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. `(rho g hr_(1) r_(2))/(2(r_(2) -r_(1)))`
B. `(rho gh (r_(2)-r_(1)))/(2r_(1)r_(2))`
C. `(2 (r_(2) - r_(1)))/(rho gh r_(1) r_(2))`
D. `(rho gh)/(2(r_(2)-r_(1)))`

Answer

A

Explanation

Here, `h_(1) = (2S cos 0^(@))/(r_(1) rho g) and h_(2) = (2S cos 0^(@))/(r_(2) rho g)`
`:. h_(1) -h_(2) = (2S cos 0^(@))/(rho g) [(1)/(r_(1)) - (1)/(r_(2))]= (2S(r_(2) - r_(1)))/(r_(1) r_(2) rho g)`
But `h_(1) - h_(2) = h` (given)
`:. h= ((2S (r_(2) - r_(1))))/(r_(1) r_(2)rho g) or S = (r_(1) r_(2) h rho g)/(2(r_(2) -r_(1)))`