relaxation improves ductility
ⓓ. Stress relaxation increases hardness
Correct Answer: Stress relaxation can reduce clamping force, leading to loosening
Explanation: Bolts under constant strain lose stress with time, which may reduce clamping force. This must be accounted for in design.
496. If stress relaxation follows $\sigma = \sigma_0 e^{-t/\tau}$, what is the stress after $3\tau$?
ⓐ. $\sigma_0$
ⓑ. $0.37 \sigma_0$
ⓒ. $0.05 \sigma_0$
ⓓ. $0$
Correct Answer: $0.05 \sigma_0$
Explanation: After $t = 3\tau$, $\sigma = \sigma_0 e^{-3} \approx 0.05 \sigma_0$.
497. In material testing, stress relaxation helps to determine:
ⓐ. Young’s modulus
ⓑ. Relaxation time and viscoelastic properties
ⓒ. Fracture toughness
ⓓ. Bulk modulus
Correct Answer: Relaxation time and viscoelastic properties
Explanation: Stress relaxation tests measure how stress decreases with time under constant strain, providing data about viscoelastic properties.
498. A rubber band stretched to constant length shows decreasing force with time. This is an example of:
ⓐ. Elastic deformation
ⓑ. Stress relaxation
ⓒ. Strain hardening
ⓓ. Work hardening
Correct Answer: Stress relaxation
Explanation: The decreasing restoring force in a rubber band at constant strain is due to stress relaxation.
499. Which factor significantly accelerates stress relaxation in metals?
ⓐ. Low temperature
ⓑ. High temperature
ⓒ. Zero applied strain
ⓓ. Large grain size
Correct Answer: High temperature
Explanation: At elevated temperatures, atomic diffusion facilitates stress relaxation, especially in metals.
500. Why is stress relaxation test significant for materials used in gaskets and seals?
ⓐ. Gaskets should break under load
ⓑ. Stress relaxation can reduce sealing pressure, causing leakage
ⓒ. Stress relaxation increases hardness of seals
ⓓ. Stress relaxation reduces elasticity permanently
Correct Answer: Stress relaxation can reduce sealing pressure, causing leakage
Explanation: Gaskets and seals rely on sustained stress for tight sealing. Stress relaxation reduces sealing pressure over time, leading to potential failure.
The chapter Mechanical Properties of Solids is a vital section of NCERT/CBSE Class 11 Physics, forming the basis for many questions in board exams and competitive exams like JEE, NEET, and state-level tests. It explains concepts such as Young’s modulus, stress-strain behavior, elasticity, and fracture mechanics. Out of the total 560 MCQs with answers, Part 5 offers the fifth set of 100 MCQs, designed for revision and in-depth practice before moving to the final part of the series.
- 👉 Total MCQs in this chapter: 560.
- 👉 This page contains: Fifth set of 100 solved MCQs.
- 👉 Perfect for exam revision and JEE/NEET preparation.
- 👉 To explore other chapters and subjects, use the top navigation bar.
- 👉 Now click the Part 6 button for the final set of questions.