A glass tube of 1mm diameter is dipped vertically into a tube of mercury, with its lower end 3 cmm below the mercury surface. What must be the gauge pressure of air in the tube to blow a hemispherical bubble at its lower and ? Given, density of mercury `= 13.6 g//c c` and surface tension of mercury `= 0.540 Nm^(-1). g =10 ms^(-2).`

Answer

4.59 cm of Hg

Explanation

Here radius of glass tube, `r = (1)/(2)mm = 0.5 xx10^(-3)m, h =3 cm = 3xx10^(-2) m,`
`rho = 13.6 g//c c = 13.6 xx 10^3 kg//m^3, S = 0.540 Nm^(-1),` Excess of pressure inside the air bubble blown at the lower end of the tube is `P = (2S)/(r ) = (2xx 0.540)/(0.5 xx 10^(-3)) = 2160 Nm^(-2)` Gauge pressure of air inside the air bubble ` =p + h rho g`
`=2160 + (3xx10^(-2))xx(13.6xx10^3)xx10`
`=(2160)/((13.6 xx 10^3)xx10) + (3xx10^(-2)xx(13.6 xx 10^3)xx10)/((13.6 xx 10^3)xx10)`
= 0.0159 + 0.03 = 0.0459m of Hg = 4.59 cm of Hg