Question

A body of volume \(V\) floats on water with \(\frac{1}{3}\) of its volume above the surface. Find the volume of the object above the surface when floating on a liquid of specific gravity 1.5.

• A. \(\frac{3V}{8}\)
• B. \(\frac{4V}{9}\)
• C. \(\frac{5V}{9}\)
• D. \(\frac{2V}{3}\)

Hint:

For floating bodies, fraction submerged \(V_s/V = \rho_{\text{body}}/\rho_{\text{liquid}}\). Volume above surface = total volume minus submerged volume.

Answer 1

The Correct Option is C

Step 1: Pin down the object's own specific gravity first, using the water data.
When a body floats, the fraction of its volume submerged equals the ratio of its own density to that of the liquid it floats in. Since $\frac{1}{3}$ of the volume is above water, $\frac{2}{3}$ is submerged, so directly:
\[ \text{SG}_{\text{body}} = \frac{2}{3}\times\text{SG}_{\text{water}} = \frac{2}{3}\times1 = \frac{2}{3} \]

Step 2: Apply the same submerged-fraction rule to the new liquid.
\[ \text{fraction submerged} = \frac{\text{SG}_{\text{body}}}{\text{SG}_{\text{liquid}}} = \frac{2/3}{1.5} = \frac{2}{3}\times\frac{2}{3} = \frac{4}{9} \]

Step 3: Subtract from the total volume to get the part above the surface.
\[ V_{\text{above}} = V - \frac{4}{9}V = \frac{5}{9}V \]

Step 4: Sanity check.
The new liquid is denser than water, so it should support the body's weight with less volume submerged, and indeed $\frac{4}{9} < \frac{2}{3}$, confirming the answer moves in the right direction.

Step 5: Conclusion.
\[ \boxed{\frac{5V}{9}} \]