Physics of Solids

Physics, HSSC-KPK Second Year. This chapter is split into 33 topics and has 859 practice questions. The topics covered are listed below, followed by 10 free sample questions from the chapter.

Topics in this chapter

1. Young's modulus, stress and strain 26 questions
2. Shear modulus, bulk modulus and Hooke's law 27 questions
3. Superconductors 30 questions
4. Energy band theory 23 questions
5. Magnetic hysteresis and hysteresis loop 26 questions
6. Classification of magnetic materials and Curie point 30 questions
7. Classification of solids and crystal structure 28 questions
8. Theory of magnetism 26 questions
9. Stress-strain curve, mechanical properties and strain energy 30 questions
10. Tensile stress: numerical problems 30 questions
11. Tensile strain and extension: numerical problems 30 questions
12. Young's modulus from experimental data 26 questions
13. Extension of wires and rods using Young's modulus 24 questions
14. Load, force and mass supported by a stretched wire 25 questions
15. Shear modulus and angle of shear: numerical problems 26 questions
16. Bulk modulus and volume change: numerical problems 26 questions
17. Strain energy stored in a stretched wire: numerical problems 28 questions
18. Strain energy per unit volume: numerical problems 27 questions
19. Springs, load-extension and Hooke's law: numerical problems 26 questions
20. Stress, strain and elastic moduli: concepts and proportional reasoning 30 questions
21. Elastic moduli table: comparing materials 30 questions
22. Stress-strain curve: numerical and graph-based problems 24 questions
23. Energy bands: numerical comparisons of band gap and thermal energy 21 questions
24. Hysteresis loss: numerical problems 24 questions
25. Superconductors: critical temperatures and conversions 22 questions
26. Crystalline, polycrystalline and amorphous solids: comparisons and reasoning 20 questions
27. Lattice, basis, unit cell and crystal structure: reasoning 20 questions
28. Mechanical properties of materials: definitions and examples 30 questions
29. Stress-strain curve and strain energy: concepts 20 questions
30. Energy bands: comparing insulators, conductors and semiconductors 20 questions
31. Hysteresis loop: path, points and properties 27 questions
32. Magnetic materials: soft and hard types, examples and applications 27 questions
33. Assertion-reason questions on Physics of Solids 30 questions

Sample questions: Physics of Solids

10 sample questions with the correct answer and an explanation. Try to answer each one before reading the explanation.

Q1. Discuss crystalline, amorphous and polymer solids. Which is the most complete correct distinction between them?

• A. Amorphous solids have MORE order than crystalline solids, contrary to their defining characteristic
• B. Crystalline solids have both long-range and short-range atomic order; amorphous solids have only short-range order with randomly oriented particles; polymeric (polymer) solids consist of long chain molecules that may show varying degrees of local order depending on how the chains are packed(correct answer)
• C. Polymeric solids are always perfectly crystalline, with no amorphous character at all
• D. Crystalline, amorphous, and polymeric solids are all identical in atomic arrangement, with no meaningful distinction

Answer: B.Crystalline solids exhibit both long-range and short-range atomic order (regular, periodic 3D arrangement); amorphous solids have only short-range order with randomly oriented particles and no defined lattice; polymeric solids consist of long chain molecules whose packing can range from largely disordered to partially ordered regions, distinguishing them from simple crystalline or amorphous classifications.

Q2. Is there any difference in the length of a 20-metre steel girder when it stands vertically and when it lies horizontally?

• A. Yes, a very slight difference — when vertical, the girder's own weight causes a small additional compressive/tensile stress along its length (depending on support), whereas when horizontal, gravity instead causes bending (sagging) rather than a length change along the girder's axis(correct answer)
• B. No, there is absolutely no difference under any circumstances
• C. Orientation has no effect on any physical property of the girder
• D. The girder's length changes only when it is painted a different color

Answer: A.When vertical, the girder's own weight adds a small compressive or tensile stress distributed along its length (elongating or compressing it very slightly via Y=stress/strain); when horizontal, gravity instead causes the girder to sag/bend (a different type of deformation, not a simple axial length change) — so orientation does subtly affect its deformation behavior, though the effect is generally very small for a rigid steel girder.

