A solid sphere of volume V and density ρ floats at the interface of two immiscible liquids of densities ρ 1 and ρ 2 respectively. If ρ 1

Question

A solid sphere of volume V and density ρ floats at the interface of two immiscible liquids of densities ρ 1 and ρ 2 respectively. If ρ 1<ρ <ρ 2, then the ratio of volume of the parts of the sphere in upper and lower liquids is: (A)  (ρ -ρ 2/ρ 2-ρ )  (B)  (ρ 2-ρ /ρ -ρ 1)  (C)  (ρ +ρ 1/ρ +ρ 2)  (D)  (ρ +ρ 2/ρ +ρ 1)

Tardigrade Admin · Accepted Answer

Let V1 and V2 be the volumes, then V1+V2=V As ball is floating. Weight of ball = upthrust on ball due to two liquids Vρ g=V1ρ 1g+V2ρ 2g ⇒ Vρ =V1ρ 1+(V-V1)ρ 2 ⇒ V1=( (ρ -ρ 2/ρ 1-ρ 2) )V Fraction in upper part =(V1/V)=(ρ -ρ 2/ρ 1-ρ 2) Fraction in lower part =1-(V1/V)=1-(ρ -ρ 2/ρ 1-ρ 2) =(ρ 1-ρ /ρ 1-ρ 2) ∴ Ratio of lower and upper parts =(ρ -ρ 2/ρ 1-ρ )

Q. A solid sphere of volume $V$ and density $ρ$ floats at the interface of two immiscible liquids of densities $ρ_{1}$ and $ρ_{2}$ respectively. If $ρ_{1} < ρ < ρ_{2},$ then the ratio of volume of the parts of the sphere in upper and lower liquids is:

$\frac{ρ - ρ _{2}}{ρ _{2} - ρ}$

$\frac{ρ _{2} - ρ}{ρ - ρ _{1}}$

$\frac{ρ + ρ _{1}}{ρ + ρ _{2}}$

$\frac{ρ + ρ _{2}}{ρ + ρ _{1}}$

$\frac{ρ _{1} ρ _{2}}{ρ}$

Q. A solid sphere of volume V and density ρ floats at the interface of two immiscible liquids of densities ρ1​ and ρ2​ respectively. If ρ1​<ρ<ρ2​, then the ratio of volume of the parts of the sphere in upper and lower liquids is:

ρ2​−ρρ−ρ2​​

ρ−ρ1​ρ2​−ρ​

ρ+ρ2​ρ+ρ1​​

ρ+ρ1​ρ+ρ2​​