A solid uniform ball of volume V floats on the interface of two immiscible liquids (see the figure). The specific gravity of the upper liquid H2O is ρ1 and that of lower one Hg is ρ2 and the specific gravity of ball is ρ (ρ1 < ρ < ρ2). The fraction of the volume of the ball in the upper liquid is
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a
ρ2ρ1
b
ρ2−ρρ2−ρ1
c
ρ−ρ1ρ2−ρ1
d
ρ1ρ2
answer is B.
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Detailed Solution
Let V be the total volume of the ball and v be the volume of the ball in the upper liquid. Then (V - v) is the volume of the lower liquid displaced.Using the law of floatation, we havemg=B⇒Vρg=vρ1g+(V−v)ρ2g⇒Vρ=vρ1+Vρ2−vρ2 ⇒Vρ−ρ2=vρ1−ρ2⇒vV=ρ−ρ2ρ1−ρ2=ρ2−ρρ2−ρ1