A solid sphere of radius 15 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 9 cm and its height is 100 cm, find the uniform thickness of the cylinder.

A solid sphere is melted and reshaped into a hollow cylinder. The volume remains constant during this process.

Given:

• Solid sphere radius = 15 cm
• Hollow cylinder external radius = 9 cm
• Hollow cylinder height = 100 cm
• Uniform thickness = $t$ cm

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Volume of the sphere with radius 15 cm:

$V = \dfrac{4}{3}\pi r^3$

$V = \dfrac{4}{3} \times \pi \times (15)^3$

$V = \dfrac{4}{3} \times \pi \times 3375$

$V = 4500\pi$ cm³

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The hollow cylinder has:

• External radius $R = 9$ cm
• Internal radius $r = (9 - t)$ cm
• Height $h = 100$ cm

Volume of hollow cylinder:

$V = \pi(R^2 - r^2)h$

$V = \pi[9^2 - (9-t)^2] \times 100$

$(9 - t)^2 = 81 - 18t + t^2$

$V = \pi[81 - 81 + 18t - t^2] \times 100$

$V = 100\pi(18t - t^2)$ cm³

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Since the volumes are equal:

$4500\pi = 100\pi(18t - t^2)$

$4500 = 100(18t - t^2)$

$45 = 18t - t^2$

$t^2 - 18t + 45 = 0$

$(t - 3)(t - 15) = 0$

$t = 3$ or $t = 15$

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If $t = 15$ cm, then internal radius $= 9 - 15 = -6$ cm (not possible)

If $t = 3$ cm, then internal radius $= 9 - 3 = 6$ cm (valid)

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The uniform thickness of the cylinder is 3 cm.