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A hemispherical bowl just floats without sinking in a liquid of density $1.2 \times 10^{3} kg / m^{3}$ If the outer diameter and the density of the bowl are 1m and $2 \times 10^{4} kg / m^{3}$ respectively, then the inner diameter of the bowl will be

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0.94 m

0.97 m

0.98 m

0.99 m

answer is C.

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Detailed Solution

Let $d$ be the inner diameter of the hemispherical bowl. Then, B = mg

$\frac{4}{3} π \times {(\frac{1}{2})}^{3} \times (1.2 \times 10^{3}) g$ $= \frac{4}{3} π [{(\frac{1}{2})}^{3} - {(\frac{d}{2})}^{3}] \times (2 \times 10^{4}) g$

Solving, we get d = 0.98 m

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