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Question

# Find the density of a metallic body which floats at the interface of mercury of specific gravity 13.6 and water such that 31.75% of its volume is submerged in mercury and 68.25% in water.

A
6000 kg/m3
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B
5000 kg/m3
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C
4000 kg/m3
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D
7000 kg/m3
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Solution

## The correct option is B 5000 kg/m3Given Specific gravity of mercury, sm=13.6 Volume submerged in the mercury =31.75% Volume submerged in water =68.25% Let the volume of the body =V m3 Volume of the body submerged in mercury =31.75100V =0.3175V m3 and volume of body submerged in water = 68.25100V =0.6825V m3 As we know, Density of liquid = specific gravity × ρw where ρw= density of water So, density of mercury (ρm) =13.6×1000=13600 kg/m3 Total upward buoyant force (B) = Buoyant force due to water (B1) + Buoyant force due to mercury (B2) B1 = ρw×g×0.6825V = 682.5gV B2 = ρm×g×0.3175V = 4318gV So, B = B1 + B1 = 5000.5gV For the equilibrium of the body, Total buoyant force (upward force) = Weight of the body Assume that the density of the body is ρb ⇒5000.5gV = ρb×g×V ⇒ρb = 5000.5 kg/m3 ∴ Density of body (ρb)≈5000 kg/m3

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