A wooden block floating in a bucket of water has 4/5 of its volume submerged. When certain amount of an oil is poured into the bucket, it is found that the block is just under the oil surface with half of its volume under water and half in oil. The density of oil relative to that of water is :-

Properties of Matter PYQ — JEE Main 2019 · 9 Apr · Shift 2 · Q50 with Solution

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Correct answer: C

For a floating body, $\text{weight of block} = \text{buoyant force}.$

If the block has volume $V$ and density $\rho_b$ , then initially $\frac{4}{5}$ of its volume is submerged in water.

So, $\rho_b V g = \rho_w \left(\frac{4V}{5}\right) g$

Hence, $\rho_b = \frac{4}{5}\rho_w.$

Now the block is just under the oil surface, with:

So total buoyant force is due to both liquids: $\text{buoyant force} = \rho_w\left(\frac{V}{2}\right)g + \rho_o\left(\frac{V}{2}\right)g$

Since the block is floating, $\rho_b V g = \rho_w\left(\frac{V}{2}\right)g + \rho_o\left(\frac{V}{2}\right)g.$

Cancel $Vg$ : $\rho_b = \frac{\rho_w + \rho_o}{2}.$

Substitute $\rho_b = \frac{4}{5}\rho_w$ : $\frac{4}{5}\rho_w = \frac{\rho_w + \rho_o}{2}.$

Multiply by 2: $\frac{8}{5}\rho_w = \rho_w + \rho_o.$

Therefore, $\rho_o = \frac{8}{5}\rho_w - \rho_w = \frac{3}{5}\rho_w.$

So the relative density of oil with respect to water is $\frac{\rho_o}{\rho_w} = \frac{3}{5} = 0.6.$

Therefore, the correct option is C.