Question

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 1

Answer:-

Let the thickness of the cylinder be "x" cm.

Given:

Radius of the sphere = 6 cm

Height of the cylinder = 32 cm

External radius of the cylinder = 5 cm.

We know that,

Internal radius = External radius - thickness

Internal radius of the cylinder = (5 - x) cm.

According to the question,

Volume of Sphere = Volume of Cylinder.

→ 4/3 * π r³ = π*r²*h

→ 4/3 * (6)³ = (5 - x)² * (32)

→ [ 1.33*(6)³ ] / 32 = (5 - x)²

→ 9 = (5 - x)²

→ √9 = 5 - x

→ ± 3 = 5 - x

→ x = 5 - 3 [ Thickness of a cylinder can't be greater than External radius]

→ x = 2 cm.

Hence, the thickness of the Cylinder is 2 cm.

Answer 2

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$\huge\tt{GIVEN:}$

• A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness.
• external radius of the base of the cylinder is 5 cm and its height is 32 cm

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$\huge\tt{TO~FIND:}$

• thickness of the cylinder

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$\huge\tt{SOLUTION:}$

➩Let the thickness of the cylinder be n cm

We know that,

➩Internal radius = External radius - Thickness

➩Internal radius = (5-n) cm

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We also know that ,

➩Volume of sphere = Volume of Cylinder

➩4/3×πr³ = πr²h

➩4/3×(6³)=(5-n²)×32

➩1.33×6³/32 = (5-n²)

➩9 =(5-n²)

➩ √9 = 5-n

➩±3=5-n

➩n = 5-3

➩n = 2 cm

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