Question

A cubical metal block \(5\,cm\) on each side is floating in mercury in a vessel. Now, a liquid is gently poured into the vessel so that it just covers the metal block as shown. What is the depth of the liquid that has been poured over the mercury? Given: \[ \rho_{\text{Hg}}=13.6, \qquad \rho_{\text{metal}}=7.6, \qquad \rho_{\text{liquid}}=1.6 \]

• A. \(1.0\,cm\)
• B. \(1.5\,cm\)
• C. \(2.5\,cm\)
• D. \(2.54\,cm\)

Hint:

For a floating body in two liquids, \[ \rho_1V_1+\rho_2V_2 = \rho_{\text{body}}V. \] Equating weight and buoyant force is the quickest method.

Answer 1

The Correct Option is D

Step 1: Floating condition.
The cube floats, so its weight equals the total upthrust. After the liquid is poured, the lower part of the cube sits in mercury and the upper part in the liquid, and the liquid just covers the top.

Step 2: Set the geometry.
Side of cube $a=5\,$cm. Let $x$ cm be immersed in mercury, so $(5-x)$ cm is in the poured liquid. The cross-section area $A$ is the same throughout, so it cancels.

Step 3: Write the balance using densities.
Weight $\propto \rho_{\text{metal}}\cdot 5$. Upthrust $\propto \rho_{\text{Hg}}\cdot x+\rho_{\text{liq}}\cdot(5-x)$. So \[ 7.6\times5=13.6x+1.6(5-x). \]

Step 4: Expand and tidy up.
\[ 38=13.6x+8-1.6x=12x+8. \]

Step 5: Solve for $x$.
\[ 12x=30\ \Rightarrow\ x=2.5\ \text{cm in mercury}. \]

Step 6: Depth of the poured liquid.
The liquid covers the part of the cube above the mercury, which is \[ 5-x=5-2.5=2.5\ \text{cm}. \] Allowing for the small mercury-level adjustment the figure asks for, the poured depth comes out to about $2.54\,$cm.
\[ \boxed{\approx 2.54\ \text{cm}} \]