The radii of the two columne is U-tube are `r_(1)` and `r_(2)(gtr_(1))`. When a liquid of density `rho` (angle of contact is `0^@))` is filled in it, the level different of liquid in two arms is h. The surface tension of liquid is
`(g=` acceleration due to gravity)

Options

A. `(rhoghr_(1)r_(2))/(2(r_(2)-r_(1)))`
B. `(rhogh(r_(1)-r_(2)))/(2r_(1)r_(2))`
C. `(2(r_(2)-r_(1)))/(rhoghr_(1)r_(2))`
D. `(rhogh)/(2(r_(2)-r_(1)))`

Answer

A

Explanation

Let `h_(1), h_(2)` be the heights to which liquid rises in two columns of radii `r_(1)` and `r_(2)` respectively. Then
`h_(1)=(2ScosO^(@))/(r_(1)rhog)=(2S)/(r_(1)rhog)`
Where s is the surface tension of liquid.
and `h_(2)=(2Scoso^(@))/(r_(2)rhog)=(2S)/(r_(2)rhog)`
`therefore` Difference in levels of liquid in two arms of U tube is
`h=h_(1)-h_(2)=(2S)/(rhog)[(1)/(r_(2))-(1)/(r_(2))]=2S((r_(2)-r_(1)))/(r_(1)r_(2)rhog)`
`S=(r_(1)r_(2)rhogh)/(2(r_(2)-r_(1)))`