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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

A
0.92 m
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B
0.94 m
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C
0.96 m
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D
0.98 m
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Solution

The correct option is D 0.98 m
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

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