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Drops of liquid of density d are floating half immersed in a l

Question

Drops of liquid of density $d$ are floating half immersed in a liquid of density $ρ$ . If the surface tension of liquid is $T$ then the radius of the drop will be $(d =$ density of liquid drop).

\frac{3 T}{g ( 2 d - ρ )}

\frac{6 T}{g ( 2 d - ρ )}

\frac{2 T}{g ( 2 d - ρ )}

\frac{3 T}{g ( 4 d - 3 ρ )}

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Samuel Discussed

Drops of liquid of density

d

are floating half immersed in a liquid of density

ρ

. If the surface tension of liquid is

T

then the radius of the drop will be

(d =

density of liquid drop).

11 mins ago

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11 mins ago

Text solutionVerified

Force due to surface tension

= T \times 2 π r

T \times 2 π r + \frac{2}{3} π r^{3} \times ρ g = \frac{4}{3} π r^{3} \times d g 2 T + \frac{2}{3} r^{2} ρ g = \frac{4}{3} r^{2} d g r^{2} (\frac{4}{3} d g - \frac{2}{3} ρ g) = 2 T r = \frac{2 T}{\frac{2}{3} g ( 2 d - ρ )} r = \frac{3 T}{g ( 2 d - ρ )}

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Question Text	Drops of liquid of density $d$ are floating half immersed in a liquid of density $ρ$ . If the surface tension of liquid is $T$ then the radius of the drop will be $(d =$ density of liquid drop).
Answer Type	Text solution:1
Upvotes	150