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AIIMS 2017
Exam
AIIMS
Year
2017
Subject
Physics
Chapter
Mechanical Properties of Fluids

AIIMS 2017 · Physics · Mechanical Properties of Fluids

A solid sphere of volume V and density floats at the interface of...

A solid sphere of volume VV and density ρ\rho floats at the interface of two immiscible liquids of densities ρ1\rho_1 and ρ2\rho_2 respectively. If ρ1<ρ<ρ2\rho_1 < \rho < \rho_2, then the ratio of volume of the parts of the sphere in upper and lower liquid is

Options

  1. Aρ−ρ1ρ2−ρ\frac{\rho-\rho_1}{\rho_2-\rho}
  2. Bρ2−ρρ−ρ1\frac{\rho_2-\rho}{\rho-\rho_1}
  3. Cρ+ρ1ρ+ρ2\frac{\rho+\rho_1}{\rho+\rho_2}
  4. Dρ+ρ2ρ+ρ1\frac{\rho+\rho_2}{\rho+\rho_1}

Correct answer

B. ρ2−ρρ−ρ1\frac{\rho_2-\rho}{\rho-\rho_1}

Step-by-step solution

V=V= Volume of solid sphere. Let V1=V_1= Volume of the part of the sphere immersed in a liquid of density ρ1andV2=\rho_1 and V_2= Volume of the part of the sphere immersed in liquid of density ρ2\rho_2. According to law of floatation, Hence from eqns (i) and (ii), V1ρg+V2ρg=V1ρ1g+V2ρ2g or V1(ρ−ρ1)g=V2(ρ2−ρ)g or V1V2=ρ2−ρρ−ρ1\begin{aligned} & V_1 \rho g+V_2 \rho g=V_1 \rho_1 g+V_2 \rho_2 g \\ & \text{ or } V_1\left(\rho-\rho_1\right) g=V_2\left(\rho_2-\rho\right) g \\ & \text{ or } \frac{V_1}{V_2}=\frac{\rho_2-\rho}{\rho-\rho_1}\end{aligned}
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