Question

# A hemispherical bowl just floats without sinking in a liquid of density 1.2×103 kg/m3. If the outer diameter and the density of the bowl are 1 m and 2×104 kg/m3 respectively, then the inner diameter of the bowl will be (in m)  [Answer upto two decimal places]

Solution

## Weight of the bowl =mg  =Vρg=23π[(D2)3−(d2)3]ρg Where D= Outer diameter  d= Inner diameter, ρ= Density of bowl  Now, weight of the liquid displaced by the bowl  =Vσg=23π(D2)3σg where σ is the density of the liquid  By law of floatation, 23π(D2)3σg=23π[(D2)3−(d2)3]ρg ⇒(12)3×1.2×103=[(12)3−(d2)3]2×104 By solving we get, d=0.98 m

Suggest corrections