Question

A bar of iron is 10 cm at ${20}^{o}C$. At ${19}^{o}C$ it will be (a of iron = 11 X ${10}^{-6}/{}^{o}C$)

• A. 11 ´ 10^-6 cm longer
• B. 11 ´ 10^-6 cm shorter
• C. 11 ´ 10^-5 cm shorter
• D. 11 ´ 10^-5 cm longer

Answer 1

The correct option is C

11 ´ 10^-5 cm shorter

$L=L_0(1+\alpha △\theta )\Rightarrow \frac{L_1}{L_2}=\frac{1+\alpha (△\theta {)}_1}{1+\alpha (△\theta {)}_2}$
$\Rightarrow \frac{10}{L_2}=\frac{1+11\times {10}^{-6}\times 20}{1+11\times {10}^{-6}\times 19}\Rightarrow L_2=9.99989$
$\Rightarrow$ Length is shorten by
$10-9.99989=0.00011=11\times {10}^{-5}cm$