Question

A steel tape is calibrated at ${20}^{\circ }C$. On a cold day when the temperature is $-{15}^{\circ }C$, the percentage error in the tape is $-\frac{x}{1000}\%$. Find $x$.

Coefficient of linear thermal expansion for steel $\alpha =12\times {10}^{-6}$ $K^{-1}$

Answer 1

Coefficient of linear thermal expansion for steel $\alpha =12\times {10}^{-6}$ $K^{-1}$

Let the length of the tape at ${20}^{o}C$ be $L$.

New length of the steel tape after heating $L^{\prime }=L(1+\alpha \Delta T)$

$\Delta L=L^{\prime }-L=L\alpha \Delta T$

$\therefore$ $\Delta L=L\alpha (-15-20)=L\alpha (-35)$

Percentage error $\frac{\Delta L}{L}\times 100=\frac{-L\times 35\times 12\times {10}^{-6}}{L}\times 100=-42\times {10}^{-3}$ %

$⟹$ $x=42$