If the volume of a sphere increases at constant rate `((dv)/(dt)=4)`. If radius of the sphere is denoted by `r` then surface area of the sphere inceases at the rate`:-`

Options

A. `(4)/(r )`
B. `(8)/(r )`
C. `(12)/(r )`
D. `(16)/(r )`

Explanation

To solve the problem, we need to find the rate of change of the surface area of a sphere when the volume is increasing at a constant rate. Here are the steps to derive the solution:

### Step 1: Write the formula for the volume of a sphere.
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]

### Step 2: Differentiate the volume with respect to time.
To find how the volume changes with respect to time, we differentiate \( V \) with respect to \( t \):
\[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) = 4 \pi r^2 \frac{dr}{dt} \]
We know from the problem statement that \( \frac{dV}{dt} = 4 \).

### Step 3: Set the derivative equal to the given rate of change of volume.
Now we can set the expression we derived equal to the given rate of change of volume:
\[ 4 \pi r^2 \frac{dr}{dt} = 4 \]

### Step 4: Solve for \( \frac{dr}{dt} \).
Rearranging the equation gives:
\[ \frac{dr}{dt} = \frac{4}{4 \pi r^2} = \frac{1}{\pi r^2} \]

### Step 5: Write the formula for the surface area of a sphere.
The surface area \( S \) of a sphere is given by:
\[ S = 4 \pi r^2 \]

### Step 6: Differentiate the surface area with respect to time.
Now we differentiate \( S \) with respect to \( t \):
\[ \frac{dS}{dt} = \frac{d}{dt} (4 \pi r^2) = 8 \pi r \frac{dr}{dt} \]

### Step 7: Substitute \( \frac{dr}{dt} \) into the surface area derivative.
Substituting \( \frac{dr}{dt} = \frac{1}{\pi r^2} \) into the equation for \( \frac{dS}{dt} \):
\[ \frac{dS}{dt} = 8 \pi r \left( \frac{1}{\pi r^2} \right) = \frac{8}{r} \]

### Conclusion
Thus, the rate at which the surface area of the sphere increases is:
\[ \frac{dS}{dt} = \frac{8}{r} \]