Question

A glass flask is filled upto a mark with $50{\text{cm}}^3$ 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 $10\times {{10}^{-6}}^{\circ }{C}^{-1}$ and coefficient of volumetric expansion of mercury is $180\times {{10}^{-6}}^{\circ }{C}^{-1}$

• A. $0.25{\text{cm}}^3$
• B. $0.5{\text{cm}}^3$
• C. $0.35{\text{cm}}^3$
• D. $0.15{\text{cm}}^3$

Answer 1

The correct option is D $0.15{\text{cm}}^3$

Volume of mercury at ${18}^{\circ }C,V_0=50{\text{cm}}^3$
Volume of mercury at ${38}^{\circ }C,V=V_0(1+{\gamma }_m\Delta T)$
$\therefore \Delta V_m=V-V_0=V_0{\gamma }_m\Delta T$

Similarly
Change in volume of flask upto mark due to change in temperature from ${18}^{\circ }C$ to ${38}^{\circ }C$;
$\Delta V_f=V_0{\gamma }_f\Delta T$
$\therefore$Volume of mercury at ${38}^{\circ }C$ above the mark
$=\Delta V_m-\Delta V_f$
$=V_0({\gamma }_m-{\gamma }_f)\Delta T$
$=50(180\times {{10}^{-6}}^{\circ }{C}^{-1}-3\times {\alpha }_f)\times \Delta T$
($\because \gamma =3\times \alpha$)
$=50(180\times {{10}^{-6}}^{\circ }{C}^{-1}-3\times 10\times {{10}^{-6}}^{\circ }{C}^{-1})\times (38-18)$
$=0.15{\text{cm}}^3$