Question

A glass flask is filled up to a mark with $50\, c c$ of mercury at $18^{\circ} C$. If the flask and contents are heated to $38^{\circ} C$, how much mercury will be above the mark? ( $\alpha$ for glass is $9 \times 10^{-6},{ }^{\circ} C$ and coefficient of real expansion of mercury is $180 \times 10^{-6} /{ }^{\circ} C$ )

• A. 0.85 c c
• B. 0.46 c c
• C. 0.153 c c
• D. 0.05 c c

Answer 1

Answer: (C)

Due to volume expansion of both mercury and flask, the change in volume of mercury relative to flask is given by
$\Delta V =V_{0}\left[\gamma_{L}-\gamma_{g}\right] \Delta \theta=V\left[\gamma_{L}-3 \alpha_{g}\right] \Delta \theta $
$=50\left[180 \times 10^{-6}-3 \times 9 \times 10^{-6}\right](38-18) $
$=0.153 \,c c$