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 $1m$ 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 $=mg$
$=V\rho g=\frac{4}{3}\pi \left[{\left(\frac{D}{2}\right)}^3-{\left(\frac{d}{2}\right)}^3\right]\rho g$
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 $d=0.98m$.