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A bakelite beaker has volume capacity of $500\, cc$ at $30^{\circ} C$. When it is partially filled with $V _{ m }$ volume (at $30^{\circ}$ ) of mercury, it is found that the unfilled volume of the beaker remains constant as temperature is varied. If $\gamma_{\text {(beaker) }}=6 \times 10^{-6}{ }^{\circ} C ^{-1}$ and $\gamma_{(\text {mercury })}=1.5 \times 10^{-4}{ }^{\circ} C ^{-1},$ where $\gamma$ is the coefficient of volume expansion, then $V _{ m }($in $cc )$ is close to
$20$
$25$
$35$
$27$
Solution
$\Delta V =\left( V _{0}- V _{ m }\right)$
After increasing temperature
$\Delta V ^{\prime}=\left( V _{0}^{\prime}- V _{ m }^{\prime}\right)$
$\Delta V ^{\prime}=\Delta V$
$V _{0}- V _{ m }= V _{0}\left(1+\gamma_{ b } \Delta T \right)- V _{ m }\left(1+\gamma_{ M } \Delta T \right)$
$V _{{0} \gamma_b} = V _{ m } \gamma_{ m }$
$V _{ m }=\frac{ V _{0} \gamma_{ b }}{\gamma_{ m }}=\frac{(500)\left(6 \times 10^{-6}\right)}{\left(1.5 \times 10^{-4}\right)}$
$20\,CC$
