This problem involves the concept of volume conservation. When a solid object like a sphere is melted and recast into another shape, like a wire (which is essentially a long cylinder), its volume remains the same. We need to calculate the volume of the original sphere and equate it to the volume of the wire to find its length.
The question gives the diameter of the sphere as 8 cm.
The radius of the sphere ($r_s$) is half of its diameter:
$r_s = \frac{\text{Diameter}}{2} = \frac{8 \text{ cm}}{2} = 4 \text{ cm}$
The wire is cylindrical. Its diameter is given as 3 mm.
First, let's convert the wire's diameter to centimeters to maintain consistent units:
$1 \text{ cm} = 10 \text{ mm}$
Wire Diameter = $3 \text{ mm} = \frac{3}{10} \text{ cm} = 0.3 \text{ cm}$
The radius of the wire ($r_w$) is half of its diameter:
$r_w = \frac{\text{Diameter}}{2} = \frac{0.3 \text{ cm}}{2} = 0.15 \text{ cm}$
The formula for the volume of a sphere ($V_s$) is:
$V_s = \frac{4}{3}\pi r_s^3$
Substituting the radius of the sphere:
$V_s = \frac{4}{3}\pi (4 \text{ cm})^3 = \frac{4}{3}\pi (64) \text{ cm}^3 = \frac{256}{3}\pi \text{ cm}^3$
The wire is treated as a cylinder. The formula for the volume of a cylinder ($V_w$) is:
$V_w = \pi r_w^2 h$
where $h$ is the length of the wire (height of the cylinder).
Substituting the radius of the wire:
$V_w = \pi (0.15 \text{ cm})^2 h = \pi (0.0225) h \text{ cm}^3$
Since the volume of the metal remains constant:
$V_s = V_w$
$\frac{256}{3}\pi \text{ cm}^3 = \pi (0.0225) h \text{ cm}^3$
Now, we solve for $h$:
$h = \frac{\frac{256}{3}\pi}{0.0225\pi} \text{ cm}$
Cancel out $\pi$:
$h = \frac{256}{3 \times 0.0225} \text{ cm} = \frac{256}{0.0675} \text{ cm}$
$h \approx 3792.59 \text{ cm}$
The options are given in meters, so we need to convert the calculated length from centimeters to meters.
$1 \text{ m} = 100 \text{ cm}$
$h = \frac{3792.59}{100} \text{ m} \approx 37.9259 \text{ m}$
Rounding to one decimal place, the length of the wire is approximately 37.9 m.
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