Lower end of a capillary tube of radius `10^(-3)` m is dipped vertically into a liquid. Surface tension of liquid is 0.5 N/m and specific gravity of liquid is 5. Contact angle between liquid and material of capillary tube is `120^(@)`. Choose the correct options (use g = 10m/`s^(2)`)

Options

A. Maximum possible depression of liquid column in the capillary tube is 1 cm.
B. Maximum possible depression of mercury column in the capillary tube is 2 cm.
C. If the length of the capillary tube dipped inside mercury is half of the maximum possible depression of mercury column in the capilary tube, angle made by the mercury surface at the end of the capillary tube with the verticle, is `cos^(-1)(-1/4)`.
D. If the length of the capillary tube dipped inside mercury is one third of the maximum possible depression of mercury column in the capilary tube, angle made by the mercury surface at the end of the capillary tube with the vertical, is `cos^(-1)(-1/6)`

Explanation

`S=0.5 N//m=r =10^(-3) m theta_(c)=120^(@) rho=5xx10^(3) kg//m^(3)`
`h_(max)=(2Scos theta_(c))/(r rho g) =((2)(1/2)(-1/2))/((10^(-3))(5xx10^(-3))(10))=10^(-2) m =1 cm `
If `h=(h_(max))/2`
`(2S cos theta)/(r rhog)=1/2(2Scos theta_(c))/(r rhog)`
`rArr cos theta=-1/4`
`theta=cos^(-1)(-1/4)`
If `h=(h_(max))/3`
`(2S cos theta)/(rrhog)=1/3 (2S cos theta_(c))/(r rhog)`
`rArr cos theta=-1/6, theta=cos^(-1)(-1/6)`