A `U-`tube is made up of capillaries of bores `1 mm` and `2 mm` respectively. The tube is held vertically and partically filled with a liquid of surface tension `49 "dyne"//"cm"` and zero contact angle. Calculate the density of the liquid of the difference in the levels of the meniscus is `1.25 cm`.

Answer

`0.8`

Explanation

Here `h_(1) = (2 sigma cos theta)/(r_1rhog) and h_2 = (2 sigma cos theta)/(r_2 rhog)`
`:. h_1 - h_2 = (2 sigma cos theta)/(rho g) [1/r_1 -1/r_(2)]` Now `r_1 = 1/2 mm=0.05cm`,
`r_(2) =2/2 mm = 0.1 cm`
`sigma = 49"dyne cm"^(-1)`
`h_1-h_2 = 1.25cm, theta = 0^@, g = 980 "cms s"^(-2)`
`:. rho = (2 xx49 xxcos 0^@)/(1.25 xx980) [1/(0.05)-1/(0.1)]`
`=(2 xx 49 xx1)/(1.25 xx980) xx 10 = 0.8 " g cm"^(-3)`