A cylindrical block floats vertically in a li | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $V$ be the total volume of the cylindrical block and $\rho$ be its density.In the first case,the block floats in a liquid of density $\rho_{1}

Vedclass · Accepted Answer

$\frac{1-x_{2}}{x_{1}-x_{2}}$

Vedclass · Answer

(A) Let $V$ be the total volume of the cylindrical block and $\rho$ be its density.In the first case,the block floats in a liquid of density $\rho_{1}$. According to the law of floatation,the weight of the block equals the buoyant force:$V \rho g = V x_{1} \rho_{1} g$$\Rightarrow \rho = x_{1} \rho_{1} \quad ... (1)$In the second case,the block is fully immersed in two immiscible liquids of densities $\rho_{1}$ and $\rho_{2}$. The volume fraction $x_{2}$ is in liquid $\rho_{1}$ and the remaining fraction $(1 - x_{2})$ is in liquid $\rho_{2}$. The total buoyant force equals the weight of the block:$V \rho g = V x_{2} \rho_{1} g + V (1 - x_{2}) \rho_{2} g$$\Rightarrow \rho = x_{2} \rho_{1} + (1 - x_{2}) \rho_{2} \quad ... (2)$Equating $(1)$ and $(2)$:$x_{1} \rho_{1} = x_{2} \rho_{1} + (1 - x_{2}) \rho_{2}$$x_{1} \rho_{1} - x_{2} \rho_{1} = (1 - x_{2}) \rho_{2}$$\rho_{1} (x_{1} - x_{2}) = \rho_{2} (1 - x_{2})$$\frac{\rho_{1}}{\rho_{2}} = \frac{1 - x_{2}}{x_{1} - x_{2}}$

Similar Questions

$A$ container of liquid is in free fall. Without any liquid spilling out, does this container obey the principle of Archimedes?

$A$ boat carrying steel balls is floating on the surface of water in a tank. If the balls are thrown into the tank one by one,how will it affect the level of water?

The density of ice is $0.9 \, g/cm^3$. What percentage by volume of the block of ice floats outside the water is ..........$\%$

At what depth does a body remain in equilibrium in a liquid? Explain.

Vedclass Test Series

Exam Paper Generator

Online Exam Module