A wooden block floating in a bucket of water | vedclass

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(C) In the first situation,the block is floating in water. By the law of floatation,the weight of the block equals the weight of the displaced water:$

Vedclass · Accepted Answer

$0.6$

Vedclass · Answer

(C) In the first situation,the block is floating in water. By the law of floatation,the weight of the block equals the weight of the displaced water:$V_b \rho_b g = V_s \rho_w g$$\frac{V_s}{V_b} = \frac{\rho_b}{\rho_w} = \frac{4}{5} \quad ... (i)$Here,$V_b$ is the volume of the block,$V_s$ is the submerged volume,$\rho_b$ is the density of the block,and $\rho_w$ is the density of water.In the second situation,the block is floating in a mixture of oil and water. The total weight of the block is balanced by the buoyant force from both the oil and the water:$V_b \rho_b g = \left(\frac{V_b}{2}\right) \rho_o g + \left(\frac{V_b}{2}\right) \rho_w g$Dividing by $V_b g$,we get:$\rho_b = \frac{\rho_o}{2} + \frac{\rho_w}{2}$$2 \rho_b = \rho_o + \rho_w$Substitute $\rho_b = \frac{4}{5} \rho_w$ from equation $(i)$:$2 \left(\frac{4}{5} \rho_w\right) = \rho_o + \rho_w$$\frac{8}{5} \rho_w - \rho_w = \rho_o$$\rho_o = \frac{3}{5} \rho_w = 0.6 \rho_w$Therefore,the relative density of oil with respect to water is $\frac{\rho_o}{\rho_w} = 0.6$.

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