A hole is drilled in a copper sheet. The diam | vedclass

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(D) Initial temperature $T_{1} = 27.0 \; ^{\circ}C$.Initial diameter $d_{1} = 4.24 \; cm$.Final temperature $T_{2} = 227 \; ^{\circ}C$.Change in tempe

Vedclass · Accepted Answer

$1.44 	imes 10^{-2} \; cm$

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(D) Initial temperature $T_{1} = 27.0 \; ^{\circ}C$.Initial diameter $d_{1} = 4.24 \; cm$.Final temperature $T_{2} = 227 \; ^{\circ}C$.Change in temperature $\Delta T = T_{2} - T_{1} = 227 - 27 = 200 \; K$.Coefficient of linear expansion $\alpha = 1.70 	imes 10^{-5} \; K^{-1}$.Thermal expansion of a hole in a material follows the same rule as the expansion of a solid disc of the same material.The change in diameter $\Delta d$ is given by the formula $\Delta d = d_{1} \alpha \Delta T$.Substituting the values:$\Delta d = 4.24 	imes (1.70 	imes 10^{-5}) 	imes 200$.$\Delta d = 4.24 	imes 3.40 	imes 10^{-3}$.$\Delta d = 14.416 	imes 10^{-3} \; cm$.$\Delta d = 1.4416 	imes 10^{-2} \; cm \approx 1.44 	imes 10^{-2} \; cm$.

$A$ hole is drilled in a copper sheet. The diameter of the hole is $4.24 \; cm$ at $27.0 \; ^{\circ}C$. What is the change in the diameter of the hole when the sheet is heated to $227 \; ^{\circ}C$? (Coefficient of linear expansion of copper $\alpha = 1.70 \times 10^{-5} \; K^{-1}$)

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