A cylindrical block floats vertically in a liquid of density ρ1 kept in a container such that the fraction of volume of the cylinder inside the liquid is x1. then some amount of another immiscible liquid of density ρ2 (ρ2

Question

A cylindrical block floats vertically in a liquid of density ρ1 kept in a container such that the fraction of volume of the cylinder inside the liquid is x1. then some amount of another immiscible liquid of density ρ2 (ρ2 < ρ1) is added to the liquid in the container so that the cylinder now floats just fully immersed in the liquids with x2 fraction of volume of the cylinder inside the liquid of density ρ1. The ratio ρ1/ρ2 will be (A) (1-x2/x1-x2) (B) (1-x1/x1-x2) (C) (x1-x2/x1+x2) (D) (x2/x1)-1

Tardigrade Admin · Accepted Answer

The given situation is shown as <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/495df22ad044689aacadfbf15375fa7b-.png" /> As the Bouyant force in both the cases are same ∴ ρ1 x1 g=ρ1 x2 g+ρ2(1-x2) g On solving (ρ1/ρ2)=((1-x2)/(x1-x2))

$\frac{1 - x _{2}}{x _{1} - x _{2}}$

$\frac{1 - x _{1}}{x _{1} - x _{2}}$

$\frac{x _{1} - x _{2}}{x _{1} + x _{2}}$

$\frac{x _{2}}{x _{1}} - 1$

The given situation is shown as

As the Bouyant force in both the cases are same
$∴ ρ_{1} x_{1} g = ρ_{1} x_{2} g + ρ_{2} (1 - x_{2}) g$
On solving $\frac{ρ _{1}}{ρ _{2}} = \frac{( 1 - x _{2} )}{( x _{1} - x _{2} )}$

x1​−x2​1−x2​​

x1​−x2​1−x1​​

x1​+x2​x1​−x2​​

x1​x2​​−1

The given situation is shown as As the Bouyant force in both the cases are same ∴ρ1​x1​g=ρ1​x2​g+ρ2​(1−x2​)g On solving ρ2​ρ1​​=(x1​−x2​)(1−x2​)​

$\frac{1 - x _{2}}{x _{1} - x _{2}}$

$\frac{1 - x _{1}}{x _{1} - x _{2}}$

$\frac{x _{1} - x _{2}}{x _{1} + x _{2}}$

$\frac{x _{2}}{x _{1}} - 1$

The given situation is shown as

As the Bouyant force in both the cases are same
$∴ ρ_{1} x_{1} g = ρ_{1} x_{2} g + ρ_{2} (1 - x_{2}) g$
On solving $\frac{ρ _{1}}{ρ _{2}} = \frac{( 1 - x _{2} )}{( x _{1} - x _{2} )}$