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 are 7.7g/cm³ and 13.6g/cm³ respectively

Dear Sohan

Suppose x₁ length is in water and x₂ in mercury

and x₁+x₂=h=60mm...............(1)

here equilibrium equation will be

Weight = total buoyant force for each medium,  W=B=B_w + B_hg...........(2)

weight=W= density of cubex volumexg= d_cx(Axh)xg,   where A is h²

B_w=d_w*(A*x₁)*g

B_hg=d_hg*(A*x₂)*g

use equation 2

after cancelling terms on rhs and lhs you will get

d_c*h=d_w*x₁+d_hg*x₂

put values of density

7.7h=x₁+13.6x₂..............(3)

solve for x₁ and x₂ using 1 and 3

x₂=6.7h/12.6= 6.7*60/12.6 = 31.9 mm in mercury

in water=60-31.9=28.1 mm

 All the best.                                                          

AKASH GOYAL

AskiitiansExpert-IIT Delhi

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