Two narrow bores of diameters 3.0mm and 6.0 mm are joined together to form a U-shaped tube open at both ends. If th U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is `7.3xx10^(-2)Nm^(-1)`. Take the angle of contact to be zero. and density of water to be `1.0xx10^(3)kg//m^(3)`.
`(g=9.8 ms^(-2))`

Explanation

Here , `S = 7.3 xx 10^(-2) Nm^(-1),rho = 1.0 xx 10^(3), theta = 0^@)`
For narrow tube, `2r_(1) = 3.00mm = 3xx10^(-3)m or r_(1)=1.5xx10^(-3)m`
For wider tube ,`2r_(2) = 6.00 mm = 6xx10^(-3)m` or `r_(2) = 3 xx10^(-3)m`
Let `h_(1),h_(2)` be the heights to which water rises in narrow tube and wider tube respectively.
Then, `h_(1) = (2S cos theta)/(r_(1)rho g)` and `h_(2) = (2S cos theta)/(r_(2)rho g)`
`:.` Differnece in levels of water in two limbs of U tube is , `h_(1)-h_(2) = (2S cos theta)/(rho g) [(1)/(r_1)-(1)/(r_2)]`
`=(2 xx 7.3 xx 10^(-2)xxcos 0^(@))/(10^(3)xx9.8) xx [(1)/(1.5xx10^(-3))-(1)/(3xx10^(-3)) = 4.97 xx 10^(-3)m]`.