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1. What branch of physics does the chapter “Mechanical Properties of Solids” mainly deal with?
ⓐ. Motion of planets
ⓑ. Behavior of rigid bodies under forces
ⓒ. Nature of light and optics
ⓓ. Nuclear reactions
Correct Answer: Behavior of rigid bodies under forces
Explanation: The chapter “Mechanical Properties of Solids” studies how solids deform under applied forces. It focuses on stress, strain, elasticity, and related properties. Planetary motion belongs to gravitation, optics deals with light, and nuclear reactions are studied in atomic/nuclear physics.
2. Which of the following best defines elasticity?
ⓐ. Property of solids to resist deformation permanently
ⓑ. Property of solids to regain shape after deformation
ⓒ. Property of fluids to flow indefinitely
ⓓ. Property of gases to expand continuously
Correct Answer: Property of solids to regain shape after deformation
Explanation: Elasticity is the ability of a material to regain its original shape and size after the deforming force is removed. Plasticity is the opposite, where deformation is permanent. Flowing is a property of fluids, and gases expand due to low intermolecular forces.
3. Who gave the fundamental law relating stress and strain for solids?
ⓐ. Archimedes
ⓑ. Hooke
ⓒ. Newton
ⓓ. Pascal
Correct Answer: Hooke
Explanation: Robert Hooke proposed Hooke’s Law, which states that within elastic limits, stress is directly proportional to strain ($\sigma \propto \epsilon$). Archimedes worked on buoyancy, Newton on mechanics and gravitation, and Pascal on fluid pressure.
4. Which physical quantity is defined as force per unit area?
ⓐ. Strain
ⓑ. Stress
ⓒ. Modulus of elasticity
ⓓ. Work
Correct Answer: Stress
Explanation: Stress is defined as the restoring force per unit area applied on a body, given by $\sigma = \frac{F}{A}$. Strain is a ratio of deformation, modulus of elasticity is the ratio of stress to strain, and work involves force and displacement.
5. What is the SI unit of stress?
ⓐ. Newton (N)
ⓑ. Joule (J)
ⓒ. Pascal (Pa)
ⓓ. Watt (W)
Correct Answer: Pascal (Pa)
Explanation: Stress is force per unit area ($\text{N}/\text{m}^2$), which is also the unit of pressure. This unit is called Pascal (Pa). 1 Pa = 1 N/m². Joule is energy, Watt is power, and Newton is force.
6. Which type of strain is produced when a wire is stretched by a force along its length?
ⓐ. Shear strain
ⓑ. Volumetric strain
ⓒ. Longitudinal strain
ⓓ. Angular strain
Correct Answer: Longitudinal strain
Explanation: When a wire is stretched, its length changes. The fractional change in length is called longitudinal strain ($\frac{\Delta L}{L}$). Shear strain involves angular deformation, volumetric strain involves change in volume, and angular strain is not a standard term.
7. What does Young’s modulus measure?
ⓐ. Resistance of a solid to volume change
ⓑ. Resistance of a solid to shearing force
ⓒ. Resistance of a solid to change in length
ⓓ. Resistance of a fluid to flow
Correct Answer: Resistance of a solid to change in length
Explanation: Young’s modulus $Y$ is defined as the ratio of longitudinal stress to longitudinal strain. It measures stiffness of a material against stretching or compression. Bulk modulus measures resistance to volume change, shear modulus to shearing, and viscosity relates to fluids.
8. A solid cube is subjected to equal compressive forces from all sides. What strain is produced?
ⓐ. Longitudinal strain
ⓑ. Shear strain
ⓒ. Volumetric strain
ⓓ. No strain
Correct Answer: Volumetric strain
Explanation: Equal compressive forces from all directions reduce the cube’s volume without changing its shape. This leads to volumetric strain, which is the fractional change in volume. Longitudinal strain occurs in stretching, shear strain in angular distortion.
9. Which modulus is associated with volumetric strain?
ⓐ. Young’s modulus
ⓑ. Bulk modulus
ⓒ. Shear modulus
ⓓ. Modulus of rigidity
Correct Answer: Bulk modulus
Explanation: Bulk modulus $K$ is defined as the ratio of bulk stress (hydrostatic pressure) to volumetric strain. It measures how incompressible a material is. Young’s modulus relates to length, shear modulus to shape distortion, and rigidity is another name for shear modulus.
10. Which of the following is a dimensionless quantity?
ⓐ. Stress
ⓑ. Strain
ⓒ. Young’s modulus
ⓓ. Bulk modulus
Correct Answer: Strain
Explanation: Strain is the ratio of change in dimension to original dimension ($\Delta L/L$ or $\Delta V/V$), making it a pure number without units. Stress and all elastic moduli (Young’s, Bulk, Shear) have units of pressure (Pa).
11. Why is it important to study the mechanical properties of solids?
ⓐ. To understand planetary motion
ⓑ. To design safe and durable structures
ⓒ. To calculate chemical reactions
ⓓ. To measure electrical conductivity
Correct Answer: To design safe and durable structures
Explanation: Mechanical properties like elasticity, plasticity, and strength help engineers design buildings, bridges, and machines that can withstand forces without failure. Planetary motion belongs to gravitation, chemistry deals with reactions, and conductivity is related to electricity.
12. Which of the following fields most directly benefits from knowledge of mechanical properties of solids?
ⓐ. Astronomy
ⓑ. Engineering and Material Science
ⓒ. Music
ⓓ. Medicine only
Correct Answer: Engineering and Material Science
Explanation: Engineering uses stress, strain, and elasticity concepts to design strong materials and structures. While medicine uses biomechanics, the primary field of application is engineering. Astronomy and music are not directly related to material strength.
13. The importance of stress-strain study in construction lies in:
ⓐ. Predicting material’s behavior under load
ⓑ. Measuring electric potential difference
ⓒ. Understanding chemical bonding
ⓓ. Studying planetary gravitation
Correct Answer: Predicting material’s behavior under load
Explanation: Stress-strain analysis helps determine how materials behave when subjected to different loads. This ensures buildings, machines, and tools are safe and efficient. The other options are unrelated to mechanical properties of solids.
14. Why is elasticity an important property in bridges and buildings?
ⓐ. It allows permanent deformation to absorb stress
ⓑ. It allows temporary deformation and recovery
ⓒ. It prevents all kinds of motion
ⓓ. It reduces the mass of the structure
Correct Answer: It allows temporary deformation and recovery
Explanation: Bridges and buildings experience loads such as wind, traffic, or earthquakes. Elasticity allows them to bend slightly and return to original shape, preventing cracks or collapse. Permanent deformation would be unsafe, and mass reduction isn’t the purpose of elasticity.
