A bakelite beaker has volume capacity of 500 cc at 30 degree C.
Question:

A bakelite beaker has volume capacity of $500 \mathrm{cc}$ at $30^{\circ} \mathrm{C}$. When it is partially filled with $\mathrm{V}_{\mathrm{m}}$ volume (at $30^{\circ}$ ) of mercury, it is found that the unfilled volume of the beaker remains constant as temperature is varied. If $\gamma_{\text {(beaker) }}=6 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}$ and $\gamma_{\text {(mercury) }}=1.5 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$, where $\gamma$ is the coefficient of volume expansion, then $\mathrm{V}_{\mathrm{m}}$ (in $\mathrm{cc}$ ) is close to_______.

Solution:

$\Delta \mathrm{V}=\left(\mathrm{V}_{0}-\mathrm{V}_{\mathrm{m}}\right)$

After increasing temperature

$\Delta \mathrm{V}^{\prime}=\left(\mathrm{V}_{0}^{\prime}-\mathrm{V}_{\mathrm{m}}^{\prime}\right)$

$\Delta \mathrm{V}^{\prime}=\Delta \mathrm{V}$

$\mathrm{V}_{0}-\mathrm{V}_{\mathrm{m}}=\mathrm{V}_{0}\left(1+\gamma_{\mathrm{b}} \Delta \mathrm{T}\right)-\mathrm{V}_{\mathrm{m}}\left(1+\gamma_{\mathrm{M}} \Delta \mathrm{T}\right)$

$\mathrm{V}_{0} \gamma_{\mathrm{b}}=\mathrm{V}_{\mathrm{m}} \gamma_{\mathrm{m}}$

$\mathrm{V}_{\mathrm{m}}=\frac{\mathrm{V}_{0} \gamma_{\mathrm{b}}}{\gamma_{\mathrm{m}}}=\frac{(500)\left(6 \times 10^{-6}\right)}{\left(1.5 \times 10^{-4}\right)}$

$=20 \mathrm{CC}$

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