A hollow cone (vertex down, axis vertical) floats with its vertex submerged, sinking up to one-third of its height in a liquid of relative density 0.8. When a second liquid of relative density s is poured into the cone up to one-third of its height, the cone sinks until exactly half its he

Question

A hollow cone (vertex down, axis vertical) floats with its vertex submerged, sinking up to one-third of its height in a liquid of relative density 0.8. When a second liquid of relative density s is poured into the cone up to one-third of its height, the cone sinks until exactly half its height is submerged. The cone has height 0.10 m and base radius 0.05 m. Find the specific gravity s of the poured liquid.

Accepted Answer

1.71. Let total cone height H, base radius proportional to height. A sub-cone of fractional height k has volume V*k³ where V is full cone volume. Case 1 (empty cone): submerged to k=1/3, so buoyant volume = V*(1/3)³ = V/27. Weight of cone W_cone = 0.8*rho_w*g*(V/27). Case 2: cone holds liquid (RD s) up to height 1/3, so contained liquid volume = V*(1/3)³ = V/27, weight = s*rho_w*g*(V/27). Now submerged to k=1/2, buoyant volume = V*(1/2)³ = V/8, buoyant force = 0.8*rho_w*g*(V/8). Balance: 0.8*(V/8) = 0.8*(V/27) + s*(V/27) (dividing rho_w*g). So 0.8/8 = 0.8/27 + s/27 -> 0.1 = (0.8 + s)/27 -> 0.8 + s = 2.7 -> s = 1.9. Re-evaluating with poured liquid filling to one-third of internal height give

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