15. In material testing, why is it crucial to know the breaking point of a solid?
ⓐ. To measure electrical resistance
ⓑ. To prevent sudden structural failure
ⓒ. To improve optical properties
ⓓ. To determine melting point
Correct Answer: To prevent sudden structural failure
Explanation: The breaking point tells engineers the maximum stress a material can endure before it fractures. This is critical in designing aircrafts, bridges, and machinery. Electrical resistance, optics, and melting are unrelated to breaking strength.
16. How does knowledge of mechanical properties help in sports equipment design?
ⓐ. It makes equipment lighter and more elastic
ⓑ. It increases the cost of production
ⓒ. It reduces the lifespan of equipment
ⓓ. It prevents equipment from being used
Correct Answer: It makes equipment lighter and more elastic
Explanation: Sports equipment like tennis rackets, bats, and helmets are designed using elastic materials for better performance and safety. Elasticity ensures shock absorption and flexibility. Increasing cost or reducing lifespan is not the purpose.
17. Which mechanical property is most important for designing earthquake-resistant buildings?
ⓐ. Plasticity
ⓑ. Elasticity
ⓒ. Viscosity
ⓓ. Magnetism
Correct Answer: Elasticity
Explanation: Earthquake-resistant structures need to bend temporarily without breaking, which requires elasticity. Plasticity means permanent deformation, viscosity refers to fluids, and magnetism has no role in structural safety.
18. Why is studying plastic deformation important in material science?
ⓐ. It helps predict permanent changes in shape under high loads
ⓑ. It explains why materials regain their original shape
ⓒ. It determines how transparent a material is
ⓓ. It shows how sound travels in solids
Correct Answer: It helps predict permanent changes in shape under high loads
Explanation: Plastic deformation describes irreversible changes in material shape. Understanding this helps prevent permanent damage in engineering structures. The other options are unrelated to mechanical properties.
19. Which real-life example shows the importance of knowing tensile strength?
ⓐ. Designing suspension bridge cables
ⓑ. Measuring boiling point of water
ⓒ. Observing planets with a telescope
ⓓ. Studying sound waves in air
Correct Answer: Designing suspension bridge cables
Explanation: Suspension bridge cables must withstand huge tensile forces. Engineers select materials with high tensile strength for safety. Boiling points, planetary observation, and sound waves are unrelated to tensile strength.
20. Why is it important to study mechanical properties for medical devices?
ⓐ. To improve their color and texture
ⓑ. To ensure strength, flexibility, and biocompatibility
ⓒ. To reduce their chemical reactivity
ⓓ. To increase their electrical conductivity
Correct Answer: To ensure strength, flexibility, and biocompatibility
Explanation: Medical devices like stents, implants, and prosthetics require materials that are strong yet flexible and compatible with the human body. This ensures safety and performance. Color, reactivity, or conductivity are not the key focus here.
21. Which of the following is an example of elasticity?
ⓐ. Permanent bending of a metal rod
ⓑ. Stretching of a rubber band and returning to original length
ⓒ. Breaking of glass into pieces
ⓓ. Flow of honey under gravity
Correct Answer: Stretching of a rubber band and returning to original length
Explanation: Elasticity is the property of a body to regain its original shape after deformation. A rubber band stretches and returns to its original length when the force is removed. Permanent bending indicates plasticity, breaking is fracture, and honey flow relates to viscosity.
22. Plasticity in solids refers to:
ⓐ. Temporary deformation under load
ⓑ. Permanent deformation after load removal
ⓒ. Resistance to flow of fluids
ⓓ. Ability to conduct electricity
Correct Answer: Permanent deformation after load removal
Explanation: Plasticity is the property of a material to undergo permanent deformation even after the force is removed. Elasticity is temporary deformation, viscosity refers to fluids, and electrical conduction is unrelated to plasticity.
23. Which property of solids explains why railway tracks have gaps left between them?
ⓐ. Ductility
ⓑ. Elasticity
ⓒ. Thermal expansion
ⓓ. Plasticity
Correct Answer: Thermal expansion
Explanation: Railway tracks expand in hot weather and contract in cold weather. Gaps are left to allow expansion and prevent buckling. Ductility is ability to draw into wires, elasticity is temporary deformation, and plasticity is permanent deformation.
24. The ability of a material to be drawn into wires is called:
ⓐ. Elasticity
ⓑ. Plasticity
ⓒ. Ductility
ⓓ. Malleability
Correct Answer: Ductility
Explanation: Ductility is the property of a material that allows it to be stretched into thin wires, e.g., copper and aluminum. Malleability refers to beating into thin sheets. Elasticity and plasticity describe deformation behavior, not shaping.
25. Which of the following materials shows high malleability?
ⓐ. Glass
ⓑ. Gold
ⓒ. Rubber
ⓓ. Wood
Correct Answer: Gold
Explanation: Malleability is the ability to be hammered or rolled into thin sheets. Gold is highly malleable and can be beaten into sheets as thin as a few atoms. Glass breaks, rubber is elastic, and wood cracks under hammering.
26. Which property is most useful in making springs?
ⓐ. Elasticity
ⓑ. Plasticity
ⓒ. Malleability
ⓓ. Ductility
Correct Answer: Elasticity
Explanation: Springs must regain their original shape after deformation. Hence, elasticity is essential. Plasticity would lead to permanent deformation, malleability allows shaping into sheets, and ductility allows wire drawing but not spring action.
27. What does brittleness of a material mean?
ⓐ. Ability to deform without breaking
ⓑ. Sudden breaking without significant deformation
ⓒ. Ability to conduct heat
ⓓ. Resistance to flow of fluids
Correct Answer: Sudden breaking without significant deformation
Explanation: Brittle materials like glass and ceramics break suddenly without appreciable deformation. Tough materials like metals can deform before breaking. Heat conduction and viscosity are unrelated to brittleness.
28. Which property is tested in tensile strength experiments?
ⓐ. Resistance to stretching
ⓑ. Resistance to shearing
ⓒ. Resistance to rolling
ⓓ. Resistance to flow
Correct Answer: Resistance to stretching
Explanation: Tensile strength is the maximum stress a material can withstand while being stretched before breaking. It indicates resistance to stretching. Shear relates to sliding layers, rolling refers to malleability, and flow relates to viscosity.
29. Which of the following is the property of being permanently stretched without rupture?
ⓐ. Ductility
ⓑ. Malleability
ⓒ. Elasticity
ⓓ. Rigidity
Correct Answer: Ductility
Explanation: Ductility is the ability of a material to be permanently stretched into wires without breaking. Malleability is for sheets, elasticity is temporary deformation, and rigidity is resistance to deformation.
