A vessel contains oil (density $= 0.8\text{ gm}/\text{cm}^3$) over mercury (density $= 13.6\text{ gm}/\text{cm}^3$). A homogeneous sphere floats with half of its volume immersed in mercury and the other half in oil. The density of the material of the sphere in $\text{gm}/\text{cm}^3$ is

Question

A piece of granite floats at the interface of mercury and water contained in a beaker as shown in the figure. If the densities of granite, water, and mercury are $ \rho $, $ \rho_1 $, and $ \rho_2 $ respectively, the ratio of the volume of granite in water to the volume of granite in mercury is:

Answer 1

Step 1: Compare per unit volume:
Think per unit volume of the sphere. Its weight per unit volume is $\rho g$.

Step 2: Upthrust per unit volume:
Half the volume feels upthrust from mercury and half from oil, so the average upthrust per unit volume is $\frac12(13.6+0.8)g = 7.2g$ (densities in g/cm$^3$).

Step 3: Equilibrium:
Floating means weight equals upthrust, so $\rho g = 7.2g$ and $\rho = 7.2$ g/cm$^3$.

Step 4: Answer:
This is option (B).

Final Answer:
Option (B). \[ \boxed{7.2\ \text{g/cm}^3} \]

A vessel contains oil (density \(= 0.8\text{ gm}/\text{cm}^3\)) over mercury (density \(= 13.6\text{ gm}/\text{cm}^3\)). A homogeneous sphere floats with half of its volume immersed in mercury and the other half in oil. The density of the material of the sphere in \(\text{gm}/\text{cm}^3\) is

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The Correct Option is B

Solution and Explanation

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