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Heat and Thermodynamics question

2025 · 4 Apr · Shift 2 · Q58
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  5. /2025 · 4 Apr · Shift 2 · Q58

Heat and Thermodynamics question

2025 · 4 Apr · Shift 2 · Q58

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
Consider a rectangular sheet of solid material of length l=9 cml=9 \mathrm{~cm}l=9 cm and width d=4 cm\mathrm{d}=4 \mathrm{~cm}d=4 cm. The coefficient of linear expansion is α=3.1×10−5 K−1\alpha=3.1 \times 10^{-5} \mathrm{~K}^{-1}α=3.1×10−5 K−1 at room temperature and one atmospheric pressure. The mass of sheet m=0.1 kgm=0.1 \mathrm{~kg}m=0.1 kg and the specific heat capacity Cv=900 J kg−1 K−1C_{\mathrm{v}}=900 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}Cv​=900 J kg−1 K−1. If the amount of heat supplied to the material is 8.1×102 J8.1 \times 10^2 \mathrm{~J}8.1×102 J then change in area of the rectangular sheet is :
  1. A
    2.0×10−6 m22.0 \times 10^{-6} \mathrm{~m}^22.0×10−6 m2
  2. B
    6.0×10−7 m26.0 \times 10^{-7} \mathrm{~m}^26.0×10−7 m2
  3. C
    3.0×10−7 m23.0 \times 10^{-7} \mathrm{~m}^23.0×10−7 m2
  4. D
    4.0×10−7 m24.0 \times 10^{-7} \mathrm{~m}^24.0×10−7 m2
View written solutionFree

Correct answer: A

  1. Given data
  • Length: l=9 cm=0.09 ml = 9\text{ cm} = 0.09\text{ m}l=9 cm=0.09 m
  • Width: d=4 cm=0.04 md = 4\text{ cm} = 0.04\text{ m}d=4 cm=0.04 m
  • Coefficient of linear expansion: α=3.1×10−5 K−1\alpha = 3.1 \times 10^{-5}\,\text{K}^{-1}α=3.1×10−5K−1
  • Mass: m=0.1 kgm = 0.1\,\text{kg}m=0.1kg
  • Specific heat capacity: Cv=900 J kg−1K−1C_v = 900\,\text{J kg}^{-1}\text{K}^{-1}Cv​=900J kg−1K−1
  • Heat supplied: Q=8.1×102=810 JQ = 8.1 \times 10^2 = 810\,\text{J}Q=8.1×102=810J

We need the change in area of the sheet.

  1. Find rise in temperature

Using

Q=mCvΔTQ = m C_v \Delta TQ=mCv​ΔT

so,

ΔT=QmCv=8100.1×900=81090=9 K\Delta T = \frac{Q}{mC_v} = \frac{810}{0.1 \times 900} = \frac{810}{90} = 9\,\text{K}ΔT=mCv​Q​=0.1×900810​=90810​=9K
  1. Initial area of the sheet
A0=l×d=0.09×0.04=3.6×10−3 m2A_0 = l \times d = 0.09 \times 0.04 = 3.6 \times 10^{-3}\,\text{m}^2A0​=l×d=0.09×0.04=3.6×10−3m2
  1. Use area expansion relation

For an isotropic solid, coefficient of area expansion is approximately

β=2α\beta = 2\alphaβ=2α

Thus,

ΔA=A0βΔT=A0(2α)ΔT\Delta A = A_0 \beta \Delta T = A_0 (2\alpha) \Delta TΔA=A0​βΔT=A0​(2α)ΔT

Substitute values:

ΔA=3.6×10−3×2×3.1×10−5×9\Delta A = 3.6 \times 10^{-3} \times 2 \times 3.1 \times 10^{-5} \times 9ΔA=3.6×10−3×2×3.1×10−5×9

Now calculate:

2×3.1×10−5=6.2×10−52 \times 3.1 \times 10^{-5} = 6.2 \times 10^{-5}2×3.1×10−5=6.2×10−5 6.2×10−5×9=55.8×10−5=5.58×10−46.2 \times 10^{-5} \times 9 = 55.8 \times 10^{-5} = 5.58 \times 10^{-4}6.2×10−5×9=55.8×10−5=5.58×10−4 ΔA=3.6×10−3×5.58×10−4\Delta A = 3.6 \times 10^{-3} \times 5.58 \times 10^{-4}ΔA=3.6×10−3×5.58×10−4 ΔA=20.088×10−7=2.0088×10−6 m2\Delta A = 20.088 \times 10^{-7} = 2.0088 \times 10^{-6}\,\text{m}^2ΔA=20.088×10−7=2.0088×10−6m2

So,

ΔA≈2.0×10−6 m2\boxed{\Delta A \approx 2.0 \times 10^{-6}\,\text{m}^2}ΔA≈2.0×10−6m2​
  1. Check options
  • A: 2.0×10−6 m22.0 \times 10^{-6}\,\text{m}^22.0×10−6m2 ✅
  • B: 6.0×10−7 m26.0 \times 10^{-7}\,\text{m}^26.0×10−7m2 ❌
  • C: 3.0×10−7 m23.0 \times 10^{-7}\,\text{m}^23.0×10−7m2 ❌
  • D: 4.0×10−7 m24.0 \times 10^{-7}\,\text{m}^24.0×10−7m2 ❌

Hence the correct option is A.

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