30. Which property is most desirable in making kitchen utensils?
ⓐ. Malleability and thermal conductivity
ⓑ. Ductility and brittleness
ⓒ. Elasticity and viscosity
ⓓ. Rigidity and magnetism
Correct Answer: Malleability and thermal conductivity
Explanation: Kitchen utensils need to be shaped into sheets and conduct heat efficiently. Malleable metals like aluminum and copper are commonly used. Brittleness is undesirable, elasticity is not needed, and magnetism has no role.
31. Which property allows a metal to be hammered into thin sheets without breaking?
ⓐ. Elasticity
ⓑ. Malleability
ⓒ. Ductility
ⓓ. Rigidity
Correct Answer: Malleability
Explanation: Malleability is the property that allows a material to be beaten or rolled into sheets. Gold and silver are highly malleable. Ductility refers to drawing into wires, elasticity to temporary deformation, and rigidity to resistance against deformation.
32. Which of the following materials is highly ductile?
ⓐ. Copper
ⓑ. Glass
ⓒ. Wood
ⓓ. Rubber
Correct Answer: Copper
Explanation: Copper can be drawn into very thin wires without breaking, making it highly ductile. Glass and wood are brittle, while rubber is elastic but not ductile.
33. Which pair of properties are considered opposite to each other in solids?
ⓐ. Elasticity and Plasticity
ⓑ. Malleability and Ductility
ⓒ. Rigidity and Toughness
ⓓ. Brittleness and Strength
Correct Answer: Elasticity and Plasticity
Explanation: Elasticity refers to regaining the original shape after deformation, while plasticity refers to permanent deformation. They represent opposite behaviors in materials. Malleability and ductility are shaping properties, not opposites.
34. What does toughness of a material represent?
ⓐ. Ability to deform permanently
ⓑ. Ability to absorb energy before fracture
ⓒ. Ability to resist volume change
ⓓ. Ability to conduct electricity
Correct Answer: Ability to absorb energy before fracture
Explanation: Toughness is the ability of a material to absorb energy and withstand shock loads before fracturing. Plasticity is permanent deformation, bulk modulus resists volume change, and conductivity relates to electricity.
35. Which property is most important for designing wires used in electrical transmission?
ⓐ. Elasticity
ⓑ. Ductility
ⓒ. Brittleness
ⓓ. Viscosity
Correct Answer: Ductility
Explanation: Transmission wires must be long, thin, and flexible without breaking, requiring high ductility. Elasticity ensures recovery but does not allow permanent shaping into wires. Brittleness would cause breakage, and viscosity is unrelated.
36. A material that is both malleable and ductile is:
ⓐ. Glass
ⓑ. Gold
ⓒ. Rubber
ⓓ. Wood
Correct Answer: Gold
Explanation: Gold is highly malleable (can form sheets) and ductile (can form wires). Glass is brittle, rubber is elastic, and wood is rigid but not malleable or ductile.
37. Which property describes the resistance of a solid body to change in its shape or volume?
ⓐ. Rigidity
ⓑ. Plasticity
ⓒ. Malleability
ⓓ. Ductility
Correct Answer: Rigidity
Explanation: Rigidity is the property that measures resistance to deformation. The greater the rigidity, the less the material changes its shape or volume under applied forces. Plasticity, malleability, and ductility are deformation properties.
38. Which of the following is an example of a brittle material?
ⓐ. Steel
ⓑ. Copper
ⓒ. Glass
ⓓ. Gold
Correct Answer: Glass
Explanation: Glass breaks suddenly without significant deformation, making it brittle. Steel and copper can deform plastically before breaking, and gold is malleable and ductile.
39. What property of steel makes it more useful than iron in construction?
ⓐ. Higher brittleness
ⓑ. Higher ductility and toughness
ⓒ. Lower elasticity
ⓓ. Lower strength
Correct Answer: Higher ductility and toughness
Explanation: Steel combines strength, ductility, and toughness, making it ideal for construction. It can withstand shock loads and deform without sudden fracture. Iron is more brittle in comparison.
40. Which property of solids is responsible for the permanent bending of a metal rod after applying a strong force?
ⓐ. Elasticity
ⓑ. Plasticity
ⓒ. Malleability
ⓓ. Rigidity
Correct Answer: Plasticity
Explanation: Plasticity causes permanent deformation when the applied stress exceeds the elastic limit. The rod bends and does not return to its original shape. Elasticity is temporary deformation, malleability is sheet formation, and rigidity resists deformation.
41. Why is steel preferred over iron in building bridges?
ⓐ. It is heavier than iron
ⓑ. It has higher ductility and toughness
ⓒ. It has lower elasticity
ⓓ. It is cheaper than all materials
Correct Answer: It has higher ductility and toughness
Explanation: Steel combines high strength, ductility, and toughness, making it resistant to breaking under heavy loads and vibrations. Iron is more brittle and can fail suddenly. Elasticity is similar in both, and cost alone is not the reason for preference.
42. Which mechanical property is most crucial in designing aircraft bodies?
ⓐ. Elasticity and light weight
ⓑ. High viscosity
ⓒ. High brittleness
ⓓ. Low conductivity
Correct Answer: Elasticity and light weight
Explanation: Aircraft bodies must withstand stress and vibrations while remaining light for efficiency. Aluminum alloys are chosen for their strength, elasticity, and low density. High viscosity, brittleness, or low conductivity are irrelevant to aircraft safety.
43. In material science, why is Young’s modulus important for engineers?
ⓐ. It measures resistance to electrical flow
ⓑ. It determines resistance to change in length under load
ⓒ. It measures resistance to thermal expansion
ⓓ. It defines density of a material
Correct Answer: It determines resistance to change in length under load
Explanation: Young’s modulus is the ratio of stress to strain in the elastic region. It helps engineers choose materials that can sustain load without large deformations. Electrical resistance, thermal expansion, and density are different concepts.
44. Which property makes copper suitable for making electrical wires?
ⓐ. Brittleness and low cost
ⓑ. High ductility and good conductivity
ⓒ. High viscosity and malleability
ⓓ. Elasticity and opacity
Correct Answer: High ductility and good conductivity
Explanation: Copper can be drawn into thin wires due to ductility and conducts electricity efficiently. Brittleness would cause failure, viscosity applies to fluids, and opacity is not relevant for electrical wiring.
45. Why are alloys like brass and bronze used in engineering instead of pure metals?
ⓐ. They are lighter than pure metals
ⓑ. They improve strength, durability, and resistance to corrosion
ⓒ. They are easier to break
ⓓ. They are cheaper to produce
Correct Answer: They improve strength, durability, and resistance to corrosion
Explanation: Alloys combine metals to improve mechanical properties like hardness, toughness, and resistance to wear. Brass (copper + zinc) and bronze (copper + tin) are stronger and more durable than pure copper. The main purpose is property enhancement, not just cost.
