Thermal Properties of Matter NCERT Solutions Class 11 Physics - Solved Exercise Question 11.8

Question 11.8:

A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C? Coefficient of linear expansion of copper = 1.70 × 10^–5 K^–1.

Solution:

Initial temperature, T₁ = 27.0°C
Diameter of the hole at T₁, d₁ = 4.24 cm
Final temperature, T₂ = 227°C
Diameter of the hole at T₂ = d₂
Co-efficient of linear expansion of copper, α_Cu= 1.70 × 10^–5 K^–1
For co-efficient of superficial expansion β,and change in temperature ΔT, we have the relation:
Change in area (∆)  /  Original area (A)  =  β∆T
[ (πd₂²/ 4) - (πd₁² / 4) ] / (πd₁¹ / 4)  =  ∆A / A
∴ ∆A / A = (d₂² - d₁²) / d₁²
But β = 2α
∴ (d₂² - d₁²) / d₁² = 2α∆T
(d₂² / d₁²) - 1  =  2α(T₂ - T₁)
d₂² / 4.24² = 2 X 1.7 X 10^-5 (227 - 27) +1
d₂² = 17.98 X 1.0068  =  18.1
∴ d₂ = 4.2544 cm
Change in diameter = d₂ – d₁ = 4.2544 – 4.24 = 0.0144 cm
Hence, the diameter increases by 1.44 × 10^–2 cm.