Q3. Explain the Curie point (a stated chapter learning objective). If a piece of iron is heated above its Curie temperature and then allowed to cool back below it, does it automatically regain its previous magnetization?

• A. The material remains paramagnetic forever, even after cooling below the Curie point
• B. The Curie point has no relationship to a material's magnetization at all
• C. No — above the Curie point, the material becomes paramagnetic and loses its ferromagnetic domain alignment; upon cooling back below the Curie point, it becomes ferromagnetic again, but the domains re-form in random orientations unless an external magnetizing field is reapplied, so it does not automatically regain its PREVIOUS specific magnetization(correct answer)
• D. Yes, it automatically and perfectly regains its exact previous magnetization with no external field needed

Answer: C.Above the Curie point, thermal agitation overcomes the domain-aligning interaction, and the material becomes paramagnetic (domains effectively randomized); upon cooling back below Tc, the material regains its ferromagnetic character (domains can align again), but the domains re-form in essentially random orientations unless a magnetizing field is applied to realign them — so the specific PREVIOUS magnetization is not automatically restored.

Q4. Doping a pure (intrinsic) semiconductor with certain impurity atoms can dramatically increase its conductivity, even though the overall band structure remains largely the same. How might this be explained using the energy band picture?

• A. Doping always converts a semiconductor into a perfect insulator
• B. Doping has no effect on a semiconductor's energy bands or conductivity whatsoever
• C. Doping eliminates the forbidden energy gap entirely, converting the semiconductor into a superconductor
• D. Impurity atoms can introduce additional, localized energy levels within the previously forbidden gap (closer to either the conduction or valence band), making it much easier for electrons (or holes) to be excited into the conduction (or valence) band, since the effective energy 'jump' needed is now much smaller than the full intrinsic band gap(correct answer)

Answer: D.Doping introduces impurity atoms whose energy levels sit within the forbidden gap, much closer to the conduction band (donor levels, providing easily excitable extra electrons) or valence band (acceptor levels, providing easily created extra holes) than the full intrinsic band gap — dramatically increasing the number of available charge carriers and hence conductivity, without requiring the full original band-gap energy for excitation.

Q5. Classify insulators, conductors, and semiconductors on the basis of energy bands (chapter learning objective). If you were given only a graph of a material's conductivity versus temperature, how might you distinguish a metal (conductor) from a semiconductor?

• A. A metal's conductivity typically DECREASES slightly as temperature rises (due to increased lattice vibration scattering electrons), while a semiconductor's conductivity typically INCREASES markedly as temperature rises (due to more electrons being thermally excited across the band gap)(correct answer)
• B. Both metals and semiconductors show identical conductivity-temperature behavior, with no distinguishing trend
• C. A metal's conductivity always increases dramatically with temperature, exactly like a semiconductor
• D. Temperature has no measurable effect on either a metal's or a semiconductor's conductivity

Answer: A.A metal's conductivity typically decreases somewhat with rising temperature (increased thermal vibrations scatter the already-abundant free electrons more), while a semiconductor's conductivity typically increases markedly with temperature (since more electrons gain enough thermal energy to cross the band gap into the conduction band) — this contrasting temperature dependence is a classic experimental way to distinguish the two.

Q6. Photovoltaic (solar) cells are made from semiconductor materials. How does the energy band picture explain how sunlight can generate electricity in such a cell?

• A. Sunlight has no relationship to the energy band structure of a semiconductor
• B. Photovoltaic cells require the semiconductor to have zero band gap, exactly like a metal
• C. Solar cells work by heating the semiconductor to extremely high temperatures using focused sunlight, unrelated to any band-gap process
• D. Photons of sunlight with sufficient energy (greater than or equal to the semiconductor's band gap) can excite electrons from the valence band directly into the conduction band, creating free electron-hole pairs that can then be collected as an electric current(correct answer)

Answer: D.In a photovoltaic cell, incoming photons with energy at least equal to the semiconductor's band gap can directly excite electrons from the valence band into the conduction band, creating mobile electron-hole pairs; these charge carriers are then separated and collected by the cell's internal structure to produce a usable electric current — a direct application of energy band theory (and connects to amorphous silica's photovoltaic use mentioned earlier in the chapter).

