Question

A hemispherical bowl just floats without sinking in a liquid of density 1.2 $\times$ ${10}^3$ kg/ $m^3$.. If 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

The correct option is C

0.98 m

$weightofthebowl=mg=Vpg=\frac{4}{3}\pi [{\left(\frac{D}{2}\right)}^3-{\left(\frac{d}{2}\right)}^3]pgwhereDistheouterdiameter,distheinnerdiameterandisthedensityofbowlWeightoftheliquiddisplacedbythebowl=V\sigma g=\frac{4}{3}\pi {\left(\frac{D}{2}\right)}^3\sigma gwhere\sigma isthedensityoftheliquid.Fortheflotation\frac{4}{3}\pi {\left(\frac{D}{2}\right)}^3\sigma g=\frac{4}{3}\pi [{\left(\frac{D}{2}\right)}^3-{\left(\frac{d}{2}\right)}^3]pg{\left(\frac{1}{2}\right)}^3\times 1.2\times {10}^3=[{\left(\frac{1}{2}\right)}^3-{\left(\frac{d}{2}\right)}^3]2\times {10}^4$

By solving we get d = 0.98 m.