Question

A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm; find the height of the cylinder.

Hint:

In recasting problems, always equate volumes and cancel \( \pi \) early to simplify calculations.

Answer 1

Concept: When a solid is melted and recast into another shape:

• Volume remains constant.

Formulas used: \[ \text{Volume of sphere} = \frac{4}{3}\pi r^3, \quad \text{Volume of cylinder} = \pi R^2 h. \]
Step 1: Equate the volumes.
Let height of cylinder = \( h \). Given: \[ r = 4.2 \text{ cm}, \quad R = 6 \text{ cm}. \] \[ \frac{4}{3}\pi (4.2)^3 = \pi (6)^2 h. \]
Step 2: Cancel \( \pi \) and simplify.
\[ \frac{4}{3} (4.2)^3 = 36h. \]
Step 3: Evaluate \( (4.2)^3 \).
\[ 4.2^3 = 74.088. \] \[ \frac{4}{3} \times 74.088 = 98.784. \]
Step 4: Solve for \( h \).
\[ 98.784 = 36h \Rightarrow h = \frac{98.784}{36}. \] \[ h \approx 2.744 \text{ cm}. \] Conclusion:
The height of the cylinder is approximately: \[ \boxed{2.74 \text{ cm}}. \]