A liquid drop of density `rho` is floating half immersed in a liquid of surface tension `S ` and density `rho/2`. If the surface tension `S` of the liquid is numerically equal to 10 times of acceleration due to gravity, then the diameter of the drop is :

Options

A. `sqrt( 20/rho)`
B. `sqrt( 80/rho)`
C. `sqrt( 60/rho)`
D. `sqrt( 40/rho)`

Answer

B

Explanation

A figure given here, shows a liquid drop half immersed in a liquid, where density of drop, `rho_(d) = rho, ` density of liquid, `rho_(L) = rho/2`

and surface tension, `T = 10 g`. Force acting on the drop due to surface tension,
`F = Tl = 10 g (piD)" "...(i)`
The observed weight of the drop,
`w= w_(d) - w_(L) = rho V g - rho/2 V/2 g = 3/4 rho V g" " ...(ii)`
where, V = volume of the drop `=4/3 pir^(3)`
Since, the ball is in equilibrium.
`therefore F = w`
`10g pi D = 3/4 rho g frac(pi D^(3))(6)`
`D = sqrt(80/rho ) m ` [ From Eqs. (i) and (ii) ]
Hence, the correct option is (b).