Question

An iron ring measuring $15.00cm$ in diameter is to be shrunk on a pulley which is $15.05cm$ in diameter. All measurements refer to the room temperature ${20}^{\circ }C$. To what minimum temperature should the ring be heated to make the job possible? Calculate the strain developed in the ring when it comes to the room temperature. Coefficient of linear expansion of iron $=12\times {10}^{-6}{/}^{\circ }C$.

• A. Temperature $={278}^{\circ }C$, Strain $=3.33\times {10}^{-3}$.
• B. Temperature $={298}^{\circ }C$, Strain $=3.33\times {10}^{-3}$.
• C. Temperature $={298}^{\circ }C$, Strain $=3.33\times {10}^{-5}$.
• D. Temperature $={278}^{\circ }C$, Strain $=3.33\times {10}^{-5}$.

Answer 1

The correct option is A Temperature $={278}^{\circ }C$, Strain $=3.33\times {10}^{-3}$.

The ring should be heated to increase its diameter from $15.00cmto15.05cm$
Using ${\ell }_2={\ell }_1(1+\alpha \Delta \theta )$,
$=\frac{0.05cm}{15.00cm\times 12\times {10}^{-6}{/}^{\circ }C}={278}^{\circ }C$
The temperature $={20}^{\circ }C+{278}^{\circ }C={298}^{\circ }C$.
The strain developed $=\frac{{\ell }_2-{\ell }_1}{{\ell }_1}=3.33\times {10}^{-3}$.