A 1-L flask contains some mercury. It is found that at different temperature, the volume of air inside the flask remains the same. What is the volume of mercury in the flask, given that the coefficient of linear expansion of glass`=9xx10^(-6)//^(@)C` and the coefficient of volume expansion of `Hg=1.8xx10^(-4)//^(@)C` ?

Answer

20

Explanation

Volume of beaker at `t_1""^@C=V_1`
Volume of mercury at `t_2 """^@C=V_2`
`therefore` Volume of empty space at `t_1""^@C=V_1-V_2`
On heating to `t_2^@C`, volume of beaker =`V_1{1+gamma_1(t_2-t_1)}`
Volume of mercury at `=V_2[1+gamma_2(t_2-t_1)}`
`therefore` Volume of empty space at `t_2""^@C`
`V_1{1+gamma_1(t_2-t_1)}-V_2{1+gamma_2(t_2-t_1)}=V_1-V_2+(V_1gamma_1-V_2gamma_2)(t_2-t_1)`
Since the volumes of empty space must be the same at all the temperatures,
`V_1-V_2=V_1-V_2+(V_1gamma_1-V_2 gamma_2)(t_2-t_1)`
`therefore V_1 gamma_1-V_2 gamma_2=0`
`therefore V_1gamma_1=V_2 gamma_2 implies 140 xx 3 xx alpha=V_2xx0..000189`
`therefore V_2=(140xx3xx0.00009)/(0.000189)=20 cc`
Volume of mercury in beaker = 20 cc.