A spherical balloon is being inflated so that its volume increase uniformly at the rate of `40(cm)^3/min`. How much the radius will increases during the next 1/2 minute,if initial radius is 8cm. ?

Options

A. `0.025 cm`
B. `0.050 cm`
C. `0.075 cm`
D. `0.01 cm`

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the initial volume of the balloon
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
Given that the initial radius \( r = 8 \) cm, we can substitute this value into the formula:
\[ V = \frac{4}{3} \pi (8)^3 \]
Calculating \( 8^3 \):
\[ 8^3 = 512 \]
Now substituting back into the volume formula:
\[ V = \frac{4}{3} \pi (512) = \frac{2048}{3} \pi \approx 2144.66 \text{ cm}^3 \]

### Step 2: Determine the volume increase in half a minute
The problem states that the volume increases at a rate of \( 40 \text{ cm}^3/\text{min} \). Therefore, in half a minute, the volume increase will be:
\[ \Delta V = 40 \times \frac{1}{2} = 20 \text{ cm}^3 \]

### Step 3: Calculate the final volume of the balloon
Now, we can find the final volume after the increase:
\[ V_{\text{final}} = V_{\text{initial}} + \Delta V = 2144.66 + 20 = 2164.66 \text{ cm}^3 \]

### Step 4: Calculate the final radius using the final volume
We will use the volume formula again to find the final radius \( r_{\text{final}} \):
\[ V_{\text{final}} = \frac{4}{3} \pi r_{\text{final}}^3 \]
Rearranging to solve for \( r_{\text{final}} \):
\[ r_{\text{final}}^3 = \frac{3 V_{\text{final}}}{4 \pi} \]
Substituting \( V_{\text{final}} = 2164.66 \):
\[ r_{\text{final}}^3 = \frac{3 \times 2164.66}{4 \pi} \]
Calculating the right-hand side:
\[ r_{\text{final}}^3 \approx \frac{6493.98}{12.5664} \approx 517.64 \]
Now take the cube root:
\[ r_{\text{final}} \approx \sqrt[3]{517.64} \approx 8.0247 \text{ cm} \]

### Step 5: Calculate the increase in radius
Now we find the increase in radius:
\[ \Delta r = r_{\text{final}} - r_{\text{initial}} = 8.0247 - 8 = 0.0247 \text{ cm} \]
Rounding this to three decimal places gives:
\[ \Delta r \approx 0.025 \text{ cm} \]

### Final Answer
The radius of the balloon increases by approximately \( 0.025 \text{ cm} \) during the next half minute.

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