Q. A bar of iron is $10\ cm$ at $20℃$. At $19℃$ it will be ($\alpha$ of iron $= 11 \times 10^{- 6}/℃$)
A) $11 \times 10^{- 6}cm\ longer$
B) $11 \times 10^{- 6}cm\ shorter$
C) $11 \times 10^{- 5}cm\ shorter$
D) $11 \times 10^{- 5}cm\ longer$
Explanation
(c) $L = L_{0}(1 + \alpha\mathrm{\Delta}\theta) \Rightarrow \frac{L_{1}}{L_{2}} = \frac{1 + \alpha(\mathrm{\Delta}\theta)_{1}}{1 + \alpha(\mathrm{\Delta}\theta)_{2}}$
$\Rightarrow \frac{10}{L_{2}} = \frac{1 + 11 \times 10^{- 6} \times 20}{1 + 11 \times 10^{- 6} \times 19} \Rightarrow L_{2} = 9.99989$
$\Rightarrow$ Length is shorten by
$10 - 9.99989 = 0.00011 = 11 \times 10^{- 5}cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.