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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.
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Solution 1
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(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 = d2 – d1 = 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`
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