46. Which property is most critical in designing railway tracks?
ⓐ. Elasticity and toughness
ⓑ. High brittleness
ⓒ. Low conductivity
ⓓ. High viscosity
Correct Answer: Elasticity and toughness
Explanation: Railway tracks must withstand continuous heavy loads and vibrations. Toughness prevents sudden fracture, while elasticity allows slight bending without damage. Brittleness would cause cracks, conductivity is not crucial, and viscosity is irrelevant.
47. In material testing, what does fracture toughness measure?
ⓐ. Ability to deform elastically
ⓑ. Resistance to crack propagation
ⓒ. Resistance to volume change
ⓓ. Ability to conduct heat
Correct Answer: Resistance to crack propagation
Explanation: Fracture toughness indicates a material’s ability to resist failure when a crack is present. High fracture toughness ensures safety in structures like airplanes and bridges. Elasticity, bulk modulus, and thermal conductivity are different properties.
48. Which mechanical property is most important in making automobile tires?
ⓐ. Elasticity and resilience
ⓑ. Brittleness and ductility
ⓒ. Viscosity and malleability
ⓓ. Rigidity and hardness
Correct Answer: Elasticity and resilience
Explanation: Tires need to absorb shocks and return to their original shape after deformation. Rubber’s elasticity and resilience make it suitable. Brittleness or excessive hardness would cause cracking, while viscosity and malleability are irrelevant.
49. Why is glass not preferred in structural engineering for load-bearing purposes?
ⓐ. It is too flexible
ⓑ. It is brittle and lacks toughness
ⓒ. It is very ductile
ⓓ. It is too elastic
Correct Answer: It is brittle and lacks toughness
Explanation: Glass breaks suddenly without significant deformation, making it unsafe for heavy load applications. Metals like steel are used instead. Flexibility and elasticity are not the reasons for glass’s unsuitability, and ductility does not apply to glass.
50. Which property is most important in designing helmets and protective gear?
ⓐ. Toughness and impact resistance
ⓑ. Malleability and brittleness
ⓒ. Ductility and viscosity
ⓓ. Rigidity and brittleness
Correct Answer: Toughness and impact resistance
Explanation: Helmets and safety gear must absorb large amounts of energy without cracking. Tough materials resist sudden impact and protect users. Brittleness would make them unsafe, and viscosity applies only to fluids, not solids.
51. What is the correct definition of elasticity in solids?
ⓐ. Ability of a body to deform permanently under stress
ⓑ. Ability of a body to regain its original shape after stress is removed
ⓒ. Ability of a body to resist flow of fluids
ⓓ. Ability of a body to resist conduction of electricity
Correct Answer: Ability of a body to regain its original shape after stress is removed
Explanation: Elasticity is the property of a solid to return to its original shape and size when the deforming force is removed. Permanent deformation is plasticity, resistance to flow is viscosity, and conduction relates to electricity.
52. Which of the following is the best everyday example of elasticity?
ⓐ. Stretching of a rubber band and its return to original length
ⓑ. Melting of ice into water
ⓒ. Permanent bending of a plastic ruler
ⓓ. Flow of honey on a spoon
Correct Answer: Stretching of a rubber band and its return to original length
Explanation: A rubber band regains its original length after stretching, which demonstrates elasticity. Melting ice is a phase change, bending a ruler is plastic deformation, and honey flow is viscosity.
53. When does a body show perfectly elastic behavior?
ⓐ. When it retains a small permanent deformation
ⓑ. When it returns completely to its original state after deformation
ⓒ. When it deforms and does not recover
ⓓ. When it resists all external forces without deformation
Correct Answer: When it returns completely to its original state after deformation
Explanation: Perfect elasticity means no permanent deformation remains once the stress is removed. In practice, ideal elasticity is rare, but materials like steel approximate it. A small permanent deformation indicates plasticity.
54. Which property is opposite to elasticity?
ⓐ. Ductility
ⓑ. Plasticity
ⓒ. Malleability
ⓓ. Rigidity
Correct Answer: Plasticity
Explanation: Elasticity is the recovery of shape, while plasticity refers to permanent deformation after force removal. Ductility and malleability describe shaping properties, and rigidity refers to resistance to deformation.
55. Elasticity of a material depends on:
ⓐ. Stress only
ⓑ. Strain only
ⓒ. Both stress and strain within elastic limit
ⓓ. Volume of the material
Correct Answer: Both stress and strain within elastic limit
Explanation: Elasticity is quantified as the ratio of stress to strain within the elastic limit. Stress alone or strain alone cannot define elasticity. Volume has no direct role in defining elasticity.
56. What is meant by a perfectly plastic body?
ⓐ. It completely regains its shape after deformation
ⓑ. It cannot be deformed by any force
ⓒ. It undergoes permanent deformation after smallest force
ⓓ. It stores maximum elastic potential energy
Correct Answer: It undergoes permanent deformation after smallest force
Explanation: A perfectly plastic body does not regain its shape after stress removal; even the smallest deforming force causes permanent change. In reality, such ideal plasticity is rare.
57. Which of the following is an almost perfectly elastic material?
ⓐ. Glass
ⓑ. Clay
ⓒ. Steel
ⓓ. Rubber
Correct Answer: Steel
Explanation: Steel exhibits nearly perfect elasticity within its elastic limit. Rubber is highly elastic but not perfectly proportional to stress and strain. Glass is brittle, and clay is plastic.
58. Elasticity plays a key role in which of the following phenomena?
ⓐ. Bouncing of a ball
ⓑ. Melting of metals
ⓒ. Flow of liquids
ⓓ. Absorption of heat
Correct Answer: Bouncing of a ball
Explanation: A ball bounces because it deforms temporarily when hitting the ground and then regains its shape due to elasticity. Melting is a phase change, flow is viscosity, and heat absorption is thermal property.
59. Which of the following statements is correct about elastic bodies?
ⓐ. They regain shape only partially after deformation
ⓑ. They regain shape fully after stress is removed (within elastic limit)
ⓒ. They never regain original shape after stress
ⓓ. They permanently deform under small forces
Correct Answer: They regain shape fully after stress is removed (within elastic limit)
Explanation: Elastic bodies recover their original shape completely when stress is within the elastic limit. Beyond that limit, plastic deformation occurs.
60. Why is elasticity important in engineering materials?
ⓐ. To allow materials to melt easily
ⓑ. To ensure materials regain original shape under load cycles
ⓒ. To make materials heavier and stronger
ⓓ. To reduce their electrical resistance
Correct Answer: To ensure materials regain original shape under load cycles
Explanation: Engineering materials such as beams, bridges, and springs must regain their shape after loads are applied and removed. This prevents permanent damage and ensures durability. Other options are unrelated to elasticity.
