A long horizontal glass capillary tube open at both ends contains a mercury thread 1 m long at `0^@C`. Find the length of the mercury thread, as read on this scale, at `100^@C`.

Explanation

`(gamma_("mercury")=18.2xx10^-5K^-1alpha_("glass")=9xx10^-6K^-1)`
let `V_0`,`V_t=` volume of mercury at `0^@C` and `t^@C`, respectively. `A_0`,`A_t=` area of cross-section of capillary tube at `0^@C`, respectively.
Then `v_0=I_0A_0V_t=I_tA_t V_t=V_0(1+gamma_(r)t)`
`I_tA_t=I_0A_0(1+gamma_(r)t)`
or `I_tA_0(1+2alpha_gt)=I_0A_0(1+gamma_(r)t)`
or `I_t=(1+gamma_r t)/(1+2alpha_gt)`
expanding and neglecting terms
`I_t=I_0[1+(gamma_r-2alpha_g)t]`
`I_(100)=1[1+(182xx10^-6-2xx9xx10^-6)100]`
`=1.0164m`
let L= required reading of thread, on glass scale, at temperature t. Then this section of the glass (scale) has length L at `0^@C` and if at `t^@C`
`It=L(1+alphagt)`
or `L=(I_0[1+(gamma_r-2alpha_g)t])/(1+alpha_gt)`
`=I_0[1+(gamma_r-30_g)t]`
or `L=1[1+(182xx10^-6-3xx9xx10^-6)100]`
`=1.0155m`