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

Let rx be the radius of one bore and `r_(2)` e the radius of second bore of the U-tube the if `h_(1)` and `h_(2)` are the heights of water on two sides, then
`h_(1)=(2Scostheta)/(r_(1)rhog) and h_(2)=(2Scostheta)/(r_(2)rhog)`
On substraction, we get
`h_(1)-h_(2)=(2Scostheta)/(r_(1)rhog)-(2Scostheta)/(r_(2)rhog)=(2Scostheta)/(rhog)[(1)/(r_(1))-(1)/(r_(2))]`
here, `S=7.3xx10^(-2)Nm^(-1),theta0,rho=1.0xx10^(3)kgm^(-3)`
`g=9.8ms^(-2),r_(1)=(3)/(2)min=1.5xx10^(-3)m and r_(2)=(6)/(2)mm`
`=3xx10^(-3)m`
`thereforeh_(1)-h_(2)=(2xx7.3xx10^(-2)xxcostheta)/(1xx10^(3)xx9.8)[(1)/(1.5xx10^(-3))-(1)/(3xx10^(-3))]`
`=1.49xx10^(-5)xx(1)/(3xx10^(-3))=4.97xx10^(-3)m=4.97mm`