Question

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

• A. $0.92\text{m}$
• B. $0.94\text{m}$
• C. $0.96\text{m}$
• D. $0.98\text{m}$

Answer 1

The correct option is D $0.98\text{m}$

Weight of the bowl $=mg$
$=V\rho g=\frac{2}{3}\pi [{\left(\frac{D}{2}\right)}^3-{\left(\frac{d}{2}\right)}^3]\rho g$
Where $D=$ Outer diameter
$d=$ Inner diameter, $\rho =$ Density of bowl

Now, weight of the liquid displaced by the bowl
$=V\sigma g=\frac{2}{3}\pi {\left(\frac{D}{2}\right)}^3\sigma g$
where $\sigma$ is the density of the liquid
By law of floatation,
$\frac{2}{3}\pi {\left(\frac{D}{2}\right)}^3\sigma g=\frac{2}{3}\pi [{\left(\frac{D}{2}\right)}^3-{\left(\frac{d}{2}\right)}^3]\rho g$
$\Rightarrow {\left(\frac{1}{2}\right)}^3\times 1.2\times {10}^3=[{\left(\frac{1}{2}\right)}^3-{\left(\frac{d}{2}\right)}^3]2\times {10}^4$
By solving we get, $d=0.98\text{m}$