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

Answer

D

Explanation

When the capillary tube is broken at a height of 6 cm , the height of water column will be only 6 cm .
` :' T = (rhrhog)/(2 cos theta ) " " :. h = ( 2T cos theta )/(r rho g) `
` :. h/(cos theta ) = (2T)/(rrhog) = ` constant
` :. (h_(1))/(cos theta _(1)) =(h_(2))/(cos theta _(2))`
` :. 8/(cos 0^(@)) = 6/(cos theta_(2))`
` :. cos theta_(2) = (6 cos 0^(@))/8 = 6/8 = 3/4`
` :. theta _(2) = cos^(-1) (3/4)`