Answer to Question #134111 in Molecular Physics | Thermodynamics for Sridhar

Question

A steel rod at 25°C is observed to be 1m long when measured by another metal scale which is correct at 0°C .The exact length of steel rod at 0°C is
( α steel )=12×10^(-6)/°C
(α metal )=20×10^(-6)/°C)
Ans:1.0002m

Answer 1

First calculate the Actual length measured by steel scale at 25℃

$L=1(1+t\times\alpha)$

$L=1(1+20\times10^{-6}\times 25)=1.0005m$

L=1.0005m

Now find the length of the steel rod at $0\degree$ by letting its length be $L_0$

$L=1.0005=L_0(1+12\times10^{-6}\times 25)$

$L_0=\frac{1.0005}{(1+12\times 10^{-6}\times25)}=1.0002m$

Answer= 1.0002m