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A hollow cone floats with its axis vertical upto one third of its height in a liquid in a liquid of relative density 0.8 and with its vertex submerged. When another liquid of relative density rho is filled in it upto one third of its height, the cone floats upto half its vertical height. The height of the cone is 0.10m and the radius of the circular base is 0.05m. Find the specific gravity rhois given. |
Answer»  `(r)/(x)=(R)/(H)implies r=(R)/(h)x` `r_(1)=(R)/(h)xx(h)/(3) implies r_(1)=(R)/(3)W=B` `mg=(1)/(3)pi((R )/(3))^(2)xx(h)/(3)xx(4)/(5) g` `m=(1)/(27)xx(1)/(3)piR^(2)HXX(4)/(5)"".....(1)` when liquid is fille in cone. `W=B` `(m+(1)/(3)pi((R)/(3))^(2)(h)/(3)rho]g` `=(1)/(3) pi ((R)/(2))^(2)(h)/(2)xx((4)/(5))g` `[m+(1)/(27)xx(1)/(3)piR^(2)hrho]=(1)/(10)xx(1)/(3)piR^(2)h"".....(2)` From (1) and (2) `(1)/(27) [rho+(4)/(5)]=10implies rho=1.9` 
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