Question

The volume of the sphere and cube of its diameter are in direct proportion. Two solid spheres of diameter 3 cm and 4 cm are melted together and cast into a hollow sphere of diameter 6 cm. Find the thickness of the outer cover of the sphere.

Answer 1

$$\begin{gathered} \mathrm{Let}V\mathrm{denote}\mathrm{the}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{sphere}\mathrm{and}d\mathrm{denote}\mathrm{the}\mathrm{diameter}\mathrm{of}\mathrm{the}\mathrm{sphere}. \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ V\alpha d^3 \\ V=k{d}^3(k\mathrm{is}\mathrm{a}\mathrm{constant}) \\ {V}_1=k.3^3=27k\left(\mathrm{Solid}\mathrm{sphere}\mathrm{of}\mathrm{radius}3\mathrm{cm}\right) \\ {V}_2=k.4^3=64k\left(\mathrm{Solid}\mathrm{sphere}\mathrm{of}\mathrm{radius}4\mathrm{cm}\right) \\ \mathrm{Total}\mathrm{volume}=27k+64k=91k \\ \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{question}, \\ \mathrm{Two}\mathrm{solid}\mathrm{spheres}\mathrm{of}\mathrm{diameter}3\mathrm{cm}\mathrm{and}4\mathrm{cm}\mathrm{are}\mathrm{melted}\mathrm{together}\mathrm{and}\mathrm{cast}\mathrm{into}\mathrm{hollow}\mathrm{sphere}\mathrm{of}\mathrm{diameter}6\mathrm{cm} \\ \mathrm{Let}\mathrm{the}\mathrm{inner}\mathrm{diameter}\mathrm{of}\mathrm{hollow}\mathrm{sphere}\mathrm{be}\mathit{}x\mathrm{cm} \\ k{\left(6\right)}^3-k{x}^3=91k \\ 216-x^3=91 \\ 216-91=x^3 \\ {x}^3=125 \\ \therefore x=5\mathrm{cm} \\ \mathrm{Thickness}=\frac{\mathrm{Outer}\mathrm{diameter}-\mathrm{Inner}\mathrm{diameter}}{2}=\frac{6-5}{2}=0.5\mathrm{cm} \end{gathered}$$