The barometric height at a certain place is 0.75 m of mercury. What will be the barometric height ifa liquid of density `3.4xx 10^3kg/m^3` is used to llll the barometric tube?
`[P_("mercury")=13.6xx10^3kg//m^3`]

Explanation

Data : `T = 0.472 N//m, theta = 148^(@), rho = 13.6 xx 10^(3) (kg)/m^(3) ,` barometric height
`(H) = 0.76 m , g = 9.8 m//s^(2),|h| lt 0.5%" of "H`
` cos 148^(@) = cos (180^(@) - 32^(@)) = - cos 32^(@) = - 0.8480`
For mercury , there will be capillary depression given by
` h = (2 T cos theta)/(R rho g) `
` :. ` A correction `|h|` due to capillarity must be added to the barometric height.
For `|h_("max")| = 0.005` H,
` R_("min") = (2T|cos theta|)/(rho g |h_("max")|)`
`= (2(0.472)| - 0.8480|)/((13.6 xx 10^(3))(9.8)(0.005 xx 0.76))`
` = (0.944 xx 0.8480 xx 10^(-3))/(13.6 xx 9.8 xx 3.80 xx 10^(-3))`
` = 1.581 xx 10 ^(-3) m`
` = 1.581` mm
This gives the minimum value of R .
`{:(log 13.6,," "1.1335),(log 9.8,," "0.9912),(log 3.8,,ul(+0.5798)),(,,ul(" "2.7045)):}`
`{:(log 0.944,," "bar(1).9750),(log 0.8480,,ul(+bar(1).9284)),(,," "bar(1).9034),(,,ul(-2.7045)),(,,ul(" "bar(3).1989)):}`
AL ` bar(3) .1989 = 1.581 xx 10^(-3)`