Question

A metal rod has a length of $1m$ at ${30}^{o}C$. $\alpha$ of metal is $2.5\times {10}^{-5}{/}^{o}C$. The temperature at which it will be shortened by $1mm$ is:

• A. $-{30}^o$
• B. $-{40}^o$
• C. $-{10}^o$
• D. ${10}^o$

Answer 1

The correct option is C $-{10}^o$

$l=1mT=30°C{\alpha }_{metal}=2.5\times {10}^{-5}°C\Delta l=1mm{l}_f=l_i(1+\alpha \Delta T)\frac{l_f-l_i}{l_i}=\alpha \Delta T\Rightarrow \frac{-1mm}{1mm}=\alpha \Delta T\Rightarrow \Delta T=\frac{-2.5\times {10}^{-5}\times 1}{1\times {10}^{-3}}\Rightarrow \Delta T=\frac{-1\times {10}^{-3}}{1\times 2.5\times {10}^{-5}}=-\left(\frac{100}{2.5}\right)=-40\Rightarrow T_f-T_i=-40\Rightarrow T_f=T_i-40=30-40=-10°C$