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 1

Initial temperature, *T*_{1} = 27.0°C

Diameter of the hole at *T*_{1}, *d*_{1} = 4.24 cm

Final temperature, *T*_{2} = 227°C

Diameter of the hole at *T*_{2 }= *d*_{2}

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 Δ*T*, we have the relation:

`"Change in area(ΔA)"/"Original area(A)" = betaΔT`

`(((pi (d_2^2))/4 - pi (d_1^2)/4))/((pi d_1^2/4)) = (triangleA)/A`

`:.(triangleA)/A = (d_2^2 - d_1^2)/d_1^2`

But `beta = 2 alpha`

`:.(d_2^2-d_1^2)/d_1^2 = 2alpha triangleT`

`d_2^2/d_1^2 - 1= 2alpha (T_2 - T_1)`

`d_2^2/(4.24)^2 = 2xx 1.7 xx10^(-5)(227- 27) + 1`

`d_2^2 = 17.98 xx 1.0068 = 18.1`

`:. d_2 = 4.2544 cm`

Change in diameter = *d*_{2} – *d*_{1 }= 4.2544 – 4.24 = 0.0144 cm

Hence, the diameter increases by 1.44 × 10^{–2} cm.

#### Solution 2

In this superficial expansion of copper sheet will be involved on heating

Here, area of hole at `27^@, A_1 = piD_1^2/4 = pi/4 xx(4.24)^2 cm^2`

if `D_2` cm is the diameter of the hole at `227^@C` then area of the hole at `227^@C`

`A_2 = (piD_2^2)/4 cm^2`

Coefficient of superficial expansion of copper is

`beta = 2 alpha = 2xx1.70 xx 10^(-5) = 3.4 xx 10^(-5)"'^@C^(-1)`

Increase in area = `A_2 - A_1 = betaA_1triangleT or A_2 = A_1 +betaA_1 triangle T = A_1(1 + beta triangleT)`

`(piD_2^2)/4 = pi/4(4.24)^2 [1+3.4xx10^(-5)(228 - 27)]`

`=> D_2^2 = (4.24)^2xx 1.0068` or

`D_2 = 4.2544 cm`

Change4 in diameter = `D_2 - D_1 = 4.2544 - 4.24 = 0.0114 cm`