Three liquids of densities `rho_(1), rho_(2) " and " rho_(3)` (with `rho_(1) gt rho_(2) gt rho_(3))`, having the same value of surface tension T, rise to the same height in three identical capillaries. The angles of contact `theta_(1), theta_(2) " and " theta_(3)` obey

Options

A. `(pi)/(2) lt theta_(1) lt theta_(2) lt theta_(3) lt pi`
B. `pi gt theta_(1) gt theta_(2) gt theta_(3) gt (pi)/(2)`
C. `(pi)/(2) gt theta_(1) gt theta_(2) gt theta_(3) ge 0`
D. `0 le theta_(1) lt theta_(2) lt theta_(3) lt (pi)/(2)`

Answer

D

Explanation

`h= (2T cos theta)/(rho gr)`
As r, h, T are same,`(cos theta)/(rho)`= constant
`rArr (cos theta_(1))/(rho_(1))= (cos theta_(2))/(rho_(2))= (cos theta_(3))/(rho_(3))`
As `rho_(1) gt rho_(2) gt rho_(3)`
`rArr cos theta_(1) gt cos theta_(2) gt cos theta_(3) rArr theta_(1) lt theta_(2) lt theta_(3)`
As water rises so `theta` must be acute
So, `0 le theta_(1) lt theta_(2) lt theta_(3) lt pi//2`