  Question

A layer of oil with density $$724\ kg/m^{3}$$ floats on water of density $$1000\ kg/m^{2}$$. A block floats at the oil-water interface with $$1/6$$ of its volume in oil and $$5/6$$ of its volume in water, as shown in the figure. What is the density of the block? A
1024 kg/m3  B
1276 kg/m3  C
776 kg/m3  D
954 kg/m3  Solution

The correct option is C $$954\ kg/m^{3}$$Given,Density of oil $$(\rho_{1}) = 724\ kg/m^{3}$$Density of water $$(\rho_{2}) = 1000\ kg/m^{3}$$According to Archemedes' principle,Upthrust $$=$$ Weight of the liquid displaced$$V_{\rho_{b}g} = V_{0\rho_{1}g} + V_{w\rho_{2}g}$$($$V_{0}$$ and $$V_{w}$$ are volume of oil and water displaced, respectively)$$V_{\rho_{b}g} = \dfrac {V}{6} (724)g + \dfrac {5V}{6}(1000)g$$where, $$\rho_{b} =$$ density of blockand $$V_{0} = \dfrac {V}{6}$$ and $$V_{w} = \dfrac {5V}{6}$$$$\Rightarrow V_{\rho_{b}g} = \left (\dfrac {724V}{6} + \dfrac {5000V}{6}\right )g$$$$\Rightarrow V_{\rho_{b}g} = \left (\dfrac {724V + 5000V}{6}\right )g$$$$\Rightarrow V_{\rho_{b}g} = \dfrac {5724}{6}Vg$$$$\rho_{b} = 954\ kg/m^{3}$$Physics

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