A long capillary tube of radius 0.2 mm is placed vertically inside a beaker of water. Water rises to a hight of 7.05 cm in the tube making zero angle os contact. Now if tube is pushed into water so that only 5 cm of its length is above the surface, What is the new angle of contact in degrees between water and glass surface (Take `sqrt(2)=1.41`, surface tension of water 0.07 N/m)

Answer

`45.00`

Explanation

If two points are just above and below the
meniscus then ` rho gh_(0)=(2T)/R`…….i
When `theta=0,R=r` and `h_(0)=h`
`:.rho gh=(2T)/r`……..ii
From equation I and ii
`impliesr/R=(h_(0))/h`
New angle of contact
`theta=cos^(-1)(r/R)=cos^(-1)((h_(0))/h)`
`=cos^(-1)(5/7.05)`
`=cos^(-1)(5/(5xxsqrt(2)))=cos^(-1)(1/(sqrt(2)))`
`:.theta=45^(@)`