A solid sphere of radius 6 cm is melted into a hollow cylinder of unif

1.	A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness .If the external radius of the base of the cylinder is 5 cm and its height is 32 cm, then find the thickness of the cylinder.
Answer» GIVEN: Radius of sphere = 6 CM External radius of cylinder = 5 cm Height of the cylinder = 32 cm TO FIND: What is the THICKNESS of the cylinder ? SOLUTION: LET the internal radius of the cylinder be 'r' cm and external radius be 'R' cm We know that the FORMULA for finding the Volume of sphere is:- $\large{\boxed{\bf{\star \: VOLUME = \dfrac{4}{3} \pi r^3 \: \star}}}$ We know that the formula for finding the Volume of hollow cylinder is:- $\large{\boxed{\bf{\star \: VOLUME = \pi (R^2 - r^2) h \: \star}}}$ According to question:- $\large\bf{\star \: \dfrac{4}{3} \pi r^3 = \pi (R^2 - r^2 ) h\: \star}$ On putting the given values in the formula, we get $\sf{\longmapsto \dfrac{4}{3} \cancel{\pi} r^3 = \cancel{\pi }(R^2 - r^2 ) h}$ $\sf{\longmapsto \dfrac{4}{3} \times (6)^3 = (5^2 - r^2 ) \times 32 }$ $\sf{\longmapsto \dfrac{4}{\cancel{3}} \times \cancel{216} = (25-r^2 ) \times 32 }$ $\sf{\longmapsto \dfrac{288}{32} = 25-r^2 }$ $\sf{\longmapsto 9-25 = -r^2 }$ $\sf{\longmapsto \cancel{-}16 = \cancel{-} r^2 }$ $\sf{\longmapsto \sqrt{16} = r }$ $\bf{\longmapsto 4 \: cm = r }$ Internal radius = r = 4 cm ❍ Thickness of cylinder = (R–r) = (5–4) = 1 cm ❝ Hence, the thickness of the cylinder is 1 cm ❞ ______________________