Two narrow bores of diameters 3.0 mm and 6.0mm are joined together to form a U- tube open at both ends . If the 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)kgm^(-3)(g=9.8ms^(-2))`.

Explanation

Here surface tension of water
`T=7.3xx10^(-2)Nm^(-1)`
Density of water `rho=1xx10^(3)kgm^(-3)`
angle of contact `theta=0^(@)`
Gravitation `g=9.8ms^(-1)`
Radius of first bore ,
`r_(1)=(D_(1))/(2)=(3.0)/(2)=1.5mm`
`=1.5pi16^(-3)m`
Radius of second bore , `r_(2)=(D_(2))/(2)=(6.0)/(2)=3.0mm`
`=3.0-10^(-3)m`
Suppose height of water in first tube is `h_(1)and` in second tube is `h_(2)`.
but `h_(1)=(2Tcostheta)/(r_(1)rhog)andh_(2)=(2Tcostheta)/(r_(2)rhog)`
Here `r_(2)gtr_(1)thereforeh_(1)gth_(2)`
`therefore` Height difference, `h_(1)-h_(2)=(2Tcostheta)/(r_(1)rhog)-(2Tcostheta)/(r_(2)rhog)`
`=(2Tcostheta)/(rhog)[(1)/(r_(1))-(1)/(r_(2))]`
`=(2xx7.3xx10^(-2)xxcos0^(@))/(1xx10^(3)xx9.8)`
`[(1)/(1.5xx10^(-3))-(1)/(3.0xx10^(-3))]`
`=(14.6xx10^(-2))/(9.8xx10^(3))[(1.5xx10^(3))/(3xx1.5)]`
`=(14.6xx1.5xx10)/(9.8xx3xx1.5xx10^(3))`
`=0.49659xx10^(-2)m`
`=4.96xx10^(-3)=5.0mm`