A glass capillary of length l and inside radius `r(r lt lt l)` is submerged vertically into water. The upper end of the capillary is scaled. The atmospheric pressure is `p_(0)`. To what length h has the capillary to be submerged to make the water levels inside and outside the capillary coincide. Assume that temperature of air in the capillary remains constant. (given, surface tension of water = T, angle of contact between glass water interface `= 0^(@)`)

Options

A. `(l)/((1 + (P_0 r)/(4 gamma)))`
B. `l (1 - (p_0 r)/(4 gamma) )`
C. `l (1 - (p_0 r)/(2 gamma) ) `
D. `(l)/((1 + (p_0 r)/(2 gamma) ))`

Answer

D

Explanation

Let p. be the pressure of air inside the sealed capillary tube
here `p. = (p_0 + (2 gamma)/(r ))`
Using boyle.s law `p_0 V_0 = p.V.`
` therefore p_0 (lA) = p. (l-x)A rArr p. = (p_0 l)/(l-x)` ….(i)
Where A is the cross sectional area of capillary tube.
Now since level of water inside capillary coincides with outside ,
` therefore p. -p_0 = (2 gamma)/(r )`
` therefore (p_0 l)/((l-x)) -p_0 = (2 gamma)/(r ) ` (from eqn (i))
`p_0 ( (l)/(l-x) - 1) = (2 gamma)/(r ) rArr p_0 x (2 gamma)/( r) (l-x)`
` rArr x = (1 + ( (l)/(p_0 x) )/(2 gamma)) `