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 liquid `=10^(3)kg//m^(3)`, surface tension `=72xx10^(-3)N//m` and `g=9.8ms^(-12)`.

Explanation

Let `R_(1)` and `R_(2)` be the radii of curvature of the liquid meniscus in wide and narrow limbs of barometer. Let `P_(1)` and `P_(2)` be the pressure in wide and narrow limbs respectively. Let `r_(1), r_(2)` be the radii of wide and narrow limbs of barometer respectively and S be the surface tension of liquid (i.e., water). For water-glass surface, taking angle of contact `theta` to be zero, we have
`R_(1) = (r_1)/(cos theta) ~= r_(1) and R_(2) = (r_2)/(cos theta) ~= r_(2)`
Pressure just below the liquid meniscus of wide limb `=P_(1) -(2S)/(R_1)`
Pressure just below the liquid meniscus of narrow limb = `P_(2)-(2S)/(R_2)`
Differnece of thes pressure:
`(P_(1) -(2S)/(R_1))-(P_(2)-(2S)/(R_2)) = h rho g`
or `P_(1)-P_(2) = h rho g - 2S ((1)/(R_2) -(1)/(R_1))`
`h rho g - 2S ((1)/(r_2) -(1)/(r_1))`
`= 0.2 xx 10^(3) xx 9.8 - 2 xx 972 xx 10^(-3)`
`xx[(1)/(7.2 xx 10^(-4))- (1)/(1.44 xx 10^(-3))]`
`=1860 Pa`.