Question

A steel scale measures the length of a copper rod as $l_0$ when both are at ${20}^{o}C$, the calibration temperature for the scale. The scale reading when both are at ${40}^{o}C$, is

• A. $(1+20{\alpha }_C)l_0$
• B. $(1+20{\alpha }_S)l_0$
• C. $\left(\frac{1+20{\alpha }_S}{1+20{\alpha }_C}\right)l_0$
• D. $\left(\frac{1+20{\alpha }_C}{1+20{\alpha }_S}\right)l_0$

Answer 1

The correct option is D $\left(\frac{1+20{\alpha }_C}{1+20{\alpha }_S}\right)l_0$

$l_o$ is the reading at ${20}^{o}C$.
Let $l_1$ be the scale reading at ${20}^{o}C$ which gets aligned with the scale at ${40}^{o}C$.

At ${40}^{o}C$, length of the scale is $l_o(1+20{\alpha }_c)$

The aligned reading on scale is $l_1(1+20{\alpha }_s)$

So $l_o(1+20{\alpha }_c)=l_1(1+20{\alpha }_s)$

$\Rightarrow l_1=\left(\frac{1+20{\alpha }_c}{1+20{\alpha }_s}\right)l_o$