A rectangular tube of uniform cross section has three liquids of densities `rho_(1), rho_(2)` and `rho_(3)`. Each liquid column has length `l` equal to length of sides of the equilateral triangle. Find the length `x` of the liquid of density `rho_(1)` in the horizontal limb of the tube, if the triangular tube is kept in the vertical plane.

Options

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

Answer

B

Explanation

Ascent formula for capillary tube,
`h=(2Tcostheta)/(rhogr)`
`therefore (costheta_(1))/(rho_(1)) = (costheta_(2))/(rho_(2))=(costheta_(3))/(rho_(3))`
Thus, `costheta propto rho`
`therefore costheta_(1) gt costheta_(2) gt costheta_(3)`
`0lt=theta_(1) lt theta_(2) lt theta_(3) lt pi/2`