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 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
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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.
Updated On: Jun 26, 2026
- \(800\,\text{kg m}^{-3}\)
- \(600\,\text{kg m}^{-3}\)
- \(550\,\text{kg m}^{-3}\)
- \(625\,\text{kg m}^{-3}\)
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The Correct Option is D
Solution and Explanation
Step 1: Use the floating condition in water.
For a floating body, \[ \text{Weight of body}=\text{Buoyant force} \] Let the density of wood be \[ \rho_w \] In water, half of the volume is submerged.
So, \[ \rho_w Vg=\rho_{\text{water}}\left(\frac{V}{2}\right)g \] Cancel \(Vg\): \[ \rho_w=\frac{\rho_{\text{water}}}{2} \] Given, \[ \rho_{\text{water}}=1000\,\text{kg m}^{-3} \] Therefore, \[ \rho_w=500\,\text{kg m}^{-3} \]
Step 2: Use the floating condition in oil.
In oil, submerged volume is \[ 0.8V \] Let the density of oil be \[ \rho_o \] Again, by floating condition: \[ \rho_w Vg=\rho_o(0.8V)g \] Cancel \(Vg\): \[ \rho_w=0.8\rho_o \]
Step 3: Find the density of oil.
Substitute \[ \rho_w=500\,\text{kg m}^{-3} \] \[ 500=0.8\rho_o \] \[ \rho_o=\frac{500}{0.8} \] \[ \rho_o=625\,\text{kg m}^{-3} \]
Step 4: Final conclusion.
Hence, the density of the oil is \[ \boxed{625\,\text{kg m}^{-3}} \]
For a floating body, \[ \text{Weight of body}=\text{Buoyant force} \] Let the density of wood be \[ \rho_w \] In water, half of the volume is submerged.
So, \[ \rho_w Vg=\rho_{\text{water}}\left(\frac{V}{2}\right)g \] Cancel \(Vg\): \[ \rho_w=\frac{\rho_{\text{water}}}{2} \] Given, \[ \rho_{\text{water}}=1000\,\text{kg m}^{-3} \] Therefore, \[ \rho_w=500\,\text{kg m}^{-3} \]
Step 2: Use the floating condition in oil.
In oil, submerged volume is \[ 0.8V \] Let the density of oil be \[ \rho_o \] Again, by floating condition: \[ \rho_w Vg=\rho_o(0.8V)g \] Cancel \(Vg\): \[ \rho_w=0.8\rho_o \]
Step 3: Find the density of oil.
Substitute \[ \rho_w=500\,\text{kg m}^{-3} \] \[ 500=0.8\rho_o \] \[ \rho_o=\frac{500}{0.8} \] \[ \rho_o=625\,\text{kg m}^{-3} \]
Step 4: Final conclusion.
Hence, the density of the oil is \[ \boxed{625\,\text{kg m}^{-3}} \]
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