(iii) The diameter of a sphere is 6 cm . It is melted and drawn into a wire with a diameter of 0.2 cm . Calculate the wire's length. (i) Three solid spheres of lead are melted into a single solid sphere. If the radii of the three spheres are $$ 1 \mathrm{~cm}, 6 \mathrm{~cm} $$ and 8 cm respectively, find the radius of the new sphere.

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### Problem (iii): Length of the Wire

1. **Calculate the volume of the sphere:**
   The formula for the volume $$ V $$ of a sphere is given by:
   $$
   V = \frac{4}{3} \pi r^3
   $$
   where $$ r $$ is the radius of the sphere. The diameter of the sphere is 6 cm, so the radius $$ r $$ is:
   $$
   r = \frac{6}{2} = 3 \text{ cm}
   $$

Now, substituting $$ r $$ into the volume formula:
   $$
   V = \frac{4}{3} \pi (3)^3
   $$

2. **Calculate the volume of the wire:**
   The volume of the wire can also be calculated using the formula for the volume of a cylinder:
   $$
   V = \pi r^2 h
   $$
   where $$ r $$ is the radius of the wire and $$ h $$ is its height (length). The diameter of the wire is 0.2 cm, so the radius $$ r $$ is:
   $$
   r = \frac{0.2}{2} = 0.1 \text{ cm}
   $$

Setting the volume of the sphere equal to the volume of the wire:
   $$
   \frac{4}{3} \pi (3)^3 = \pi (0.1)^2 h
   $$

3. **Solve for $$ h $$:**
   We can cancel $$ \pi $$ from both sides:
   $$
   \frac{4}{3} (3)^3 = (0.1)^2 h
   $$

Now, calculate $$ (3)^3 $$:
   $$
   (3)^3 = 27
   $$
   Thus, we have:
   $$
   \frac{4}{3} \cdot 27 = 0.01 h
   $$

Now, calculate $$ \frac{4 \cdot 27}{3} $$:
   $$
   \frac{108}{3} = 36
   $$
   So, we have:
   $$
   36 = 0.01 h
   $$

Finally, solve for $$ h $$:
   $$
   h = \frac{36}{0.01} = 3600 \text{ cm}
   $$

### Problem (i): Radius of the New Sphere

1. **Calculate the volume of the three spheres:**
   Using the same volume formula for each sphere:
   - For the first sphere with radius $$ r_1 = 1 \text{ cm} $$:
     $$
     V_1 = \frac{4}{3} \pi (1)^3 = \frac{4}{3} \pi
     $$
   - For the second sphere with radius $$ r_2 = 6 \text{ cm} $$:
     $$
     V_2 = \frac{4}{3} \pi (6)^3 = \frac{4}{3} \pi (216) = 288 \pi
     $$
   - For the third sphere with radius $$ r_3 = 8 \text{ cm} $$:
     $$
     V_3 = \frac{4}{3} \pi (8)^3 = \frac{4}{3} \pi (512) = \frac{2048}{3} \pi
     $$

2. **Total volume of the new sphere:**
   The total volume $$ V_{total} $$ is:
   $$
   V_{total} = V_1 + V_2 + V_3 = \frac{4}{3} \pi + 288 \pi + \frac{2048}{3} \pi
   $$

To combine these, we can express $$ 288 \pi $$ in terms of thirds:
   $$
   288 \pi = \frac{864}{3} \pi
   $$
   Thus:
   $$
   V_{total} = \left(\frac{4}{3} + \frac{864}{3} + \frac{2048}{3}\right) \pi = \frac{4 + 864 + 2048}{3} \pi = \frac{2916}{3} \pi = 972 \pi
   $$

3. **Find the radius of the new sphere:**
   Let $$ R $$ be the radius of the new sphere. The volume of the new sphere is:
   $$
   V_{new} = \frac{4}{3} \pi R^3
   $$
   Setting this equal to the total volume:
   $$
   \frac{4}{3} \pi R^3 = 972 \pi
   $$

Cancel $$ \pi $$ from both sides:
   $$
   \frac{4}{3} R^3 = 972
   $$

Multiply both sides by $$ \frac{3}{4} $$:
   $$
   R^3 = 972 \cdot \frac{3}{4} = 729
   $$

Finally, take the cube root:
   $$
   R = \sqrt[3]{729} = 9 \text{ cm}
   $$

### Summary of Results
- The length of the wire is $$ 3600 \text{ cm} $$.
- The radius of the new sphere is $$ 9 \text{ cm} $$.
- The wire is 3600 cm long. - The new sphere has a radius of 9 cm.

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