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The volume of the bulb of a mercury thermometer at $0^o C$ is $V_0$and cross section of the capillary is $A_0$. The coefficient of linear expansion of glass is $a_g$ $per ^o C$ and the cubical expansion of mercury $\gamma_m$ $per ^o C$. If the mercury just fills the bulb at $0^o C$, what is the length of mercury column in capillary at $T^o C.$
$\frac{{{V_0}T\left( {{\gamma _m}\, - \,3{a_g}} \right)}}{{{A_0}\left( {1\, + \,2{a_g}T} \right)}}$
$\frac{{{V_0}T\left( {{\gamma _m}\, - \,3{a_g}} \right)}}{{{A_0}\left( {1\, + \,2{a_g}T} \right)}}$
$\frac{{{V_0}T\left( {{\gamma _m}\, + \,2{a_g}} \right)}}{{{A_0}\left( {1\, + \,3{a_g}T} \right)}}$
$\frac{{{V_0}T\left( {{\gamma _m}\, + \,2{a_g}} \right)}}{{{A_0}\left( {1\, + \,3{a_g}T} \right)}}$
Solution
Expansion in the volume of mercury $=V_{0} \gamma_{m} T$
Expansion in the volume of bulbe $V_{0}\left(3 a_{g}\right) T$
Now after this expansion, the mercury fills both the bulb and the column. Hence the volume that goes in the column is $V_{0}\left(\gamma_{m}-3 a_{g}\right) T$
Area of the column at that temperature= $A_{0}\left(1+2 a_{g} T\right)$
Hence $h A_{0}\left(1+2 a_{g} T\right)=V_{0}\left(\gamma_{(m)}-3 a_{g}\right) T$
$\Longrightarrow h=\frac{V_{0}T\left(\gamma_{m}-3 a _{g}\right) }{A_{0}\left(1+2 a _{g} T\right)}$
