Find the % Volume left when the largest possible sphere is kept in a cube

Dear Student,

Please find below the solution to the asked query:

When a largest possible sphere kept in a cube , So 

Side of cube =  Diameter of sphere  (  As we shoe in following diagram )



Let , Sid e of cube  =  a  , So

Diameter of sphere = a  , Then Radius of sphere  = $\frac{a}{2}$

We know volume of cube =  ( Side )³ , So

Volume of given cube =  a³  

And

We know Volume of sphere = $\frac{4 \mathrm{\pi } {\mathrm{r}}^3}{3}$ , So

Volume of given sphere = $\frac{4 \times {\displaystyle \frac{22}{7}} \times {\left({\displaystyle \frac{\mathrm{a}}{2}}\right)}^3}{3} = \frac{88 \times {\displaystyle \frac{{\mathrm{a}}^3}{8}}}{21}= \frac{\mathbf{11}\mathbf{ }{\mathbf{a}}^{\mathbf{3}}}{\mathbf{21}}$

So,

Volume left when the largest possible sphere is kept in a cube =  Volume of cube -  Volume of sphere = a³ - $\frac{11 a^3}{21} = \frac{{\displaystyle 21 a^3 - 11 a^3}}{{\displaystyle 21}} = \frac{{\displaystyle 10 a^3}}{{\displaystyle 21}}$

Therefore,

Percentage of volume left when the largest possible sphere is kept in a cube = $\frac{{\displaystyle \frac{10 a^3}{21}}}{a^3}\times 100 = \frac{{\displaystyle 10 a^3}}{21 a^3}\times 100 = \frac{{\displaystyle 1000}}{21} =47.619\% \approx \mathbf{47}\mathbf{.}\mathbf{62}\mathbf{\%}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{Ans}\mathbf{ }\mathbf{)}$
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