Question

A black smith fixes ring on the rime of the wooden wheel of a bullock cart the diameter of the rime and iron ring arc $5.243$ and $5.231$ resp. at ${27}^{\circ }C$ to what temp, should the ring be heated. So as to fit the rime of the wheel $\alpha =1.20\times {10}^{-5}$.

Answer 1

Given initial temperature $T_1=27{}^{\circ }C$

Initial length $l_1=5.231m$

Final length $l_2=5.243m$

Now, ${\alpha }_1=\frac{\Delta l}{l_1\Delta T}=\frac{l_2-l_1}{l_1\Delta T}$

$\Rightarrow l_2=l_1[1+{\alpha }_1(T_2-T_1\left)\right]$

i.e $5.243m=5.231m[1+1.20\times {10}^{-5}{k}^{-1}(T_2-27{}^{\circ }C\left)\right]$

This gives $T_2=218{}^{\circ }C$