Question

A steel scale measures the length of a copper rod as $L\text{cm}$ when both are at ${20}^{\circ }\text{C}$, the calibration temperature for the scale. If the coefficients of linear expansion for steel and copper are ${\alpha }_s$ and ${\alpha }_c$ respectively, what would be the scale reading (in cm) when both are at ${21}^{\circ }\text{C}$?

• A. $L\frac{(1+{\alpha }_c)}{(1+{\alpha }_s)}$
• B. $L\frac{{\alpha }_c}{{\alpha }_s}$
• C. $L\frac{{\alpha }_s}{{\alpha }_c}$
• D. $L$

Answer 1

The correct option is A $L\frac{(1+{\alpha }_c)}{(1+{\alpha }_s)}$

Given, the length of steel scale and copper rod at ${20}^{\circ }\text{C}$ $=L$ .
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
$L_{steel}=L(1+{\alpha }_s\Delta T)......\left(1\right)$
For copper rod,
New length due to change in temperature is given by
$L_{copper}=L(1+{\alpha }_c\Delta T)......\left(2\right)$
Hence, the relative expansion of copper rod w.r.t. steel scale after the rise in temperature is given by
$L_{\text{app}}=L_{copper}-L_{steel}=\left[L\right(1+{\alpha }_c\Delta T)-L(1+{\alpha }_s\Delta T\left)\right]$
$\left[\text{Using}\right(1\left)\text{and}\right(2\left)\right]$
$\Rightarrow L_{\text{app}}=L({\alpha }_c-{\alpha }_s)\Delta T.....\left(3\right)$
Given,
Initial temperature $T_1={20}^{\circ }\text{C}$
Final temperature $T_2={21}^{\circ }\text{C}$
$\therefore$ Change in temperature $\left(\Delta T\right)=1^{\circ }\text{C}$
From $\left(3\right)$, we can write that
$L_{\text{app}}=L({\alpha }_c-{\alpha }_s)$
Therefore, measured length at $21{}^{\circ }\text{C}$
$=L+L_{\text{app}}=L(1+{\alpha }_c-{\alpha }_s)$

By applying binomial expansion on option (a),
$L\frac{(1+{\alpha }_c)}{(1+{\alpha }_s)}=L(1+{\alpha }_c)(1+{\alpha }_s{)}^{-1}$
$=L(1+{\alpha }_c)(1-{\alpha }_s)$
$=L(1+{\alpha }_c-{\alpha }_s-{\alpha }_c{\alpha }_s)$
[$\alpha$ value is very small, hence we can neglect ${\alpha }_c{\alpha }_s$]
$\approx L(1+{\alpha }_c-{\alpha }_s)=L_{\text{app}}$
Thus, option (a) is correct.