Initially the diameter of a balloon is 28 cm. It can explode when the diameter becomes 5/2 times of the initial diameter . Air is brown at 156 cc/s. It is known that the shape of balloon always remains spherical . In how many seconds the balloon will explode ?

Options

A. A)`1078 s`
B. B)`1368 s `
C. C)`1087 s`
D. D)none of these

Answer

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

### Step 1: Determine the initial diameter and volume of the balloon.
- The initial diameter of the balloon is given as 28 cm.
- The formula for the volume \( V \) of a sphere is:
\[ V = \frac{\pi}{6} \times d^3 \]
where \( d \) is the diameter.

Calculating the initial volume:
\[ V_{\text{initial}} = \frac{\pi}{6} \times (28)^3 \]
Calculating \( 28^3 \):
\[ 28^3 = 21952 \]
Thus,
\[ V_{\text{initial}} = \frac{\pi}{6} \times 21952 \]

### Step 2: Calculate the final diameter and volume of the balloon.
- The balloon explodes when the diameter becomes \( \frac{5}{2} \) times the initial diameter:
\[ d_{\text{final}} = \frac{5}{2} \times 28 = 70 \text{ cm} \]

Calculating the final volume:
\[ V_{\text{final}} = \frac{\pi}{6} \times (70)^3 \]
Calculating \( 70^3 \):
\[ 70^3 = 343000 \]
Thus,
\[ V_{\text{final}} = \frac{\pi}{6} \times 343000 \]

### Step 3: Determine the change in volume.
The change in volume \( \Delta V \) is given by:
\[ \Delta V = V_{\text{final}} - V_{\text{initial}} \]
Substituting the values we calculated:
\[ \Delta V = \left(\frac{\pi}{6} \times 343000\right) - \left(\frac{\pi}{6} \times 21952\right) \]
Factoring out \( \frac{\pi}{6} \):
\[ \Delta V = \frac{\pi}{6} \times (343000 - 21952) \]
Calculating \( 343000 - 21952 \):
\[ 343000 - 21952 = 321048 \]
Thus,
\[ \Delta V = \frac{\pi}{6} \times 321048 \]

### Step 4: Relate the volume change to the air blown into the balloon.
The volume of air blown into the balloon per second is given as 156 cc/s. If \( t \) is the time in seconds until the balloon explodes, then:
\[ \Delta V = 156 \times t \]
Setting the two expressions for \( \Delta V \) equal:
\[ 156t = \frac{\pi}{6} \times 321048 \]

### Step 5: Solve for \( t \).
Rearranging the equation to solve for \( t \):
\[ t = \frac{\frac{\pi}{6} \times 321048}{156} \]
Calculating \( \frac{\pi}{6} \) using \( \pi \approx \frac{22}{7} \):
\[ \frac{\pi}{6} \approx \frac{22}{7 \times 6} = \frac{22}{42} = \frac{11}{21} \]
Now substituting this back into the equation for \( t \):
\[ t = \frac{\frac{11}{21} \times 321048}{156} \]

Calculating \( \frac{11 \times 321048}{21 \times 156} \):
Calculating \( 21 \times 156 = 3276 \):
\[ t = \frac{11 \times 321048}{3276} \]
Calculating \( 11 \times 321048 = 3531528 \):
\[ t = \frac{3531528}{3276} \approx 1078 \]

### Final Answer:
Thus, the balloon will explode in approximately **1078 seconds**.

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