The surface area of a solid metallic sphere is $5024 \; cm^2$. It is melted and recast into solid cones of radius $5 \; cm$ and height $10 \; cm$. Calculate the radius of the sphere and the number of cones recast. Take $\pi = 3.14$
Let radius of sphere $= R$
Given: Surface area of sphere $= 5024 \; cm^2$
i.e. $\;$ $4 \pi R^2 = 5024$
i.e. $\;$ $R^2 = \dfrac{5024}{4 \pi} = \dfrac{1256}{3.14} = 400$
$\implies$ $R = \sqrt{400} = 20$ $\;\;\;$ [negative value discarded since radius of sphere cannot be negative]
$\therefore \;$ Radius of sphere $= R = 20 \; cm$
Volume of sphere $= V_S = \dfrac{4}{3} \pi R^3$
Given: Radius of cone $= r = 5 \; cm$; $\;$ Height of cone $= h = 10 \; cm$
Volume of cone $= V_C = \dfrac{1}{3} \pi r^2 h$
Let the number of cones made $= n$
$\because \;$ The sphere is melted and recast into a number of cones, we have
$V_S = n \times V_C$
$\implies$ $n = \dfrac{V_S}{V_C}$
i.e. $\;$ $n = \dfrac{\dfrac{4}{3}\pi R^3}{\dfrac{1}{3}\pi r^2 h}$
i.e. $\;$ $n = \dfrac{4 R^3}{r^2 h} = \dfrac{4 \times 20^3}{5^2 \times 10} = 128$
i.e. $\;$ Number of cones made $= n = 128$