Solutions
Volume of sphere = \(\frac{4}{3}\pi {r^3}\)
Where, r is radius of sphere.
Volume of the given sphere = (4/3) × π × 63
This volume is converted to the hollow cylinder.
Let, the thickness of cylinder be x.
External radius = 5 cm
Internal radius = (5 – x) cm
Area of circular base = πr2, where r is radius of base.
Area of hollow base = (π × 52) – π (5 – x)2
= π [52 – (5 – x)2]
= π (5 + 5 – x) (5 – 5 + x)
= πx(10 – x)
Volume of cylinder = Area of base × height
∴(4/3) × π × 63 = 32 × π (10 – x)x
⇒ 9 = 10x – x2
⇒ x2 – 10x + 9 = 0
⇒ x2 – 9x – x + 9 = 0
⇒ x(x – 9) – 1(x – 9) = 0
⇒ (x – 1) (x – 9) = 0
⇒ x = 1 or x = 9
x = 9 not possible, since the thickness cannot be greater than 5, which is the external radius.
So, x = 1