The lower end of a capillary tube is dipped in water. Water rises to a height of 8 cm. The tube is then broken at a height of 6 cm. The height of water colume and angle of contact will be

Options

A. 6 cm `sin^(-1)""(3)/4`
B. 6 cm `cos^(-1)""(3)/4`
C. 4 cm `sin^(-1)""(1)/2`
D. 4 cm `cos^(-1)""(1)/2`

Answer

B

Explanation

`h=(2sigmacos theta)/(rhogr)impliesh/(costheta)` = constant
`:. 8/(cos0^@)=6/(costheta)`
`implies costheta=3/4`
`theta=cos^(-1)(3/4)`