61. What is the main difference between elastic and plastic deformation?
ⓐ. Elastic deformation is permanent, plastic deformation is temporary
ⓑ. Elastic deformation is temporary, plastic deformation is permanent
ⓒ. Both are permanent changes
ⓓ. Both are temporary changes
Correct Answer: Elastic deformation is temporary, plastic deformation is permanent
Explanation: Elastic deformation disappears when the force is removed, allowing the material to return to its original shape. Plastic deformation remains even after the force is removed, leading to a permanent change in the material’s shape.
62. Which of the following is an example of plastic deformation?
ⓐ. Stretching of a rubber band within limits
ⓑ. Permanent bending of a metal rod after exceeding elastic limit
ⓒ. Compression of a spring within its elastic limit
ⓓ. Restoring shape of a tennis ball after squeezing
Correct Answer: Permanent bending of a metal rod after exceeding elastic limit
Explanation: Once stress exceeds the elastic limit, the material undergoes plastic deformation and cannot return to its original form. Rubber bands, springs, and balls show elastic deformation within limits.
63. In stress-strain behavior, where does plastic deformation start?
ⓐ. At the proportional limit
ⓑ. At the elastic limit
ⓒ. At the breaking point
ⓓ. At the origin of the curve
Correct Answer: At the elastic limit
Explanation: Plastic deformation begins once the stress crosses the elastic limit. Up to the elastic limit, deformation is reversible. Beyond it, the material undergoes permanent deformation until fracture.
64. Why is elastic deformation preferable in construction materials?
ⓐ. It reduces the cost of production
ⓑ. It ensures recovery of shape under stress
ⓒ. It prevents all kinds of deformation
ⓓ. It increases weight of the structure
Correct Answer: It ensures recovery of shape under stress
Explanation: Elastic materials can bear loads and return to their original form, maintaining structural safety. Permanent deformation (plasticity) would weaken the structure over time.
65. Which statement about plastic deformation is correct?
ⓐ. It occurs only within the elastic limit
ⓑ. It is always reversible
ⓒ. It remains even after stress is removed
ⓓ. It does not depend on load magnitude
Correct Answer: It remains even after stress is removed
Explanation: Plastic deformation is permanent and irreversible. Elastic deformation occurs within the elastic limit, but once exceeded, plastic deformation sets in and does not reverse.
66. A copper wire is stretched beyond its elastic limit. What type of deformation occurs?
ⓐ. Elastic only
ⓑ. Plastic only
ⓒ. Both elastic and plastic
ⓓ. No deformation
Correct Answer: Both elastic and plastic
Explanation: First, the wire undergoes elastic deformation, but beyond the elastic limit, plastic deformation occurs. The elastic part recovers after force removal, while the plastic part remains permanent.
67. Which of the following is NOT a characteristic of elastic deformation?
ⓐ. Temporary change in shape
ⓑ. Proportional relation between stress and strain
ⓒ. Complete recovery on stress removal
ⓓ. Permanent change in structure of the material
Correct Answer: Permanent change in structure of the material
Explanation: Permanent change in structure is a property of plastic deformation. Elastic deformation is temporary, proportional (Hooke’s law), and reversible.
68. In forging metals, which type of deformation is primarily involved?
ⓐ. Elastic deformation
ⓑ. Plastic deformation
ⓒ. Vibrational deformation
ⓓ. Elastic and plastic deformation equally
Correct Answer: Plastic deformation
Explanation: Forging involves shaping metals by applying high forces beyond the elastic limit, resulting in permanent deformation. Elastic deformation plays no significant role here.
69. Why does a spring return to its original length after compression?
ⓐ. Because of plastic deformation
ⓑ. Because of brittleness
ⓒ. Because of elasticity
ⓓ. Because of malleability
Correct Answer: Because of elasticity
Explanation: Springs are designed to operate within their elastic limit. When compressed or stretched, they store potential energy and return to their original shape once the load is removed.
70. Which of the following best represents a case of elastic deformation?
ⓐ. Clay being molded into shapes
ⓑ. Glass shattering under impact
ⓒ. A rubber ball bouncing back after hitting the ground
ⓓ. A metal rod permanently bent after overload
Correct Answer: A rubber ball bouncing back after hitting the ground
Explanation: The ball undergoes temporary deformation and regains shape due to elasticity. Clay molding and metal bending involve plasticity, while glass shattering is brittleness, not elasticity.
71. What is meant by linear elastic deformation?
ⓐ. Deformation where stress is proportional to strain
ⓑ. Deformation that is permanent
ⓒ. Deformation where stress and strain are unrelated
ⓓ. Deformation that occurs only in fluids
Correct Answer: Deformation where stress is proportional to strain
Explanation: In linear elastic deformation, Hooke’s law holds true ($\sigma \propto \epsilon$). Stress and strain increase proportionally, and the material regains its original shape after load removal. Permanent deformation is plastic, not linear elastic.
72. In which region of a stress-strain curve does linear elastic deformation occur?
ⓐ. Proportional limit region
ⓑ. Plastic region
ⓒ. Yield region
ⓓ. Breaking point
Correct Answer: Proportional limit region
Explanation: Linear elastic deformation is observed up to the proportional limit where stress is directly proportional to strain. Beyond this limit, non-linear behavior and eventually plastic deformation occur.
73. Which law governs linear elastic deformation?
ⓐ. Newton’s law
ⓑ. Hooke’s law
ⓒ. Pascal’s law
ⓓ. Archimedes’ principle
Correct Answer: Hooke’s law
Explanation: Hooke’s law states that within the elastic limit, stress is directly proportional to strain. This is the defining feature of linear elastic deformation. Other laws deal with mechanics and fluids, not elasticity.
74. What is non-linear elastic deformation?
ⓐ. Deformation where stress is proportional to strain
ⓑ. Deformation where material regains shape but stress is not proportional to strain
ⓒ. Deformation where material breaks instantly
ⓓ. Permanent deformation beyond yield point
Correct Answer: Deformation where material regains shape but stress is not proportional to strain
Explanation: Non-linear elastic deformation occurs when stress and strain are not proportional but the material still regains its shape once the load is removed. Rubber is a good example of non-linear elastic material.
75. Which of the following is an example of a linear elastic material?
ⓐ. Steel within elastic limit
ⓑ. Rubber band under all conditions
ⓒ. Clay
ⓓ. Glass
Correct Answer: Steel within elastic limit
Explanation: Steel obeys Hooke’s law very well up to its elastic limit, showing linear elastic behavior. Rubber exhibits non-linear elasticity, clay is plastic, and glass is brittle.
