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Q.

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)

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a

3Tg(2d−ρ)

b

6Tg′(2d−ρ)

c

2Tg(2d−ρ)

d

3Tg(4d−3ρ)

answer is A.

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Detailed Solution

FB=FT=Fg m′g2+2πrT=mg ρ43πr3g2+2πrT=d43πr3g 2πrρ23πr2g2+T=43πr3gd ρr2g3+T=23r2gd⇒T=r2g3[2d−ρ] r=3Tg(2d−ρ)
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