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The coefficient of apparent expansion of mercury in a glass vessel is $153\times 10^{-6}/\,^oC$ and in a steel vessel is $144\times 10^{-6}/\,^oC.$ If $\alpha $ for steel is $12 \times 10^{-6}/\,^oC,$ then $\alpha $ that of glass is
$9 \times 10^{-6}/\,^oC$
$6 \times 10^{-6}/\,^oC$
$36 \times 10^{-6}/\,^oC$
$27 \times 10^{-6}/\,^oC$
Solution
$\gamma_{\mathrm{real}}=\gamma_{\mathrm{app}}+\gamma_{\mathrm{vessel}}$
So $\left(\gamma_{\text {app}}, \gamma_{\text {vessel }}\right)_{\text {gass }}=\left(\gamma_{\text {app }}+\gamma_{\text {vessel}}\right)_{steel}$
$\Rightarrow 153 \times 10^{-6}+\left(Y_{\text {vessel}}\right)_{\text {glass }}=\left(144 \times 10^{-6}+\gamma_{\text {vessel }}\right)_{\text {steel }}$
Further. $\left(y_{\text {vessel}}\right)_{\text {steel}}=3 a=3 \times\left(12 \times 10^{-6}\right)=36 \times 10^{-6}/^\circ {C}$
$\Rightarrow 153 \times 10^{-6}+\left(\gamma_{\text {vessel }}\right)_{\text {glass}}=144 \times 10^{-6}+36 \times 10^{-6}$
$\Rightarrow\left(\gamma_{\text {vessel }}\right)_{\text {glass }}=3 \mathrm{a}=27 \times 10^{-6}/^\circ {C}$
$\Rightarrow a=9 \times 10^{-6} /^{\circ} \mathrm{C}$
