By melting two solid metallic spheres of redii 1 cm and 6 cm, a hollow sphere of thickness 1 cm is made. The external radios of the hollow sphere will be

Given $$r_{1} = 1$$ cm and $$r_{2}$$ = 6 cm

Volume of 1st sphere = $$\frac{4}{3} \pi r_{1}^{3}$$ = $$\frac{4}{3} \pi 1^{3}$$ = $$\frac{4}{3}\pi$$

Volume of 2nd sphere = $$\frac{4}{3} \pi r_{2}^{3}$$ = $$\frac{4}{3} \pi (6)^{3}$$ 

Combined volume of two spheres = $$\frac{4}{3} \pi (217)$$ 

Let outer radius of hollow sphere = $$x$$ then inner radius = $$x - 1$$ 

Volume of the hollow sphere is given by,

$$\frac{4}{3} \pi (x^{3} - (x - 1)^{3})$$ = $$\frac{4}{3} \pi (217)$$ 

$$x^{3}-(x - 1)^{3} = 217$$ 

$$x^{3}-x^{3} - 1 -3x^{2} + 3x = 217$$ 

$$3x^{2} - 3x - 216 = 0$$ 

x = 9 or -8 (which cannot be a solution)

Hence, option C is the correct answer.