First slide

Thermal expansion

Question

The coefficient of apparent expansion of mercury in a glass vessel is 153 x 10^{- 6}/⁰C and in a steel vessel is 144 x 10^{- 6}/^°C. If $α$ for steel is 12 x 10^-6/°C, then that of glass is

Difficult

A

9 x 10^-6/°C
B

6 x 10^-6/°C
C

36 x 10^-6/°C
D

27 x 10^-6/°C

Solution

We know that
$γ_{real} = γ_{apparent} + γ_{vessel}$
So,
${(γ_{app} + γ_{vessel})}_{glass} = {(γ_{app} + γ_{vessel})}_{steel}$
$(∵ γ_{real} is same in both cases)$
Or
$\begin{aligned} 153 \times 10^{- 6} + {(γ_{vessel})}_{glass} \\ = 144 \times 10^{- 6} + {(γ_{vessel})}_{steel} \end{aligned}$
$Further {(γ_{vessel})}_{steel}$
$= 3 α = 3 \times (12 \times 10^{- 6}) = 36 \times 10^{- 6} /^{\circ} C$
$\begin{aligned} ∴ 153 \times 10^{- 6} + {(γ_{vessel})}_{glass} \\ = 144 \times 10^{- 6} + 36 \times 10^{- 6} \end{aligned}$
Solving, we get
${(γ_{vessel})}_{glass} = 27 \times 10^{- 6} /^{\circ} C$
or $α = \frac{γ_{galss}}{3} = 9 \times 10^{- 6} /^{\circ} C$

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