Question

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

• A. $80.0096\text{cm}$
• B. $80.0272\text{cm}$
• C. $1\text{cm}$
• D. $25.2\text{cm}$

Answer 1

The correct option is A $80.0096\text{cm}$

Let the length of steel scale and copper wire at ${20}^{\circ }\text{C}$ be $L_0$ .
For a change in temperature $\left(\Delta T\right)$, the length of a material changes to the new length given by, $L=L_0(1+\alpha \Delta T)$
For steel scale,
New length due to change in temperature is given by
$L_{steel}=L_0(1+{\alpha }_{steel}\Delta T)......\left(1\right)$
For copper wire,
New length due to change in temperature is given by
$L_{Cu}=L_0(1+{\alpha }_{cu}\Delta T)......\left(2\right)$

The relative expansion of the copper wire w.r.t. steel scale after rise in temperature is given by
$L_{app}=L_{Cu}-L_{steel}=\left[L_0\right(1+{\alpha }_{Cu}\Delta T)-L_0(1+{\alpha }_{steel}\Delta T\left)\right]$
$\left[\text{Using}\right(1\left)\text{and}\right(2\left)\right]$
$\Rightarrow L_{app}=L_0({\alpha }_{cu}-{\alpha }_{steel})\Delta T$
Change in temperature $\left(\Delta T\right)=40-20={20}^{\circ }\text{C}$
$\Rightarrow L_{app}=80\times (17\times {10}^{-6}-11\times {10}^{-6})\times 20=0.0096\text{cm}$
Hence, the new length of the wire measured at ${40}^{\circ }\text{C}$
$=L_0+L_{app}$
$=80.0096\text{cm}$
Hence, option (a) is the correct answer.