76. Which of the following is an example of non-linear elastic material?
ⓐ. Copper wire within elastic limit
ⓑ. Steel rod under small stress
ⓒ. Rubber band stretched significantly
ⓓ. Glass rod under compression
Correct Answer: Rubber band stretched significantly
Explanation: Rubber regains its shape but does not show a proportional relationship between stress and strain, hence it exhibits non-linear elasticity. Metals like copper and steel are linear elastic within their limits.
77. In linear elastic deformation, the stress-strain graph is:
ⓐ. A straight line through the origin
ⓑ. A parabola
ⓒ. A horizontal line
ⓓ. A random curve
Correct Answer: A straight line through the origin
Explanation: In the proportional limit region, stress is directly proportional to strain, producing a straight-line graph through the origin. Beyond this region, non-linear behavior begins.
78. In non-linear elastic deformation, the stress-strain graph is:
ⓐ. Perfectly straight
ⓑ. Curved but material still recovers
ⓒ. Horizontal with no slope
ⓓ. Always discontinuous
Correct Answer: Curved but material still recovers
Explanation: In non-linear elasticity, stress-strain relationship is not proportional, producing a curved graph. However, the material still returns to its original shape once the force is removed.
79. Why is steel considered nearly perfectly linear elastic?
ⓐ. It can deform permanently under all conditions
ⓑ. It obeys Hooke’s law closely up to high stress values
ⓒ. It shows plastic deformation from the beginning
ⓓ. It has no elastic properties at all
Correct Answer: It obeys Hooke’s law closely up to high stress values
Explanation: Steel maintains a direct proportionality between stress and strain up to its elastic limit, making it almost perfectly linear elastic. Plastic deformation occurs only beyond this limit.
80. Which of the following best represents non-linear elastic behavior?
ⓐ. Stretching of a steel wire within elastic limit
ⓑ. Compression of a spring within Hooke’s law
ⓒ. Stretching of a rubber sheet
ⓓ. Breaking of a glass rod
Correct Answer: Stretching of a rubber sheet
Explanation: Rubber regains shape after deformation but the stress-strain relation is non-linear. Steel and springs are linear elastic, while glass breaking is brittle failure, not elasticity.
81. In linear elastic deformation, which mathematical relation holds true?
ⓐ. $\sigma = k \epsilon^2$
ⓑ. $\sigma = E \cdot \epsilon$
ⓒ. $\sigma = \frac{1}{\epsilon}$
ⓓ. $\sigma = \epsilon + E$
Correct Answer: $\sigma = E \cdot \epsilon$
Explanation: In linear elasticity, stress ($\sigma$) is directly proportional to strain ($\epsilon$) with the constant of proportionality being Young’s modulus ($E$). Thus, $\sigma = E \cdot \epsilon$. Options A, C, and D do not represent Hooke’s law.
82. Which graph correctly represents linear elastic deformation?
ⓐ. Stress vs. strain is a straight line through the origin
ⓑ. Stress vs. strain is a horizontal line
ⓒ. Stress vs. strain is a vertical line
ⓓ. Stress vs. strain is a parabola
Correct Answer: Stress vs. strain is a straight line through the origin
Explanation: Linear elastic deformation obeys Hooke’s law ($\sigma \propto \epsilon$), so the graph is a straight line. Non-linear elasticity shows a curved graph. Horizontal or vertical lines do not represent real material behavior.
83. A steel wire of length $L = 2 \, m$, area of cross-section $A = 2 \times 10^{-6} \, m^2$, and Young’s modulus $Y = 2 \times 10^{11} \, Pa$ is subjected to a tensile force $F = 200 \, N$. What is the longitudinal strain?
ⓐ. $1.0 \times 10^{-4}$
ⓑ. $2.0 \times 10^{-4}$
ⓒ. $5.0 \times 10^{-4}$
ⓓ. $1.0 \times 10^{-5}$
Correct Answer: $1.0 \times 10^{-5}$
Explanation: Stress $\sigma = \frac{F}{A} = \frac{200}{2 \times 10^{-6}} = 1 \times 10^{8} \, Pa$.
Strain $\epsilon = \frac{\sigma}{Y} = \frac{1 \times 10^{8}}{2 \times 10^{11}} = 1 \times 10^{-3} / 200 = 1.0 \times 10^{-5}$. Hence correct answer is D.
84. Which material typically shows non-linear elastic deformation even at small strains?
ⓐ. Steel
ⓑ. Rubber
ⓒ. Copper
ⓓ. Glass
Correct Answer: Rubber
Explanation: Rubber regains its original shape after stretching but stress is not proportional to strain. Its stress-strain graph is curved (non-linear). Steel shows linear elasticity up to elastic limit, copper is ductile, and glass is brittle.
85. For a linear elastic material, if stress is doubled, strain becomes:
ⓐ. Half
ⓑ. Same
ⓒ. Double
ⓓ. Zero
Correct Answer: Double
Explanation: From Hooke’s law, $\sigma = E \cdot \epsilon$. For constant $E$, if $\sigma$ doubles, then $\epsilon$ also doubles. Thus strain is directly proportional to applied stress in linear elastic region.
86. A steel rod obeys Hooke’s law up to stress $\sigma = 2 \times 10^{8} \, Pa$. If Young’s modulus is $2 \times 10^{11} \, Pa$, what is the maximum elastic strain?
ⓐ. $1.0 \times 10^{-2}$
ⓑ. $1.0 \times 10^{-3}$
ⓒ. $1.0 \times 10^{-4}$
ⓓ. $1.0 \times 10^{-5}$
Correct Answer: $1.0 \times 10^{-3}$
Explanation: $\epsilon = \frac{\sigma}{Y} = \frac{2 \times 10^{8}}{2 \times 10^{11}} = 1.0 \times 10^{-3}$. This represents the maximum elastic strain before plastic deformation begins.
87. Which of the following stress-strain relations corresponds to non-linear elasticity?
ⓐ. $\sigma \propto \epsilon$
ⓑ. $\sigma \propto \epsilon^2$ or other non-linear relation
ⓒ. $\sigma = 0$ always
ⓓ. $\sigma$ independent of $\epsilon$
Correct Answer: $\sigma \propto \epsilon^2$ or other non-linear relation
Explanation: In non-linear elasticity, stress and strain are not proportional, but the material still recovers after stress removal. Linear elasticity corresponds to option A. Options C and D are unrealistic.
