Question

A hole is drilled in a copper sheer. 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 ? Coeffecient of linear expansion of copper = $1.70 \times 10^{-5}\,K^{-1}$

• A. $1.44 \times 10^{-2}\,cm$
• B. $14.4 \times 10^{-2}\,cm$
• C. $144 \times 10^{-2}\,cm$
• D. $0.144 \times 10^{-2}\,cm$

Answer 1

Answer: (A)

$1.44 \times 10^{-2}\,cm$

Answer 2

Initial temperature, $T _{1}=27.0^{\circ} C$ Diameter of the hole at $T _{1}, d _{1}=4.24 cm$ Final temperature, $T _{2}=227^{\circ} C$ Diameter of the hole at $T _{2}= D _{2}$ Co-efficient of linear expansion of copper, $\alpha_{ Cu }=1.70 \times 10^{-5} K ^{-1}$ For co-efficient of superficial expansion $\beta$, and change in temperature $\triangle T$, we have the relation: Change in area $(\triangle) /$ Original area $(A)=\beta \triangle T$ $\left[\left(\pi d _{2}^{2} / 4\right)-\left(\pi d _{2}^{2} / 4\right)\right] /\left(\pi d _{1}^{1} / 4\right)=\triangle A / A$ $\therefore \triangle A / A =\left( d _{2}^{2}- d _{1}^{2}\right) / d _{1}^{2}$ But $\beta=2 \alpha$ $\therefore\left( d _{2}^{2}- d _{1}^{2}\right) / d _{1}^{2}=2 \alpha \Delta T$