Volume is conserved. Set (1/3)πr₁²h = (4/3)πr₂³ and solve for r₂. Class 10 conversion problem.

A Cone is Melted and Recast into a Sphere — Find the Radius

Question

A cone of height 24 cm and base radius 6 cm is made of modelling clay. A child reshapes it into a sphere. Find the radius of the sphere.

(NCERT Class 10, Chapter 13, Example 1)

Solution — Step by Step

We need two formulas:

V_{\text{cone}} = \frac{1}{3}\pi r^2 h \qquad V_{\text{sphere}} = \frac{4}{3}\pi R^3

Here, $r = 6$ cm, $h = 24$ cm for the cone, and $R$ is what we want.

When a solid is melted and recast, no material is lost. The total volume stays the same. So:

\frac{1}{3}\pi r^2 h = \frac{4}{3}\pi R^3

$\pi$ appears on both sides — cancel it immediately. This is a step most students skip, which makes the arithmetic messier than it needs to be.

\frac{1}{3} r^2 h = \frac{4}{3} R^3

\frac{1}{3} \times 36 \times 24 = \frac{4}{3} R^3

\frac{864}{3} = \frac{4}{3} R^3

288 = \frac{4}{3} R^3

Multiply both sides by $\frac{3}{4}$ :

R^3 = 288 \times \frac{3}{4} = 216

R = \sqrt[3]{216} = 6 \text{ cm}

The radius of the sphere is 6 cm.

Why This Works

The key principle here is conservation of volume. When you melt and recast, you’re just rearranging the same amount of material into a new shape. No clay is added, none is lost.

This is different from surface area — surface area is not conserved when you change shape. (The cone and sphere here have very different surface areas.) Only volume is conserved, so we always equate volumes in melting/recasting problems.

The reason the answer comes out so cleanly ( $R^3 = 216 = 6^3$ ) is by design — NCERT picked these numbers to give a nice cube root. In exams, if your $R^3$ is not a perfect cube, double-check your arithmetic before assuming the answer is irrational.

Alternative Method

Instead of setting up the equation symbolically, you can compute the cone’s volume numerically first, then equate.

V_{\text{cone}} = \frac{1}{3} \times \pi \times 36 \times 24 = 288\pi \text{ cm}^3

Now set this equal to the sphere volume:

\frac{4}{3}\pi R^3 = 288\pi

R^3 = \frac{288 \times 3}{4} = 216 \implies R = 6 \text{ cm}

Same answer, slightly different path. The symbolic method (Step 3 above) is faster because you cancel $\pi$ before substituting — useful when $r$ , $h$ are not nice numbers.

Common Mistake

Students often write $(4/3)\pi R^3 = (1/3)\pi r^2 h$ correctly, but then forget to multiply by $3/4$ when isolating $R^3$ . They accidentally write $R^3 = 288$ instead of $R^3 = 216$ , getting $R \approx 6.6$ cm. Always multiply both sides by the reciprocal of $4/3$ , which is $3/4$ .

This type — cone/cylinder/hemisphere melted into sphere — is a guaranteed 3-mark question in CBSE Class 10 boards. The working is always the same: equate volumes, cancel $\pi$ , isolate the unknown. Practice 5-6 variations and you’ll never lose marks here.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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