**Question:**

A bakelite beaker has volume capacity of $500 \mathrm{cc}$ at $30^{\circ} \mathrm{C}$. When it is partially filled with $V_{m}$ volume (at $30^{\circ} \mathrm{C}$ ) 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 $V_{m}$ (in cc) is close to__________

**Solution:**

$(\mathbf{2 0 . 0 0})$

Volume capacity of beaker, $V_{0}=500 \mathrm{cc}$

$V_{b}=V_{0}+V_{0} \gamma_{\text {beaker }} \Delta T$

When beaker is partially filled with $V_{m}$ volume of mercury,

$V_{b}^{1}=V_{m}+V_{m} \gamma_{m} \Delta T$

Unfilled volume $\left(V_{0}-V_{m}\right)=\left(V_{b}-V_{m}^{1}\right)$

$\Rightarrow V_{0} \gamma_{\text {beaker }}=V_{m} \gamma_{M}$

$\therefore V_{m}=\frac{V_{0} \gamma_{\text {beaker }}}{\gamma_{M}}$

or, $V_{m}=\frac{500 \times 6 \times 10^{-6}}{1.5 \times 10^{-4}}=20 \mathrm{cc}$

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