The coefficient of apparent expansion of merc | vedclass

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Question

(A) The relationship between real expansion,apparent expansion,and vessel expansion is given by: $\gamma_{real} = \gamma_{app} + \gamma_{vessel}$.Sinc

Vedclass · Accepted Answer

$9 	imes 10^{-6}/^{\circ}C$

Vedclass · Answer

(A) The relationship between real expansion,apparent expansion,and vessel expansion is given by: $\gamma_{real} = \gamma_{app} + \gamma_{vessel}$.Since the real expansion coefficient of mercury $(\gamma_{real})$ is constant,we have: $(\gamma_{app} + \gamma_{vessel})_{glass} = (\gamma_{app} + \gamma_{vessel})_{steel}$.Given $\gamma_{app, glass} = 153 	imes 10^{-6}/^{\circ}C$ and $\gamma_{app, steel} = 144 	imes 10^{-6}/^{\circ}C$.For a solid vessel,the volume expansion coefficient is $\gamma_{vessel} = 3\alpha$.For steel,$\gamma_{vessel, steel} = 3 	imes (12 	imes 10^{-6}) = 36 	imes 10^{-6}/^{\circ}C$.Substituting these values into the equation: $153 	imes 10^{-6} + \gamma_{vessel, glass} = 144 	imes 10^{-6} + 36 	imes 10^{-6}$.$153 	imes 10^{-6} + \gamma_{vessel, glass} = 180 	imes 10^{-6}$.$\gamma_{vessel, glass} = (180 - 153) 	imes 10^{-6} = 27 	imes 10^{-6}/^{\circ}C$.Since $\gamma_{vessel, glass} = 3\alpha_{glass}$,we have $3\alpha_{glass} = 27 	imes 10^{-6}/^{\circ}C$.Therefore,$\alpha_{glass} = 9 	imes 10^{-6}/^{\circ}C$.

$x$	$1$	$2$	$3$	$4$
$y$	$0.7111$	$0.7222$	$0.7333$	$0.7444$

The coefficient of apparent expansion of mercury in a glass vessel is $153 \times 10^{-6}/^{\circ}C$ and in a steel vessel is $144 \times 10^{-6}/^{\circ}C$. If $\alpha$ for steel is $12 \times 10^{-6}/^{\circ}C$,then $\alpha$ for glass is:

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