A U-tube is supported with its limbs vertical andis partly filled with water. If internal diameters of the limbs are 1xx10^(-2)m and 1xx10^(-4)m respectively, wht will be the difference in heights of water columns in the two limbs (Surface tension of water is 0.07 Nm^(-1))

A U-tube is supported with its limbs vertical andis partly filled with

1.	A U-tube is supported with its limbs vertical andis partly filled with water. If internal diameters of the limbs are 1xx10^(-2)m and 1xx10^(-4)m respectively, wht will be the difference in heights of water columns in the two limbs (Surface tension of water is 0.07 Nm^(-1))
Answer» Solution :Surface tension, `S=0.07Nm^(-1)` Density, `rho=1000k gm^(-3),g=9.8ms^(-2)` ANGLE of CONTACT `theta=0^(@)`, RADIUS, `r_(1)=0.5xx10^(-2)m`, Radius, `r_(2)=0.5xx10^(-4)m` Let `h_(1)` be the height of water in the limb of radius `r_(1)`. Then, `h_(1)=(2Scostheta)/(r_(1)rhog)=(2xx0.07xxcos0^(@))/(0.5xx10^(-2)xx1000xx9.8)m` `=2.86xx10^(-3)m` similarly `h_(2)=2.86xx10^(-1)m`. Difference in HEIGHTS `=h_(2)-h_(1)=2.86xx10^(-1)m-2.86xx10^(-3)m` `=(0.286-0.00286)m=0.283m`