Question

A steel rod at $25^{\circ} C$ is observed to be $1\, m$ long when measured by another metal scale which is correct at $0{ }^{\circ} C$. The exact length of steel rod at $0{ }^{\circ} C$ is
$\left(\alpha_{\text {steel }}=12 \times 10^{-6} /{ }^{\circ} C , \alpha_{\text {metal }}=20 \times 10^{-6} /{ }^{\circ} C \right)$

• A. 1.00002 m
• B. 1.0002 m
• C. 0.998 m
• D. 0.9998 m

Answer 1

Answer: (B)

A s both steel rod and metal scale expands at $25^{\circ} C$,
we use True value of length of rod at $25^{\circ} C$
$=$ Scale reading $\times\left(1+\alpha\left(\theta'-\theta\right)\right)$
where, $\alpha=$ coefficient of linear expansion of scale
( $=20 \times 10^{-6} /{ }^{\circ} C$ )
$\theta'=$ temperature at which observation is taken $\left(=25^{\circ} C _{r}\right)$ and $\theta=$ temperature at which metre scalc reads correctly $\left(=0^{\circ} C \right)$
Substituting values in Eq. (i), we get
Length of rod at $25^{\circ} C$
$=1 \times\left[1+20 \times 10^{-6}(25-0)\right] $
$=1+5 \times 10^{-4}=1.0005\, m$
Now, if $l_{2}=$ length of steel rod at $25^{\circ} C$ and
$l_{1}=$ length of steel rod at $0^{\circ} C$, then we have
$l_{2}=l_{1}(1+\alpha \Delta \theta)$
Substituting values, we have
$1.0005=l_{1}\left[1+12 \times 10^{-6} \times(25-0)\right]$
$\Rightarrow l_{1}=\frac{1.0005}{1.0003}=1.000199=1.0002 \,m$