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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?
640311_f5439d1a40e347a5a2308827f5a749a9.png


A
1024 kg/m3
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B
1276 kg/m3
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C
776 kg/m3
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D
954 kg/m3
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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 block
and $$V_{0} = \dfrac {V}{6}$$ 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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