Question

A steel scale measures the length of a copper wire as $80.0cm$, when both are at ${20}^{o}C$ (the calibration temperature for scale). What would be the scale read for the length of the wire when both are at ${40}^{o}C$? (Given ${\alpha }_{steel}=11\times {10}^{-6}pe{r}^{o}C$ and ${\alpha }_{copper}=17\times {10}^{-6}pe{r}^{o}C)$

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

Answer 1

Answer: (A)

$80.0096cm$

According to the question

the length of the copper wire at the temperature of 20°C is given, $L=80cm$

Now when the temperature increased the both the things will expand.

Let the length of the wire after expanding is $L^{\prime }\left(c\right)$

$L^{\prime }\left(c\right)=L+(1+\alpha (c\left)\Delta T\right)$

Let the length of the scale after expanding is L'(s)

$L^{\prime }\left(s\right)=L+(1+\alpha (s\left)\Delta T\right)$

Now the apparent length of the wire,

$L_{apparent}=L^{\prime }\left(c\right)-L^{\prime }\left(s\right)$

$=(L+(1+\alpha \left(c\right)\Delta T\left)\right)-(L+(1+\alpha \left(s\right)\Delta T\left)\right)$

$=L\left(\alpha \right(c)-\alpha (s\left)\right)\Delta T$

$=80\left(\right(17\times {10}^{(}-6)-11\times {10}^{(}-6))\times 20$

$=80.0096cm$