Respuesta :
Given data:
* The actual length of the iron rod is 30 cm.
* The initial temperature is 20 degree celsius.
* The final temperature is 80 degree celsius.
* The mass of the iron rod is 3 kg.
* The value of constants are,
[tex]\begin{gathered} \alpha=1.2\times10^{-5\circ}C^{-1} \\ c=450Jkg^{-1\circ}C^{-1} \end{gathered}[/tex]Solution:
(a). The increase in the length of the iron rod with the change in the temperature is,
[tex]\alpha=\frac{dL}{LdT}[/tex]where L is the actual length, and dL is the change in the length,
Substituting the known values,
[tex]\begin{gathered} 1.2\times10^{-5}=\frac{dL^{}}{30\times10^{-2}\times(80^{\circ}-20^{\circ})} \\ 1.2\times10^{-5}=\frac{dL^{}}{1800\times10^{-2}} \\ 1.2\times10^{-5}=\frac{dL^{}}{18} \\ dL=1.2\times10^{-5}\times18 \\ dL=21.6\times10^{-5}\text{ m} \\ dL=0.2\times0^{-3}\text{ m} \\ dL=0.2\text{ mm} \end{gathered}[/tex]Thus, the change in the length of the iron rod is 0.2 mm.
Hence, first option is the correct answer.
