Question

A bakelite beaker has a volume capacity of $500cc$ at $30°C$. When it is partially filled with $V_m$ volume (at $30°C$) of mercury, it is found that the unfilled volume of the beaker remains constant as the temperature is varied. If ${\gamma }_{\left(beaker\right)}=6\times {10}^{-6}°C^{-1}$ and ${\gamma }_{\left(mercury\right)}=1.5\times {10}^{-4}°C^{-1}$where $\gamma$ is the coefficient of volume expansion, then $V_m\left(cc\right)$ is close to

Answer 1

Step 1. Given Data,

Volume capacity of the beaker at $30°C$,$V_b=500cc$

Expansion coefficient of the beaker ${\gamma }_{\left(beaker\right)}=6\times {10}^{-6}°C^{-1}$

Expansion coefficient of mercury ${\gamma }_{\left(mercury\right)}=1.5\times {10}^{-4}°C^{-1}$

Volume of partially field mercury (at $30°C$) $=V_m$,

Step 2. Calculate the Volume of partially field mercury (at $30°C$),

For the volume to remain constant at all temperature

Expansion of the liquid $=$Expansion of solid

$$\begin{gathered} V_b{\gamma }_{\left(beaker\right)}=V_m{\gamma }_{\left(mercury\right)} \\ \Rightarrow 500\times \left(6\times {10}^{-6}\right)=V_m\times \left(1.5\times {10}^{-4}\right) \\ \Rightarrow V_m=\frac{500\times \left(6\times {10}^{-6}\right)}{\left(1.5\times {10}^{-4}\right)} \\ \Rightarrow V_m=\frac{30}{1.5} \\ \Rightarrow V_m=20cc \end{gathered}$$

Hence, the volume of partially field mercury (at $30°C$), $V_m=20cc$.