Question

A metallic solid cylinder of diameter 8√21 cm is melted and recast in the form of a solid sphere of diameter 42 cm. The height of the metallic cylinder
is​

Answer 1

Answer: 36.75 cm

Step-by-step explanation:

The diameter of the solid cylinder, d₁ = 8√21 cm

∴ Radius, r₁ = d₁/2 = (8√21) / 2 = 4√21 cm

And,

The diameter of the solid sphere, d₂ = 42 cm

∴ Radius, r₂ = d₂/2 = 42 / 2 = 21 cm

Now, the required formulas are:

The volume of a solid cylinder, V₁ = π*(r₁)²*h ….. (i)…where h = height of the solid cylinder

And,

The volume of the solid sphere, V₂ = 4/3 * π*(r₂)³ …. (ii)

Since we are given that the solid cylinder is melted and recast into a solid sphere, therefore, the volume of the cylinder & the sphere remains the same.

So, we can write

V₁ = V₂

⇒ [π * (r₁)² * h] = [4/3 * π * (r₂)³ ]…. [from (i) & (ii)]

Cancelling the similar terms and substituting the given values of r₁ & r₂

⇒ h = [4/3*21*21*21] / [(4√21)²]

⇒ h = [4/3 * 21 * 21 * 21] / [16 * 21]

⇒ h = [7 * 21] / 4

⇒ h = 36.75 cm

Thus, the height of the metallic cylinder is 36.75 cm.