A barometer contains two uniform capillaries of radii `1.4xx10^(-3)m` and `7.2xx10^(-4)m`. If the height of liquid in narrow tube is `0.2m` more than that in wide tube, calculate the true pressure difference. Density of liqid `=10^(3)kg//m^(3)`, surface tension `=72xx10^(-3)N//m` and `g=9.8ms^(-12)`.

Explanation

Let the presures in wide and narrow limbs of `P_(1)` and `P_(2)`, respectively. If `R_(1)` and `R_(2)` be the radii of meniscus in wide and narrow limb pressure just below hhe meniscus of wide tube `=P_(1)- (2T)/(R_(1))` and pressure just below the meniscus of narrow limb `=P_(2)=(2T)/(R_(2))`.
Therefore, difference of these pressures
`(P_(1)-(2T)/(R_(1)))-(P_(2)-(2T)/(R_(2)))=hrhog`
Therefore, true pressure difference,
`P_(1)-P_(2)=hrhog-2T(1/(R_(2))-1/(R_(1)))`
For the water and glass surface, takingg in the angle of contact `theta` to be zero, we have `R_(1)=(r_(1))/(costheta)~~r_(1)` and `R_(2)=(r_(2))/(costheta)~~r_(2)` where `r_(1)` and `r_(2)` are radii of wide and narrow limbs, respectively.
`:.P_(1)-P_(2)=hrhog-2T(1/(r_(2))-1/(r_(1)))=0.2xx10^(3)xx9.8-2xx72`
`xx10^(-3)xx(1/(7.2xx10^(-4))-1/(1.44xx10^(-3)))`
`=1.96xx10^(3)-0.10xx10^(3)=1.86xx10^(3)N//m^(2)=1860N//m^(2)`