Class 11 Physics Chapter 10: Thermal Properties of Matter MCQs with answers
10 multiple-choice questions on Thermal Properties of Matter, each with its answer and the reason, written from the NCERT chapter. For school exams on the CBSE pattern, JEE Main and NEET, which are set from the same NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.
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Answer: (d) −40°
Put tF = tC = t in tF = (9/5)tC + 32: t = 9t/5 + 32, so −4t/5 = 32 and t = −40. The same number is read at −40 °C = −40 °F.
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Answer: (a) 25 K
The Kelvin and Celsius scales have the same size of unit (T = tC + 273.15), so a rise of 25 °C is a rise of 25 K. 298.15 K and 316.15 K are temperatures, not a rise; 45 is the rise in °F.
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Answer: (c) 109.2 kPa
At constant volume P ∝ T, and the straight P–t line reaches zero pressure at −273 °C. From −273 °C to 0 °C the pressure rises by 0.40 kPa per degree over 273 degrees: P = 0.40 × 273 = 109.2 kPa.
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Answer: (b) aluminium
Δl = α_l l ΔT, so with l and ΔT the same, the rod with the largest α_l lengthens the most: aluminium, 2.5 × 10⁻⁵ K⁻¹.
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Answer: (c) 5.0 mm
Δl = α_l l ΔT = 2.5 × 10⁻⁵ × 2.0 × 100 = 5.0 × 10⁻³ m = 5.0 mm. 2.5 mm forgets the length of 2.0 m; 0.5 mm and 50 mm slip a power of ten.
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Answer: (a) 1.2 × 10⁻⁵ K⁻¹
α_v = (ΔV/V)/ΔT = 0.0036/100 = 3.6 × 10⁻⁵ K⁻¹. Since α_v = 3α_l, α_l = 1.2 × 10⁻⁵ K⁻¹. 3.6 × 10⁻⁵ K⁻¹ is α_v itself; 1.8 divides by 2 (the area rule); 10.8 multiplies by 3.
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Answer: (d) first increases, then decreases
Water contracts on heating from 0 °C to 4 °C, so its density rises; above 4 °C it expands and the density falls. Water is densest at 4 °C.
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Answer: (b) water near 4 °C, being densest, stays at the bottom while ice forms at the top and floats
Water has its maximum density at 4 °C. As the pond cools, water near 4 °C sinks and stays at the bottom; colder, less dense water stays on top and freezes there, so the pond freezes from the top and the water below stays liquid.
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Answer: (d) 3.3 × 10⁻³ K⁻¹
For an ideal gas at constant pressure α_v = 1/T, with T in kelvin: 1/300 K = 3.3 × 10⁻³ K⁻¹. 3.7 × 10⁻³ K⁻¹ is the value at 0 °C; 3.7 × 10⁻² K⁻¹ uses 1/27, the Celsius number.
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Answer: (c) 90 cm
The difference stays fixed only if both rods lengthen by the same amount for any ΔT: liron α_iron ΔT = lbrass α_brass ΔT, so liron = 60 × 1.8/1.2 = 90 cm. 40 cm inverts the ratio; 150 cm adds the lengths.
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