question_answer

A conical vessel of radius 12 cm and height 16 cm is completely filled with water. A sphere is lowered into the water and its sized is such that, when it touches the sides, it is just immersed. What fraction of the water overflows?

A)  \[\frac{3}{8}\]                                   

B)  \[\frac{4}{7}\]                       

C)  \[\frac{1}{2}\]                       

D)  \[\frac{5}{9}\]                                                           

Correct Answer: A

Solution :

Let radius of the sphere be r cm Radius of Conical Vessel (R) = 12 cm Height of Conical Vessel (h) = 16 cm Slant height of the cone \[(AO)\]                                     \[=\sqrt{{{16}^{2}}+{{12}^{2}}}=20\,cm\] Now,  \[\Delta \,OBA\tilde{\ }\Delta OCD\] \[\frac{AB}{CD}=\frac{OA}{OD}\,\,\Rightarrow \frac{12}{r}=\frac{20}{16-r}\]                                     \[[\therefore \,\,OD=OB-BD]\] \[\Rightarrow \]\[192-12\,r=20r\,\,\,\Rightarrow \,\,32\,r=192\Rightarrow \,r=6\,cm\] Volume of water in the vessel \[=\frac{1}{3}\pi {{R}^{2}}h\] \[=\frac{1}{3}\pi \times {{12}^{2}}\times 16=768\pi \,c{{m}^{3}}\] Volume of water that over flows = Volume of the sphere \[=\frac{4}{3}\pi {{r}^{3}}=\frac{4}{3}\times \pi \times {{6}^{3}}=288\pi \,c{{m}^{3}}\] Fraction of volume of water over flows \[=\frac{288\,\pi }{768\,\pi }=\frac{3}{8}\]