Q7. Liquid nitrogen (boiling point 77K) is much cheaper and easier to produce/handle than liquid helium (boiling point ~4K). Why is the discovery of superconductors with Tc above 77K considered such a major practical breakthrough?

• A. Liquid nitrogen and liquid helium are equally expensive and convenient, so there is no practical advantage to a higher Tc
• B. Materials with Tc above 77K can be cooled into their superconducting state using the relatively cheap and convenient liquid nitrogen, rather than requiring the far more expensive and technically demanding liquid helium needed for lower-Tc superconductors(correct answer)
• C. The choice of coolant has no bearing on the practicality of using a superconductor
• D. A Tc above 77K would actually make a material harder to use as a superconductor, not easier

Answer: B.As implied by the chapter's mention of the '77-K temperature of liquid nitrogen,' achieving superconductivity above this temperature is a major practical breakthrough because it allows the relatively cheap, safe, and widely available liquid nitrogen to be used for cooling, instead of the far more expensive and technically demanding liquid helium needed for lower-temperature ('conventional') superconductors.

Q8. Why might MRI machines specifically benefit from superconducting magnets rather than conventional electromagnets, given the very high, stable currents (and hence strong magnetic fields) that MRI requires?

• A. Superconducting magnets cannot sustain any current at all, making them unsuitable for MRI
• B. MRI machines specifically require magnets that DO waste large amounts of energy as heat
• C. The very high currents needed for MRI's strong magnetic field would, in a conventional (resistive) electromagnet, generate enormous, continuous heating (I²R losses) that would be costly and difficult to manage; a superconducting magnet sustains this same high current with zero resistive heating, making it far more practical for continuous operation(correct answer)
• D. MRI machines could use conventional electromagnets with no meaningful difference in performance or cost

Answer: C.MRI machines require very strong, stable magnetic fields sustained continuously — achievable via high current; a conventional resistive electromagnet carrying such current would generate substantial, continuous I²R heating (costly to manage and potentially unsafe near a patient), while a superconducting magnet sustains the same current with zero resistive loss, exactly why the chapter specifically cites MRI machines as a superconductor application.

Q9. The chapter states that 'the spinning motion of electrons in particular is responsible for magnetism of a substance,' more so than the orbital motion. What does this suggest about materials whose electrons have paired-up (opposite) spins within each atomic orbital?

• A. Materials with paired electron spins are always the strongest ferromagnets
• B. Such materials would tend to have their spin-related magnetic effects cancel out (since paired electrons typically have opposite, cancelling spins), leaving orbital motion and any unpaired electrons as the main potential source of any remaining magnetism(correct answer)
• C. Paired electron spins would double the material's magnetism, rather than cancelling it
• D. Electron spin pairing has no relevance to a material's magnetic behavior

Answer: B.Since spin motion is highlighted as the primary contributor to magnetism, and electrons that are 'paired' in an orbital conventionally have opposite (cancelling) spins, such paired electrons would largely cancel their spin-magnetic contributions — consistent with why materials with mostly paired electrons (many common materials) tend to be only weakly magnetic (diamagnetic), while materials with UNPAIRED electron spins (like iron's) can show much stronger magnetic behavior.

Q10. A compass needle (a small permanent magnet) is made from a hard magnetic material rather than a soft one. Why is this choice essential for the compass to function reliably over time?

• A. A soft magnetic material would work exactly as well as a hard one for a compass needle
• B. The choice between hard and soft magnetic material has no bearing on a compass's reliability
• C. A hard magnetic material has high retentivity, retaining its magnetization strongly and stably over long periods without an external field; a soft material would lose its magnetization too easily (from stray fields, shocks, or simply low retentivity), making it unreliable as a persistent direction-indicator(correct answer)
• D. Hard magnetic materials cannot hold any magnetization at all, making them unsuitable for compasses

Answer: C.A compass needle needs to retain a strong, stable, PERMANENT magnetization over long periods to reliably indicate direction; a hard magnetic material's high retentivity (much residual magnetism remaining after the initial magnetizing field is removed) makes it ideal for this, whereas a soft magnetic material's low retentivity would cause the needle to lose its magnetization too readily to be a reliable, persistent compass.