A spherical ball of lead 6 cm in radius is melted and recast into three spherical balls. The radii of two of these balls are 3cm and 4 cm. What is the radius of the third sphere ?

Options

A. 4.5 cm
B. 5 cm
C. 6 cm
D. 7 cm

Answer

To solve the problem of finding the radius of the third sphere after melting a lead ball and recasting it into three smaller spheres, we can follow these steps:

### Step 1: Calculate the volume of the original sphere
The formula for the volume \( V \) of a sphere is given by:

\[ V = \frac{4}{3} \pi r^3 \]

For the original sphere with a radius of 6 cm:

\[ V = \frac{4}{3} \pi (6)^3 \]

Calculating \( (6)^3 \):

\[ (6)^3 = 216 \]

So, the volume becomes:

\[ V = \frac{4}{3} \pi \times 216 = \frac{864}{3} \pi = 288 \pi \, \text{cm}^3 \]

### Step 2: Calculate the volumes of the two smaller spheres
For the first smaller sphere with a radius of 3 cm:

\[ V_1 = \frac{4}{3} \pi (3)^3 \]

Calculating \( (3)^3 \):

\[ (3)^3 = 27 \]

So, the volume becomes:

\[ V_1 = \frac{4}{3} \pi \times 27 = \frac{108}{3} \pi = 36 \pi \, \text{cm}^3 \]

For the second smaller sphere with a radius of 4 cm:

\[ V_2 = \frac{4}{3} \pi (4)^3 \]

Calculating \( (4)^3 \):

\[ (4)^3 = 64 \]

So, the volume becomes:

\[ V_2 = \frac{4}{3} \pi \times 64 = \frac{256}{3} \pi \, \text{cm}^3 \]

### Step 3: Calculate the total volume of the two smaller spheres
Now, we add the volumes of the two smaller spheres:

\[ V_{\text{total}} = V_1 + V_2 = 36 \pi + \frac{256}{3} \pi \]

To add these, we need a common denominator. The common denominator is 3:

\[ V_{\text{total}} = \frac{108}{3} \pi + \frac{256}{3} \pi = \frac{364}{3} \pi \, \text{cm}^3 \]

### Step 4: Calculate the volume of the third sphere
The volume of the third sphere \( V_3 \) can be found by subtracting the total volume of the two smaller spheres from the volume of the original sphere:

\[ V_3 = V - V_{\text{total}} = 288 \pi - \frac{364}{3} \pi \]

To perform this subtraction, we convert \( 288 \pi \) to a fraction with a denominator of 3:

\[ 288 \pi = \frac{864}{3} \pi \]

Now we can subtract:

\[ V_3 = \left(\frac{864}{3} - \frac{364}{3}\right) \pi = \frac{500}{3} \pi \, \text{cm}^3 \]

### Step 5: Find the radius of the third sphere
Using the volume formula for a sphere, we can set \( V_3 \) equal to the volume formula and solve for the radius \( r_3 \):

\[ \frac{4}{3} \pi r_3^3 = \frac{500}{3} \pi \]

We can cancel \( \pi \) and \( \frac{4}{3} \):

\[ r_3^3 = \frac{500}{4} = 125 \]

Taking the cube root:

\[ r_3 = \sqrt[3]{125} = 5 \, \text{cm} \]

### Final Answer
The radius of the third sphere is **5 cm**.

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