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10-1.Thermometry, Thermal Expansion and Calorimetry
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A bar of iron is $10\, cm$ at $20°C$. At $19°C$ it will be ($\alpha$ of iron $= 11 \times 10^{-6}/°C$)
A
$11 \times 10^{-6} cm$ longer
B
$11 \times 10^{-6} cm$ shorter
C
$11 \times 10^{-5} cm$ shorter
D
$11 \times 10^{-6} cm$ longer
Solution
(c) $L = {L_0}(1 + \alpha \Delta \theta )$ $⇒$ $\frac{{{L_1}}}{{{L_2}}} = \frac{{1 + \alpha {{(\Delta \theta )}_1}}}{{1 + \alpha {{(\Delta \theta )}_2}}}$
$\Rightarrow$ $\frac{{10}}{{{L_2}}} = \frac{{1 + 11 \times {{10}^{ – 6}} \times 20}}{{1 + 11 \times {{10}^{ – 6}} \times 19}}$ $⇒$ ${L_2} = 9.99989$
$\Rightarrow$ Length is shorten by
$10 – 9.99989 = 0.00011 = 11 \times {10^{ – 5}}\,cm$
Standard 11
Physics
