A copper sphere of diameter $12\text{cm}$ is drawn into a wire of diameter $4\text{mm}$. What is the length (in $\text{cm}$) of the wire? (Where $\pi =\frac{22}{7}$)

When a solid object is reshaped into another solid object, its total volume remains unchanged. Here, the copper sphere is transformed into a long cylindrical wire.
1. Volume of the Copper Sphere:

* Diameter of the sphere ($D$) = $12\text{cm}$
* Radius of the sphere ($R$) = $\frac{12}{2}=6\text{cm}$

Formula for the volume of a sphere:

$$V_{\text{sphere}}=\frac{4}{3}\pi R^3$$

$$V_{\text{sphere}}=\frac{4}{3}\times \pi \times (6{)}^3=\frac{4}{3}\times \pi \times 216=288\pi {\text{cm}}^3$$

2. Volume of the Cylindrical Wire:

* Diameter of the wire ($d$) = $4\text{mm}=0.4\text{cm}$
* Radius of the wire ($r$) = $\frac{0.4}{2}=0.2\text{cm}$
* Let the length (height) of the wire be $h\text{cm}$.

Formula for the volume of a cylinder:

$$V_{\text{wire}}=\pi r^{2}h$$

$$V_{\text{wire}}=\pi \times (0.2{)}^2\times h=0.04\pi h{\text{cm}}^3$$

3. Equating both volumes:

$$V_{\text{wire}}=V_{\text{sphere}}$$

$$0.04\pi h=288\pi$$

Canceling $\pi$ from both sides:

$$0.04h=288$$

$$h=\frac{288}{0.04}=\frac{28800}{4}=7200\text{cm}$$

Therefore, the length of the wire is $7200\text{cm}$.