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, αCu= 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 = d2d1 = 4.2544 – 4.24 = 0.0144 cm
Hence, the diameter increases by 1.44 × 10–2 cm.

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