50 Physics Objective Questions on Linear, Area and Volume Expansivity (With Answers and Solutions)

This set of 50 multiple-choice questions covers the core topics of thermal expansion in solids: linear expansivity (α), area/superficial expansivity (β), and volume/cubic expansivity (γ). Based on senior secondary physics curricula and suitable for JAMB, WAEC, NABTEB and similar examinations, the questions test understanding of definitions, formulas, experimental determination, calculations involving length, area and volume changes, and practical consequences and applications such as bimetallic strips, thermostats, expansion gaps and sagging of wires. Detailed solutions are provided for calculation-based questions. Use this resource to master thermal expansivity.

Questions

1. The linear expansivity of a substance is defined as the increase in length per unit length perA. unit mass

   B. degree rise in temperature

   C. unit area

   D. unit volume

2. The SI unit of linear expansivity isA. K

   B. m

   C. K⁻¹

   D. mK⁻¹

3. In the formula α = e / (l₁θ), the symbol ‘e’ represents theA. original length

   B. final length

   C. increase in length

   D. temperature change

4. Which of the following equations correctly defines linear expansivity α?A. α = (l₂ – l₁) / (l₁ × θ)

   B. α = l₁θ / (l₂ – l₁)

   C. α = (l₂ + l₁) / (l₁θ)

   D. α = l₁ / (θ × l₂)

5. A metal rod expands when heated. The magnitude of this expansion depends on:I. the temperature change

   II. the nature of the substance

   III. the original size of the rod

   A. I and II only

   B. II and III only

   C. I and III only

   D. I, II and III

6. In the experiment to determine linear expansivity of a metal rod, the micrometer screw gauge measures theA. temperature of the steam

   B. length of the rod

   C. expansion of the rod

   D. pressure of the steam

7. During the linear expansivity experiment, cold water is first passed through the rod toA. clean the rod

   B. determine the initial temperature θ₁ and initial micrometer reading

   C. lubricate the rod

   D. prevent rust

8. A rod of initial length 2 m at 25°C is heated to 80°C. If its linear expansivity is 4.0 × 10⁻³ K⁻¹, the increase in length isA. 0.26 m

   B. 0.44 m

   C. 0.53 m

   D. 0.84 m

9. A brass rod is 10 m long at 41°C. What will be its length at 30°C? (α_brass = 2.0 × 10⁻⁵ K⁻¹)A. 9.9978 m

   B. 9.9997 m

   C. 10.0002 m

   D. 10.0003 m

10. Brass of length 100 cm at 50°C changes to 100.054 cm when heated. If α_brass = 0.000018 K⁻¹, the final temperature isA. 51°C

    B. 54°C

    C. 72°C

    D. 80°C

11. An iron rod is 2.58 m long at 0°C. If the difference between the lengths of the iron rod and a brass rod is to remain constant at all temperatures, the length of the brass rod at 0°C should be(α_iron = 1.2×10⁻⁵ K⁻¹, α_brass = 1.9×10⁻⁵ K⁻¹)

    A. 1.63 m

    B. 2.58 m

    C. 3.00 m

    D. 4.00 m

12. Superficial (area) expansivity β is related to linear expansivity α byA. β = α

    B. β = 2α

    C. β = 3α

    D. β = α/2

13. The cubic expansivity γ of a solid is related to its linear expansivity α byA. γ = α

    B. γ = 2α

    C. γ = 3α

    D. γ = 3α²

14. A square metal sheet of area 600 mm² is heated through 15 K. If the linear expansivity of the metal is 1.9×10⁻⁵ K⁻¹, its new area isA. 600.057 mm²

    B. 600.114 mm²

    C. 600.171 mm²

    D. 600.342 mm²

15. A metal cube of linear expansivity α and initial volume V is heated through a temperature rise t. The increase in volume isA. (1/3)αVt

    B. (1/2)αVt

    C. αVt

    D. 3αVt

16. When a metal ball is heated through 30°C, its final volume becomes 1.0018 cm³. If its linear expansivity is 2.0×10⁻⁵ K⁻¹, the original volume wasA. 1.0000 cm³

    B. 1.0020 cm³

    C. 1.0036 cm³

    D. 1.0180 cm³

17. A brass cube of side 10 cm is heated through 30°C. (α_brass = 2.0×10⁻⁵ K⁻¹). The increase in its volume isA. 0.06 cm³

    B. 0.18 cm³

    C. 0.60 cm³

    D. 1.80 cm³

18. The statement “the linear expansivity of brass is 2.0×10⁻⁵ K⁻¹” means thatA. a brass rod of any length will expand by 2.0×10⁻⁵ m for every 1°C rise

    B. a unit length of brass will increase by 2.0×10⁻⁵ unit length per degree rise

    C. the volume of brass increases by 2.0×10⁻⁵ m³ per Kelvin

    D. the mass of brass changes by that amount

19. The S.I unit of area expansivity isA. m²

    B. K

    C. m

    D. K⁻¹

20. Why is the term α²θ² neglected when deriving the relationship between area and linear expansivity?A. Because α is large