88. Which one of the following statements is true about linear and non-linear elastic deformation?
ⓐ. Both are proportional to stress
ⓑ. Both lead to permanent deformation
ⓒ. Both are reversible, but proportionality exists only in linear elasticity
ⓓ. Both are irreversible
Correct Answer: Both are reversible, but proportionality exists only in linear elasticity
Explanation: Elastic deformation (linear or non-linear) is always reversible, but Hooke’s law applies only in linear elasticity. Non-linear elastic materials like rubber do not follow proportionality.
89. If a wire of length $1 \, m$ and cross-sectional area $1 \, mm^2$ (Young’s modulus $2 \times 10^{11} \, Pa$) is elongated by $0.5 \, mm$, what is the tensile force applied?
ⓐ. 50 N
ⓑ. 100 N
ⓒ. 200 N
ⓓ. 500 N
Correct Answer: 100 N
Explanation: Strain $\epsilon = \frac{\Delta L}{L} = \frac{0.5 \times 10^{-3}}{1} = 5 \times 10^{-4}$.
Stress $\sigma = Y \cdot \epsilon = (2 \times 10^{11})(5 \times 10^{-4}) = 1 \times 10^{8} \, Pa$.
Force $F = \sigma \cdot A = (1 \times 10^{8})(1 \times 10^{-6}) = 100 \, N$.
Actually $A = 1 \, mm^2 = 1 \times 10^{-6} \, m^2$. So, $F = 100 \, N$. Correct answer: B. 100 N.
90. Which example best shows linear elasticity in daily life?
ⓐ. Stretching of a steel spring within limits
ⓑ. Stretching of a rubber band to large lengths
ⓒ. Permanent bending of a plastic ruler
ⓓ. Shattering of glass
Correct Answer: Stretching of a steel spring within limits
Explanation: A steel spring follows Hooke’s law closely, making it a perfect example of linear elasticity. Rubber shows non-linear elasticity, plastic ruler bending is plastic deformation, and glass shattering is brittle failure.
91. A steel wire of length $2 \, m$ and radius $1 \, mm$ is subjected to a tensile force of $100 \, N$. If Young’s modulus of steel is $2 \times 10^{11} \, Pa$, what is the elongation of the wire?
ⓐ. $0.16 \, mm$
ⓑ. $0.32 \, mm$
ⓒ. $0.64 \, mm$
ⓓ. $1.28 \, mm$
Correct Answer: $0.32 \, mm$
Explanation: Area $A = \pi r^2 = \pi (1 \times 10^{-3})^2 = 3.14 \times 10^{-6} \, m^2$.
Stress $ \sigma = \frac{F}{A} = \frac{100}{3.14 \times 10^{-6}} \approx 3.18 \times 10^{7} \, Pa$.
Strain $ \epsilon = \frac{\sigma}{Y} = \frac{3.18 \times 10^{7}}{2 \times 10^{11}} \approx 1.59 \times 10^{-4}$.
Elongation $ \Delta L = \epsilon L = 1.59 \times 10^{-4} \times 2 = 3.18 \times 10^{-4} \, m = 0.32 \, mm$.
92. For a linear elastic material, the potential energy stored per unit volume (elastic energy density) is:
ⓐ. $\frac{1}{2} \sigma \epsilon$
ⓑ. $\sigma \epsilon$
ⓒ. $\frac{\sigma}{\epsilon}$
ⓓ. $\frac{1}{2} E \sigma$
Correct Answer: $\frac{1}{2} \sigma \epsilon$
Explanation: Elastic potential energy per unit volume is given by
$$ U = \int_0^\epsilon \sigma \, d\epsilon = \frac{1}{2} \sigma \epsilon $$
using Hooke’s law ($\sigma = E \epsilon$). This formula holds for linear elastic deformation.
93. A copper wire of length $1.5 \, m$ and cross-sectional area $2 \times 10^{-6} \, m^2$ is stretched by $1 \, mm$ under a load. If Young’s modulus of copper is $1.1 \times 10^{11} \, Pa$, what is the applied force?
ⓐ. 150 N
ⓑ. 220 N
ⓒ. 330 N
ⓓ. 440 N
Correct Answer: 150 N
Explanation: Strain $ \epsilon = \frac{\Delta L}{L} = \frac{1 \times 10^{-3}}{1.5} \approx 6.67 \times 10^{-4}$.
Stress $ \sigma = Y \cdot \epsilon = (1.1 \times 10^{11})(6.67 \times 10^{-4}) = 7.34 \times 10^{7} \, Pa$.
Force $ F = \sigma \cdot A = (7.34 \times 10^{7})(2 \times 10^{-6}) \approx 146.8 \, N$.
Let’s recheck → $F = 7.34 \times 10^{7} \times 2 \times 10^{-6} = 146.8 \, N$.
Correct Answer: A. 150 N (approx).
94. Which of the following correctly represents the condition for linear elasticity?
ⓐ. $\sigma \propto \epsilon^2$
ⓑ. $\sigma \propto \sqrt{\epsilon}$
ⓒ. $\sigma \propto \epsilon$
ⓓ. $\sigma$ independent of $\epsilon$
Correct Answer: $\sigma \propto \epsilon$
Explanation: In linear elastic deformation, stress is directly proportional to strain (Hooke’s law). Non-linear materials like rubber do not follow this proportionality.
95. A wire of length $3 \, m$ and area $1 \times 10^{-6} \, m^2$ is subjected to a tensile force of $60 \, N$. If its elongation is $0.3 \, mm$, what is the Young’s modulus of the material?
ⓐ. $2 \times 10^{10} \, Pa$
ⓑ. $6 \times 10^{10} \, Pa$
ⓒ. $2 \times 10^{11} \, Pa$
ⓓ. $6 \times 10^{11} \, Pa$
Correct Answer: $6 \times 10^{11} \, Pa$
Explanation: Stress $ \sigma = \frac{F}{A} = \frac{60}{1 \times 10^{-6}} = 6 \times 10^{7} \, Pa$.
Strain $ \epsilon = \frac{\Delta L}{L} = \frac{0.3 \times 10^{-3}}{3} = 1 \times 10^{-4}$.
$ Y = \frac{\sigma}{\epsilon} = \frac{6 \times 10^{7}}{1 \times 10^{-4}} = 6 \times 10^{11} \, Pa$.
Correct Answer: D. $6 \times 10^{11} \, Pa$.
96. In non-linear elastic deformation, which of the following is true?
ⓐ. Stress is proportional to strain
ⓑ. Stress-strain curve is a straight line
ⓒ. Stress-strain curve is curved but reversible
ⓓ. Deformation is always permanent
Correct Answer: Stress-strain curve is curved but reversible
Explanation: Non-linear elasticity (e.g., rubber) shows a curved stress-strain relation but deformation is reversible after removal of stress. This differentiates it from plasticity.
97. Which formula relates the elastic strain energy stored in a stretched wire of length $L$, cross-sectional area $A$, Young’s modulus $Y$, and elongation $\Delta L$?
