Question

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 $1\, m$ and $2 \times 10^{4} kg / m ^{3}$, respectively, then the inner diameter of the bowl will be

• A. 0.94 m
• B. 0.97 m
• C. 0.98 m
• D. 0.99 m

Answer 1

Answer: (C)

Weight of the bowl $=m g$
$=V \rho g=\frac{4}{3} \pi\left[\left(\frac{D}{2}\right)^{3}-\left(\frac{d}{2}\right)^{3}\right] \rho g$
where $D=$ Outer diameter
$d=$ Inner diameter
$\rho=$ Density of bowl
Weight of the liquid displaced by the bowl
$V \sigma g=\frac{4}{3} \pi\left(\frac{D}{2}\right)^{3} \sigma g$
where $\sigma$ is the density of the liquid.
For floatation: $\frac{4}{3} \pi\left(\frac{D}{2}\right)^{3} \sigma g=\frac{4}{3} \pi\left[\left(\frac{D}{2}\right)^{3}-\left(\frac{d}{2}\right)^{3}\right] \rho g$
$\Rightarrow\left(\frac{1}{2}\right)^{3} \times 1.2 \times 10^{3}=\left[\left(\frac{1}{2}\right)^{3}-\left(\frac{d}{2}\right)^{3}\right] 2 \times 10^{4}$
By solving, we get $d=0.98\, m$.