Question

The coefficient of thermal expansion of a rod is temperature-dependent and is given by the formula $\alpha =aT$ , where $a$ is $3.465\times 10^{- 5} \,{}^\circ C^{- 2}$ and $T$ in $^{o}C$ . If the length of the rod is $l$ at temperature $0^{o}C$ , then the temperature at which the length will be $2l$ is [Given, $ln\left(2\right)=0.693$ ]

• A. $12 \,{}^{o}C \, $
• B. $20 \,{}^{o}C$
• C. $200 \,{}^{o}C$
• D. $100 \,{}^{o}C$

Answer 1

Answer: (C)

As; $dl=\alpha ldT$
$\therefore \displaystyle \int _{l}^{2 l}\frac{d l}{l}=a \, \displaystyle \int _{0}^{T}TdT$
$ln2=a\frac{T^{2}}{2}$
$\therefore T=\left[\frac{ln 4}{a}\right]^{\frac{1}{2}}=\left[\frac{2 \times 0.693}{3.465 \times 10^{- 5}}\right]^{\frac{1}{2}}=\sqrt{\frac{2}{5} \times 10^{5}}=200 \,{}^{o}C$