ⓐ. $U = \frac{1}{2} \frac{F \Delta L}{A}$
ⓑ. $U = \frac{1}{2} \frac{Y A (\Delta L)^2}{L}$
ⓒ. $U = \frac{F}{A} \cdot L$
ⓓ. $U = \frac{1}{2} \frac{(\Delta L)^2}{Y}$
Correct Answer: $U = \frac{1}{2} \frac{Y A (\Delta L)^2}{L}$
Explanation: Elastic potential energy stored in a stretched wire is
$$ U = \frac{1}{2} \cdot \frac{F^2 L}{A Y} = \frac{1}{2} \frac{Y A (\Delta L)^2}{L} $$
using Hooke’s law and strain relations.
98. A spring of spring constant $k = 200 \, N/m$ is stretched by $5 \, cm$. What is the strain energy stored?
ⓐ. $0.125 \, J$
ⓑ. $0.25 \, J$
ⓒ. $0.5 \, J$
ⓓ. $1.0 \, J$
Correct Answer: $0.125 \, J$
Explanation: Elastic potential energy $ U = \frac{1}{2} k x^2 = \frac{1}{2}(200)(0.05^2) = 0.25 \times 0.01 = 0.125 \, J$.
99. Which material shows nearly perfect linear elasticity?
ⓐ. Steel
ⓑ. Rubber
ⓒ. Glass
ⓓ. Clay
Correct Answer: Steel
Explanation: Steel follows Hooke’s law very closely up to the elastic limit, showing linear elasticity. Rubber is non-linear elastic, glass is brittle, and clay is plastic.
100. Which everyday object best demonstrates non-linear elastic deformation?
ⓐ. A steel spring
ⓑ. A stretched rubber band
ⓒ. A glass rod under pressure
ⓓ. A metal wire under small stress
Correct Answer: A stretched rubber band
Explanation: Rubber regains its shape after stretching but the stress-strain curve is non-linear. Steel springs and wires follow linear elasticity, while glass breaking is brittle fracture.
Welcome to Class 11 Physics MCQs – Chapter 9: Mechanical Properties of Solids (Part 1). This page is a chapter-wise question bank for the NCERT/CBSE Class 11 Physics syllabus—built for quick revision and exam speed. Practice MCQs / objective questions / Physics quiz items with solutions and explanations, ideal for CBSE Boards, JEE Main, NEET, competitive exams, and Board exams.
These MCQs are suitable for international competitive exams—physics concepts are universal.
Navigation & pages: The full chapter has 560 MCQs in 6 parts (100 + 100 + 100 + 100 + 100 + 60). Part 1 contains 100 MCQs split across 10 pages—you’ll see 10 questions per page. Use the page numbers above to view the remaining questions.
What you will learn & practice
- Introduction to Mechanical Properties of Solids
- Elastic behaviour of solids and deformation basics
- Stress and strain (normal, shear, volumetric) and Hooke’s law
- Stress–strain curve: elastic limit, yield point, ultimate strength
- Elastic moduli: Young’s modulus, shear modulus, bulk modulus; Poisson’s ratio
- Applications of elastic behaviour of materials
- Plastic deformation and strain hardening
- Fracture mechanics (brittle vs ductile failure—intro)
- Creep and stress relaxation (time-dependent effects)
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The full chapter includes 560 solved multiple-choice questions in 6 parts. This page (Part 1) gives the first 100 MCQs with answers and explanations to build your foundations. Explore more with: Class 11 Physics – Chapter Index, Class 11 – Subject Index, All Classes – MCQ Library.
- 👉 Total MCQs in this chapter: 560 (100 + 100 + 100 + 100 + 100 + 60)
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FAQs on Mechanical Properties of Solids ▼
▸ What are Mechanical Properties of Solids MCQs in Class 11 Physics?
These are multiple-choice questions from Chapter 9 of NCERT Class 11 Physics – Mechanical Properties of Solids. They test concepts like stress, strain, elastic modulus, Hooke’s law, and stress-strain curves.
▸ How many MCQs are available in this chapter?
There are a total of 560 MCQs from Mechanical Properties of Solids. They are divided into 6 sets – five sets of 100 questions each and one set of 60 questions.
▸ Are these MCQs useful for NCERT, CBSE, and state board exams?
Yes, these MCQs are based on the NCERT/CBSE Class 11 Physics syllabus and are equally useful for state board exams. They help students build a strong foundation in mechanics and problem-solving.
▸ Are Mechanical Properties of Solids MCQs important for JEE and NEET?
Yes, this chapter is highly important for JEE, NEET, and other competitive exams. Topics like Young’s modulus, bulk modulus, shear modulus, and elasticity frequently appear in entrance tests.
▸ Do these MCQs include answers and explanations?
Yes, every MCQ comes with the correct answer along with detailed explanations wherever required. This ensures students not only practice but also understand the concepts thoroughly.
▸ Which subtopics are covered in these MCQs?
The MCQs cover all major subtopics of this chapter, including stress and strain, elastic behavior, Hooke’s law, stress-strain curve, Young’s modulus, bulk modulus, shear modulus, and applications of elasticity in daily life.
▸ Why is the stress-strain curve important in this chapter?
The stress-strain curve is an important concept as it explains the behavior of materials under different loads. MCQs based on this topic strengthen understanding of elasticity and fracture points, which are frequently asked in exams.
▸ Who should practice Mechanical Properties of Solids MCQs?
These MCQs are ideal for Class 11 students, CBSE/state board aspirants, and candidates preparing for JEE, NEET, NDA, UPSC, and other competitive exams.
▸ Can I practice these MCQs online for free?
Yes, all Mechanical Properties of Solids MCQs on GK Aim are available online for free. Students can practice them anytime on mobile, tablet, or desktop.
▸ Are these MCQs useful for quick revision before exams?
Yes, practicing these MCQs regularly helps in quick revision, improves memory recall, and boosts exam performance by enhancing accuracy and speed.
▸ Do these MCQs cover both basic and advanced level problems?
Yes, the MCQs range from simple conceptual questions to advanced application-based problems, covering the entire depth of the chapter as per NCERT and competitive exam standards.
▸ Why are the 560 MCQs divided into 6 parts?
The MCQs are divided into 6 sets to make practice more organized and manageable. This allows students to learn step by step and avoid information overload.
▸ Can teachers and coaching institutes use these MCQs?
Yes, teachers and coaching centers can use these MCQs as readymade assignments, quizzes, and practice material for board exam and competitive exam preparation.
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Yes, the Mechanical Properties of Solids MCQs pages are fully optimized for smartphones and tablets so students can practice anytime, anywhere.
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