A capillary tube of radius 1 mm is kept vertical with the lower end in water. (a) Find the height of water raised in the capillary. (b) If the length of the capillary tube is half the answer of part (a), find the angle `theta` made by the water surface in the capillary with the wall.

Answer

A, B, C

Explanation

`a. r=1mm=10^-3m`
`h=(2Tcostheta)/(rrhog)`
`=(2xx(0.076))/(10^-3xx10xx100)`
`=1.52cm`
`=1.52xx10^-2m`
b. `h'=(2Tcostheta)/(r rhog)`
`costheta=(h'r rhog)/(2T)`
`costheta=((1.52xx10^2)xx(10^-3)xx(10)xx(10))/(2xx0.076)`
`[because h'=(h/2)=(1.52xx10^-2)/2]`
`=1/2`
`So, theta=cos^-1(1/2)=60^@`