Question

The coefficient of linear expansion of glass is ${\alpha }_g$ per ${}^{0}C$ and the cubical expansion of mercury is ${\gamma }_{m}per{}^{o}C$ . The volume of the bulb of a mercury thermometer at $0^{o}C$ is $V_0$ and cross- section of the capillary at $T^{o}C$ is $A_0$ , if the mercury just fills the bulb at $0^{0}C$, height of the column is

• A. $\frac{V_{0}T({\gamma }_m+3{\alpha }_g)}{A_0(1+2{\alpha }_{g}T)}$
• B. $\frac{V_{0}T({\gamma }_m-3{\alpha }_g)}{A_0(1+2{\alpha }_{g}T)}$
• C. $\frac{V_{0}T({\gamma }_m+2{\alpha }_g)}{A_0(1+3{\alpha }_{g}T)}$
• D. $\frac{V_{0}T({\gamma }_m-2{\alpha }_g)}{A_0(1+3{\alpha }_{g}T)}$

Answer 1

The correct option is B $\frac{V_{0}T({\gamma }_m-3{\alpha }_g)}{A_0(1+2{\alpha }_{g}T)}$

Expansion in the volume of mercury=$V_0{\gamma }_{m}T$

Expansion in the volume of bulb=$V_0\left(3{\alpha }_g\right)T$

Now after this expansion, the mercury fills both the bulb and the column. Hence the volume that goes in the column is $V_0({\gamma }_m-3{\alpha }_g)T$.

Area of the column at that temperature=$A_0(1+2{\alpha }_{g}T)$

Hence $h{A}_0(1+2{\alpha }_{g}T)=V_0({\gamma }_m-3{\alpha }_g)T$

$⟹h=\frac{V_0({\gamma }_m-3{\alpha }_g)T}{A_0(1+2{\alpha }_{g}T)}$