(i) The radius of a metallic sphere is 3 cm .It melted and recast in to wire of diameter 0.4 cm Find the length of wire.
(ii) An iron ball oif radius 4 cm is melted .How many small spheres of radus 2 cm can be formed from the material?

Answer

To solve the given problems step by step, we will break them down into two parts as mentioned in the question.

### Part (i): Finding the Length of the Wire

1. **Calculate the Volume of the Sphere:**
The formula for the volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Here, the radius \( r \) of the sphere is 3 cm.
\[ V = \frac{4}{3} \pi (3)^3 = \frac{4}{3} \pi (27) = 36\pi \, \text{cm}^3 \]

2. **Calculate the Radius of the Wire:**
The diameter of the wire is given as 0.4 cm, so the radius \( r \) of the wire is:
\[ r = \frac{0.4}{2} = 0.2 \, \text{cm} \]

3. **Set Up the Volume of the Cylinder (Wire):**
The volume \( V \) of the cylinder (wire) is given by:
\[ V = \pi r^2 h \]
where \( h \) is the height (or length) of the wire. We know the volume of the sphere is equal to the volume of the wire:
\[ 36\pi = \pi (0.2)^2 h \]

4. **Solve for the Height (Length of the Wire):**
Simplifying the equation:
\[ 36 = (0.2)^2 h \]
\[ 36 = 0.04 h \]
\[ h = \frac{36}{0.04} = 900 \, \text{cm} \]

5. **Convert Length to Meters:**
Since 1 meter = 100 cm,
\[ h = \frac{900}{100} = 9 \, \text{m} \]

### Part (ii): Finding the Number of Small Spheres

1. **Calculate the Volume of the Large Sphere:**
The radius of the iron ball is 4 cm. Using the volume formula:
\[ V_1 = \frac{4}{3} \pi (4)^3 = \frac{4}{3} \pi (64) = \frac{256}{3} \pi \, \text{cm}^3 \]

2. **Calculate the Volume of the Small Sphere:**
The radius of the small sphere is 2 cm. Using the volume formula:
\[ V_2 = \frac{4}{3} \pi (2)^3 = \frac{4}{3} \pi (8) = \frac{32}{3} \pi \, \text{cm}^3 \]

3. **Find the Number of Small Spheres:**
To find how many small spheres can be formed from the large sphere, we divide the volume of the large sphere by the volume of a small sphere:
\[ \text{Number of small spheres} = \frac{V_1}{V_2} = \frac{\frac{256}{3} \pi}{\frac{32}{3} \pi} = \frac{256}{32} = 8 \]

### Final Answers:
(i) The length of the wire is **9 meters**.
(ii) The number of small spheres that can be formed is **8**.

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