The surface area of a ballon of spherical shape being inflated , increases at a constant rete. If initially , the radius of balloon is 3 units and after 5 second , it becomes 7 units , then its radius after 9 seconds is :

Options

A. 9
B. 10
C. 11
D. 12

Answer

To solve the problem step-by-step, we will use the relationship between the surface area of a sphere and its radius, and the fact that the surface area is increasing at a constant rate.

### Step 1: Understand the relationship between surface area and radius
The surface area \( S \) of a sphere is given by the formula:
\[ S = 4\pi r^2 \]
where \( r \) is the radius of the sphere.

### Step 2: Differentiate the surface area with respect to time
To find how the surface area changes with respect to time, we differentiate both sides of the equation with respect to \( t \):
\[ \frac{dS}{dt} = 8\pi r \frac{dr}{dt} \]
This equation shows that the rate of change of surface area is proportional to the radius and the rate of change of the radius.

### Step 3: Establish the constant rate of increase
Since the surface area is increasing at a constant rate, we can denote this constant rate as \( k \):
\[ \frac{dS}{dt} = k \]
Thus, we can equate the two expressions:
\[ k = 8\pi r \frac{dr}{dt} \]

### Step 4: Set up the data points
We know the radius at two different times:
- At \( t = 0 \) seconds, \( r = 3 \) units.
- At \( t = 5 \) seconds, \( r = 7 \) units.

### Step 5: Calculate the change in radius over time
The change in radius from \( t = 0 \) to \( t = 5 \) seconds is:
\[ \Delta r = 7 - 3 = 4 \text{ units} \]
The time interval is \( 5 \) seconds, so the rate of change of the radius is:
\[ \frac{dr}{dt} = \frac{\Delta r}{\Delta t} = \frac{4}{5} \text{ units/second} \]

### Step 6: Find the radius after 9 seconds
We need to find the radius after \( 9 \) seconds. The time from \( t = 0 \) to \( t = 9 \) seconds is \( 9 \) seconds. The radius increases at the same rate:
\[ \text{Total increase in radius} = \frac{dr}{dt} \times \Delta t = \frac{4}{5} \times 9 = \frac{36}{5} \text{ units} \]
Now, add this increase to the initial radius:
\[ r(9) = r(0) + \text{Total increase} = 3 + \frac{36}{5} = 3 + 7.2 = 10.2 \text{ units} \]

### Step 7: Conclusion
Thus, the radius of the balloon after \( 9 \) seconds is:
\[ \boxed{10.2 \text{ units}} \]