A tank contains water on top of mercury. A cube of Iron, $60 m m$ along each edge, is sitting upright in, equilibrium in the liquids. Find how much of it is in each liquid, The densities of iron and mercury are $7.7 \times 10^{3} k g / m^{3}$ and $13.6 \times 10^{3} k g / m^{3}$ respectively,

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Solution

Let $(x) m m$ is in water and $(60 - x) m m$ is in mercury.
Total upthrust=weight

$= {(0.06)}^{2} (x \times 10^{- 3}) (10^{3}) g + {(0.06)}^{2} (60 - x) (10^{- 3}) (13.6 \times 10^{3}) g$
$= {(0.06)}^{3} (7.7 \times 10^{- 3}) (g)$
Solving this equation we get,
$x = 28 m m$
and $60 - x = 32 m m$

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A tank contains water on top of mercury. A cube of Iron, 60mm along each edge, is sitting upright in, equilibrium in the liquids. Find how much of it is in each liquid, The densities of iron and mercury are7.7×103kg/m3 and 13.6×103kg/m3respectively,

Let (x)mm is in water and (60−x)mm is in mercury.Total upthrust=weight =(0.06)2(x×10−3)(103)g+(0.06)2(60−x)(10−3)(13.6×103)g =(0.06)3(7.7×10−3)(g)Solving this equation we get,x=28mmand 60−x=32mm

A tank contains water on top of mercury. A cube of Iron, $60 m m$ along each edge, is sitting upright in, equilibrium in the liquids. Find how much of it is in each liquid, The densities of iron and mercury are $7.7 \times 10^{3} k g / m^{3}$ and $13.6 \times 10^{3} k g / m^{3}$ respectively,

Let $(x) m m$ is in water and $(60 - x) m m$ is in mercury.
Total upthrust=weight

$= {(0.06)}^{2} (x \times 10^{- 3}) (10^{3}) g + {(0.06)}^{2} (60 - x) (10^{- 3}) (13.6 \times 10^{3}) g$
$= {(0.06)}^{3} (7.7 \times 10^{- 3}) (g)$
Solving this equation we get,
$x = 28 m m$
and $60 - x = 32 m m$