Question

A block of wood of volume \(V\) floats in water with half of its volume submerged. The same block floats in an oil with \(0.8V\) volume submerged. If the density of water is \(1000\,\text{kg m}^{-3}\), then the density of the oil is

• A. \(800\,\text{kg m}^{-3}\)
• B. \(600\,\text{kg m}^{-3}\)
• C. \(550\,\text{kg m}^{-3}\)
• D. \(625\,\text{kg m}^{-3}\)

Hint:

For a floating body, use \(\rho_{\text{body}}Vg=\rho_{\text{liquid}}V_{\text{submerged}}g\). The fraction submerged depends on the ratio of densities.

Answer 1

The Correct Option is D

Step 1: State the law of floatation.
A floating body displaces fluid equal to its own weight. When a body floats: \[ \text{Weight of body} = \text{Weight of fluid displaced} = \rho_{\text{fluid}} \times V_{\text{submerged}} \times g \] This is Archimedes principle applied to floating objects.
Step 2: Apply the floating condition in water to find the density of wood.
Let $\rho_w$ be the density of wood and $V$ be its total volume. Half the volume is submerged in water ($\rho_{\text{water}} = 1000 \, \text{kg m}^{-3}$): \[ \rho_w \cdot V \cdot g = \rho_{\text{water}} \cdot \frac{V}{2} \cdot g \] Cancelling $Vg$ from both sides: \[ \rho_w = \frac{\rho_{\text{water}}}{2} = \frac{1000}{2} = 500 \, \text{kg m}^{-3} \]
Step 3: Understand why only half is submerged.
The fraction submerged equals the ratio of the body's density to the fluid's density: \[ \text{fraction submerged} = \frac{\rho_{\text{body}}}{\rho_{\text{fluid}}} = \frac{500}{1000} = \frac{1}{2} \] A body with half the fluid's density floats with exactly half its volume submerged. This is a general and very useful rule.
Step 4: Apply the floating condition in oil.
In oil, $0.8V$ of the wood's volume is submerged. Let $\rho_o$ be the density of the oil: \[ \rho_w \cdot V \cdot g = \rho_o \cdot 0.8V \cdot g \] Cancelling $Vg$: \[ \rho_w = 0.8 \rho_o \]
Step 5: Solve for the density of oil.
\[ 500 = 0.8 \rho_o \] \[ \rho_o = \frac{500}{0.8} = 625 \, \text{kg m}^{-3} \]
Step 6: Verify using the fraction-submerged rule.
Fraction submerged in oil $= \rho_w / \rho_o = 500/625 = 0.8$, which matches the given condition that $0.8V$ is submerged. The oil is less dense than water (625 vs 1000), so a greater fraction of the wood must submerge in oil to displace enough weight, which is physically consistent. \[ \boxed{\rho_{\text{oil}} = 625 \, \text{kg m}^{-3}} \]