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

`r_(1)=3/2 =1.5 mm=1.5 xx 10^(-3)m`
`r_(2)=6/2 =3mm =3 xx 10^(-3)m`
`p=10^(3) kg//m^(3), theta=0, S=7.3 xx 10^(-2)N m^(-1)`
Let `h_(1) and h_(2)` be the heights to which water will rise in the two tubes. `h_(1)=(2 S cos theta)/(r_(1)pg), h_(2)=(2S cos theta)/(r_(2)pg)`
`h_(1)-h_(2)=(2S cos theta)/(pg) (1/r_(1)-1/r_(2))`
`=(2 xx 7.3 xx 10^(-2) xx cos 0^(@))/(10^(3) xx 9.8) [(1)/(1.5 xx 10^(-3)) -1/(3 xx 10^(-3))]`
On solving we get `h_(1)-h_(2)=5.01m`