Question

A metallic sphere of radius \(9\) cm is melted and recast to form a cylinder of radius \(3\) cm. Find the curved surface area of the cylinder.

Hint:

When one solid is melted and recast into another, their volumes remain equal. First find the missing dimension using volume, then calculate the required surface area.

Answer 1

Step 1: Use the concept of conservation of volume.}
Since the metallic sphere is melted and recast into a cylinder, the volume of the sphere will be equal to the volume of the cylinder.

Step 2: Write the volume of the sphere.}
Radius of the sphere is \(9\) cm.
So, volume of the sphere is:
\[ V_{\text{sphere}}=\frac{4}{3}\pi r^3 \] \[ V_{\text{sphere}}=\frac{4}{3}\pi (9)^3 \] \[ V_{\text{sphere}}=\frac{4}{3}\pi \cdot 729 \] \[ V_{\text{sphere}}=972\pi \]
Step 3: Write the volume of the cylinder.}
Let the height of the cylinder be \(h\).
Radius of the cylinder is \(3\) cm.
So, volume of the cylinder is:
\[ V_{\text{cylinder}}=\pi r^2 h \] \[ V_{\text{cylinder}}=\pi (3)^2 h \] \[ V_{\text{cylinder}}=9\pi h \]
Step 4: Equate the two volumes to find the height.}
Since both volumes are equal:
\[ 972\pi=9\pi h \] Cancel \(\pi\) from both sides:
\[ 972=9h \] \[ h=108 \]
Step 5: Find the curved surface area of the cylinder.}
Curved surface area of a cylinder is given by:
\[ \text{CSA}=2\pi rh \] Substituting \(r=3\) cm and \(h=108\) cm:
\[ \text{CSA}=2\pi(3)(108) \] \[ \text{CSA}=648\pi \]
Step 6: State the final answer.}
Hence, the curved surface area of the cylinder is:
\[ \boxed{648\pi\ \text{cm}^2} \] If expressed numerically,
\[ 648\pi \approx 2036.57\ \text{cm}^2 \]