    B. Because α is extremely small and α² is negligible

    C. Because θ is zero

    D. Because the material does not expand in two dimensions

21. An iron rod and a brass rod have different linear expansivities. For the difference in their lengths to remain constant at all temperatures, the product of linear expansivity and initial length must beA. different for both

    B. equal for both

    C. zero

    D. multiplied by the temperature

22. In bridges, one end of a steel girder is fixed and the other rests on rollers in an expansion gap. This is toA. increase the weight of the bridge

    B. allow free expansion and contraction, preventing bending

    C. reduce friction

    D. make the bridge look modern

23. Gaps are left between railway lines toA. allow for expansion on hot days, preventing buckling

    B. improve the appearance

    C. reduce noise

    D. make repairs easier

24. Electric transmission cables sag in hot weather becauseA. they become heavier

    B. the metal expands and becomes longer

    C. the poles shrink

    D. the wires contract

25. To prevent snapping in cold weather, transmission wires are givenA. a larger diameter

    B. an initial sag

    C. more insulation

    D. higher tension

26. A bimetallic strip consists of two metals with differentA. densities

    B. specific heat capacities

    C. linear expansivities

    D. melting points

27. When a brass-iron bimetallic strip is heated, it bends withA. brass on the outside of the curve

    B. iron on the outside of the curve

    C. both metals expanding equally

    D. no bending

28. A bimetallic strip is used in a thermostat toA. measure pressure

    B. control temperature by breaking or making an electric circuit

    C. increase current

    D. store heat

29. In an electric fire alarm, a bimetallic strip expands and makes contact to complete a circuit whenA. it gets cold

    B. the room is dark

    C. a fire heats it

    D. someone touches it

30. The balance wheel of some clocks uses a bimetallic strip toA. keep time by using expansion/contraction

    B. generate electricity

    C. sound an alarm

    D. store energy

31. Which of the following is a disadvantage of thermal expansion?A. Thermostat operation

    B. Bimetallic strip in fire alarm

    C. Cracking of building walls due to uneven expansion

    D. Use in clock balance wheels

32. The type of expansion considered for a solid rod that is long and thin is mainlyA. volume expansion

    B. superficial expansion

    C. linear expansion

    D. apparent expansion

33. If a metal rod does not expand when heated, its linear expansivity isA. zero

    B. 1

    C. infinite

    D. negative

34. A metal sheet has an area of 100 cm² at 20°C. If the temperature rises to 90°C and the linear expansivity is 0.000017 K⁻¹, the new area isA. 100.06 cm²

    B. 100.12 cm²

    C. 100.24 cm²

    D. 100.36 cm²

35. A brass cube has a volume of 100 cm³ at 25°C. Its volume at 0°C, if α_brass = 2.0×10⁻⁵ K⁻¹, isA. 85.00 cm³

    B. 99.85 cm³

    C. 99.95 cm³

    D. 100.05 cm³

36. In the formula for cubic expansivity γ = ΔV / (V₁θ), ΔV stands forA. final volume

    B. initial volume

    C. increase in volume

    D. average volume

37. The expansion of solids when heated is due toA. increase in the size of the molecules

    B. decrease in intermolecular forces

    C. increased vibration amplitude of molecules, increasing average distance

    D. creation of new molecules

38. If the linear expansivity of a material is 1.0×10⁻⁵ K⁻¹, its area expansivity isA. 1.0×10⁻⁵ K⁻¹

    B. 2.0×10⁻⁵ K⁻¹

    C. 3.0×10⁻⁵ K⁻¹

    D. 1.5×10⁻⁵ K⁻¹

39. A metal cube expands when heated. The change in length of each side depends onA. the original volume only

    B. the linear expansivity, original length, and temperature change

    C. the final temperature only

    D. the mass of the cube

40. Why is the term α²θ² neglected in the derivation of β = 2α?A. α is negative

    B. α² is extremely small compared to α, making the term negligible

    C. θ is always zero

    D. the material does not expand

41. A metal rod of length 50 cm is heated from 40°C to 80°C. The increase in length in terms of its linear expansivity α isA. 20α

    B. 200α

    C. 2000α

    D. 20000α

42. A zinc rod has length 200 m at 23°C. If its temperature rises to 33°C and α_zinc = 2.6×10⁻⁵ K⁻¹, the increase in length isA. 0.052 m

    B. 0.52 m

    C. 5.2 m

    D. 52 m

43. A wire of length 35 m is heated from 10°C to 50°C. (α = 2.0×10⁻⁶ K⁻¹). The change in length isA. 1.4×10⁻³ m

    B. 2.8×10⁻³ m

    C. 3.5×10⁻³ m

    D. 4.2×10⁻³ m

44. A metal rod of length 100 cm is heated through 100°C. (α = 3×10⁻⁵ K⁻¹). The change in length isA. 4 mm

    B. 3 mm

    C. 2 mm

    D. 1 mm

45. A cube made of metal of linear expansivity α is heated through temperature rise θ. If initial volume is V₀, the expression for increase in volume isA. (1/3)αV₀θ

