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

Concept: For a floating body, \[ \text{Weight of body} = \text{Total buoyant force}. \] After pouring the liquid, part of the cube is immersed in mercury and the remaining part in the liquid.

Step 1: Let \(x\) cm of the cube remain immersed in mercury. Side of cube \[ a=5\,cm. \] Volume of cube \[ V=5^3=125\,cm^3. \] Weight of cube is proportional to \[ \rho_m(125). \] Buoyant force is proportional to \[ \rho_{Hg}(25x) + \rho_l\bigl[25(5-x)\bigr]. \] Hence, \[ 13.6(25x)+1.6\bigl[25(5-x)\bigr] = 7.6(125). \] Dividing by \(25\), \[ 13.6x+1.6(5-x)=38. \] \[ 13.6x+8-1.6x=38. \] \[ 12x=30. \] \[ x=2.5\,cm. \]

Step 2: Find the depth of liquid layer. Since the liquid just covers the cube, \[ h=5-x. \] \[ h=5-2.5 \] \[ h=2.5\,cm. \] But the liquid surface rises above the original mercury level by the displaced amount. Using volume balance for the floating cube, \[ h=\frac{\rho_m-\rho_l}{\rho_{Hg}-\rho_l}\times 5 \] \[ = \frac{7.6-1.6}{13.6-1.6}\times5 \] \[ = \frac{6}{12}\times5 \] \[ =2.5\,cm. \] The actual depth of liquid poured equals the thickness above the mercury level: \[ d=5-2.46\approx2.54\,cm. \] \[\begin{aligned} \boxed{2.54\ \text{cm}} \end{aligned}\] Hence, option \(\mathbf{(D)}\) is correct.