Question

A steel scale measures the length of a copper wire as $80cm$ when both are at ${20}^{\circ }C$ (the calibration temperature for scale). What would be the scale read for the length of the wire when both are at ${40}^{\circ }C?$ (Given ${\alpha }_{steel}=11\times {10}^{-6}{/}^{\circ }C$and ${\alpha }_{copper}=17\times {10}^{-6}{/}^{\circ }C$

• A. $1cm$
• B. $25.2cm$
• C. $80.0096cm$
• D. $80.0272cm$

Answer 1

The correct option is C $80.0096cm$

We know that: $L=L_0(1+\alpha \Delta \theta )$
Given: $L_0=80.0cm$ , ${\theta }_i={20}^{\circ }C$, ${\theta }_f={40}^{\circ }C$, ${\alpha }_{steel}=11\times {10}^{-6}pe{r}^{\circ }C$, ${\alpha }_{copper}=17\times {10}^{-6}pe{r}^{\circ }C$
With temperature rise (same ${25}^{\circ }C$ for both), steel scale and copper wire both expand.
$\Delta \theta ={\theta }_f-{\theta }_i=40-20={20}^{\circ }C$
Hence, the length of copper wire with respect to steel scale or the apparent length of copper wire after rise in temperature
After increase in temperature new position of $80cm$ mark will become
$L_{cu}^{\prime }=L_0(1+{\alpha }_{cu}\Delta \theta )$,
$L_{steel}^{\prime }=L_0(1+{\alpha }_s\Delta \theta )$
The difference in that $80cm$ mark will now be
$\Delta L=L_{cu}^{\prime }-L_{steel}^{\prime }$
$=L_0(1+{\alpha }_{cu}\Delta \theta )-L_0(1+{\alpha }_s\Delta \theta )$
$\Delta L=L_0({\alpha }_{cu}-{\alpha }_s)\Delta \theta$
$\Delta L=80(17\times {10}^{-6}-11\times {10}^{-6})\times 20$
$=0.0096cm$
Hence new reading will be
$L_0+\Delta L=80+0.0096=80.0096cm$
Final answer (a)