    B. (1/2)αV₀θ

    C. 2αV₀θ

    D. 3αV₀θ

46. The experiment to determine linear expansivity of a metal rod uses steam toA. cool the rod

    B. heat the rod uniformly

    C. measure pressure

    D. clean the micrometer

47. In the linear expansivity experiment, the initial temperature of the cold rod is read after passing cold water. This temperature isA. θ₁

    B. θ₂

    C. the steam temperature

    D. room temperature

48. The bimetallic strip thermometer operates on the principle ofA. change in resistance

    B. differential expansion of two metals

    C. change in colour

    D. liquid expansion

49. Why does a creaking noise come from galvanized iron roofing sheets on a sunny day?A. The wind blows them

    B. Expansion and contraction of the sheets

    C. Birds walking on them

    D. The paint melting

50. In an electric thermostat, the bimetallic strip bends as temperature rises and eventuallyA. melts

    B. breaks the circuit, stopping current

    C. increases the current

    D. changes colour

Answer Key

1. B
2. C
3. C
4. A
5. D
6. C
7. B
8. B
9. A
10. D
11. A
12. B
13. C
14. D
15. D
16. A
17. D
18. B
19. D
20. B
21. B
22. B
23. A
24. B
25. B
26. C
27. A
28. B
29. C
30. A
31. C
32. C
33. A
34. C
35. B
36. C
37. C
38. B
39. B
40. B
41. C
42. A
43. B
44. B
45. D
46. B
47. A
48. B
49. B
50. B

Solutions to Calculation Questions

Q8: e = α l₁ (θ₂ – θ₁) = 4.0×10⁻³ × 2 × (80 – 25) = 0.008 × 55 = 0.44 m. Answer B.

Q9: l₂ = l₁ + α l₁ (θ₂ – θ₁). Temperature falls, so (θ₂ – θ₁) = 30 – 41 = -11°C. l₂ = 10 + 2.0×10⁻⁵×10×(-11) = 10 – 0.0022 = 9.9978 m. Answer A.

Q10: α = (l₂ – l₁) / [l₁ (θ₂ – θ₁)] → 0.000018 = 0.054 / [100 (θ₂ – 50)] → θ₂ – 50 = 0.054 / (0.000018×100) = 30 → θ₂ = 80°C. Answer D.

Q11: For constant difference, α_i l_i = α_b l_b → l_b = (1.2×10⁻⁵ × 2.58) / (1.9×10⁻⁵) = 3.096/1.9 ≈ 1.63 m. Answer A.

Q14: β = 2α = 3.8×10⁻⁵ K⁻¹. A₂ = A₁ (1 + βθ) = 600 (1 + 3.8×10⁻⁵×15) = 600 (1 + 0.00057) = 600.342 mm². Answer D.

Q16: γ = 3α = 6.0×10⁻⁵ K⁻¹. V₁ = V₂ / (1 + γθ) = 1.0018 / (1 + 0.0018) = 1.0018 / 1.0018 = 1.0000 cm³. Answer A.

Q17: V₁ = 10³ = 1000 cm³. γ = 3α = 6.0×10⁻⁵ K⁻¹. ΔV = γ V₁ θ = 6×10⁻⁵ × 1000 × 30 = 1.8 cm³. Answer D.

Q34: β = 2α = 3.4×10⁻⁵ K⁻¹. A₂ = A₁ (1 + βθ) = 100 (1 + 3.4×10⁻⁵×70) = 100 (1 + 0.00238) ≈ 100.24 cm². Answer C.

Q35: Cooling from 25°C to 0°C, Δθ = -25°C. γ = 6.0×10⁻⁵ K⁻¹. V₂ = V₁ (1 + γ Δθ) = 100 (1 – 0.0015) = 99.85 cm³. Answer B.

Q41: e = α l₁ (θ₂ – θ₁) = α × 0.5 m × 40 = 20α m. But options in α multiples? 20α (if l in m) but rod is 50 cm = 0.5 m. α × 0.5 × 40 = 20α. However the option likely expects cm unit: 50 cm × 40 = 2000 cm = 20 m? Let’s check: if α unit K⁻¹, e in same unit as l. For l=50 cm, e = α × 50 × 40 = 2000α cm. The answer is 2000α. So Answer C.

Q42: e = 2.6×10⁻⁵ × 200 × (33-23) = 2.6×10⁻⁵ × 200 × 10 = 5.2×10⁻² m = 0.052 m. Answer A.

Q43: e = 2.0×10⁻⁶ × 35 × (50-10) = 2.0×10⁻⁶ × 35 × 40 = 2.8×10⁻³ m. Answer B.

Q44: e = 3×10⁻⁵ × 100 cm × 100 = 0.3 cm = 3 mm. Answer B.

Q45: ΔV = 3α V₀ θ. Answer D.