Question

A piece of granite floats at the interface of mercury and water. If the densities of granite, water and mercury are \( \rho, \rho_1, \rho_2 \) respectively, the ratio of volume of granite in water to that in mercury is 

• A. \( \frac{\rho_2 - \rho}{\rho - \rho_1} \)
• B. \( \frac{\rho_2 + \rho}{\rho_1 + \rho} \)
• C. \( \frac{\rho_1\rho_2}{\rho} \)
• D. \( \frac{\rho_1}{\rho_2} \)

Hint:

Two-fluid flotation: Balance total buoyant force with weight.

Answer 1

The Correct Option is A

Concept: Floating equilibrium: \[ \text{Weight} = \text{Buoyant forces} \] Step 1: {\color{red}Let volumes.} Let volume in water = \( V_1 \), in mercury = \( V_2 \). Step 2: {\color{red}Force balance.} \[ \rho g(V_1 + V_2) = \rho_1 g V_1 + \rho_2 g V_2 \] Step 3: {\color{red}Simplify.} \[ \rho V_1 + \rho V_2 = \rho_1 V_1 + \rho_2 V_2 \] \[ (\rho - \rho_1)V_1 = (\rho_2 - \rho)V_2 \] Step 4: {\color{red}Ratio.} \[ \frac{V_1}{V_2} = \frac{\rho_2 - \rho}{\rho - \rho_1} \]