Question

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.

Answer 1

Here, diameter of hole ( D1) = 4.24 cm
So, initial area of hole (Ao)= πr² = 22/7 ( 4.24/2)²
= 4.494π cm²
intial temperature ( T1) = 27°C = 27+ 273 = 300 K
Final temperature ( T2) = 227°C = 227+ 273 = 500K
coefficient of linear expansion (a) = 1.7 × 10^-5/°C
coefficient of superficial expansion (b) = 2× linear expansion
= 2 × 1.7 × 10^-5/°C
= 3.4 × 10^-5 /°C

Use formula,
A = Ao( 1 + b∆T)
A = 4.494π [1 + 3.4 × 10^-5 × (500-300)]
= 4.494π[ 1 + 3.4 × 10^-5 × 200]
= 4.494π [ 1 + 6.8 × 10^-3 ]
= 4.494π [ 1 + 0.0068]
= 4.494π × 1.00068
= 4.525π cm² = πD2²/4
D2² = 4.525 × 4
D2 = 4.2544 cm
Change in diameter (∆D) = D2 - D1
= 4.2544 - 4.24
= 0.0144 cm

Answer 2

delta l=alpha l delta t

=1.7*10^-5*4.24*200

=0.014

hope this helps