Question

The temperature of a thin wire of lead of length $5\text{cm}$ is raised by $150{}^{\circ }\text{C}$ which increases its length by about $0.2\text{mm}$. What is the coefficient of cubical expansion of lead?

• A. $54\times {10}^{-6}{}^{\circ }{\text{C}}^{-1}$
• B. $27\times {10}^{-6}{}^{\circ }{\text{C}}^{-1}$
• C. $9\times {10}^{-6}{}^{\circ }{\text{C}}^{-1}$
• D. $81\times {10}^{-6}{}^{\circ }{\text{C}}^{-1}$

Answer 1

The correct option is D $81\times {10}^{-6}{}^{\circ }{\text{C}}^{-1}$

Given:
The initial length of the lead wire, $\text{l}=5\text{cm}$
Change in the temperature, $\Delta \text{T}=150{}^{\circ }\text{C}$
Increase in the length of the wire, $\Delta \text{l}=0.2\text{mm}=0.02\text{cm}$
If $\alpha$ is the coefficient of linear expansion of lead, then:
$\Delta \text{l}=\text{l}\alpha \Delta \text{T}$
$⟹\alpha =\frac{\Delta \text{l}}{\text{l}\Delta \text{T}}=\frac{0.02}{5\times 150}$
$=26.67\times {10}^{-6}{}^{\circ }{\text{C}}^{-1}$
$\approx 27\times {10}^{-6}{}^{\circ }{\text{C}}^{-1}$
The ratio of the coefficients of linear and cubical expansions is $\alpha :\gamma =1:3$
$⟹\gamma =3\alpha =3\times 27\times {10}^{-6}$
$=81\times {10}^{-6}{}^{\circ }{\text{C}}^{-1}$