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Q.

Drops of liquid of density d are floating half immersed in a liquid of density p. If the surface tension of liquid is T, then radius of the drop will be

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a

3Tg(2d−ρ)

b

3Tg(d−ρ)

c

6Tg(2d−ρ)

d

6Tg(d−ρ)

answer is A.

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Detailed Solution

The weight of the drop is equal to the sum of force due to surface tension on drop and force of buoyancy. Thus   43πr3dg=2πrT+12×43πr3ρg or  2r2dg=3T+r2ρg r=3Tg(2d−ρ)
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Drops of liquid of density d are floating half immersed in a liquid of density p. If the surface tension of liquid is T, then radius of the drop will be