Question

When a liquid in a glass vessel is heated, its apparent expansion coefficient is $10.30\times 10^{- 4} \, {}^{\circ}C^{- 1}$ . When the same liquid is heated in a metal vessel, its apparent expansion coefficient is $10.06\times 10^{- 4} \,{}^{\circ}C^{- 1}$ . If the coefficient of linear expansion of glass = $9\times 10^{- 6} {}^{\circ}C^{- 1}$ . What is the coefficient of linear expansion of metal?

• A. $51 \, \times 10^{- 6} {}^{\circ}C^{- 1}$
• B. $17 \, \times 10^{- 6} \, {}^{\circ}C^{- 1}$
• C. $25 \times 10^{-6}\,{}^{\circ}C^{- 1}$
• D. $43 \, \times 10^{- 6} \, {}^{\circ}C^{- 1}$

Answer 1

Answer: (B)

Here, $\gamma_{ ag }=10.30 \times 10^{-4}{ }^{\circ} C ^{-1}$
$\gamma_{ am }=10.06 \times 10^{-4} { }^{\circ} C ^{-1} $
$\alpha_{ g }=9 \times 10^{-6} { }^{\circ} C ^{-1}, \alpha_{ m }=?$
Now, $\gamma_{ r }=\gamma_{ ag }+\gamma_{ g }=\gamma_{ am }+\gamma_{ m }$
$\therefore 10.30 \times 10^{-4}+3 \times 9 \times 10^{-6}=10.06 \times 10^{-4}+\gamma_{ m } \left[\because \gamma_{ g }=3 \times \alpha_{ a }\right]$
$\therefore \gamma_{ m }=(10.30+0.27-10.06)10^{-4}=0.51 \times 10^{-4} $
$\alpha_{ m }=\frac{1}{3} \gamma_{ m }=\frac{0.51 \times 10^{-4}}{3}$
$=0.17 \times 10^{-4}=17 \times 10^{-6} { }^{\circ} C ^{-1}$